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int64 219
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob got a job as a system administrator in X corporation. His first task was to connect n servers with the help of m two-way direct connection so that it becomes possible to transmit data from one server to any other server via these connections. Each direct connection has to link two different servers, each pair of servers should have at most one direct connection. Y corporation, a business rival of X corporation, made Bob an offer that he couldn't refuse: Bob was asked to connect the servers in such a way, that when server with index v fails, the transmission of data between some other two servers becomes impossible, i.e. the system stops being connected. Help Bob connect the servers.
Input
The first input line contains 3 space-separated integer numbers n, m, v (3 ≤ n ≤ 105, 0 ≤ m ≤ 105, 1 ≤ v ≤ n), n — amount of servers, m — amount of direct connections, v — index of the server that fails and leads to the failure of the whole system.
Output
If it is impossible to connect the servers in the required way, output -1. Otherwise output m lines with 2 numbers each — description of all the direct connections in the system. Each direct connection is described by two numbers — indexes of two servers, linked by this direct connection. The servers are numbered from 1. If the answer is not unique, output any.
Examples
Input
5 6 3
Output
1 2
2 3
3 4
4 5
1 3
3 5
Input
6 100 1
Output
-1
Submitted Solution:
```
#! /usr/bin/python3
# SUBMISSION 1 (wrong answer on test 1): big dumb forgot to print answer
# SUBMISSION 2 (runtime error on test 2): big dumb again forget to handle if m out of bounds
# SUBMISSION 3 (TLE on test 10): need to optimize, maybe change handling of calculating factorial? SOLUTION: tried eliminating factorial calculation
import sys
import math
import copy
big = set()
small = []
def main():
input_list = get_input()
line = [int(x) for x in input_list[0].split(" ")]
solution = solve(line[0], line[1], line[2])
if solution != -1:
for i in solution:
print(i)
else:
print(-1)
def get_input():
input_list = []
for line in sys.stdin:
input_list.append(line.rstrip("\n"))
return input_list
def generate_sets(n, m, v):
global big
global small
if v == 1:
small = [2, 1]
for i in range(1, n + 1):
big.add(i)
big.remove(2)
big = list(big)
else:
small = [1, v]
for i in range(1, n + 1):
big.add(i)
big.remove(1)
big = list(big)
return None
def generate_basic_network(n, m, v):
global big
global small
edges = set()
temp = copy.deepcopy(big)
temp = set(temp)
temp.remove(v)
temp = list(temp)
basic = small + temp
for i in range(len(basic) - 1):
edges.add(str(basic[i]) + " " + str(basic[i + 1]))
return edges
def solve(n, m, v):
global big
global small
big = set()
small = []
edges = set()
# m_max = 1 + (math.factorial(n - 1) / (2 * math.factorial(n - 3)))
m_min = n - 1
if m >= m_min:
generate_sets(n, m, v)
edges = generate_basic_network(n, m, v)
for j in range(len(big)):
if len(edges) == m:
break
for k in range(len(big[j + 1:])):
if len(edges) == m:
break
edge = str(big[j]) + " " + str(big[j + 1:][k])
reverse_edge = str(big[j + 1:][k]) + " " + str(big[j])
if not edge in edges and not reverse_edge in edges:
edges.add(edge)
if len(edges) != m:
return -1
return edges
else:
return -1
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 11,923
| 11
| 23,846
|
Yes
|
output
| 1
| 11,923
| 11
| 23,847
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
def main():
n = int(input())
w = list(map(int, input().split()))
bits = [0] * (10 ** 6 + 100)
for e in w:
bits[e] += 1
cur, res = 0, 0
for e in bits:
cur += e
if cur % 2:
res += 1
cur //= 2
print(res)
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 12,065
| 11
| 24,130
|
Yes
|
output
| 1
| 12,065
| 11
| 24,131
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
bit = [0]*(10**6+101)
for i in l:
bit[i] = bit[i]+1
#print(bit[0],bit[1],bit[2],bit[3])
ans = 0
s = 0
for i in bit:
s = s+i
ans = ans+(s%2)
s = s//2
print(ans)
```
|
instruction
| 0
| 12,067
| 11
| 24,134
|
Yes
|
output
| 1
| 12,067
| 11
| 24,135
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from collections import Counter
n = int(input())
weights = sorted(list(map(int,input().split())))
counter = Counter(weights)
minus_counter = Counter()
ans = 0
for weight, count in sorted(counter.items()):
count -= max(0, minus_counter[2**weight])
if count > 0 and count // 2 > 0:
minus_counter[2**weight*(count//2)*2] += 1
count -= (count//2)*2
ans += 1
ans += count
print(ans)
```
|
instruction
| 0
| 12,068
| 11
| 24,136
|
No
|
output
| 1
| 12,068
| 11
| 24,137
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.
<image>
Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.
Input
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Output
Print the minimum number of steps in a single line.
