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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fighting with Zmei Gorynich — a ferocious monster from Slavic myths, a huge dragon-like reptile with multiple heads!
<image>
Initially Zmei Gorynich has x heads. You can deal n types of blows. If you deal a blow of the i-th type, you decrease the number of Gorynich's heads by min(d_i, curX), there curX is the current number of heads. But if after this blow Zmei Gorynich has at least one head, he grows h_i new heads. If curX = 0 then Gorynich is defeated.
You can deal each blow any number of times, in any order.
For example, if curX = 10, d = 7, h = 10 then the number of heads changes to 13 (you cut 7 heads off, but then Zmei grows 10 new ones), but if curX = 10, d = 11, h = 100 then number of heads changes to 0 and Zmei Gorynich is considered defeated.
Calculate the minimum number of blows to defeat Zmei Gorynich!
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.
The first line of each query contains two integers n and x (1 ≤ n ≤ 100, 1 ≤ x ≤ 10^9) — the number of possible types of blows and the number of heads Zmei initially has, respectively.
The following n lines of each query contain the descriptions of types of blows you can deal. The i-th line contains two integers d_i and h_i (1 ≤ d_i, h_i ≤ 10^9) — the description of the i-th blow.
Output
For each query print the minimum number of blows you have to deal to defeat Zmei Gorynich.
If Zmei Gorynuch cannot be defeated print -1.
Example
Input
3
3 10
6 3
8 2
1 4
4 10
4 1
3 2
2 6
1 100
2 15
10 11
14 100
Output
2
3
-1
Note
In the first query you can deal the first blow (after that the number of heads changes to 10 - 6 + 3 = 7), and then deal the second blow.
In the second query you just deal the first blow three times, and Zmei is defeated.
In third query you can not defeat Zmei Gorynich. Maybe it's better to convince it to stop fighting?
Submitted Solution:
```
def solve():
(n,x) = (map(int, input().split()))
(maxr, maxd) = (-1,-1)
for _ in range(n):
(d,h) = (map(int, input().split()))
maxr = max(maxr, d - h)
maxd = max(maxd, d)
if x <= maxd:
return 1
elif maxr <= 0:
return -1
else:
x -= maxd
return (x+maxr-1)//maxr + 1
t = int(input())
while t:
print(solve())
t = t - 1
```
|
instruction
| 0
| 23,931
| 2
| 47,862
|
Yes
|
output
| 1
| 23,931
| 2
| 47,863
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fighting with Zmei Gorynich — a ferocious monster from Slavic myths, a huge dragon-like reptile with multiple heads!
<image>
Initially Zmei Gorynich has x heads. You can deal n types of blows. If you deal a blow of the i-th type, you decrease the number of Gorynich's heads by min(d_i, curX), there curX is the current number of heads. But if after this blow Zmei Gorynich has at least one head, he grows h_i new heads. If curX = 0 then Gorynich is defeated.
You can deal each blow any number of times, in any order.
For example, if curX = 10, d = 7, h = 10 then the number of heads changes to 13 (you cut 7 heads off, but then Zmei grows 10 new ones), but if curX = 10, d = 11, h = 100 then number of heads changes to 0 and Zmei Gorynich is considered defeated.
Calculate the minimum number of blows to defeat Zmei Gorynich!
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.
The first line of each query contains two integers n and x (1 ≤ n ≤ 100, 1 ≤ x ≤ 10^9) — the number of possible types of blows and the number of heads Zmei initially has, respectively.
The following n lines of each query contain the descriptions of types of blows you can deal. The i-th line contains two integers d_i and h_i (1 ≤ d_i, h_i ≤ 10^9) — the description of the i-th blow.
Output
For each query print the minimum number of blows you have to deal to defeat Zmei Gorynich.
If Zmei Gorynuch cannot be defeated print -1.
Example
Input
3
3 10
6 3
8 2
1 4
4 10
4 1
3 2
2 6
1 100
2 15
10 11
14 100
Output
2
3
-1
Note
In the first query you can deal the first blow (after that the number of heads changes to 10 - 6 + 3 = 7), and then deal the second blow.
In the second query you just deal the first blow three times, and Zmei is defeated.
In third query you can not defeat Zmei Gorynich. Maybe it's better to convince it to stop fighting?
Submitted Solution:
```
import operator
from math import *
t = int(input())
for y in range(t):
n,x = map(int,input().split())
l = []
c = 0
for z in range(n):
d,h = map(int,input().split())
if(h >= d):
c += 1
l.append([d-h,d])
if(c == n):
print(-1)
else:
l.sort()
s = sorted(l, key = operator.itemgetter(1))
#print(s[-1],l[-1])
d = l[-1][0]
s1 = s[-1][1]
if(s1 >= x):
print("1")
else:
count = (x - s1)//d
#print(count)
if((x-s1)%d != 0):
count += 1
print(count+1)
```
|
instruction
| 0
| 23,932
| 2
| 47,864
|
No
|
output
| 1
| 23,932
| 2
| 47,865
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fighting with Zmei Gorynich — a ferocious monster from Slavic myths, a huge dragon-like reptile with multiple heads!
<image>
Initially Zmei Gorynich has x heads. You can deal n types of blows. If you deal a blow of the i-th type, you decrease the number of Gorynich's heads by min(d_i, curX), there curX is the current number of heads. But if after this blow Zmei Gorynich has at least one head, he grows h_i new heads. If curX = 0 then Gorynich is defeated.
You can deal each blow any number of times, in any order.
For example, if curX = 10, d = 7, h = 10 then the number of heads changes to 13 (you cut 7 heads off, but then Zmei grows 10 new ones), but if curX = 10, d = 11, h = 100 then number of heads changes to 0 and Zmei Gorynich is considered defeated.
Calculate the minimum number of blows to defeat Zmei Gorynich!
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.
The first line of each query contains two integers n and x (1 ≤ n ≤ 100, 1 ≤ x ≤ 10^9) — the number of possible types of blows and the number of heads Zmei initially has, respectively.
The following n lines of each query contain the descriptions of types of blows you can deal. The i-th line contains two integers d_i and h_i (1 ≤ d_i, h_i ≤ 10^9) — the description of the i-th blow.
Output
For each query print the minimum number of blows you have to deal to defeat Zmei Gorynich.
If Zmei Gorynuch cannot be defeated print -1.
Example
Input
3
3 10
6 3
8 2
1 4
4 10
4 1
3 2
2 6
1 100
2 15
10 11
14 100
Output
2
3
-1
Note
In the first query you can deal the first blow (after that the number of heads changes to 10 - 6 + 3 = 7), and then deal the second blow.
