message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
import math
n, k = map(int, input().split())
a = 0
x, y = map(int, input().split())
for i in range(n-1):
p, q = map(int, input().split())
a += math.sqrt((x - p)* (x - p) + (y - q) * (y - q))
print(a * k / 50)
``` | instruction | 0 | 91,048 | 3 | 182,096 |
No | output | 1 | 91,048 | 3 | 182,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
import math
def dist(x0, y0, x1, y1):
return math.sqrt(math.pow((x1-x0), 2) + math.pow((y1-y0), 2))
tmp = input().split()
n, k = int(tmp[0]), float(tmp[1])
list_l = []
tmp = input().split()
list_l.append((float(tmp[0]), float(tmp[1])))
dist_sum = 0
for i in range(1, int(n/2)):
tmp = input().split()
list_l.append((float(tmp[0]), float(tmp[1])))
dist_sum += dist(list_l[i-1][0], list_l[i-1][1], list_l[i][0], list_l[i][1])
dist_sum /= 50
dist_sum *= k
print(dist_sum)
``` | instruction | 0 | 91,049 | 3 | 182,098 |
No | output | 1 | 91,049 | 3 | 182,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,098 | 3 | 182,196 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
for _ in range(int(input())):
N, K, L = map(int, input().split())
A = list(map(int, input().split()))
A.append(-K)
if max(A) > L:
print("No")
continue
ok = 1
k = -K-1
for a in A:
k += 1
if k < 0:
if a + K <= L:
k = -K-1
if a - k > L:
k = -(L - a)
else:
if a + k > L:
ok = 0
break
elif a + K <= L:
k = -K-1
if ok:
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
``` | output | 1 | 91,098 | 3 | 182,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,099 | 3 | 182,198 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
def main():
t = int(input())
for i in range(t):
n, k, l = map(int, input().split())
d = tuple(map(int, input().split()))
output = "Yes" if _can_get_to_finish(d, k, l) else "No"
print(output)
def _can_get_to_finish(depths, max_tide_height, max_height_can_swim):
tide_heights_allowed = [max_height_can_swim - depth for depth in depths]
if any(allowed_height < 0 for allowed_height in tide_heights_allowed):
return False
sea_size = len(tide_heights_allowed)
def is_checkpoint(pos):
return tide_heights_allowed[pos] >= max_tide_height
get_tide_height = lambda time: _get_tide_height_by_time(time, max_tide_height)
next_pos = 0
curr_checkpoint = -1
while next_pos < sea_size:
# Time when we will be at the next position.
# Because at the checkpoint we can spend as much time as we need,
# initially it is time of the first moment when can visit next position.
# It is negative because position becomes accessible before new period starts.
time_of_next_pos = -tide_heights_allowed[next_pos]
while next_pos < sea_size and not is_checkpoint(next_pos):
max_allowed_height = tide_heights_allowed[next_pos]
if get_tide_height(time_of_next_pos) > max_allowed_height:
how_much_must_wait = get_tide_height(time_of_next_pos) - max_allowed_height
if time_of_next_pos + how_much_must_wait > 0:
return False
time_of_next_pos += how_much_must_wait
next_pos += 1
time_of_next_pos += 1
while next_pos < sea_size and is_checkpoint(next_pos):
curr_checkpoint = next_pos
next_pos += 1
return True
def _get_tide_height_by_time(time, max_tide_height):
time %= 2*max_tide_height
if time < max_tide_height:
result = time
else:
time_since_max = time - max_tide_height
result = max_tide_height - time_since_max
return result
if __name__ == '__main__':
main()
``` | output | 1 | 91,099 | 3 | 182,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,100 | 3 | 182,200 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from bisect import bisect_left as lower_bound, bisect_right as upper_bound
def so(): return int(input())
def st(): return input()
def mj(): return map(int,input().strip().split(" "))
def msj(): return map(str,input().strip().split(" "))
def le(): return list(map(int,input().split()))
def lebe():return list(map(int, input()))
def dmain():
sys.setrecursionlimit(1000000)
threading.stack_size(1024000)
thread = threading.Thread(target=main)
thread.start()
def joro(L):
return(''.join(map(str, L)))
def decimalToBinary(n): return bin(n).replace("0b","")
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def tr(n):
return n*(n+1)//2
def iu():
n,p,q=mj()
L=le()
a,b=p,-p
g=1
for i in range(n):
if(q<L[i]):
g=0
break
a=min(q-L[i],p)
if(a!=p):
b=max(1+b,-a)
else:
b=-q
if(a<b):
g=0
break
if(g==1):
print("Yes")
else:
print("No")
def main():
for i in range(so()):
iu()
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read()
``` | output | 1 | 91,100 | 3 | 182,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,101 | 3 | 182,202 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def check(k,l,d,st,en):
ad = k-1
for i in range(st+1,en):
if d[i]+abs(ad) <= l:
ad -= 1
else:
if ad <= 0:
return 0
ad = l-d[i]-1
return 1
def solve(n,k,l,d):
stops = [-1]
for ind,i in enumerate(d):
if i+k<=l:
stops.append(ind)
elif i>l:
return 'NO'
stops.append(n)
for i in range(1,len(stops)):
if not check(k,l,d,stops[i-1],stops[i]):
return 'NO'
return 'YES'
def main():
for _ in range(int(input())):
n,k,l = map(int,input().split())
d = list(map(int,input().split()))
print(solve(n,k,l,d))
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 91,101 | 3 | 182,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,102 | 3 | 182,204 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
import sys
inpy = [int(x) for x in sys.stdin.read().split()]
t = inpy[0]
index = 1
for _ in range(t):
n, k, l = inpy[index], inpy[index+1], inpy[index+2]
index += 3
d = inpy[index:index+n]
index += n
x, m = k, True
flag = True
for i in range(n):
diff = l - d[i]
if diff < 0:
flag = False
break
if diff >= k:
x = k
m = True
else:
if m:
x = min(x - 1, diff)
if x == 0:
m = False
else:
x = x + 1
if x > diff:
flag = False
break
if flag :
print('Yes')
else:
print('No')
``` | output | 1 | 91,102 | 3 | 182,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,103 | 3 | 182,206 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
# input = open('file.txt').readline
for _ in range( int(input()) ):
n , k , l = map( int , input().strip().split(" ") )
arr = list(map( int , input().strip().split(" ") ))
goods = []
bad = False
for i , a in enumerate( arr ):
if a + k <= l:
goods.append(i)
if a > l:
bad = True
break
if bad:
print('No')
continue
goods.append(n)
prev = -1
for g in goods:
st = prev
en = g
# print(st , en)
if st + 1 == en:
prev = g
continue
tk = k
while st < en-1 and tk > 0:
st += 1
tk -= 1
plc = arr[st] + tk
if plc > l:
tk -= ( plc - l )
if tk < 0:
bad = True
# print(st, en , tk , 'after')
if tk == 0:
while st < en-1:
st += 1
tk += 1
plc = arr[st] + tk
# print('inside',st,tk,plc)
if plc > l:
bad = True
break
if bad:
break
prev = g
if bad:
print('No')
else:
print('Yes')
``` | output | 1 | 91,103 | 3 | 182,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,104 | 3 | 182,208 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
#
# ------------------------------------------------
# ____ _ Generatered using
# / ___| | |
# | | __ _ __| | ___ _ __ ______ _
# | | / _` |/ _` |/ _ \ '_ \|_ / _` |
# | |__| (_| | (_| | __/ | | |/ / (_| |
# \____\____|\____|\___|_| |_/___\____|
#
# GNU Affero General Public License v3.0
# ------------------------------------------------
# Author : prophet
# Created : 2020-07-24 11:19:20.229238
# UUID : dFs0Ek0q78tkOXbf
# ------------------------------------------------
#
production = True