Examples
Input
5
1 1 2 3 3
Output
2
Input
4
0 1 2 3
Output
4
Note
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from collections import Counter
n = int(input())
weights = sorted(list(map(int,input().split())))
counter = [0] * 10** 7
for weight, count in Counter(weights).items():
counter[weight] = count
ans = 0
for i in range(10**7):
weight, count = (i, counter[i])
if count > 0 and count // 2 > 0:
counter[weight+(count//2)] += 1
count -= (count//2)*2
ans += count
print(ans)
```
|
instruction
| 0
| 12,071
| 11
| 24,142
|
No
|
output
| 1
| 12,071
| 11
| 24,143
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
import math
import sys
while True:
try:
a,b,c,d,e,f = map(float, input().split())
if a*e == b*d:
continue
x = (c*e - b*f)/(a*e - b*d)
y = (c*d - a*f)/(b*d - a*e)
if x == 0:
x = 0
if y == 0:
y = 0
print('{0:.3f} {1:.3f}'.format(x,y))
except EOFError:
break
```
|
instruction
| 0
| 12,398
| 11
| 24,796
|
Yes
|
output
| 1
| 12,398
| 11
| 24,797
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
while(True):
try:
a,b,c,d,e,f = map(float, input().split())
y = (a*f-c*d)/(a*e-b*d)
x = (c-b*y)/a
print("%.3f %.3f"%(x,y))
except:
break
```
|
instruction
| 0
| 12,399
| 11
| 24,798
|
Yes
|
output
| 1
| 12,399
| 11
| 24,799
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
# coding: utf-8
import sys
for line in sys.stdin:
a, b, c, d, e, f = map(float, line.strip().split())
y = (c*d-a*f)/(b*d-a*e)
x = (c - b*y) / a
print("%.3lf %.3lf" % (x, y))
```
|
instruction
| 0
| 12,401
| 11
| 24,802
|
Yes
|
output
| 1
| 12,401
| 11
| 24,803
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
import sys
[print("{0[0]:.3f} {0[1]:.3f}".format([round(y, 3) for y in [(x[4] * x[2] - x[1] * x[5]) / (x[0] * x[4] - x[1] * x[3]), (x[0] * x[5] - x[2] * x[3]) / (x[0] * x[4] - x[1] * x[3])]])) for x in [[float(y) for y in x.split()] for x in sys.stdin]]
```
|
instruction
| 0
| 12,402
| 11
| 24,804
|
No
|
output
| 1
| 12,402
| 11
| 24,805
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
print('ok')
```
|
instruction
| 0
| 12,403
| 11
| 24,806
|
No
|
output
| 1
| 12,403
| 11
| 24,807
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
def jisuan(lis):
ns = lis.split(" ")
for i in range(6):
ns[i] = float(ns[i])
a, b, c, d, e, f = ns[0], ns[1], ns[2], ns[3], ns[4], ns[5]
x = (c * e - b * f) / (a * e - b * d)
y = (c * d - a * f) / (b * d - a * e)
return [x, y]
while 1:
try:
inpu = str(input())
except:
break
nums = inpu.split("\n")
for l in range(len(nums)):
k0 = jisuan("1 2 3 4 5 6")[0]
k1 = jisuan("1 2 3 4 5 6")[1]
print("%.3f" % k0, "%.3f" % k1)
```
|
instruction
| 0
| 12,404
| 11
| 24,808
|
No
|
output
| 1
| 12,404
| 11
| 24,809
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which solve a simultaneous equation:
ax + by = c
dx + ey = f
The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
Input
The input consists of several data sets, 1 line for each data set. In a data set, there will be a, b, c, d, e, f separated by a single space. The input terminates with EOF.
Output
For each data set, print x and y separated by a single space. Print the solution to three places of decimals. Round off the solution to three decimal places.
Examples
Input
1 2 3 4 5 6
2 -1 -2 -1 -1 -5
Output
-1.000 2.000
1.000 4.000
Input
2 -1 -3 1 -1 -3
2 -1 -3 -9 9 27
Output
0.000 3.000
0.000 3.000
Submitted Solution:
```
# coding: utf-8
# Your code here!
import numpy as np
while True:
try:
s=input().split()
l=[int(s[0]),int(s[1]),int(s[3]),int(s[4])]
lb=[int(s[2]),int(s[5])]
l=np.dot(np.linalg.inv(np.array(l).reshape(2,2)),np.array(lb))
print('%.3f %.3f' % (l[0],l[1]))
except:
break
```
|
instruction
| 0
| 12,405
| 11
| 24,810
|
No
|
output
| 1
| 12,405
| 11
| 24,811
|
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
3
NNN
NNN
NNN
Output
Taro
|
instruction
| 0
| 12,442
| 11
| 24,884
|
"Correct Solution:
```
from collections import deque
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N = int(readline())
G = [[] for i in range(N)]
for i in range(N):
*vs, = readline()
for j in range(N):
if vs[j] == "Y":
G[i].append(j)
d = 0
C = [0, 0]
used = [0]*N
for i in range(N):
d += len(G[i])
if used[i]:
continue
used[i] = 1
que = deque([i])
c = 0
while que:
v = que.popleft()
c += 1
for w in G[v]:
if used[w]:
continue
used[w] = 1
que.append(w)
C[c % 2] += 1
r0 = (N*(N-1)//2 - d//2) & 1
memo = {}
def dfs(i, p, q):
key = (p, q)
if key in memo:
return memo[key]
if p+q == 2:
r = (q == 2) ^ r0
memo[key] = e = r ^ (i & 1)
return e
r = 0
if p > 1 or (p and q):
if dfs(i+1, p-1, q) == 0:
r = 1
if q > 1:
if dfs(i+1, p+1, q-2) == 0:
r = 1
memo[key] = r
return r
if dfs(0, C[0], C[1]):
write("Taro\n")
else:
write("Hanako\n")
solve()
```
|
output
| 1
| 12,442
| 11
| 24,885
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,569
| 11
| 25,138
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
c,l=0,[]
for i in range(n):