In the second query you just deal the first blow three times, and Zmei is defeated.
In third query you can not defeat Zmei Gorynich. Maybe it's better to convince it to stop fighting?
Submitted Solution:
```
import math
for _ in range(int(input())):
n, x = map(int, input().split())
a = []
md = -1
for i in range(n):
d, h = map(int, input().split())
md = max(md, d)
a.append(d-h)
a = max(a)
x = x - md
if a <= 0:
print(-1)
else:
if x > 0:
s = (x+a-1)//a + 1
s = 1
print(s)
```
|
instruction
| 0
| 23,933
| 2
| 47,866
|
No
|
output
| 1
| 23,933
| 2
| 47,867
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fighting with Zmei Gorynich — a ferocious monster from Slavic myths, a huge dragon-like reptile with multiple heads!
<image>
Initially Zmei Gorynich has x heads. You can deal n types of blows. If you deal a blow of the i-th type, you decrease the number of Gorynich's heads by min(d_i, curX), there curX is the current number of heads. But if after this blow Zmei Gorynich has at least one head, he grows h_i new heads. If curX = 0 then Gorynich is defeated.
You can deal each blow any number of times, in any order.
For example, if curX = 10, d = 7, h = 10 then the number of heads changes to 13 (you cut 7 heads off, but then Zmei grows 10 new ones), but if curX = 10, d = 11, h = 100 then number of heads changes to 0 and Zmei Gorynich is considered defeated.
Calculate the minimum number of blows to defeat Zmei Gorynich!
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.
The first line of each query contains two integers n and x (1 ≤ n ≤ 100, 1 ≤ x ≤ 10^9) — the number of possible types of blows and the number of heads Zmei initially has, respectively.
The following n lines of each query contain the descriptions of types of blows you can deal. The i-th line contains two integers d_i and h_i (1 ≤ d_i, h_i ≤ 10^9) — the description of the i-th blow.
Output
For each query print the minimum number of blows you have to deal to defeat Zmei Gorynich.
If Zmei Gorynuch cannot be defeated print -1.
Example
Input
3
3 10
6 3
8 2
1 4
4 10
4 1
3 2
2 6
1 100
2 15
10 11
14 100
Output
2
3
-1
Note
In the first query you can deal the first blow (after that the number of heads changes to 10 - 6 + 3 = 7), and then deal the second blow.
In the second query you just deal the first blow three times, and Zmei is defeated.
In third query you can not defeat Zmei Gorynich. Maybe it's better to convince it to stop fighting?
Submitted Solution:
```
def main():
t=int(input())
allans=[]
for _ in range(t):
n,x=readIntArr()
d=[]
h=[]
for _ in range(n):
dd,hh=readIntArr()
d.append(dd)
h.append(hh)
maxDamage=max(d)
if x-maxDamage<=0:
allans.append(0)
continue
imaxdiff=0
for i in range(n):
if d[i]-h[i]>d[imaxdiff]-h[imaxdiff]:
imaxdiff=i
if d[imaxdiff]-h[imaxdiff]<=0:
allans.append(-1)
continue
dd=d[imaxdiff]
hh=h[imaxdiff]
b=1000
k=0
while b>0:
while (dd-hh)*(k+b)+maxDamage<x:
k+=b
b//=2
k+=2
allans.append(k)
multiLineArrayPrint(allans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(i,j):
print('? {} {}'.format(i,j))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(' '.join([str(x) for x in ans])))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
```
|
instruction
| 0
| 23,934
| 2
| 47,868
|
No
|
output
| 1
| 23,934
| 2
| 47,869
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fighting with Zmei Gorynich — a ferocious monster from Slavic myths, a huge dragon-like reptile with multiple heads!
<image>
Initially Zmei Gorynich has x heads. You can deal n types of blows. If you deal a blow of the i-th type, you decrease the number of Gorynich's heads by min(d_i, curX), there curX is the current number of heads. But if after this blow Zmei Gorynich has at least one head, he grows h_i new heads. If curX = 0 then Gorynich is defeated.
You can deal each blow any number of times, in any order.
For example, if curX = 10, d = 7, h = 10 then the number of heads changes to 13 (you cut 7 heads off, but then Zmei grows 10 new ones), but if curX = 10, d = 11, h = 100 then number of heads changes to 0 and Zmei Gorynich is considered defeated.
Calculate the minimum number of blows to defeat Zmei Gorynich!
You have to answer t independent queries.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of queries.
The first line of each query contains two integers n and x (1 ≤ n ≤ 100, 1 ≤ x ≤ 10^9) — the number of possible types of blows and the number of heads Zmei initially has, respectively.
The following n lines of each query contain the descriptions of types of blows you can deal. The i-th line contains two integers d_i and h_i (1 ≤ d_i, h_i ≤ 10^9) — the description of the i-th blow.
Output
For each query print the minimum number of blows you have to deal to defeat Zmei Gorynich.
If Zmei Gorynuch cannot be defeated print -1.
Example
Input
3
3 10
6 3
8 2
1 4
4 10
4 1
3 2
2 6
1 100
2 15
10 11
14 100
Output
2
3
-1
Note
In the first query you can deal the first blow (after that the number of heads changes to 10 - 6 + 3 = 7), and then deal the second blow.
In the second query you just deal the first blow three times, and Zmei is defeated.
In third query you can not defeat Zmei Gorynich. Maybe it's better to convince it to stop fighting?
Submitted Solution:
```
import math
t = int(input())
for i in range(t):
delta = input()
delta = delta.split()
for i in range(2):
delta[i]=int(delta[i])
heads = delta[1]
m = 0
s = []
for i in range(delta[0]):
gamma = input()
gamma = gamma.split()
for i in range(2):
gamma[i]=int(gamma[i])
s.append(gamma[0]-gamma[1])
if(gamma[0]>m):
m = gamma[0]
s.sort()
val = s[-1]
if(heads<m):
print(1)
continue
if val<=0:
print(-1)
continue
# moves = 0
heads -=m
moves = math.ceil(heads/m)+1
print(moves)
```
|
instruction
| 0
| 23,935
| 2
| 47,870
|
No
|
output
| 1
| 23,935
| 2
| 47,871
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,048
| 2
| 48,096
|
Tags: data structures, greedy
Correct Solution:
```
import os, sys
from io import BytesIO, IOBase
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")
mod = 998244353
from math import ceil, gcd
from bisect import bisect_right,bisect_left
from _collections import deque
import heapq
n=int(input())
a=list(map(int,input().split()))
ans=[]
heapq.heapify(ans)
tot=0
ct=0
for i in a:
tot+=i
ct+=1
if i<0:
heapq.heappush(ans,i)
while tot<0:
t=heapq.heappop(ans)
tot+=abs(t)
ct-=1
print(ct)
```
|
output
| 1
| 24,048
| 2
| 48,097
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,049
| 2
| 48,098
|
Tags: data structures, greedy
Correct Solution:
```
#Fast I/O
import sys,os
import math
# To enable the file I/O i the below 2 lines are uncommented.
# read from in.txt if uncommented
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
# will print on Console if file I/O is not activated
#if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
# inputs template
from io import BytesIO, IOBase
def main():
import heapq
n=int(input())
arr=list(MI())
count=0
heap=[]
s=0
for i in arr:
if i>=0:
s+=i
count+=1
elif s+i>=0:
s+=i
heapq.heappush(heap,i)
count+=1
else:
if heap and heap[0]<i:
s-=heapq.heappop(heap)
heapq.heappush(heap,i)
s+=i
print(count)
# Sample Inputs/Output
# region fastio
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")
#for array of integers
def MI():return (map(int,input().split()))
# endregion
#for fast output, always take string
def outP(var): sys.stdout.write(str(var)+'\n')
# end of any user-defined functions
MOD=10**9+7
mod=998244353
# main functions for execution of the program.
if __name__ == '__main__':
#This doesn't works here but works wonders when submitted on CodeChef or CodeForces
main()
```
|
output
| 1
| 24,049
| 2
| 48,099
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,050
| 2
| 48,100
|
Tags: data structures, greedy
Correct Solution:
```
import heapq
n=int(input())
arr=list(map(int,input().split()))
heap=[]
heapq.heapify(heap)
mxm=0
count=0
for i in arr:
heapq.heappush(heap,i)
mxm+=i
if mxm>=0:
count+=1
elif mxm<0:
val=heapq.heappop(heap)
mxm+=abs(val)
print(count)
```
|
output
| 1
| 24,050
| 2
| 48,101
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,051
| 2
| 48,102
|
Tags: data structures, greedy
Correct Solution:
```
import sys
input=sys.stdin.readline
import heapq
n=int(input())
a=list(map(int,input().split()))
cnt=0
hq=[]
s=0
for i in range(n):
if a[i]>=0:
s+=a[i]
cnt+=1
else:
if s+a[i]>=0:
heapq.heappush(hq,a[i])
s+=a[i]
cnt+=1
else:
if len(hq)==0:
continue
else:
x=heapq.heappop(hq)
if x<a[i]:
heapq.heappush(hq,a[i])
s=s-x+a[i]
else:
heapq.heappush(hq,x)
print(cnt)
```
|
output
| 1
| 24,051
| 2
| 48,103
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,052
| 2
| 48,104
|
Tags: data structures, greedy
Correct Solution:
```
import heapq #heapq similar to priority queue
H = [] #list h can be converted to heapq later
#heapify() function is used to convert list into heap
n=int(input())
for i in list(map(int,input().split())):
tot = (tot+i if 'tot' in locals() else i)
#here adding we are current element
#globals() gove all variables in program
tot -= ((heapq.heappush(H,i) if tot >= 0 else heapq.heappushpop(H,i)) or 0)
#if sum<0 we will pop least element uptil now
print(len(H))
```
|
output
| 1
| 24,052
| 2
| 48,105
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,053
| 2
| 48,106
|
Tags: data structures, greedy
Correct Solution:
```
#===========Template===============
from io import BytesIO, IOBase
from math import sqrt
import sys,os
from os import path
inpl=lambda:list(map(int,input().split()))
inpm=lambda:map(int,input().split())
inpi=lambda:int(input())
inp=lambda:input()
rev,ra,l=reversed,range,len
P=print
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def B(n):
return bin(n).replace("0b","")
def factors(n):
arr=[]
for i in ra(2,int(sqrt(n))+1):
if n%i==0:
arr.append(i)
if i*i!=n:
arr.append(int(n/i))
return arr
def prime_factors(n):
omap={2:0}
while n!=1 and n%2==0:
omap[2]+=1
n/=2
for i in ra(3,int(math.sqrt(n))+1,2):
while n!=1 and n%i==0:
if i not in omap:omap[i]=1
else:omap[i]+=1
n/=i
if n!=1:
omap[int(n)]=1
return omap
def dfs(arr,n):
pairs=[[i+1] for i in ra(n)]
vis=[0]*(n+1)
for i in ra(l(arr)):
pairs[arr[i][0]-1].append(arr[i][1])
pairs[arr[i][1]-1].append(arr[i][0])
comp=[]
for i in ra(n):
stack=[pairs[i]]
temp=[]
if vis[stack[-1][0]]==0:
while len(stack)>0:
if vis[stack[-1][0]]==0:
temp.append(stack[-1][0])
vis[stack[-1][0]]=1
s=stack.pop()
for j in s[1:]:
if vis[j]==0:
stack.append(pairs[j-1])
else:
stack.pop()
comp.append(temp)
return comp
#=========I/p O/p ========================================#
from bisect import bisect_left as bl
from bisect import bisect_right as br
import sys,operator,math,operator
from collections import Counter
if(path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
import random,heapq
#==============To chaliye shuru krte he ====================#
n=inpi()
li=inpl()
heap=[]
heapq.heapify(heap)
su=0
for i in ra(n):
if su+li[i]<0:
if l(heap)>0:
ele=heapq.heappop(heap)
if ele<li[i]:
su-=ele
su+=li[i]
heapq.heappush(heap,li[i])
else:
heapq.heappush(heap,ele)
else:
su+=li[i]
heapq.heappush(heap,li[i])
P(l(heap))
```
|
output
| 1
| 24,053
| 2
| 48,107
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,054
| 2
| 48,108
|
Tags: data structures, greedy
Correct Solution:
```
# aadiupadhyay
from heapq import heappop
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
from heapq import *
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def solve():
n = inp()
l = li()
h = []
cur, ans = 0, 0
for i in l:
cur += i
ans += 1
if i < 0:
heappush(h,i)
if cur < 0:
cur -= heappop(h)
ans -= 1
pr(ans)
for _ in range(1):
solve()
```
|
output
| 1
| 24,054
| 2
| 48,109
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
|
instruction
| 0
| 24,055
| 2
| 48,110
|
Tags: data structures, greedy
Correct Solution:
```
import heapq
H = []
n = int(input())
for i in list(map(int,input().split(' '))):
tot = (tot + i if 'tot' in locals() or 'tot' in globals() else i)
tot -= ((heapq.heappush(H, i) if tot >= 0 else heapq.heappushpop(H, i)) or 0)
print(len(H))
```
|
output
| 1
| 24,055
| 2
| 48,111
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
from heapq import heapify, heappush, heappop
n = int(input())
l = list(map(int,input().split()))
heap = []
heapify(heap)
ans = 0
a = 0
for i in range(n):
a+=l[i]
ans+=1
heappush(heap,l[i])
while a<0:
a-=heappop(heap)
ans-=1
print(ans)
```
|
instruction
| 0
| 24,056
| 2
| 48,112
|
Yes
|
output
| 1
| 24,056
| 2
| 48,113
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
#DaRk DeveloPer
import sys