import sys, math, collections
def input(input_format = 0, multi = 0):
if multi > 0: return [input(input_format) for i in range(multi)]
else:
next_line = sys.stdin.readline()[:-1]
if input_format >= 10:
use_list = False
input_format = int(str(input_format)[-1])
else: use_list = True
if input_format == 0: formatted_input = [next_line]
elif input_format == 1: formatted_input = list(map(int, next_line.split()))
elif input_format == 2: formatted_input = list(map(float, next_line.split()))
elif input_format == 3: formatted_input = list(next_line)
elif input_format == 4: formatted_input = list(map(int, list(next_line)))
elif input_format == 5: formatted_input = next_line.split()
else: formatted_input = [next_line]
return formatted_input if use_list else formatted_input[0]
def out(output_line, output_format = 0, newline = True):
formatted_output = ""
if output_format == 0: formatted_output = str(output_line)
elif output_format == 1: formatted_output = " ".join(map(str, output_line))
elif output_format == 2: formatted_output = "\n".join(map(str, output_line))
elif output_format == 3: formatted_output = "".join(map(str, output_line))
print(formatted_output, end = "\n" if newline else "")
def log(*args):
if not production:
print("$$$", end = "")
print(*args)
enu = enumerate
ter = lambda a, b, c: b if a else c
ceil = lambda a, b: -(-a // b)
flip = lambda a: (a + 1) & 1
def mapl(iterable, format = 0):
if format == 0: return list(map(int, iterable))
elif format == 1: return list(map(str, iterable))
elif format == 2: return list(map(list, iterable))
#
# >>>>>>>>>>>>>>> START OF SOLUTION <<<<<<<<<<<<<<
#
def solve():
n, k, l = input(1)
d = input(1)
log(k, l)
log(d)
f = [l - i for i in d]
log(f)
p = [(0, 2 * k - 1)]
for i in f:
a, b = p[-1]
if i >= k:
p.append((0, 2 * k - 1))
else:
fb = k + i
fa = max(a + 1, k - i)
log(i, fb, fa)
if fb < fa:
out("No")
return
p.append((fa, fb))
log(p)
else:
out("Yes")
log("")
return
for i in range(input(11)): solve()
#
# >>>>>>>>>>>>>>>> END OF SOLUTION <<<<<<<<<<<<<<<
#
``` | output | 1 | 91,104 | 3 | 182,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island. | instruction | 0 | 91,105 | 3 | 182,210 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n, k, l = map(int, input().split())
d = list(map(int, input().split()))
safe = [0]
for i in range(n):
if d[i]+k<=l:
safe.append(i+1)
safe.append(n+1)
ok = True
for i in range(1, len(safe)):
tide = k
down = True
for j in range(safe[i-1], safe[i]-1):
tide += -1 if down else 1
if down:
tide -= max(0, d[j]+tide-l)
if tide<0 or d[j]+tide>l:
ok = False
break
if tide==0:
down = False
if ok:
print('Yes')
else:
print('No')
``` | output | 1 | 91,105 | 3 | 182,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n,k,l=[int(x) for x in input().split()]
d=[int(x) for x in input().split()]
for i in range(n):
d[i]-=l #as long as d[i]<=0, Koa can stay at i
safeSpots=set()
ok=True
for i in range(n):
if d[i]+k<=0:
safeSpots.add(i)
if d[i]>0: #impossible no matter what
print('No')
ok=False
break
if ok==False:
continue
def getAdditionalDepth(t):
t2=t%(2*k)
if t2>k:
t2=2*k-t2
return t2
ok=True
t=k #starting on a safe spot. start moving as soon as the tide is low enough
if d[0]+getAdditionalDepth(t)>0:
t+=d[0]+getAdditionalDepth(t) #will be 0 at this time
d.append(-float('inf')) #d[n] is safe ground
for i in range(n):
if t>=2*k:t-=2*k #t is current time
if (i+1) in safeSpots:
t=k #set t to k (max depth)
continue
nextDepth=d[i+1]+getAdditionalDepth(t+1) #next depth if Koa moves over now
if nextDepth<=0:
t+=1
continue #ok to move to this index
else: #find the earliest time to move over, if possible
if t+1<=k:
ok=False #impossible because tide is increasing
break
earliestTime=t+nextDepth #after moving over now to i+1 (at t+1), must wait for nextDepth time before it's safe on i+1
t=earliestTime+1
if ok:
print('Yes')
else:
print('No')
``` | instruction | 0 | 91,106 | 3 | 182,212 |
Yes | output | 1 | 91,106 | 3 | 182,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
def main():
t = int(input())
for i in range(t):
n, k, l = map(int, input().split())
d = tuple(map(int, input().split()))
output = "Yes" if _can_get_to_finish(d, k, l) else "No"
print(output)
def _can_get_to_finish(depths, max_tide_height, max_height_can_swim):
tide_heights_allowed = [max_height_can_swim - depth for depth in depths]
if any(allowed_height < 0 for allowed_height in tide_heights_allowed):
return False
get_tide_height = lambda time: _get_tide_height_by_time(time, max_tide_height)
next_pos = 0
last_checkpoint = None
while True:
prev_checkpoint = last_checkpoint
next_time = -tide_heights_allowed[next_pos]
while next_pos < len(tide_heights_allowed):
allowed_height = tide_heights_allowed[next_pos]
if allowed_height >= max_tide_height:
last_checkpoint = next_pos
next_pos += 1
break
if get_tide_height(next_time) > allowed_height:
how_much_must_wait = get_tide_height(next_time) - allowed_height
if next_time + how_much_must_wait > 0:
return False
next_time += how_much_must_wait
next_pos += 1
next_time += 1
if next_pos >= len(tide_heights_allowed):
return True
if last_checkpoint == prev_checkpoint:
return False
def _get_tide_height_by_time(time, max_tide_height):
time %= 2*max_tide_height
if time < max_tide_height:
result = time
else:
time_since_max = time - max_tide_height
result = max_tide_height - time_since_max
return result
if __name__ == '__main__':
main()
``` | instruction | 0 | 91,107 | 3 | 182,214 |
Yes | output | 1 | 91,107 | 3 | 182,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
t = int(input())
while t>0:
n,k,l = map(int,input().split())
a = list(map(int,input().split()))
if max(a) > l:
print("No")
else:
for i in range(n):
if a[i] + k <= l:
a[i] = -1
ghata = -1
p = 1
y = 0
for i in range(n):
if a[i] == -1:
ghata = -1
else:
if ghata == -1:
ghata = l-a[i]
y = 0
if ghata == 0:
y = 1
ghata += 1
elif i != n-1 and a[i+1] != -1:
# y = 0
ghata = ghata-1
if l<ghata+a[i+1]:
ghata = l-a[i+1]
else:
if a[i]+ghata>l:
if y == 0:
if l<ghata+a[i]:
ghata = l-a[i]
if ghata == 0:
ghata += 1
y = 1
else:
ghata = ghata-1
else:
# print(i)
p = 0
break
else:
if y == 1 or ghata == 0:
ghata = ghata+1
y = 1
else:
ghata = ghata-1
# print(ghata,i)
if p == 0:
print("No")
else:
print("Yes")
t = t-1
``` | instruction | 0 | 91,108 | 3 | 182,216 |
Yes | output | 1 | 91,108 | 3 | 182,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
for h in range(int(input())):
n, k, l = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))
dp = [0 for i in range(len(arr)+1)]
dp[0] = -1
# print(p)
there = True
for i in range(len(arr)):
if arr[i] > l:
there = False
break
elif arr[i]+k > l:
dp[i+1] = arr[i]
else:
dp[i+1] = -1
if there == False:
print("No")
continue
pointer = 0
for i in range(n+1):
if dp[i] == -1:
if i == n:
continue
elif i < n:
maxi = l-dp[i+1]
pointer = -maxi-1
else:
pointer += 1
if dp[i] + abs(pointer) > l:
if pointer >= 0:
there = False
break
else:
pointer = -(l-dp[i])
if there == False:
print("No")
else:
print("Yes")
``` | instruction | 0 | 91,109 | 3 | 182,218 |
Yes | output | 1 | 91,109 | 3 | 182,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
T = int(input())
for _ in range(T):
n,k,l = map(int, input().split())
d = list(map(int, input().split()))
ranges = []
t = -102
co = 0
for i in range(n):
ch = [max(t+1, -l+d[i]), l-d[i]]