l.append(int(input()))
for i in l:
if i%2!=0:
c+=1
cnt,p=int(c/2),0
for i in l:
if i%2==0:
print(int(i/2))
elif i%2!=0 and p<cnt and i<0:
print(int(i/2))
p+=1
elif i%2!=0 and p<cnt and i>0:
print(int(i/2)+1)
p+=1
elif i%2!=0 and p>=cnt and i<0:
print(int(i/2)-1)
p+=1
else:
print(int(i/2))
```
|
output
| 1
| 12,569
| 11
| 25,139
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,570
| 11
| 25,140
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
st=1
for i in range(n):
a=int(input())
if a%2==0:
print(a//2)
continue
if a==0:
print(0)
if a<0:
if st==1:
print(a//2)
st=0
else:
print(a//2+1)
st=1
if a>0:
if st==1:
print(a//2)
st=0
else:
print(a//2+1)
st=1
```
|
output
| 1
| 12,570
| 11
| 25,141
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,571
| 11
| 25,142
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
l=[]
no=0
po=0
for i in range(n):
x=int(input())
if(x<0 and x%2==1):
no+=1
elif(x>0 and x%2==1):
po+=1
l.append(x)
ans=[]
no//=2
po//=2
for i in l:
x=i
if(x%2==1):
if(x<0 and no>0):
# print("....",x)
x-=1
no-=1
elif(x>0 and po>0):
x+=1
po-=1
# print(x,x//2)
if(x<0):
ans.append(-(abs(x)//2))
else:
ans.append(x//2)
for i in ans:
print(i)
```
|
output
| 1
| 12,571
| 11
| 25,143
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,572
| 11
| 25,144
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
l1=[]
odd=0
for i in range (n):
l1.append(int(input()))
l=[]
sum1=0
for i in range (n):
if(l1[i]%2==0):
sum1+=int(l1[i]/2)
l.append(int(l1[i]/2))
elif(l1[i]%2!=0 and odd==0):
l.append(int(l1[i]/2))
sum1+=int(l1[i]/2)
if(sum1==0):
for i in range (n):
print(l[i])
elif(sum1<0):
for i in range (n):
if(l1[i]%2!=0 and l1[i]>0):
sum1+=1
l[i]+=1
if(sum1==0):
break
for i in range (n):
print(l[i])
elif(sum1>0):
for i in range (n):
if(l1[i]%2!=0 and l1[i]<0):
sum1-=1
l[i]-=1
if(sum1==0):
break
for i in range (n):
print(l[i])
```
|
output
| 1
| 12,572
| 11
| 25,145
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,573
| 11
| 25,146
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
mod=1
for _ in range(n):
a=int(input())
if(a%2)==0:
print(a//2)
else:
print(a//2 + mod)
mod=1-mod
```
|
output
| 1
| 12,573
| 11
| 25,147
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,574
| 11
| 25,148
|
Tags: implementation, math
Correct Solution:
```
l=[]
for i in range(int(input())):
l.append(int(input()))
ans=[]
f=1
for i in l:
if i%2==0:
ans.append(i//2)
else:
ans.append((i+f)//2)
f*=-1
for i in ans:
print(i)
```
|
output
| 1
| 12,574
| 11
| 25,149
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,575
| 11
| 25,150
|
Tags: implementation, math
Correct Solution:
```
from collections import Counter
import os
import 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")
##########################################################
#for _ in range(int(input())):
#import math
import sys
# from collections import deque
#from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=sys.stdin.readline()
#n=int(n)
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=int(inaput())
#for _ in range(int(input())):
import math
n = int(input())
f=0
for i in range(n):
u= int(input())
if u%2==0:
print(u//2)
else:
if f==0:
print(math.floor(u/2))
f=1
else:
print(math.ceil(u/2))
f=0
#arr=list(map(int,input().split()))
#for _ in range(int(input())):
#n, k = map(int, input().split())
#arr=list(map(int,input().split()))
```
|
output
| 1
| 12,575
| 11
| 25,151
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
|
instruction
| 0
| 12,576
| 11
| 25,152
|
Tags: implementation, math
Correct Solution:
```
sum=0
for i in range(int(input())):
x=int(input())
y=x//2
if y==x/2:
print(y)
else:
if sum==0:
sum=x-y*2
print(y)
else:
y=y+sum
sum=0
print(y)
```
|
output
| 1
| 12,576
| 11
| 25,153
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
import sys
import math
def main():
#for _ in range(int(input())):
n = int(sys.stdin.readline())
#r, p, s= [int(x) for x in sys.stdin.readline().split()]
#l = list(set(int(x) for x in sys.stdin.readline().split()))
#str = sys.stdin.readline()
flag=0
for i in range(n):
x=int(sys.stdin.readline())
if x%2==0:
print(x//2)
else:
if flag==0:
print(x//2)
flag=1
else:
q=x//2
q+=1
print(q)
flag=0
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 12,577
| 11
| 25,154
|
Yes
|
output
| 1
| 12,577
| 11
| 25,155
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
l=[]
low=0
even=0
for i in range(0,n):
t=int(input())
if t%2==0:
even=even+t//2
l.append((t//2,0))
else:
low=low+(t//2)
l.append((t//2))
if low!=-1*even:
count=(-1*even)-low
for i in range(0,len(l)):
if count==0:
break
if type(l[i])!=tuple:
l[i]=l[i]+1
count=count-1
for i in range(0,len(l)):
if type(l[i])==tuple:
print(l[i][0])
else:
print(l[i])
```
|
instruction
| 0
| 12,578
| 11
| 25,156
|
Yes
|
output
| 1
| 12,578
| 11
| 25,157
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import *
n = int(input())
k = []
p = 0