# taking input as string
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int, sys.stdin.readline().rstrip("\r\n").split()))
mod = 10 ** 9 + 7;
Mod = 998244353;
INF = float('inf')
# ______________________________________________________________________________________________________
from math import *
from bisect import *
from heapq import *
from collections import defaultdict as dd
from collections import OrderedDict as odict
from collections import Counter as cc
from collections import deque
from itertools import groupby
sys.setrecursionlimit(20 * 20 * 20 * 20 + 10) # this is must for dfs
MAX = 10 ** 5
def solve():
n=takein()
arr=takeiar()
sum=0
heap=[]
count=0
for i in arr:
if i<0:
heappush(heap,i)
sum+=i
count+=1
while sum<0:
sum+=abs(heappop(heap))
count-=1
print(count)
return
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt", "r")
sys.stdout = open("output.txt", "w")
t = 1
#t = takein()
# t = 1
for tt in range(1, t + 1):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :", time.time() - start_time, "seconds")
sys.stdout.close()
# ---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def takein():
return (int(sys.stdin.readline().rstrip("\r\n")))
# input the string
def takesr():
return (sys.stdin.readline().rstrip("\r\n"))
# input int array
def takeiar():
return (list(map(int, sys.stdin.readline().rstrip("\r\n").split())))
# input string array
def takesar():
return (list(map(str, sys.stdin.readline().rstrip("\r\n").split())))
# innut values for the diffrent variables
def takeivr():
return (map(int, sys.stdin.readline().rstrip("\r\n").split()))
def takesvr():
return (map(str, sys.stdin.readline().rstrip("\r\n").split()))
# ------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def ispalindrome(s):
return s == s[::-1]
def invert(bit_s):
# convert binary string
# into integer
temp = int(bit_s, 2)
# applying Ex-or operator
# b/w 10 and 31
inverse_s = temp ^ (2 ** (len(bit_s) + 1) - 1)
# convert the integer result
# into binary result and then
# slicing of the '0b1'
# binary indicator
rslt = bin(inverse_s)[3:]
return str(rslt)
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return (q)
def counter_elements(a):
q = dict()
for i in range(len(a)):
if a[i] not in q:
q[a[i]] = 0
q[a[i]] = q[a[i]] + 1
return (q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return (q)
def factorial(n, m=1000000007):
q = 1
for i in range(n):
q = (q * (i + 1)) % m
return (q)
def factors(n):
q = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return (list(sorted(list(set(q)))))
def prime_factors(n):
q = []
while n % 2 == 0: q.append(2); n = n // 2
for i in range(3, int(n ** 0.5) + 1, 2):
while n % i == 0: q.append(i); n = n // i
if n > 2: q.append(n)
return (list(sorted(q)))
def transpose(a):
n, m = len(a), len(a[0])
b = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
return (b)
def power_two(x):
return (x and (not (x & (x - 1))))
def ceil(a, b):
return -(-a // b)
def seive(n):
a = [1]
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p ** 2, n + 1, p):
prime[i] = False
p = p + 1
for p in range(2, n + 1):
if prime[p]:
a.append(p)
return (a)
def isprime(n):
if (n > 2 and not n % 2) or n == 1:
return False
for i in range(3, int(n ** 0.5 + 1), 2):
if not n % i:
return False
return True
# -----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
input = sys.stdin.readline
main()
```
|
instruction
| 0
| 24,057
| 2
| 48,114
|
Yes
|
output
| 1
| 24,057
| 2
| 48,115
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
import heapq
t = int(input())
a_list = [int(k) for k in input().split()]
heap = []
health = 0
res = 0
for v in a_list:
health += v
res +=1
heapq.heappush(heap,v)
while health<0:
health -= heapq.heappop(heap)
res -= 1
print(len(heap))
```
|
instruction
| 0
| 24,058
| 2
| 48,116
|
Yes
|
output
| 1
| 24,058
| 2
| 48,117
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
import heapq, sys
input = sys.stdin.readline
n, vals, drunk, curr = int(input()), [int(i) for i in input().split()], [], 0
for i in vals:
curr += i # push to heap regardless
heapq.heappush(drunk, i)
if curr < 0: # remove most negative
curr -= heapq.heappop(drunk)
print(len(drunk))
```
|
instruction
| 0
| 24,059
| 2
| 48,118
|
Yes
|
output
| 1
| 24,059
| 2
| 48,119
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
from sys import stdin,stdout,setrecursionlimit
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C,deque
from math import ceil,gcd,sqrt,factorial,log2,floor
from bisect import bisect_right as br,bisect_left as bl
import heapq
n = it()
l = mp()
x = []
heapq.heapify(x)
ans,res = 0,0
for i in range(n):
if ans + l[i] < 0:
if not len(x):
continue
z = heapq.heappop(x)
if abs(z) > abs(l[i]):
ans += (abs(z)+l[i])
heapq.heappush(x,l[i])
else:
res += 1
ans += l[i]
if abs(l[i]) != l[i]:
heapq.heappush(x,l[i])
print(res)
```
|
instruction
| 0
| 24,060
| 2
| 48,120
|
No
|
output
| 1
| 24,060
| 2
| 48,121
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
n=int(input())
num = list(map(int,input().strip().split()))[:n]
num.sort(reverse=True)
sum=0
count=0
for i in range(n):
if(sum+num[i]>0):
sum=sum+num[i]
count=count+1
print(count)
```
|
instruction
| 0
| 24,061
| 2
| 48,122
|
No
|
output
| 1
| 24,061
| 2
| 48,123
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
n = int(input())
st = str(input())
mass = st.split(" ")
pos = []
neg = []
checker = []
for i in range(len(mass)):
mass[i] = int(mass[i])
if mass[i] >= 0: pos.append(mass[i])
else: neg.append(mass[i])
summ = 0
for i in mass:
if i >= 0: summ += i
else:
if summ >= abs(i): checker.append(True)
else: checker.append(False)
out = 0
while summ >= 0 and len(neg) != 0:
minn = neg.index(max(neg))
if summ >= abs(neg[minn]) and checker[minn] == True:
summ += neg[minn]
neg.pop(minn)
checker.pop(minn)
out += 1
else: break
print(out + len(pos))
```
|
instruction
| 0
| 24,062
| 2
| 48,124
|
No
|
output
| 1
| 24,062
| 2
| 48,125
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The only difference is that in this version n ≤ 200000. You can make hacks only if both versions of the problem are solved.