t = max(-l+d[i],t+1)
if k<=l-d[i]:
ch = [-102,102]
t = -101
ranges.append(ch)
if ranges[i][0]>ranges[i][1]:
print("NO")
co = 1
break
if co==0:
print("YES")
``` | instruction | 0 | 91,110 | 3 | 182,220 |
No | output | 1 | 91,110 | 3 | 182,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
for _ in range(int(input())):
N, K, L = map(int, input().split())
A = list(map(int, input().split()))
A.append(-K)
if max(A) > L:
print("No")
exit()
ok = 1
k = -K-1
for a in A:
if k < 0:
k += 1
if a + k > L:
k = -(L - a)
else:
k += 1
if a + k > L:
ok = 0
break
elif a + K <= L:
k = -K+1
if ok:
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
``` | instruction | 0 | 91,111 | 3 | 182,222 |
No | output | 1 | 91,111 | 3 | 182,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
from sys import stdin, stdout
# 5 2 3
# 1 2 3 2 2
#
def koa_and_beach(n, k, l, d_a):
sp_a = []
for i in range(len(d_a)):
if d_a[i] + k <= l:
sp_a.append(i)
if d_a[i] > l:
return 'No'
sp_a.append(n)
pre = -1
down = True
for i in sp_a:
ct = k
for j in range(pre + 1, i):
if down:
ct -= 1
else:
ct += 1
if d_a[j] + ct > l:
if down:
ct = d_a[j] + ct - l
else:
return 'No'
if ct == k:
down = True
elif ct == 0:
down = False
pre = i
return 'Yes'
# d + p[i] <= l
t = int(stdin.readline())
for _ in range(t):
n, k, l = map(int, stdin.readline().split())
d_a = list(map(int, stdin.readline().split()))
res = koa_and_beach(n, k, l, d_a)
stdout.write(str(res) + '\n')
``` | instruction | 0 | 91,112 | 3 | 182,224 |
No | output | 1 | 91,112 | 3 | 182,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Submitted Solution:
```
t = int(input())
while t>0:
n,k,l = map(int,input().split())
a = list(map(int,input().split()))
if max(a) > l:
print("No")
else:
for i in range(n):
if a[i] + k <= l:
a[i] = -1
ghata = -1
p = 1
for i in range(n):
if a[i] == -1:
ghata = -1
else:
if ghata == -1:
ghata = l-a[i]
if ghata == 0:
ghata += 1
elif i != n-1:
ghata = ghata-1
if l-a[i+1]<ghata:
ghata = l-a[i+1]
else:
if a[i]+((ghata-1)%k)+1>l:
p = 0
break
else:
ghata = ghata+1
if p == 0:
print("No")
else:
print("Yes")
t = t-1
``` | instruction | 0 | 91,113 | 3 | 182,226 |
No | output | 1 | 91,113 | 3 | 182,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the following concurrent program. There are N processes and the i-th process has the following pseudocode:
repeat ni times
yi := y
y := yi + 1
end repeat
Here y is a shared variable. Everything else is local for the process. All actions on a given row are atomic, i.e. when the process starts executing a row it is never interrupted. Beyond that all interleavings are possible, i.e. every process that has yet work to do can be granted the rights to execute its next row. In the beginning y = 0. You will be given an integer W and ni, for i = 1, ... , N. Determine if it is possible that after all processes terminate, y = W, and if it is possible output an arbitrary schedule that will produce this final value.
Input
In the first line of the input you will be given two space separated integers N (1 ≤ N ≤ 100) and W ( - 109 ≤ W ≤ 109). In the second line there are N space separated integers ni (1 ≤ ni ≤ 1000).
Output
On the first line of the output write Yes if it is possible that at the end y = W, or No otherwise. If the answer is No then there is no second line, but if the answer is Yes, then on the second line output a space separated list of integers representing some schedule that leads to the desired result. For more information see note.
Examples
Input
1 10
11
Output
No
Input
2 3
4 4
Output
Yes
1 1 2 1 2 2 2 2 2 1 2 1 1 1 1 2
Input
3 6
1 2 3
Output
Yes
1 1 2 2 2 2 3 3 3 3 3 3
Note
For simplicity, assume that there is no repeat statement in the code of the processes, but the code from the loop is written the correct amount of times. The processes are numbered starting from 1. The list of integers represent which process works on its next instruction at a given step. For example, consider the schedule 1 2 2 1 3. First process 1 executes its first instruction, then process 2 executes its first two instructions, after that process 1 executes its second instruction, and finally process 3 executes its first instruction. The list must consists of exactly 2·Σ i = 1...N ni numbers.
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, w = map(int, input().split())
a = [0] + list(map(int, input().split()))
total = sum(a)
def ng():
print('No')
exit()
def ok(a):
print('Yes')
print(*a)
exit()
if w < 1 or total < w:
ng()
if n == 1:
if w == a[0]:
ok([1] * (a[0] * 2))
else:
ng()
if w == 1:
if min(a[1:]) > 1:
ng()
min_i = a.index(1)
ans = []
for i in range(1, n + 1):
if i == min_i:
continue
ans += [i] * (a[i] * 2)
ok([min_i] + ans + [min_i])
ans1, ans2, ans3 = [], [], []
if w > 2:
for i in range(1, 3):
x = min(a[i] - 1, w - 2)
w -= x
a[i] -= x
ans3 += [i] * (2 * x)
for i in range(3, n + 1):
x = min(a[i], w - 2)
w -= x
a[i] -= x
ans3 += [i] * (2 * x)
ans1 = [2] * ((a[2] - 1) * 2)
for i in range(3, n + 1):
ans1 += [i] * (a[i] * 2)
ans1 = [1] + ans1 + [1]
a[1] -= 1
ans2 = [2] + [1] * (a[1] * 2) + [2]
ok(ans1 + ans2 + ans3)
``` | instruction | 0 | 91,235 | 3 | 182,470 |
No | output | 1 | 91,235 | 3 | 182,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is under siege! The attackers have blocked all the ways out and their cannon is bombarding the city. Fortunately, Berland intelligence managed to intercept the enemies' shooting plan. Let's introduce the Cartesian system of coordinates, the origin of which coincides with the cannon's position, the Ox axis is directed rightwards in the city's direction, the Oy axis is directed upwards (to the sky). The cannon will make n more shots. The cannon balls' initial speeds are the same in all the shots and are equal to V, so that every shot is characterized by only one number alphai which represents the angle at which the cannon fires. Due to the cannon's technical peculiarities this angle does not exceed 45 angles (π / 4). We disregard the cannon sizes and consider the firing made from the point (0, 0).
The balls fly according to the known physical laws of a body thrown towards the horizon at an angle:
vx(t) = V·cos(alpha) vy(t) = V·sin(alpha) – g·t x(t) = V·cos(alpha)·t y(t) = V·sin(alpha)·t – g·t2 / 2
Think of the acceleration of gravity g as equal to 9.8.
Bertown defends m walls. The i-th wall is represented as a vertical segment (xi, 0) - (xi, yi). When a ball hits a wall, it gets stuck in it and doesn't fly on. If a ball doesn't hit any wall it falls on the ground (y = 0) and stops. If the ball exactly hits the point (xi, yi), it is considered stuck.
Your task is to find for each ball the coordinates of the point where it will be located in the end.
Input
The first line contains integers n and V (1 ≤ n ≤ 104, 1 ≤ V ≤ 1000) which represent the number of shots and the initial speed of every ball. The second line contains n space-separated real numbers alphai (0 < alphai < π / 4) which represent the angles in radians at which the cannon will fire. The third line contains integer m (1 ≤ m ≤ 105) which represents the number of walls. Then follow m lines, each containing two real numbers xi and yi (1 ≤ xi ≤ 1000, 0 ≤ yi ≤ 1000) which represent the wall’s coordinates. All the real numbers have no more than 4 decimal digits. The walls may partially overlap or even coincide.