for x in range(n):
a = int(input())
if a%2 == 0:
a = int(a/2)
k.append(a)
else:
if p%2 == 0:
a = floor(a/2)
k.append(a)
else:
a = ceil(a/2)
k.append(a)
p += 1
for i in range(len(k)):
print(k[i])
```
|
instruction
| 0
| 12,579
| 11
| 25,158
|
Yes
|
output
| 1
| 12,579
| 11
| 25,159
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
import math
n = int(input())
check1 = 0
while n:
n -= 1
a = int(input())
if a % 2 == 0: print(a//2)
elif a % 2 and check1 % 2 == 0:
print(int(math.floor(a / 2)))
check1 += 1
elif a % 2 and check1 % 2:
print(int(math.ceil(a / 2)))
check1 += 1
else: print(a)
```
|
instruction
| 0
| 12,580
| 11
| 25,160
|
Yes
|
output
| 1
| 12,580
| 11
| 25,161
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
o=0
for i in range(n):
l=int(input())
if(l>=0):
if(l%2==0):
print(l//2)
else:
if(o==0):
o=o+1
print(l//2)
else:
print(l//2+1)
o+=1
else:
l=abs(l)
if(l%2==0):
print(-(l//2))
else:
if(o==0):
o=o+1
print(-(l//2))
else:
print(-(l//2+1))
o=o+1
```
|
instruction
| 0
| 12,581
| 11
| 25,162
|
No
|
output
| 1
| 12,581
| 11
| 25,163
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import ceil,floor
n=int(input())
cou=0
while n:
a=int(input())
if(cou%2==0 and a%2==1):
print(ceil(a/2))
else:
print(floor(a/2))
cou+=1
n-=1
```
|
instruction
| 0
| 12,582
| 11
| 25,164
|
No
|
output
| 1
| 12,582
| 11
| 25,165
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import *
n = int(input())
sum = 0
found = 1
for i in range(n):
x = int(input())
if x&1:
if found == 1:
print(floor(x / 2))
found = 0
else:
print(ceil(x / 2))
found = 1
else:
print(x / 2)
```
|
instruction
| 0
| 12,583
| 11
| 25,166
|
No
|
output
| 1
| 12,583
| 11
| 25,167
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
a=[]
b=[]
for i in range(n):
x=int(input())
a.append(x)
if x>=0:
b.append(x//2)
else:
b.append(0-(abs(x)//2))
if sum(b)==(-1):
for i in range(n):
if a[i]>0 and a[i]%2==1:
b[i]+=1
break
elif sum(b)==1:
for i in range(n):
if a[i]<0 and a[i]%2==1:
b[i]-=1
break
for i in b:
print(i)
```
|
instruction
| 0
| 12,584
| 11
| 25,168
|
No
|
output
| 1
| 12,584
| 11
| 25,169
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
from collections import deque, Counter
input = sys.stdin.buffer.readline
MOD = 1000000007
T = int(input())
for _ in range(T):
n, p = map(int, input().split())
ls = list(map(int, input().split()))
ls.sort(reverse=True)
if p == 1: print(1 if n%2 else 0); continue
cp, cc = 0, 0
ok, pos = 1, 0
for i, u in enumerate(ls):
if cc == 0:
cp, cc = u, 1
else:
tp, tc = cp, cc
while tp > u and tc <= n:
tp -= 1; tc *= p
if tp == u:
cp, cc = tp, tc-1
else:
ok, pos = 0, i; break
if ok:
print((pow(p, cp, MOD)*cc)%MOD)
else:
a = (pow(p, cp, MOD)*cc)%MOD
b = 0
for u in ls[pos:]:
b += pow(p, u, MOD)
b %= MOD
print((a-b)%MOD)
```
|
instruction
| 0
| 12,610
| 11
| 25,220
|
Yes
|
output
| 1
| 12,610
| 11
| 25,221
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
left = -1
i = 0
val = [0] * 21
tmp = p
val[0] = p
slot = defaultdict(int)
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
x = arr[i]
if left == -1:
left = x
else:
slot[x] += 1
tmp = x
if x == left:
left = -1
slot.pop(x)
else:
while slot[tmp] % p == 0:
slot[tmp+1] += 1
slot.pop(tmp)
tmp += 1
if tmp == left:
left = -1
slot.pop(tmp)
i+=1
if left == -1:
stdout.write("0\n")
continue
res = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= left:
left -= pww
res = (res * val[w]) % MOD7
if left == 0:
break
for x,c in slot.items():
tp = 1
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
res = (res - tp * c) % MOD7
stdout.write(str(res)+"\n")
main()
```
|
instruction
| 0
| 12,611
| 11
| 25,222
|
Yes
|
output
| 1
| 12,611
| 11
| 25,223
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from math import log2
def main():
t=int(input())
allAns=[]
MOD=10**9+7
for _ in range(t):
n,p=readIntArr()
a=readIntArr()
if p==1: #put half in each group
if n%2==1:
ans=1
else:
ans=0
else:
a.sort(reverse=True)
totalMOD=0
totalExact=0 #store exact total//p**a[i] for current i. total must be a multiple of p**a[i]
prevPow=a[0]
i=0
while i<n:
x=a[i]
#if totalExact//p**x>n, then just subtract all the remaining elements since totalExact can't reach 0
if totalExact!=0 and log2(totalExact)+(prevPow-x)*log2(p)>log2(n): #using log or else number may get too big
while i<n:
x=a[i]
totalMOD-=pow(p,x,MOD)
totalMOD=(totalMOD+MOD)%MOD
i+=1
else:
if totalExact>0:
totalExact*=(p**(prevPow-x))
prevPow=x
if totalExact==0: #add
totalMOD+=pow(p,x,MOD)
# totalMOD+=mod_pow(p,x)
totalMOD%=MOD
totalExact+=1
else:
totalMOD-=pow(p,x,MOD)
# totalMOD-=mod_pow(p,x)
totalMOD=(totalMOD+MOD)%MOD
totalExact-=1
# print('x:{} total:{}'.format(x,total))
i+=1
ans=totalMOD
allAns.append(ans)
multiLineArrayPrint(allAns)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#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()]
main()
```
|
instruction
| 0
| 12,612
| 11
| 25,224
|
Yes
|
output