There are n potions in a line, with potion 1 on the far left and potion n on the far right. Each potion will increase your health by a_i when drunk. a_i can be negative, meaning that potion will decrease will health.
You start with 0 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.
What is the largest number of potions you can drink?
Input
The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of potions.
The next line contains n integers a_1, a_2, ... ,a_n (-10^9 ≤ a_i ≤ 10^9) which represent the change in health after drinking that potion.
Output
Output a single integer, the maximum number of potions you can drink without your health becoming negative.
Example
Input
6
4 -4 1 -3 1 -3
Output
5
Note
For the sample, you can drink 5 potions by taking potions 1, 3, 4, 5 and 6. It is not possible to drink all 6 potions because your health will go negative at some point
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
l.sort()
while(sum(l)<0):
del(l[0])
print(len(l))
```
|
instruction
| 0
| 24,063
| 2
| 48,126
|
No
|
output
| 1
| 24,063
| 2
| 48,127
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,342
| 2
| 48,684
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
import sys
count = 0
r_list = []
dic = {}
for line in sys.stdin:
if count == 0:
n = int(line.strip().split(' ')[0])
k = int(line.strip().split(' ')[1])
count += 1
continue
count += 1
if count == n + 2:
answer_len = len(line.strip())
break
tmp = line.strip()
dic[len(tmp)] = dic.get(len(tmp), 0) + 1
sorted_dic = sorted(dic.items(), key=lambda x:x[1])
num = 0
for tmp in sorted_dic:
if tmp[0] < answer_len:
num += tmp[1]
min = num + 1
max = num + dic[answer_len]
min_t = ((min - 1) // k)*5 + min
max_t = ((max - 1) // k)*5 + max
print(str(min_t) + ' '+ str(max_t))
```
|
output
| 1
| 24,342
| 2
| 48,685
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,343
| 2
| 48,686
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
n, k = list(map(int, input().split()))
pwd = []
for _ in range(n):
pwd.append(input())
pwd = list(set(pwd))
corr = input()
mn = len([p for p in pwd if len(p)<len(corr)])
mx = len([p for p in pwd if len(p)<=len(corr)])
print(1+mn + (mn//k)*5, mx+((mx-1)//k)*5)
```
|
output
| 1
| 24,343
| 2
| 48,687
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,344
| 2
| 48,688
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
n, k = map(int, input().split(' '))
ss = [input() for _ in range(n)]
pl = len(input())
def time(i): return i + (i-1) // k * 5
print(time(sum(len(s) < pl for s in ss)+1), time(sum(len(s) <= pl for s in ss)))
```
|
output
| 1
| 24,344
| 2
| 48,689
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,345
| 2
| 48,690
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
def f(l,k,p):
n = len(l)
ll = [len(s) for s in l]
ll.sort()
cl = len(p)
fi = ll.index(cl)
li = fi
while li<n and ll[li]==cl:
li += 1
li = li-1
return [1+i+5*(i//k) for i in [fi,li]]
n,k = list(map(int,input().split()))
l = [input() for _ in range(n)]
p = input()
print(*f(l,k,p))
```
|
output
| 1
| 24,345
| 2
| 48,691
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,346
| 2
| 48,692
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
n,k = map(int, input().split())
lst = []
for i in range(n):
pwd = input()
lst.append(len(pwd))
lst.sort()
codehorses = len(input())
left = lst.index(codehorses)
right = left+lst.count(codehorses)-1
mi = left+1 + (left//k)*5
ma = right+1 + (right//k)*5
print(mi,ma)
```
|
output
| 1
| 24,346
| 2
| 48,693
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,347
| 2
| 48,694
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
n, k = list(map(int, input().split()))
lst = []
for i in range(n+1):
s = input()
lst.append(s)
vanya_password = lst[-1]
lst.pop(-1)
best_case = 0
worst_case = 0
count1 = 0
count2 = 0
for i in range(len(lst)):
if(len(lst[i]) < len(vanya_password)):
count1+=1
#for j in range(len(lst)):
if(len(lst[i]) <= len(vanya_password)):
count2+=1
#print(count1,count2)
best_case = count1 + (count1//k)*5 + 1
worst_case = count2 + ((count2-1)//k)*5
print('{0} {1}'.format(best_case,worst_case))
#print(s)
```
|
output
| 1
| 24,347
| 2
| 48,695
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,348
| 2
| 48,696
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
n, k = map(int, input().split())
p = [input() for i in range(n)]
L = len(input())
p.sort(key = lambda x: len(x))
c = 0
minc = 0
maxc = 0
for i in p:
le = len(i)
if le < L:
maxc+= 1
minc+= 1
if le == L:
maxc += 1
elif le > L:
break
print(minc+1 + ((minc)//k)*5, maxc + ((maxc - 1)//k)*5)
```
|
output
| 1
| 24,348
| 2
| 48,697
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords.
Output
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Examples
Input
5 2
cba
abc
bb1
abC
ABC
abc
Output
1 15
Input
4 100
11
22
1
2
22
Output
3 4
Note
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
instruction
| 0
| 24,349
| 2
| 48,698
|
Tags: implementation, math, sortings, strings
Correct Solution:
```
n, k = map(int, input().split())
L = [input() for i in range(n)]
p = input()
res, tr, MAX = 0, 0, 0
L.sort(key = lambda x: len(x))
for i in range(n):
if len(L[i]) < len(p):
res += 1
tr += 1
if tr == k:
res += 5
tr = 0
elif len(L[i]) == len(p):
MAX += 1
tr += 1
if tr == k:
MAX += 5
tr = 0
else:
break
if tr == 0: MAX -= 5
print(res + 1, res + MAX)
```
|
output
| 1
| 24,349
| 2
| 48,699
|
Provide a correct Python 3 solution for this coding contest problem.
problem
The point $ P $ is placed at the origin on the coordinate plane. I want to move the point $ P $ to a position where the Manhattan distance from the origin is as far as possible.