Output
Print n lines containing two real numbers each — calculate for every ball the coordinates of its landing point. Your answer should have the relative or absolute error less than 10 - 4.
Examples
Input
2 10
0.7853
0.3
3
5.0 5.0
4.0 2.4
6.0 1.9
Output
5.000000000 2.549499369
4.000000000 0.378324889
Input
2 10
0.7853
0.3
2
4.0 2.4
6.0 1.9
Output
10.204081436 0.000000000
4.000000000 0.378324889 | instruction | 0 | 91,291 | 3 | 182,582 |
Tags: data structures, geometry, sortings
Correct Solution:
```
import sys
from array import array # noqa: F401
from math import pi, sin, cos
from bisect import bisect_left
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, v = map(int, input().split())
v = float(v)
alpha = [float(input()) for _ in range(n)]
m = int(input())
wall = sorted(tuple(map(float, input().split())) for _ in range(m)) + [(1e9, 1e9)]
max_angle = pi / 4
eps = 1e-9
a = [0.0] * m + [max_angle + eps]
for i in range(m):
ng_angle, ok_angle = 0.0, max_angle + eps
for _ in range(50):
mid_angle = (ok_angle + ng_angle) / 2
t = wall[i][0] / (v * cos(mid_angle))
if (v * sin(mid_angle) * t - 9.8 * t**2 / 2) - eps <= wall[i][1]:
ng_angle = mid_angle
else:
ok_angle = mid_angle
a[i] = max(a[i], ng_angle)
a[i + 1] = max(a[i], a[i + 1])
ans = [''] * n
for i in range(n):
wi = bisect_left(a, alpha[i])
ok, ng = 0.0, 1e7
sin_a = sin(alpha[i])
for _ in range(50):
t = (ok + ng) / 2
if v * sin_a * t - 9.8 * t**2 / 2 >= 0.0:
ok = t
else:
ng = t
x = v * cos(alpha[i]) * ok
if x < wall[wi][0]:
ans[i] = f'{x:.8f} {0:.8f}'
else:
ok, ng = 0.0, 1e7
cos_a = cos(alpha[i])
for _ in range(50):
t = (ok + ng) / 2
if v * cos_a * t <= wall[wi][0]:
ok = t
else:
ng = t
y = v * sin_a * ok - 9.8 * ok**2 / 2
ans[i] = f'{wall[wi][0]:.8f} {y:.8f}'
print('\n'.join(ans))
``` | output | 1 | 91,291 | 3 | 182,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Today, the Earth has been attacked by the invaders from space, Invader, and the only survivors of humankind are us at the base. There is almost no force left to compete with them. But here we do not give up. Eradication of Invaders is the last resort for us humans to survive. I will explain the contents of the strategy that will be the last from now on.
First of all, our strength is too small compared to the strength of the Invader, so we will stay at this base and fight the siege. Surrounded by high mountains, the base has no choice but to take a straight road in front of the base for the invaders to invade. We will call this road a field. With this feature, it is possible to concentrate on the front and attack the invader. At the base, we attack the Invader with two types of weapons. One is a sniper rifle that snipers one invader. The other is a grenade launcher that can attack over a wide area.
As the only programmer in humanity, your job is to simulate combat based on Invaders and our action records. Action records are given in query format. Each query consists of one or more integers and is given as follows:
0 | Invaders appear on the field at a distance of L from the base.
--- | ---
1 d | All invaders currently on the field approach the base by d. After this action, the invaders that reach the base will take damage, and their enemies will disappear from the field. If it is damaged, "damage (the number of invaders that reached the base at this time)" is output on one line.
2 k | If the number of invaders currently on the field is k or more, attack the kth invader from the closest to the base with a sniper rifle. The invader disappears from the field. And output "hit" on one line. If the number of invaders on the field is less than k, print "miss" on one line.
3 x r | Attack with a grenade launcher in the range r at a position x from the base. All invaders that land at a distance of x from the base and are less than or equal to r will disappear from the field. Then output "bomb (number of invaders killed)" on one line. As a side note, the base will not be damaged by the grenade launcher.
4 k | If the number of invaders on the field is k or more, output "distance (distance between the kth invader from the closest base to the base)" on one line. If the number of invaders on the field is less than k, print "distance -1" on one line.
That is all for the explanation of the strategy. All members start the operation! ... Good luck.
Constraints
The input satisfies the following conditions.
* All values contained in the input are integers
* 1 ≤ Q ≤ 100000
* 1 ≤ L ≤ 109
* 1 ≤ d, k ≤ 109
* 0 ≤ x ≤ L
* 0 ≤ r ≤ 109
* There can never be more than one invader in the same position
* No more than 3 datasets
Input
The input consists of multiple datasets. Each dataset is represented below. The first line is given two integers Q and L separated by spaces. The following Q line is given the Q queries described above. The end of the input consists of two zeros.
Output
For each dataset, output for queries that have output instructions. Print "end" at the end of each dataset.
Example
Input
18 10
0
4 1
1 1
4 1
0
1 1
0
2 2
1 10
0
1 1
0
1 1
0
1 5
3 4 0
3 4 1
0
9 10
4 1
2 2
3 5 5
0
4 1
2 2
3 5 5
0
2 1
0 0
Output
distance 10
distance 9
hit
damage 2
bomb 1
bomb 2
end
distance -1
miss
bomb 0
distance 10
miss
bomb 1
hit
end
Submitted Solution:
```
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1515
for i in range(3):
firstInput = input()
if firstInput == "0 0":
break
else:
invader = []
Q = int(firstInput.split(" ")[0])
L = int(firstInput.split(" ")[1])
for j in range(Q):
otherInput = input()
inputList = []
for k in range(len(otherInput.split(" "))):
inputList.append(int(otherInput.split(" ")[k]))
if inputList[0] == 0:
invader.append(L)
elif inputList[0] == 1:
for l in range(len(invader)):
invader[l] -= inputList[1]
a = 0
b = []
for l in range(len(invader)):
if invader[l] <= 0:
a += 1
for l in range(a):
del invader[0]
if a > 0:
print("damage " + str(a))
elif inputList[0] == 2:
if len(invader) < inputList[1]:
print("miss")
else:
del invader[inputList[1]-1]
print("hit")
elif inputList[0] == 3:
a = 0
b = []
for l in range(len(invader)):
if invader[l] > inputList[1]:
if invader[l] - inputList[1] <= inputList[2]:
a += 1
b.append(l)
else:
if inputList[1] - invader[l] <= inputList[2]:
a += 1
b.append(l)
#print(b)
for l in range(len(b)):
del invader[b[0]]
print("bomb " + str(a))
elif inputList[0] == 4:
if len(invader) < inputList[1]:
print("distance -1")
else:
print("distance "+str(invader[inputList[1]-1]))
#print(invader)
print("end")
``` | instruction | 0 | 91,682 | 3 | 183,364 |
No | output | 1 | 91,682 | 3 | 183,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Today, the Earth has been attacked by the invaders from space, Invader, and the only survivors of humankind are us at the base. There is almost no force left to compete with them. But here we do not give up. Eradication of Invaders is the last resort for us humans to survive. I will explain the contents of the strategy that will be the last from now on.
First of all, our strength is too small compared to the strength of the Invader, so we will stay at this base and fight the siege. Surrounded by high mountains, the base has no choice but to take a straight road in front of the base for the invaders to invade. We will call this road a field. With this feature, it is possible to concentrate on the front and attack the invader. At the base, we attack the Invader with two types of weapons. One is a sniper rifle that snipers one invader. The other is a grenade launcher that can attack over a wide area.
As the only programmer in humanity, your job is to simulate combat based on Invaders and our action records. Action records are given in query format. Each query consists of one or more integers and is given as follows:
0 | Invaders appear on the field at a distance of L from the base.
--- | ---
1 d | All invaders currently on the field approach the base by d. After this action, the invaders that reach the base will take damage, and their enemies will disappear from the field. If it is damaged, "damage (the number of invaders that reached the base at this time)" is output on one line.