| 1
| 12,612
| 11
| 25,225
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from math import log
def main():
mod = 10**9+7
for _ in range(int(input())):
n,p = map(int,input().split())
k = sorted(map(int,input().split()),reverse=1)
a,b = 0,0
# power,num
sign = [-1]*n
for ind,i in enumerate(k):
if not b:
a,b = i,1
sign[ind] = 1
else:
if a-i > log(n,p)-log(b,p):
break
b,a = b*pow(p,a-i)-1,i
ans = 0
for a,b in zip(sign,k):
ans = (ans+a*pow(p,b,mod))%mod
print(ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 12,613
| 11
| 25,226
|
Yes
|
output
| 1
| 12,613
| 11
| 25,227
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
from sys import stdin, stdout
import math
def main():
MOD7 = 1000000007
t = int(stdin.readline())
pw = [0] * 21
for w in range(20,-1,-1):
pw[w] = int(math.pow(2,w))
for ks in range(t):
n,p = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
if p == 1:
if n % 2 ==0:
stdout.write("0\n")
else:
stdout.write("1\n")
continue
arr.sort(reverse=True)
res = 0
i = 0
val = [0] * 21
tmp = p
val[0] = p
for x in range(1,21):
tmp = (tmp * tmp) % MOD7
val[x] = tmp
while i < n:
if res == 0 and i + 1 < n and arr[i] == arr[i+1]:
i +=2
continue
tp = 1
x = arr[i]
for w in range(20,-1,-1):
pww = pw[w]
if pww <= x:
x -= pww
tp = (tp * val[w]) % MOD7
if x == 0:
break
if res == 0:
res = tp
else:
res = (res - tp) % MOD7
if res == 0 and t == 9999 and ks == 9998:
print(i,res,tp)
i+=1
if t != 9999:
stdout.write(str(res)+"\n")
elif ks == 9998:
print(n,p)
main()
```
|
instruction
| 0
| 12,614
| 11
| 25,228
|
No
|
output
| 1
| 12,614
| 11
| 25,229
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
from math import log
import copy
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(x) for x in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def bo(i):
return ord(i)-ord('a')
from copy import *
from bisect import *
t=fi()
while t>0:
t-=1
n,p=mi()
a=li()
mod=10**9+7
a.sort()
d=[0 for i in range(10**6+1)]
for i in range(n):
d[a[i]]+=1
if p==1:
print(0 if n%2==0 else 1)
continue
#print(len(d))
pp=[]
for i in range(len(d)):
l=p
#print(d[i],p)
if d[i]==0:
continue
z=d[i]
for j in range(int(log(z,p))+2,-1,-1):
if d[i]>p**j:
d[j+i]+=d[i]//(p**j)
d[i]-=(d[i]//(p**j))*(p**j)
if d[i]>0:
pp.append([i,d[i]])
c=pow(p,pp[-1][0],mod)*pp[-1][1]
#print(pp)
for i in range(len(pp)-1):
c-= pow(p,pp[i][0],mod)*pp[i][1]
c=(c+mod)%mod
print(c)
```
|
instruction
| 0
| 12,615
| 11
| 25,230
|
No
|
output
| 1
| 12,615
| 11
| 25,231
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys;input=sys.stdin.readline
mod = 10**9+7
T, = map(int, input().split())
for _ in range(T):
N, M = map(int, input().split())
X = list(map(int, input().split()))
sX = sorted(list(set(X)))[::-1]
CC = dict()
for x in X:
if x not in CC:
CC[x] = 0
CC[x] += 1
rmn = 0
flag = False
bi=sX[0]
for i in sX:
cc = CC[i]
if rmn:
tmp = rmn
for _ in range(bi-i):
tmp*=M
if tmp>N:
flag = True
break
if flag:
rmn *= pow(M, bi-i, mod)
else:
rmn *= pow(M, bi-i)
if flag:
rmn = (rmn-cc)%mod
else:
rmn-=cc
if rmn <= cc:
rmn %= 2
bi = i
rmn = rmn*pow(M, bi, mod)%mod
print(rmn)
```
|
instruction
| 0
| 12,616
| 11
| 25,232
|
No
|
output
| 1
| 12,616
| 11
| 25,233
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johnny has just found the new, great tutorial: "How to become a grandmaster?". The tutorial tells many strange and unexpected for Johnny things, such as you have to be patient or that very important is solving many harder and harder problems.
The boy has found an online judge with tasks divided by topics they cover. He has picked p^{k_i} problems from i-th category (p is his favorite number). He wants to solve them in two weeks (the patience condition is too hard for Johnny, so for simplicity, he looks only at easy tasks, which can be solved in such a period). Now our future grandmaster has to decide which topics to cover first and which the second week. Help him assign topics in such a way, that workload is balanced.
Formally, given n numbers p^{k_i}, the boy wants to divide them into two disjoint sets, minimizing the absolute difference between sums of numbers in each set. Find the minimal absolute difference. Output the result modulo 10^{9}+7.
Input
Input consists of multiple test cases. The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Each test case is described as follows:
The first line contains two integers n and p (1 ≤ n, p ≤ 10^6). The second line contains n integers k_i (0 ≤ k_i ≤ 10^6).
The sum of n over all test cases doesn't exceed 10^6.
Output
Output one integer — the reminder of division the answer by 1 000 000 007.
Example
Input
4
5 2
2 3 4 4 3
3 1
2 10 1000
4 5
0 1 1 100
1 8
89
Output
4
1
146981438
747093407
Note
You have to minimize the difference, not it's remainder. For example, if the minimum difference is equal to 2, but there is also a distribution where the difference is 10^9 + 8, then the answer is 2, not 1.