First, the string $ S = s_1s_2 \ cdots s_ {| S |} $ ($ | S | $ is the number of characters in $ S $) is given. The point $ P $ is moved by reading the characters one by one from the beginning of the character string $ S $. The string $ S $ consists of the letters'U',' L',' D', and'R'. When each character is read, if the coordinates of the point $ P $ before movement are $ (x, y) $, the coordinates of the point $ P $ after movement are $ (x, y + 1) and \ (, respectively. It becomes x-1, y), \ (x, y-1), \ (x + 1, y) $.
Immediately before reading each character, you can choose whether or not to cast magic. There are two types of magic, magic 1 and magic 2. Assuming that the $ i $ th character of the character string $ S $ is $ s_i $, the change when magic is applied immediately before reading $ s_i $ is as follows.
* When magic 1 is applied: For all $ s_j \ (i \ le j \ le | S |) $, replace'U'with'D'and'D' with'U'.
* When magic 2 is applied: For all $ s_j \ (i \ le j \ le | S |) $, replace'L'with'R'and'R' with'L'.
Intuitively, Magic 1 can reverse the subsequent treatment of the top and bottom, and Magic 2 can reverse the treatment of the left and right. The number of times the magic is applied before reading a certain character may be multiple times. You can also apply both spells in succession. However, the total number of times that magic can be applied before reading all the characters in the character string $ S $ is $ K $. See the sample for details.
Find the maximum value of $ | x'| + | y'| $, where $ (x', y') $ is the coordinate of the point $ P $ after reading all the characters in the string $ S $. ..
input
Input is given from standard input in the following format.
$ S $
$ K $
output
Output the maximum value of $ | x'| + | y'| $ in one line. Also, output a line break at the end.
Example
Input
RRLUDDD
2
Output
7
|
instruction
| 0
| 24,703
| 2
| 49,406
|
"Correct Solution:
```
def main():
s = input()
k = int(input())
lrs = []
uds = []
for c in s:
if c == "R":
lrs.append(1)
if c == "L":
lrs.append(-1)
if c == "U":
uds.append(-1)
if c == "D":
uds.append(1)
lr_length = len(lrs)
ud_length = len(uds)
lr_dp1 = [0] * (lr_length + 1)
lr_dp2 = [0] * (lr_length + 1)
for i in lrs:
for j in range(lr_length, 0, -1):
lr_dp1[j] = max(lr_dp1[j], lr_dp1[j - 1]) + i * (-1) ** j
lr_dp2[j] = min(lr_dp2[j], lr_dp2[j - 1]) + i * (-1) ** j
lr_dp1[0] += i
lr_dp2[0] += i
ud_dp1 = [0] * (ud_length + 1)
ud_dp2 = [0] * (ud_length + 1)
for i in uds:
for j in range(ud_length, 0, -1):
ud_dp1[j] = max(ud_dp1[j], ud_dp1[j - 1]) + i * (-1) ** j
ud_dp2[j] = min(ud_dp2[j], ud_dp2[j - 1]) + i * (-1) ** j
ud_dp1[0] += i
ud_dp2[0] += i
lr_acc = [abs(lr_dp1[0])]
for i in range(1, lr_length + 1):
lr_acc.append(max(lr_acc[-1], abs(lr_dp1[i]), abs(lr_dp2[i])))
ud_acc = [abs(ud_dp1[0])]
for i in range(1, ud_length + 1):
ud_acc.append(max(ud_acc[-1], abs(ud_dp1[i]), abs(ud_dp2[i])))
ans = 0
for i in range(min(k + 1, lr_length + 1)):
ans = max(ans, lr_acc[i] + ud_acc[min(k - i, ud_length)])
print(ans)
main()
```
|
output
| 1
| 24,703
| 2
| 49,407
|
Provide a correct Python 3 solution for this coding contest problem.
problem
The point $ P $ is placed at the origin on the coordinate plane. I want to move the point $ P $ to a position where the Manhattan distance from the origin is as far as possible.
First, the string $ S = s_1s_2 \ cdots s_ {| S |} $ ($ | S | $ is the number of characters in $ S $) is given. The point $ P $ is moved by reading the characters one by one from the beginning of the character string $ S $. The string $ S $ consists of the letters'U',' L',' D', and'R'. When each character is read, if the coordinates of the point $ P $ before movement are $ (x, y) $, the coordinates of the point $ P $ after movement are $ (x, y + 1) and \ (, respectively. It becomes x-1, y), \ (x, y-1), \ (x + 1, y) $.
Immediately before reading each character, you can choose whether or not to cast magic. There are two types of magic, magic 1 and magic 2. Assuming that the $ i $ th character of the character string $ S $ is $ s_i $, the change when magic is applied immediately before reading $ s_i $ is as follows.
* When magic 1 is applied: For all $ s_j \ (i \ le j \ le | S |) $, replace'U'with'D'and'D' with'U'.
* When magic 2 is applied: For all $ s_j \ (i \ le j \ le | S |) $, replace'L'with'R'and'R' with'L'.
Intuitively, Magic 1 can reverse the subsequent treatment of the top and bottom, and Magic 2 can reverse the treatment of the left and right. The number of times the magic is applied before reading a certain character may be multiple times. You can also apply both spells in succession. However, the total number of times that magic can be applied before reading all the characters in the character string $ S $ is $ K $. See the sample for details.
Find the maximum value of $ | x'| + | y'| $, where $ (x', y') $ is the coordinate of the point $ P $ after reading all the characters in the string $ S $. ..
input
Input is given from standard input in the following format.
$ S $
$ K $
output
Output the maximum value of $ | x'| + | y'| $ in one line. Also, output a line break at the end.
Example
Input
RRLUDDD
2
Output
7
|
instruction
| 0
| 24,704
| 2
| 49,408
|
"Correct Solution:
```
# AOJ 2809: Graduation Ceremony
# Python3 2018.7.11 bal4u
MAX = 2002
dx = (0,1,0,-1) # URDL
dy = (-1,0,1,0)
tr = {'U':0, 'R':1, 'D':2, 'L': 3}
S = list(input())
K = int(input())
u, d, l, r = [0]*MAX, [0]*MAX, [0]*MAX, [0]*MAX
for s in S:
dir = tr[s]
j2 = j = K
while j > 0:
j -= 1
u[j2] = min(-d[j], u[j2])+dy[dir]
r[j2] = max(-l[j], r[j2])+dx[dir]
d[j2] = max(-u[j], d[j2])+dy[dir]
l[j2] = min(-r[j], l[j2])+dx[dir]
j2 -= 1
r[0] = l[0] = l[0]+dx[dir]
u[0] = d[0] = d[0]+dy[dir]
ans = 0;
for i in range(K+1):
j = max(-l[i], r[i])
j2 = max(-u[K-i], d[K-i])
j += j2;
if j > ans: ans = j
print(ans)
```
|
output
| 1
| 24,704
| 2
| 49,409
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
The point $ P $ is placed at the origin on the coordinate plane. I want to move the point $ P $ to a position where the Manhattan distance from the origin is as far as possible.