2 k | If the number of invaders currently on the field is k or more, attack the kth invader from the closest to the base with a sniper rifle. The invader disappears from the field. And output "hit" on one line. If the number of invaders on the field is less than k, print "miss" on one line.
3 x r | Attack with a grenade launcher in the range r at a position x from the base. All invaders that land at a distance of x from the base and are less than or equal to r will disappear from the field. Then output "bomb (number of invaders killed)" on one line. As a side note, the base will not be damaged by the grenade launcher.
4 k | If the number of invaders on the field is k or more, output "distance (distance between the kth invader from the closest base to the base)" on one line. If the number of invaders on the field is less than k, print "distance -1" on one line.
That is all for the explanation of the strategy. All members start the operation! ... Good luck.
Constraints
The input satisfies the following conditions.
* All values contained in the input are integers
* 1 ≤ Q ≤ 100000
* 1 ≤ L ≤ 109
* 1 ≤ d, k ≤ 109
* 0 ≤ x ≤ L
* 0 ≤ r ≤ 109
* There can never be more than one invader in the same position
* No more than 3 datasets
Input
The input consists of multiple datasets. Each dataset is represented below. The first line is given two integers Q and L separated by spaces. The following Q line is given the Q queries described above. The end of the input consists of two zeros.
Output
For each dataset, output for queries that have output instructions. Print "end" at the end of each dataset.
Example
Input
18 10
0
4 1
1 1
4 1
0
1 1
0
2 2
1 10
0
1 1
0
1 1
0
1 5
3 4 0
3 4 1
0
9 10
4 1
2 2
3 5 5
0
4 1
2 2
3 5 5
0
2 1
0 0
Output
distance 10
distance 9
hit
damage 2
bomb 1
bomb 2
end
distance -1
miss
bomb 0
distance 10
miss
bomb 1
hit
end
Submitted Solution:
```
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1515
for i in range(3):
firstInput = input()
if firstInput == "0 0":
break
else:
invader = []
Q = int(firstInput.split(" ")[0])
L = int(firstInput.split(" ")[1])
for j in range(Q):
otherInput = input()
inputList = []
for k in range(len(otherInput.split(" "))):
inputList.append(int(otherInput.split(" ")[k]))
if inputList[0] == 0:
invader.append(L)
elif inputList[0] == 1:
for l in range(len(invader)):
invader[l] -= inputList[1]
a = 0
b = []
for l in range(len(invader)):
if invader[l] <= 0:
a += 1
for l in range(a):
del invader[0]
if a > 0:
print("damage " + str(a))
elif inputList[0] == 2:
if len(invader) < inputList[1]:
print("miss")
else:
del invader[inputList[1]-1]
print("hit")
elif inputList[0] == 3:
a = 0
b = []
for l in range(len(invader)):
if invader[l] > inputList[1]:
if invader[l] - inputList[1] <= inputList[2]:
a += 1
b.append(l)
else:
if inputList[1] - invader[l] <= inputList[2]:
a += 1
b.append(l)
print(b)
for l in range(len(b)):
del invader[b[0]]
print("bomb " + str(a))
elif inputList[0] == 4:
if len(invader) < inputList[1]:
print("distance -1")
else:
print("distance "+str(invader[inputList[1]-1]))
#print(invader)
print("end")
``` | instruction | 0 | 91,683 | 3 | 183,366 |
No | output | 1 | 91,683 | 3 | 183,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Today, the Earth has been attacked by the invaders from space, Invader, and the only survivors of humankind are us at the base. There is almost no force left to compete with them. But here we do not give up. Eradication of Invaders is the last resort for us humans to survive. I will explain the contents of the strategy that will be the last from now on.
First of all, our strength is too small compared to the strength of the Invader, so we will stay at this base and fight the siege. Surrounded by high mountains, the base has no choice but to take a straight road in front of the base for the invaders to invade. We will call this road a field. With this feature, it is possible to concentrate on the front and attack the invader. At the base, we attack the Invader with two types of weapons. One is a sniper rifle that snipers one invader. The other is a grenade launcher that can attack over a wide area.
As the only programmer in humanity, your job is to simulate combat based on Invaders and our action records. Action records are given in query format. Each query consists of one or more integers and is given as follows:
0 | Invaders appear on the field at a distance of L from the base.
--- | ---
1 d | All invaders currently on the field approach the base by d. After this action, the invaders that reach the base will take damage, and their enemies will disappear from the field. If it is damaged, "damage (the number of invaders that reached the base at this time)" is output on one line.
2 k | If the number of invaders currently on the field is k or more, attack the kth invader from the closest to the base with a sniper rifle. The invader disappears from the field. And output "hit" on one line. If the number of invaders on the field is less than k, print "miss" on one line.
3 x r | Attack with a grenade launcher in the range r at a position x from the base. All invaders that land at a distance of x from the base and are less than or equal to r will disappear from the field. Then output "bomb (number of invaders killed)" on one line. As a side note, the base will not be damaged by the grenade launcher.
4 k | If the number of invaders on the field is k or more, output "distance (distance between the kth invader from the closest base to the base)" on one line. If the number of invaders on the field is less than k, print "distance -1" on one line.
That is all for the explanation of the strategy. All members start the operation! ... Good luck.
Constraints
The input satisfies the following conditions.
* All values contained in the input are integers
* 1 ≤ Q ≤ 100000
* 1 ≤ L ≤ 109
* 1 ≤ d, k ≤ 109
* 0 ≤ x ≤ L
* 0 ≤ r ≤ 109
* There can never be more than one invader in the same position
* No more than 3 datasets
Input
The input consists of multiple datasets. Each dataset is represented below. The first line is given two integers Q and L separated by spaces. The following Q line is given the Q queries described above. The end of the input consists of two zeros.
Output
For each dataset, output for queries that have output instructions. Print "end" at the end of each dataset.
Example
Input
18 10
0
4 1
1 1
4 1
0
1 1
0
2 2
1 10
0
1 1
0
1 1
0
1 5
3 4 0
3 4 1
0
9 10
4 1
2 2
3 5 5
0
4 1
2 2
3 5 5
0
2 1
0 0
Output
distance 10
distance 9
hit
damage 2
bomb 1
bomb 2
end
distance -1
miss
bomb 0
distance 10
miss
bomb 1
hit
end
Submitted Solution:
```
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1515
for i in range(3):
firstInput = input()
if firstInput == "0 0":
break
else:
invader = []
Q = int(firstInput.split(" ")[0])
L = int(firstInput.split(" ")[1])
for j in range(Q):
otherInput = input()
inputList = []
for k in range(len(otherInput.split(" "))):
inputList.append(int(otherInput.split(" ")[k]))
if inputList[0] == 0:
invader.append(L)
elif inputList[0] == 1:
for l in range(len(invader)):
invader[l] -= inputList[1]
a = 0
b = []
for l in range(len(invader)):
if invader[l] <= 0:
a += 1
if a > 0:
for l in range(a):
del invader[0]
print("damage " + str(a))
elif inputList[0] == 2:
if len(invader) < inputList[1]:
print("miss")
else:
del invader[inputList[1]-1]
print("hit")
elif inputList[0] == 3:
a = 0
b = []
for l in range(len(invader)):
if invader[l] > inputList[1]:
if invader[l] - inputList[1] <= inputList[2]:
a += 1
b.append(l)
else:
if inputList[1] - invader[l] <= inputList[2]:
a += 1
b.append(l)
#print(b)
for l in range(len(b)):
del invader[b[0]]
print("bomb " + str(a))
elif inputList[0] == 4:
if len(invader) < inputList[1]:
print("distance -1")
else:
print("distance "+str(invader[inputList[1]-1]))
#print(invader)
print("end")
``` | instruction | 0 | 91,684 | 3 | 183,368 |
No | output | 1 | 91,684 | 3 | 183,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Today, the Earth has been attacked by the invaders from space, Invader, and the only survivors of humankind are us at the base. There is almost no force left to compete with them. But here we do not give up. Eradication of Invaders is the last resort for us humans to survive. I will explain the contents of the strategy that will be the last from now on.