In the first test case of the example, there're the following numbers: 4, 8, 16, 16, and 8. We can divide them into such two sets: {4, 8, 16} and {8, 16}. Then the difference between the sums of numbers in sets would be 4.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
MOD = 10 ** 9 + 7
def compress(string):
string.append(-1)
n = len(string)
begin, end, cnt = 0, 1, 1
while end < n:
if string[begin] == string[end]:
end, cnt = end + 1, cnt + 1
else:
yield string[begin], cnt
begin, end, cnt = end, end + 1, 1
t = int(input())
LIMIT = 10 ** 6
for _ in range(t):
n, p = map(int, input().split())
k = list(map(int, input().split()))
k.sort(reverse=True)
k = compress(k)
if p == 1:
print(n % 2)
continue
over_cnt = 0
over_ans = 1
while over_ans <= LIMIT:
over_ans *= p
over_cnt += 1
ans = -1
ans_cnt = -1
flag = False
res = 0
for val, cnt in k:
if flag:
res -= cnt * pow(p, val, MOD)
res %= MOD
if ans == -1 and cnt % 2 == 0:
continue
if ans == -1:
ans_cnt = 1
ans = val
continue
if ans - val >= over_cnt or ans_cnt >= LIMIT:
flag = True
res = ans_cnt * pow(p, ans, MOD) % MOD
res -= cnt * pow(p, val, MOD)
res %= MOD
continue
nokori = ans_cnt * p ** (ans - val)
nokori -= cnt
if nokori == 0:
ans = -1
ans_cnt = -1
elif nokori < 0:
ans = val
ans_cnt = -nokori % 2
else:
ans = val
ans_cnt = nokori
if res != 0:
print(res)
elif ans == -1:
print(0)
else:
print(ans_cnt * pow(p, ans, MOD) % MOD)
```
|
instruction
| 0
| 12,617
| 11
| 25,234
|
No
|
output
| 1
| 12,617
| 11
| 25,235
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 ≤ A_i ≤ 10^3) flamingoes inside (1 ≤ i ≤ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 ≤ N ≤ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 ≤ L < R ≤ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
5
15
10
Output
? 1 2
? 5 6
? 3 4
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
# DEFINING SOME GOOD STUFF
import heapq
import sys
from math import *
import threading
from heapq import *
from itertools import count
from pprint import pprint
from collections import defaultdict
from heapq import heapify, heappop, heappush
# threading.stack_size(10**8)
# sys.setrecursionlimit(300000)
'''
-> if you are increasing recursionlimit then remember submitting using python3 rather pypy3
-> sometimes increasing stack size don't work locally but it will work on CF
'''
mod = 10 ** 9+7
inf = 10 ** 15
decision = ['NO', 'YES']
yes = 'YES'
no = 'NO'
# ------------------------------FASTIO----------------------------
import os
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")
# ________________________FAST FACTORIAL______________________________#
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was "+str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n+1-len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n+1):
prev = nextArr[i-initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was "+str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n+1-len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n+1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was "+str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n+1-len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n+1):
prev = nextArr[i-initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n-k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n-k, k)) * f.invFactorial(min(k, n-k)) % self.MOD
def npr(self, n, k):
if k < 0 or n < k:
return 0
f = self.factorial
return (f.calc(n) * f.invFactorial(n-k)) % self.MOD
#_______________SEGMENT TREE ( logn range modifications )_____________#
class SegmentTree:
def __init__(self, data, default = 0, func = lambda a, b: max(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len-1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size+self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i+i], self.data[i+i+1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx+self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx+1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# ____________________MY FAVOURITE FUNCTIONS_______________________#
def lower_bound(li, num):
answer = len(li)
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] >= num:
answer = middle
end = middle-1
else:
start = middle+1
return answer # min index where x is not less than num
def upper_bound(li, num):
answer = len(li)
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] <= num:
start = middle+1
else:
answer = middle
end = middle-1
return answer # max index where x is greater than num
def abs(x):
return x if x >= 0 else -x
def binary_search(li, val):
# print(lb, ub, li)
ans = -1
lb = 0
ub = len(li)-1
while (lb <= ub):
mid = (lb+ub) // 2
# print('mid is',mid, li[mid])
if li[mid] > val:
ub = mid-1
elif val > li[mid]:
lb = mid+1
else:
ans = mid # return index
break
return ans
def kadane(x): # maximum sum contiguous subarray
sum_so_far = 0
current_sum = 0
for i in x:
current_sum += i
if current_sum < 0:
current_sum = 0
else:
sum_so_far = max(sum_so_far, current_sum)
return sum_so_far
def pref(li):
pref_sum = [0]
for i in li:
pref_sum.append(pref_sum[-1]+i)
return pref_sum
def SieveOfEratosthenes(n):
prime = [{1, i} for i in range(n+1)]
p = 2
while (p <= n):
for i in range(p * 2, n+1, p):
prime[i].add(p)
p += 1
return prime
def primefactors(n):
factors = []
while (n % 2 == 0):
factors.append(2)
n //= 2
for i in range(3, int(sqrt(n))+1, 2): # only odd factors left
while n % i == 0:
factors.append(i)
n //= i
if n > 2: # incase of prime
factors.append(n)
return factors
def prod(li):
ans = 1
for i in li:
ans *= i
return ans