First, the string $ S = s_1s_2 \ cdots s_ {| S |} $ ($ | S | $ is the number of characters in $ S $) is given. The point $ P $ is moved by reading the characters one by one from the beginning of the character string $ S $. The string $ S $ consists of the letters'U',' L',' D', and'R'. When each character is read, if the coordinates of the point $ P $ before movement are $ (x, y) $, the coordinates of the point $ P $ after movement are $ (x, y + 1) and \ (, respectively. It becomes x-1, y), \ (x, y-1), \ (x + 1, y) $.
Immediately before reading each character, you can choose whether or not to cast magic. There are two types of magic, magic 1 and magic 2. Assuming that the $ i $ th character of the character string $ S $ is $ s_i $, the change when magic is applied immediately before reading $ s_i $ is as follows.
* When magic 1 is applied: For all $ s_j \ (i \ le j \ le | S |) $, replace'U'with'D'and'D' with'U'.
* When magic 2 is applied: For all $ s_j \ (i \ le j \ le | S |) $, replace'L'with'R'and'R' with'L'.
Intuitively, Magic 1 can reverse the subsequent treatment of the top and bottom, and Magic 2 can reverse the treatment of the left and right. The number of times the magic is applied before reading a certain character may be multiple times. You can also apply both spells in succession. However, the total number of times that magic can be applied before reading all the characters in the character string $ S $ is $ K $. See the sample for details.
Find the maximum value of $ | x'| + | y'| $, where $ (x', y') $ is the coordinate of the point $ P $ after reading all the characters in the string $ S $. ..
input
Input is given from standard input in the following format.
$ S $
$ K $
output
Output the maximum value of $ | x'| + | y'| $ in one line. Also, output a line break at the end.
Example
Input
RRLUDDD
2
Output
7
Submitted Solution:
```
a = input()
b=int(input())
RL = ['R','L']
UD = ['U','D']
RL_c = [0, 0]
UD_c = [0, 0]
for l in a:
try:
RL_c[RL.index(l)]+=1
except:
UD_c[UD.index(l)]+=1
if b>0:
for _ in range(len(a)):
if RL_c[0]!=0 and RL_c[1]!=0:
if RL_c[0]>=RL_c[1]:
RL_c[0]+=1
RL_c[1]-=1
else:
RL_c[0]-=1
RL_c[1]+=1
elif UD_c[0]!=0 and UD_c[1]!=0:
if UD_c[0]>=UD_c[1]:
UD_c[0]+=1
UD_c[1]-=1
else:
UD_c[0]-=1
UD_c[1]+=1
else:
break
b-=1
if b==0:
break
print(abs(RL_c[0]-RL_c[1])+abs(UD_c[0]-UD_c[1]))
```
|
instruction
| 0
| 24,705
| 2
| 49,410
|
No
|
output
| 1
| 24,705
| 2
| 49,411
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,023
| 2
| 50,046
|
Tags: greedy, sortings
Correct Solution:
```
def beat_dragons(s, n, dragons):
for i in range(n):
if s > dragons[i][0]:
s+=dragons[i][1]
else:
return False
return True
def main():
s,n = [int(i) for i in input().split()]
dragons = []
for i in range(n):
pair = input().split()
dragons.append((int(pair[0]), int(pair[1])))
dragons.sort()
print("YES") if beat_dragons(s, n, dragons) else print("NO")
main()
```
|
output
| 1
| 25,023
| 2
| 50,047
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,024
| 2
| 50,048
|
Tags: greedy, sortings
Correct Solution:
```
M = lambda : map(int,input().split())
s,n = M()
lvl = []
for i in range(n):
d,p = M()
lvl += [[d,p]]
flag = 1
for i in sorted(lvl) :
if s-i[0] <= 0:
flag = 0
s += i[1]
if flag:
print("YES")
else:
print("NO")
```
|
output
| 1
| 25,024
| 2
| 50,049
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,025
| 2
| 50,050
|
Tags: greedy, sortings
Correct Solution:
```
import sys
line=input().split()
s=int(line[0])
n=int(line[1])
d={}
ts=0
for i in range (n):
line=input().split()
x=int(line[0])
y=int(line[1])
while (x,ts) in d:
ts=ts+1
d[(x,ts)]=y
d_sorted=sorted(d)
for j in d_sorted:
if s>j[0]:
s=s+d[j]
else:
print('NO')
sys.exit()
print('YES')
```
|
output
| 1
| 25,025
| 2
| 50,051
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,026
| 2
| 50,052
|
Tags: greedy, sortings
Correct Solution:
```
# 230A. Dragons
s, n = [int(k) for k in input().split()]
lst = [0] * n
for i in range(n):
lst[i] = [int(k) for k in input().split()]
lst.sort()
for i in range(n):
if s <= lst[i][0]:
print("NO")
exit()
else:
s += lst[i][1]
if s > 0:
print("YES")
```
|
output
| 1
| 25,026
| 2
| 50,053
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,027
| 2
| 50,054
|
Tags: greedy, sortings
Correct Solution:
```
s,n=[int(i) for i in input( ).split( )]
dic={}
ts=0
for i in range(n):
x,y=[int(i) for i in input( ).split( )]
while(x,ts)in dic:
ts+=1
dic[(x,ts)]=y
dragon=sorted(dic)
for i in dragon:
stren=s-i[0]
if stren<=0:
print('NO')
exit( )
else:
s+=dic[i]
print('YES')
```
|
output
| 1
| 25,027
| 2
| 50,055
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,028
| 2
| 50,056
|
Tags: greedy, sortings
Correct Solution:
```
h=list(map(int,input().split()))
list1=[]
x=[]
flag=0
for i in range(h[1]):
a=tuple(map(int,input().split()))
list1.append(a)
x=sorted(list1)
for i in range(h[1]):
if h[0]>x[i][0]:
h[0]=h[0]+x[i][1]
flag+=1
else:
flag=0
break
if flag==0:
print("NO")
else:
print("YES")
```
|
output
| 1
| 25,028
| 2
| 50,057
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,029
| 2
| 50,058
|
Tags: greedy, sortings
Correct Solution:
```
s, n = map(int, input().split())
drag = []
for i in range(n):
drag.append(tuple(map(int, input().split())))
drag.sort(key=lambda x: x[0])
for d, x in drag:
if s <= d:
print("NO")
exit()
s += x
print("YES")
```
|
output
| 1
| 25,029
| 2
| 50,059
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
instruction
| 0
| 25,030
| 2
| 50,060
|
Tags: greedy, sortings
Correct Solution:
```
s,n=map(int,input().split())