First of all, our strength is too small compared to the strength of the Invader, so we will stay at this base and fight the siege. Surrounded by high mountains, the base has no choice but to take a straight road in front of the base for the invaders to invade. We will call this road a field. With this feature, it is possible to concentrate on the front and attack the invader. At the base, we attack the Invader with two types of weapons. One is a sniper rifle that snipers one invader. The other is a grenade launcher that can attack over a wide area.
As the only programmer in humanity, your job is to simulate combat based on Invaders and our action records. Action records are given in query format. Each query consists of one or more integers and is given as follows:
0 | Invaders appear on the field at a distance of L from the base.
--- | ---
1 d | All invaders currently on the field approach the base by d. After this action, the invaders that reach the base will take damage, and their enemies will disappear from the field. If it is damaged, "damage (the number of invaders that reached the base at this time)" is output on one line.
2 k | If the number of invaders currently on the field is k or more, attack the kth invader from the closest to the base with a sniper rifle. The invader disappears from the field. And output "hit" on one line. If the number of invaders on the field is less than k, print "miss" on one line.
3 x r | Attack with a grenade launcher in the range r at a position x from the base. All invaders that land at a distance of x from the base and are less than or equal to r will disappear from the field. Then output "bomb (number of invaders killed)" on one line. As a side note, the base will not be damaged by the grenade launcher.
4 k | If the number of invaders on the field is k or more, output "distance (distance between the kth invader from the closest base to the base)" on one line. If the number of invaders on the field is less than k, print "distance -1" on one line.
That is all for the explanation of the strategy. All members start the operation! ... Good luck.
Constraints
The input satisfies the following conditions.
* All values contained in the input are integers
* 1 ≤ Q ≤ 100000
* 1 ≤ L ≤ 109
* 1 ≤ d, k ≤ 109
* 0 ≤ x ≤ L
* 0 ≤ r ≤ 109
* There can never be more than one invader in the same position
* No more than 3 datasets
Input
The input consists of multiple datasets. Each dataset is represented below. The first line is given two integers Q and L separated by spaces. The following Q line is given the Q queries described above. The end of the input consists of two zeros.
Output
For each dataset, output for queries that have output instructions. Print "end" at the end of each dataset.
Example
Input
18 10
0
4 1
1 1
4 1
0
1 1
0
2 2
1 10
0
1 1
0
1 1
0
1 5
3 4 0
3 4 1
0
9 10
4 1
2 2
3 5 5
0
4 1
2 2
3 5 5
0
2 1
0 0
Output
distance 10
distance 9
hit
damage 2
bomb 1
bomb 2
end
distance -1
miss
bomb 0
distance 10
miss
bomb 1
hit
end
Submitted Solution:
```
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1515
for i in range(3):
firstInput = input()
if firstInput == "0 0":
break
else:
invader = []
Q = int(firstInput.split(" ")[0])
L = int(firstInput.split(" ")[1])
for j in range(Q):
otherInput = input()
inputList = []
for k in range(len(otherInput.split(" "))):
inputList.append(int(otherInput.split(" ")[k]))
if inputList[0] == 0:
invader.append(L)
elif inputList[0] == 1:
for l in range(len(invader)):
invader[l] -= inputList[1]
a = 0
b = []
for l in range(len(invader)):
if invader[l] <= 0:
a += 1
for l in range(a):
del invader[0]
if a > 0:
print("damage " + str(a))
elif inputList[0] == 2:
if len(invader) < inputList[1]:
print("miss")
else:
del invader[inputList[1]-1]
print("hit")
elif inputList[0] == 3:
a = 0
b = []
for l in range(len(invader)):
if invader[l] > inputList[1]:
if invader[l] - inputList[1] <= inputList[2]:
a += 1
b.append(l)
else:
if inputList[1] - invader[l] <= inputList[2]:
a += 1
b.append(l)
print(b)
for l in range(len(b)):
del invader[b[0]]
print("bomb " + str(a))
elif inputList[0] == 4:
if len(invader) < inputList[1]:
print("distance -1")
else:
print("distance "+str(invader[inputList[1]-1]))
print(invader)
print("end")
``` | instruction | 0 | 91,685 | 3 | 183,370 |
No | output | 1 | 91,685 | 3 | 183,371 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,686 | 3 | 183,372 |
"Correct Solution:
```
while True:
e = int(input())
if e == 0:
break
ans = e
z = 0
while z**3<=e:
y = int((e-z**3)**0.5)
x = e-z**3-y**2 #zとyを先に定める
ans = min(x+y+z,ans)
z +=1
print(ans)
``` | output | 1 | 91,686 | 3 | 183,373 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,687 | 3 | 183,374 |
"Correct Solution:
```
while True:
e = int(input())
if e == 0:
break
mz = int( e **(1/3))+1
L = []
for k in range(mz+1):
ek = e-k**3
if ek < 0:
break
j = int(ek**0.5)
i = ek - j**2
if i >= 0 and j >= 0 and k >=0:
L.append((i+j+k))
print(min(L))
``` | output | 1 | 91,687 | 3 | 183,375 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,688 | 3 | 183,376 |
"Correct Solution:
```
import sys
inf = 1<<30
def solve():
while 1:
e = int(sys.stdin.readline().rstrip())
if e == 0:
return
ans = inf
for z in range(int(e**(1/3)) + 10):
rest = e - z**3
if rest < 0:
break
y = int((rest)**0.5)
x = rest - y**2
ans = min(ans, x + y + z)
print(ans)
if __name__ == '__main__':
solve()
``` | output | 1 | 91,688 | 3 | 183,377 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,689 | 3 | 183,378 |
"Correct Solution:
```
# ??¶?´?: x + y^2 + z^3 = e
# ????????¨??§???????°???¨?????? x + y + z ????±??????????
import math
# 1. z ????±???????
# 2. y ??? ???y**2 < (e - z**3)???????????????y????????§???
# 3. x ??? e - y**2 - z**3
# ?????????e????????????????????????101???
while True:
e = int(input())
if e == 0:
break
min_e = e
for i in range(101):
if (e - i**3) < 0:
break;
z = i
y = math.floor(math.sqrt(e - i**3))
x = e - y**2 - z**3
if ( (x + y + z) < min_e ):
min_e = x + y + z
print(min_e)
``` | output | 1 | 91,689 | 3 | 183,379 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,690 | 3 | 183,380 |
"Correct Solution:
```
import math
def solve(e):
k = 2 ** 32
for z in range(100, -1, -1):
z3 = z * z * z
if z3 > e:
continue
e2 = e - z3
ylm = int(math.sqrt(e2))
xzlm = 3 * z * z + 3 * z + 1
for y in range(ylm, -1, -1):
y2 = y * y
if e2 > (y + 1) * (y + 1):
break
e3 = e2 - y2
xylm = 2 * y + 1
x = e3
if x > xylm or x > xzlm:
continue
k = min(k, x + y + z)
return k
def main():
while True:
a = int(input())
if a == 0:
break
print(solve(a))
if __name__ == '__main__':
main()
``` | output | 1 | 91,690 | 3 | 183,381 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,691 | 3 | 183,382 |
"Correct Solution:
```
while True:
e = int(input())
if e == 0:
break
ans = float('inf')
max_z = int(e ** (1 / 3) + 1e-6)
for z in range(max_z, -1, -1):
y = int((e - z ** 3) ** 0.5 + 1e-6)
ans = min(ans, e - y ** 2 - z ** 3 + y + z)
print(ans)
``` | output | 1 | 91,691 | 3 | 183,383 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,692 | 3 | 183,384 |
"Correct Solution:
```
#!/usr/bin/env python3
from math import sqrt
while True:
e = int(input())
if e == 0: break
z, ans = 0, e
while z ** 3 <= e:
y = int(sqrt(e - z**3))
x = e - y**2 - z**3
ans = min(ans, x + y + z)
z += 1
print(ans)
``` | output | 1 | 91,692 | 3 | 183,385 |
Provide a correct Python 3 solution for this coding contest problem.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44 | instruction | 0 | 91,693 | 3 | 183,386 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:
break
m = n
z = 0
while z**3<=n:
y = int((n-z**3)**0.5)
x = n-z**3-y**2
m = min(x+y+z,m)
z +=1
print(m)
``` | output | 1 | 91,693 | 3 | 183,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
# your code goes here
#grab
import math
e=int(input())
#e=1250
while e>0:
z=0
t=z**3
f=e-t
q=math.sqrt(f)
y=int(q)
if q-y>0.999999:
y+=1
x=y*y
x=f-x
m=x+y+z
while t<=e:
f=e-t
q=math.sqrt(f)
y=int(q)
if q-y>0.999999:
y+=1
x=y*y
x=f-x
l=x+y+z
if l<m:
m=l
z+=1
t=z**3
print(m)
e=int(input())
``` | instruction | 0 | 91,694 | 3 | 183,388 |
Yes | output | 1 | 91,694 | 3 | 183,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
import math
def solve(e):
k = 2 ** 32
for z in range(100, -1, -1):
z3 = z * z * z
if z3 > e:
continue
e2 = e - z3
ylm = int(math.sqrt(e2))
for y in range(ylm, -1, -1):
y2 = y * y
if e2 > (y + 1) * (y + 1):
break
e3 = e2 - y2
x = e3
k = min(k, x + y + z)
return k
def main():
while True:
a = int(input())
if a == 0:
break
print(solve(a))
if __name__ == '__main__':
main()
``` | instruction | 0 | 91,695 | 3 | 183,390 |
Yes | output | 1 | 91,695 | 3 | 183,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
while True:
e = int(input())
if e == 0:
break
z, ans = 0, float('inf')
while z ** 3 <= e:
r = e - z ** 3
y = int(r ** 0.5)
x = r - y ** 2
ans = min(ans, x + y + z)
z += 1
print(ans)
``` | instruction | 0 | 91,696 | 3 | 183,392 |
Yes | output | 1 | 91,696 | 3 | 183,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
from math import floor, sqrt
cubics = [ x * x * x for x in range(101)]
squares = [ x * x for x in range(1000)]
while True:
m = 1000000
e = int(input())
if e == 0: break
for z in range(len(cubics)):
if e < cubics[z]: break
tmp = e - cubics[z]
y = floor(sqrt(tmp))
x = tmp - y ** 2
m = min(m, x + y + z)
print(m)
``` | instruction | 0 | 91,697 | 3 | 183,394 |
Yes | output | 1 | 91,697 | 3 | 183,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
while 1:
e = int(input().strip())
if e ==0:break
m=e
for z in range(e):
for y in range(e):
x = e - y**2 - z**3
if x < 0:break
if m > x+y+z: m = x+y+z
print(m)
``` | instruction | 0 | 91,698 | 3 | 183,396 |
No | output | 1 | 91,698 | 3 | 183,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
import math
def main():
while True:
e = int(input())
if e == 0:
break
p = 0
cs = list(reversed(range(math.floor(e**(1/3))+1)))
bs = list(reversed(range(math.floor(e**(1/2))+1)))
for c in cs:
for b in bs:
if c**3+b**2<e and c**3+b**2>p:
r = c+b
p = c**3+b**2
if c**3+b**2<p:
break
print(r+e-p)
main()
``` | instruction | 0 | 91,699 | 3 | 183,398 |
No | output | 1 | 91,699 | 3 | 183,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import heapq
import math
def solve(e):
candidates = []
for z in range(0, math.floor(math.pow(e, 1.0 / 3.0)) + 1):
if e < z * z * z:
break
else:
for y in range(0, math.floor(math.sqrt(e - z * z * z)) + 1):
x = e - z * z * z - y * y
heapq.heappush(candidates, x + y + z)
return candidates[0]
if __name__ == '__main__':
while True:
e = int(input())
if e == 0:
break
else:
print(solve(e))
``` | instruction | 0 | 91,700 | 3 | 183,400 |
No | output | 1 | 91,700 | 3 | 183,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Space Coconut Crab
Space coconut crab
English text is not available in this practice contest.
Ken Marine Blue is a space hunter who travels through the entire galaxy in search of space coconut crabs. The space coconut crab is the largest crustacean in the universe, and it is said that the body length after growth is 400 meters or more, and if you spread your legs, it will reach 1,000 meters or more. Many people have already witnessed the space coconut crab, but none have succeeded in catching it.
Through a long-term study, Ken uncovered important facts about the ecology of space coconut crabs. Surprisingly, the space coconut crab did the same thing as the latest warp technology called phase transition navigation, and lived back and forth between normal space and hyperspace. Furthermore, it was found that it takes a long time for the space coconut crab to warp out from the hyperspace to the normal space, and that it cannot move to the hyperspace for a while after the warp out.
So Ken finally decided to catch the space coconut crab. The strategy is as follows. First, we observe the energy of the space coconut crab as it plunges from normal space into hyperspace. When this energy is e, it is known that the coordinates (x, y, z) at which the space coconut crab warps out of hyperspace satisfy the following conditions.
* x, y, z are all non-negative integers.
* x + y2 + z3 = e.
* Minimize the value of x + y + z under the above conditions.
These conditions alone do not always uniquely determine the coordinates, but it is certain that the coordinates to warp out are on the plane x + y + z = m, where m is the minimum value of x + y + z. Is. Therefore, a barrier of sufficient size is placed on this plane. Then, the space coconut crab will warp out to the place where the barrier is stretched. Space coconut crabs affected by the barrier get stuck. It is a setup to capture it with the weapon breaker, which is a state-of-the-art spacecraft operated by Ken.
The barrier can only be set once, so it cannot fail. So Ken decided to use a calculator to carry out his mission. Your job is to write a program that finds the plane x + y + z = m to which the barrier should be placed when the energy for the space coconut crab to enter the hyperspace is given. Your program will be accepted when it outputs the correct results for all of the prepared test cases.
Input
The input consists of multiple datasets. Each dataset consists of only one row and contains one positive integer e (e ≤ 1,000,000). This represents the energy when the space coconut crab rushes into hyperspace. The input ends when e = 0, which is not included in the dataset.
Output
For each dataset, output the value of m on one line. The output must not contain any other characters.