def sumk(a, b):
print('called for', a, b)
ans = a * (a+1) // 2
ans -= b * (b+1) // 2
return ans
def sumi(n):
ans = 0
if len(n) > 1:
for x in n:
ans += int(x)
return ans
else:
return int(n)
def checkwin(x, a):
if a[0][0] == a[1][1] == a[2][2] == x:
return 1
if a[0][2] == a[1][1] == a[2][0] == x:
return 1
if (len(set(a[0])) == 1 and a[0][0] == x) or (len(set(a[1])) == 1 and a[1][0] == x) or (len(set(a[2])) == 1 and a[2][0] == x):
return 1
if (len(set(a[0][:])) == 1 and a[0][0] == x) or (len(set(a[1][:])) == 1 and a[0][1] == x) or (len(set(a[2][:])) == 1 and a[0][0] == x):
return 1
return 0
# _______________________________________________________________#
inf = 10**9 + 7
def main():
# karmanya = int(input())
karmanya = 1
# divisors = SieveOfEratosthenes(200010)
# print(divisors)
while karmanya != 0:
karmanya -= 1
n = int(input())
# a,b,c,d = map(int, input().split())
# s = [int(x) for x in list(input())]
# s = list(input())
# a = list(map(int, input().split()))
# b = list(map(int, input().split()))
# c = list(map(int, input().split()))
# d = defaultdict(list)
ans = [0]*n
# print(ans)
l,r = 1, n
print('?',l,r)
sys.stdout.flush()
s = int(input())
for i in range(n-2):
r -= 1
print('?',l,r)
sys.stdout.flush()
x = int(input())
ans[n-1-i] = s - x
s = x
ans[1] = s-1
ans[0] = 1
print('!', *ans)
sys.stdout.flush()
main()
# t = threading.Thread(target=main)
# t.start()
# t.join()
```
|
instruction
| 0
| 12,650
| 11
| 25,300
|
No
|
output
| 1
| 12,650
| 11
| 25,301
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 ≤ A_i ≤ 10^3) flamingoes inside (1 ≤ i ≤ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 ≤ N ≤ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 ≤ L < R ≤ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
5
15
10
Output
? 1 2
? 5 6
? 3 4
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
import sys
import time
t=int(input())
print("? 1 2",flush=True)
a=int(input())
print("? 1 3",flush=True)
b=int(input())
print("? 2 3",flush=True)
c=int(input())
l=[(a+b-c)//2,(a+c-b)//2,(b+c-a)//2]
d=(b+c-a)//2
for i in range(2,t-1):
print(f"? {i+1} {i+2} ",flush=True)
e=int(input())
l.append(e-d)
d=e-d
print("!",end= " ")
for i in range(t):
print(l[i],end=" ")
print()
```
|
instruction
| 0
| 12,651
| 11
| 25,302
|
No
|
output
| 1
| 12,651
| 11
| 25,303
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 ≤ A_i ≤ 10^3) flamingoes inside (1 ≤ i ≤ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 ≤ N ≤ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 ≤ L < R ≤ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
5
15
10
Output
? 1 2
? 5 6
? 3 4
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
''' ===============================
-- @uthor : Kaleab Asfaw
-- Handle : kaleabasfaw2010
-- Bio : High-School Student
==============================='''
import sys
def inOut(x):
print("?", x)
sys.stdout.flush()
val = int(input())
return val
n = int(input())
if n%2:
lst = []
for i in range(1, n-2, 2):
val = inOut(str(i) + " " + str(i+1))
lst.append(1)
lst.append(val-1)
val = inOut(str(i+2) + " " + str(i+3))
val1 = inOut(str(i+3) + " " + str(i+4))
val2 = inOut(str(i+2) + " " + str(i+4))
y = val+val1-val2
x = val-y
z = val1-y
lst += [x, y, z]
else:
lst = []
for i in range(1, n+1, 2):
val = inOut(str(i) + " " + str(i+1))
lst.append(1)
lst.append(val-1)
ans = "! " + " ".join(list(map(str, lst)))
print(ans)
sys.stdout.flush()
# 1 4 4 6 7 8
```
|
instruction
| 0
| 12,652
| 11
| 25,304
|
No
|
output
| 1
| 12,652
| 11
| 25,305
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. You have to use a flush operation right after printing each line. For example, in C++ you should use the function fflush(stdout), in Java — System.out.flush(), in Pascal — flush(output) and in Python — sys.stdout.flush().
Mr. Chanek wants to buy a flamingo to accompany his chickens on his farm. Before going to the pet shop, Mr. Chanek stops at an animal festival to have fun. It turns out there is a carnival game with a flamingo as the prize.
There are N mysterious cages, which are numbered from 1 to N. Cage i has A_i (0 ≤ A_i ≤ 10^3) flamingoes inside (1 ≤ i ≤ N). However, the game master keeps the number of flamingoes inside a secret. To win the flamingo, Mr. Chanek must guess the number of flamingoes in each cage.
Coincidentally, Mr. Chanek has N coins. Each coin can be used to ask once, what is the total number of flamingoes inside cages numbered L to R inclusive? With L < R.
Input
Use standard input to read the responses of your questions.
Initially, the judge will give an integer N (3 ≤ N ≤ 10^3), the number of cages, and the number of coins Mr. Chanek has.
For each of your questions, the jury will give an integer that denotes the number of flamingoes from cage L to R inclusive.
If your program does not guess the flamingoes or ask other questions, you will get "Wrong Answer". Of course, if your program asks more questions than the allowed number, your program will get "Wrong Answer".
Output
To ask questions, your program must use standard output.
Then, you can ask at most N questions. Questions are asked in the format "? L R", (1 ≤ L < R ≤ N).
To guess the flamingoes, print a line that starts with "!" followed by N integers where the i-th integer denotes the number of flamingo in cage i. After answering, your program must terminate or will receive the "idle limit exceeded" verdict. You can only guess the flamingoes once.
Examples
Input
6
5
15
10
Output
? 1 2
? 5 6
? 3 4
! 1 4 4 6 7 8
Note
In the sample input, the correct flamingoes amount is [1, 4, 4, 6, 7, 8].