D=[]
for i in range(n):
a,b=map(int,input().split())
D.append([a,b])
D.sort()
done=False
for i in range(n):
if(s>D[i][0]):
s+=D[i][1]
else:
print("NO")
done=True
break
if( not done):
print("YES")
```
|
output
| 1
| 25,030
| 2
| 50,061
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
kripower,nom=map(int,input().split())
matches=[]
order=[]
sortedorder=[]
count=0
for i in range(nom):
matches.append(list(map(int,input().split())))
order.append(matches[i][0])
for i in range(nom):
mini=order[0]
index=0
for i in range(1,nom):
if order[i]<mini:
mini=order[i]
index=i
sortedorder.append(index)
order[index]=100000
for k in range(nom):
a=sortedorder[k]
if kripower>matches[a][0]:
kripower+=matches[a][1]
else:
count=1
break
if count==1:
print ("NO")
else:
print ("YES")
```
|
instruction
| 0
| 25,031
| 2
| 50,062
|
Yes
|
output
| 1
| 25,031
| 2
| 50,063
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
s, n = map(int, input().split())
dragon=[]
flag = 1
for i in range(n):
x, y = map(int, input().split())
dragon+=[[x, y]]
#Sorting of a 2D list is done as any order Kirito will fight
dragon.sort(key=lambda x: x[0])
for i in range(n):
#Kirito's strength needs to be more
if s>dragon[i][0]:
s+=dragon[i][1]
else:
flag=0
break
if flag==0:
print("NO")
else:
print("YES")
```
|
instruction
| 0
| 25,032
| 2
| 50,064
|
Yes
|
output
| 1
| 25,032
| 2
| 50,065
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
s,n=map(int,input().split())
for x,y in sorted([list(map(int,input().split())) for i in range(n)],key=lambda x:x[0]):
if s<=x:print('NO');break
else:s+=y
else:print('YES')
```
|
instruction
| 0
| 25,033
| 2
| 50,066
|
Yes
|
output
| 1
| 25,033
| 2
| 50,067
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
s, n = map(int, input(). split())
b = True
a = [[0] * 2 for _ in range(n)]
for i in range(n):
a[i] = list(map(int, input(). split()))
a.sort()
for i in range(n):
if a[i][0] >= s:
print('NO')
b = False
break
else:
s += a[i][1]
if b:
print('YES')
```
|
instruction
| 0
| 25,034
| 2
| 50,068
|
Yes
|
output
| 1
| 25,034
| 2
| 50,069
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
s, n = [int(i) for i in input().split(" ")]
for i in range(n):
d_force, bonus = [int(i) for i in input().split(" ")]
if d_force >= s:
print("NO")
break
else:
s = s + bonus
else:
print("YES")
```
|
instruction
| 0
| 25,035
| 2
| 50,070
|
No
|
output
| 1
| 25,035
| 2
| 50,071
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
s,n = map(int,input().split())
for i in range(n):
opp, bonus = map(int,input().split())
if s>opp:
s+=bonus
else:
s=-1
break
if s>0:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 25,036
| 2
| 50,072
|
No
|
output
| 1
| 25,036
| 2
| 50,073
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
def MoveOn(strength, arr):
#Sort array
arr = sorted(arr, key=lambda x: x[0])
for i in arr:
if strength < int(i[0]):
return "NO"
else:
strength = strength + int(i[1])
return "YES"
if __name__ == "__main__":
n = input().split(" ")
arr = []
for i in range(0,int(n[1])):
arr.append(input().split(" "))
print(MoveOn(int(n[0]),arr))
```
|
instruction
| 0
| 25,037
| 2
| 50,074
|
No
|
output
| 1
| 25,037
| 2
| 50,075
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input
The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.
Output
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Examples
Input
2 2
1 99
100 0
Output
YES
Input
10 1
100 100
Output
NO
Note
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
Submitted Solution:
```
a,b=map(int,input().split())
m=[]
n=[]
for i in range(b):
x,y=map(int,input().split())
m.append(x)
n.append(y)
k=m.copy()
k.sort()
for i in k:
if a>i:
t=m.index(i)
a=a+n[t]
else:
print("NO")
break
else:
print("YES");
```
|
instruction
| 0
| 25,038
| 2
| 50,076
|
No
|
output
| 1
| 25,038
| 2
| 50,077
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have n layers. The first layer is hate, second one is love, third one is hate and so on...
For example if n = 1, then his feeling is "I hate it" or if n = 2 it's "I hate that I love it", and if n = 3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of layers of love and hate.
Output
Print Dr.Banner's feeling in one line.
Examples
Input
1
Output
I hate it
Input
2
Output
I hate that I love it
Input
3
Output
I hate that I love that I hate it
|
instruction
| 0
| 25,220
| 2
| 50,440
|
Tags: implementation
Correct Solution:
```
n = int(input(""))
a = "I hate "
b = "I love "
for i in range(1,n+1):
if i%2==1:
print(a,end = "")
if i != n:
print("that ",end = "")
if i%2==0:
print(b,end = "")
if i != n:
print("that ",end = "")
print("it")
```
|
output
| 1
| 25,220
| 2
| 50,441
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have n layers. The first layer is hate, second one is love, third one is hate and so on...
For example if n = 1, then his feeling is "I hate it" or if n = 2 it's "I hate that I love it", and if n = 3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
Input
The only line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of layers of love and hate.
Output
Print Dr.Banner's feeling in one line.
Examples
Input
1
Output
I hate it
Input
2
Output
I hate that I love it
Input
3
Output
I hate that I love that I hate it
|
instruction
| 0
| 25,221
| 2
| 50,442
|
Tags: implementation
Correct Solution:
```
n=int(input())
for i in range(1,n+1):
if(i%2):
print("I hate",end=" ")
else:
print("I love",end=" ")
if(i==n):
print("it")
else:
print("that",end=" ")
```
|
output
| 1
| 25,221
| 2
| 50,443
|
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