Sample Input
1
2
Four
27
300
1250
0
Output for the Sample Input
1
2
2
3
18
44
Example
Input
1
2
4
27
300
1250
0
Output
1
2
2
3
18
44
Submitted Solution:
```
while True:
e = int(input())
if e == 0:
break
m = e + 1
y = 0
while y <= e:
y2 = y ** 2
if y2 > e:
break
z = 0
while z <= e:
z3 = z ** 3
if z3 + y2 > e:
break
x = e - y2 - z3
m = min(m, x + y + z)
z += 1
y += 1
print(m)
``` | instruction | 0 | 91,701 | 3 | 183,402 |
No | output | 1 | 91,701 | 3 | 183,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
def command(cmd) :
t=cmd[0]
if t==1 :
pour(cmd[1],cmd[2])
else :
amount(cmd[1])
def pour(p,amount) :
Skip=[]
while p<=n and amount>0:
t=C[p-1]-W[p-1]
if amount>=t :
amount-=t
W[p-1]=C[p-1]
Skip.append(p)
else :
W[p-1]+=amount
break
p=Links[p]
for i in Skip :
Links[i]=p
#print(W)
def amount(p) :
P.append(W[p-1])
n=int(input())
C=list(map(int,input().split()))
W=[0]*n
Links=list(map(int,range(1,n+2)))
P=[]
m=int(input())
for i in range(m) :
command([int(i) for i in input().split()])
for i in P :
print(i)
``` | instruction | 0 | 92,090 | 3 | 184,180 |
Yes | output | 1 | 92,090 | 3 | 184,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = [0]*n
after = list(map(int,range(1,n+2)))
#print(a)
def add_water(p,x):
i = p-1
skip = []
while x>0 and i<n:
space = a[i]-b[i]
if x >= space:
x -= space
b[i] = a[i]
skip.append(i)
i = after[i]
else:
b[i] += x
break
for j in skip:
after[j] = i
m = int(input())
P =[]
for i in range(m):
l = [int(j) for j in input().split()]
#print(l)
if l[0] == 1:
add_water(l[1],l[2])
else:
P.append(b[l[1]-1])
for j in P:
print(j)
``` | instruction | 0 | 92,091 | 3 | 184,182 |
Yes | output | 1 | 92,091 | 3 | 184,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
from sys import stdin, stdout
def find_next(root, v):
q = v
while root[q] != q:
q = root[q]
while v != q:
root[v], v = q, root[v]
return v
input = stdin.readline
n = int(input())
a = tuple(map(int, input().split()))
indexes = [i for i in range(n + 1)]
res = [0] * (n + 1)
for _ in range(int(input())):
req = input().split()
if req[0] == '1':
p, x = int(req[1]), int(req[2])
p -= 1
while p < n and res[p] + x >= a[p]:
x = res[p] + x - a[p]
res[p] = a[p]
indexes[p] = p = find_next(indexes, p + 1)
res[p] += x
else:
k = int(req[1])
stdout.write(f'{res[k - 1]}\n')
``` | instruction | 0 | 92,092 | 3 | 184,184 |
Yes | output | 1 | 92,092 | 3 | 184,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
'''
Created on 2016-4-9
@author: chronocorax
'''
def line():
return [int(c) for c in input().split()]
n = line()[0]
cap = line()
cap0 = cap[:]
nxt = [i + 1 for i in range(n)]
result = []
m = line()[0]
for _ in range(m):
tup = line()
if tup[0] == 1:
p = tup[1] - 1
p0 = p
x = tup[2]
# flush(p - 1, x)
marker = []
while p < n and cap[p] < x:
x -= cap[p]
cap[p] = 0
marker.append(p)
p = nxt[p]
if p < n: cap[p] -= x
for i in marker:
nxt[i] = p
else:
k = tup[1]
result.append(cap0[k - 1] - cap[k - 1])
for res in result:
print(res)
``` | instruction | 0 | 92,093 | 3 | 184,186 |
Yes | output | 1 | 92,093 | 3 | 184,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
n = int(input())
string = input().split()
vessels = []
transbordaEm = 300000
positions = {}
for i in range(n):
vessels.append([0,int(string[i])])
positions[i+1] = i+1
m = int(input())
for q in range(m):
query = input().split()
# print(vessels)
if(query[0] == '2'):
position = int(query[1]) - 1
print(vessels[position][0])
elif(query[0] == '1'):
position = positions[int(query[1])] - 1
x = int(query[2])
for v in range(position, n):
# print('no for!')
c = vessels[v][1]
t = vessels[v][0]
if(x > (c-t)): #entrou aqui, vai encher!
x = x - (c-t)
vessels[v][0] = c # t = c
last = v
if(x == 0):
break
else:
vessels[v][0] = t + x # t = x
last = v
break
for i in range(int(query[1])-1, last+1):
positions[i] = last+1
# print('na query:', int(query[1]))
# print('no positions:', positions[int(query[1])])
# print(positions)
``` | instruction | 0 | 92,094 | 3 | 184,188 |
No | output | 1 | 92,094 | 3 | 184,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = [0]*n
m = int(input())
q1 = []
for l in range(m):
q = [int(i) for i in input().split()]
if q[0] == 1:
if len(q1) != 0 and q1[-1][1] == q[1]:
q1[-1][2] += q[2]
else:
q1.append(q)
elif q[0] == 2:
if len(q1) != 0:
q1.sort(key = lambda x:x[1])
v,j = 0,0
for i in range(q1[0][1]-1,n):
if j < len(q1) and q1[j][1]-1 == i:
v += q1[j][2]
j += 1
b[i] += v
if b[i] <= a[i]:
v = 0
else:
v = b[i] - a[i]
b[i] = a[i]
if j == len(q1) and v == 0:
break
print(b[q[1]-1])
q1.clear()
``` | instruction | 0 | 92,095 | 3 | 184,190 |
No | output | 1 | 92,095 | 3 | 184,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = [0]*n
m = int(input())
q1 = []
for l in range(m):
q = [int(i) for i in input().split()]
if q[0] == 1:
q1.append(q)
elif q[0] == 2:
if len(q1) != 0:
q1.sort(key = lambda x:x[1])
v,j = 0,0
for i in range(q1[0][1]-1,n):
if j < len(q1) and q1[j][1]-1 == i:
v += q1[j][2]
j += 1
b[i] += v
if b[i] <= a[i]:
v = 0
else:
v = b[i] - a[i]
b[i] = a[i]
if j == len(q1) and v == 0:
break
print(b[q[1]-1])
q1.clear()
``` | instruction | 0 | 92,096 | 3 | 184,192 |
No | output | 1 | 92,096 | 3 | 184,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a system of n vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to n, in the order from the highest to the lowest, the volume of the i-th vessel is ai liters.
<image>
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the i-th vessel goes to the (i + 1)-th one. The liquid that overflows from the n-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add xi liters of water to the pi-th vessel;
2. Print the number of liters of water in the ki-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input
The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·105). The second line contains n integers a1, a2, ..., an — the vessels' capacities (1 ≤ ai ≤ 109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer m — the number of queries (1 ≤ m ≤ 2·105). Each of the next m lines contains the description of one query. The query of the first type is represented as "1 pi xi", the query of the second type is represented as "2 ki" (1 ≤ pi ≤ n, 1 ≤ xi ≤ 109, 1 ≤ ki ≤ n).
Output
For each query, print on a single line the number of liters of water in the corresponding vessel.
Examples
Input
2
5 10
6
1 1 4
2 1
1 2 5
1 1 4
2 1
2 2
Output
4
5
8
Input
3
5 10 8
6
1 1 12
2 2
1 1 6
1 3 2
2 2
2 3
Output
7
10
5
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = [0]*n
prev ={}
after = {}
for i in range(n):
prev[i] = i-1
after[i] = i+1
#print(a)
def add_water(p,x):
if (a[p-1]-b[p-1])>x:
b[p-1] += x
else:
total = a[p-1]-b[p-1]
b[p-1] += total
after[prev[p-1]]=after[p-1]
prev[after[p-1]]=prev[p-1]
i = after[p-1]
while x-total>0 and i<n:
add = min((a[i]-b[i]),x-total)
b[i] += add
if x-total>=(a[i]-b[i]):
total += add
after[prev[i]]=after[i]
prev[after[i]]=prev[i]
i = after[i]
def print_water(k):
print(b[k-1])
m = int(input())
for i in range(m):
l = [int(j) for j in input().split()]
#print(l)
if l[0] == 1:
add_water(l[1],l[2])
else:
print_water(l[1])
``` | instruction | 0 | 92,097 | 3 | 184,194 |
No | output | 1 | 92,097 | 3 | 184,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
mod=998244353
n,m= map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
pre=[[0 for i in range (m)]for j in range (n+1)]
for i in range (n):
for j in range (m):
pre[i+1][j]=pre[i][j]
if j+1==a[i]:
pre[i+1][j]+=1
s=sum(b)
ch=0
for i in range (s,n+1):
curr=0
for j in range (m):
if pre[i][j]-pre[i-s][j]==b[j]:
curr+=1
if curr==m:
ch=1
#print(i)
break
if ch:
print("YES")
else:
print("NO")
``` | instruction | 0 | 92,361 | 3 | 184,722 |
Yes | output | 1 | 92,361 | 3 | 184,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES
Submitted Solution:
```
n, m = map(int, input().split())
c = list(map(int, input().split()))
k = list(map(int, input().split()))
s = sum(k)
for i in range(n - s + 1):
v = [0] * m
for j in range(i, i + s):
v[c[j] - 1] += 1
if v == k:
print('YES')
exit(0)
print('NO')
``` | instruction | 0 | 92,362 | 3 | 184,724 |
Yes | output | 1 | 92,362 | 3 | 184,725 |
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