Submitted Solution:
```
import sys
n=int(input())
a=[0]*n
print("1",n)
sys.stdout.flush()
x=int(input())
c=x
for j in range(1,n-1):
print("1",n-j)
sys.stdout.flush()
y=int(input())
a[n-j]=x-y
x=y
print("2",n)
sys.stdout.flush()
x=int(input())
a[0]=c-x
a[1]=y-a[0]
print("!",' '.join(map(str,a)))
```
|
instruction
| 0
| 12,653
| 11
| 25,306
|
No
|
output
| 1
| 12,653
| 11
| 25,307
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
d={}
for i in range(0,n):
if arr[i]-(i+1) not in d:
d[arr[i]-(i+1)]=1
else:
d[arr[i]-(i+1)]+=1
ans=0
for key in d.keys():
if d[key]>1:
k=d[key]
ans+=(k*(k-1))//2
print(ans)
```
|
instruction
| 0
| 12,706
| 11
| 25,412
|
Yes
|
output
| 1
| 12,706
| 11
| 25,413
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
def main():
import sys
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
s = list(map(int, sys.stdin.readline().split()))
t = dict()
c = 0
for i in range(n):
r = s[i] - i - 1
if r in t:
c += t[r]
if r in t:
t[r] += 1
else:
t[r] = 1
sys.stdout.write(str(c) + '\n')
main()
```
|
instruction
| 0
| 12,707
| 11
| 25,414
|
Yes
|
output
| 1
| 12,707
| 11
| 25,415
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
t = int(input())
while t!=0:
t-=1
n = int(input())
arr = list(map(int , input().split()))
a = arr[0]-0
'''
i=1
while i <n:
arr[i] = arr[i] + a
i+= 1
'''
ans=0
j=0
a = arr[0]
while j<n:
print( a , arr[j])
if arr[j] == a:
ans += 1
a +=1
j+=1
if ans !=1:
ans = (ans*(ans-1))//2
#print('here' , ans)
print(ans)
```
|
instruction
| 0
| 12,710
| 11
| 25,420
|
No
|
output
| 1
| 12,710
| 11
| 25,421
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
t = int(input())
for tc in range(t):
n = int(input())
arr = [int(z) for z in input().split()]
model = list(range(1, n+1))
c = []
for i in range(n):
if arr[i] == model[i]:
c.append(i)
ans = 0
for i in range(1, len(c)):
ans += len(c) - i
print(ans)
```
|
instruction
| 0
| 12,711
| 11
| 25,422
|
No
|
output
| 1
| 12,711
| 11
| 25,423
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of n integers. Count the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — array a.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case output the number of pairs of indices (i, j) such that i < j and a_j - a_i = j - i.
Example
Input
4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6
Output
1
3
3
10
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
ls = list(map(int,input().split()))
ls2 = []
for i,v in enumerate(ls):
ls2.append(v - i)
d = {}
for i in ls2:
if i in d:
d[i] += 1
else:
d[i] = 1
dd = []
for j in d:
if d[j] == 1:
dd.append(j)
for k in dd:
d.pop(k)
sov = sum(d.values())
print(((sov * (sov - 1)) // 2))
```
|
instruction
| 0
| 12,713
| 11
| 25,426
|
No
|
output
| 1
| 12,713
| 11
| 25,427
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 ≤ q ≤ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n — the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
map = {}
for i in range(0,n):
input_string = input()
input_string = input_string.split(" ")
if input_string[0] in map:
temp = map[input_string[0]]
map.pop(input_string[0])
map[input_string[1]] = temp
else:
map[input_string[1]] = input_string[0]
print(len(map))
for key in map.keys():
value = map[key]
print(f'{value} {key}')
```
|
instruction
| 0
| 12,867
| 11
| 25,734
|
Yes
|
output
| 1
| 12,867
| 11
| 25,735
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 ≤ q ≤ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n — the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
q = int(input())
old = dict()
new = dict()
for i in range(q):
name = input().split()
if new.get(name[0]) == None:
old[name[0]] = name[1]
new[name[1]] = name[0]
else:
old[new[name[0]]] = name[1]
new[name[1]] = new[name[0]]
print(len(old))
for i in old:
print(i, old[i])
```
|
instruction
| 0
| 12,868
| 11
| 25,736
|
Yes
|
output
| 1
| 12,868
| 11
| 25,737
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 ≤ q ≤ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n — the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
from collections import defaultdict
class DisjSet:
def __init__(self, n):
self.rank = defaultdict(lambda:1)
self.parent = defaultdict(lambda:None)
def sol(self,a):
if self.parent[a]==None:
self.parent[a]=a
def find(self, x):
#ans=1
if (self.parent[x] != x):
self.parent[x] = self.find(self.parent[x])
#ans+=1
return self.parent[x]
def Union(self, x, y):
xset = self.find(x)
yset = self.find(y)
if xset == yset:
return
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
n=int(input())
obj=DisjSet(n+1)
temp=[]
visited=defaultdict(lambda:None)
for p in range(n):
a,b=input().split()
temp.append([a,b])
obj.sol(a)
obj.sol(b)
obj.Union(a,b)
#print(obj.find('MikeMirzayanov'))
cont=0
ans=[]
while temp:
a,b=temp.pop()
if visited[obj.find(b)]==None:
ans.append([obj.find(b),b])
cont+=1
visited[obj.find(b)]=True
print(cont)
for val in ans:
print(*val)
```
|
instruction
| 0
| 12,869
| 11
| 25,738
|
Yes
|
output
| 1
| 12,869
| 11
| 25,739
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 ≤ q ≤ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n — the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
m, rev = {}, {}
for i in range(n):
a, b = input().split()
if a not in rev:
m[a] = b
rev[b] = a
else:
m[rev[a]] = b
rev[b] = rev[a]
print(len(m))
for key in m:
print(key, m[key])
```
|
instruction
| 0
| 12,870
| 11
| 25,740
|
Yes
|
output
| 1
| 12,870
| 11
| 25,741
|
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