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stringlengths 2
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values | message_id
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int64 219
108k
| cluster
float64 11
11
| __index_level_0__
int64 438
217k
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|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya tries to break in a safe. He knows that a code consists of n numbers, and every number is a 0 or a 1. Vasya has made m attempts to enter the code. After each attempt the system told him in how many position stand the right numbers. It is not said in which positions the wrong numbers stand. Vasya has been so unlucky that he hasn’t entered the code where would be more than 5 correct numbers. Now Vasya is completely bewildered: he thinks there’s a mistake in the system and it is self-contradictory. Help Vasya — calculate how many possible code variants are left that do not contradict the previous system responses.
Input
The first input line contains two integers n and m (6 ≤ n ≤ 35, 1 ≤ m ≤ 10) which represent the number of numbers in the code and the number of attempts made by Vasya. Then follow m lines, each containing space-separated si and ci which correspondingly indicate Vasya’s attempt (a line containing n numbers which are 0 or 1) and the system’s response (an integer from 0 to 5 inclusively).
Output
Print the single number which indicates how many possible code variants that do not contradict the m system responses are left.
Examples
Input
6 2
000000 2
010100 4
Output
6
Input
6 3
000000 2
010100 4
111100 0
Output
0
Input
6 3
000000 2
010100 4
111100 2
Output
1
|
instruction
| 0
| 95,655
| 11
| 191,310
|
Tags: brute force
Correct Solution:
```
from itertools import combinations
def calculate(s, dif):
x = int(s, 2)
for j in combinations(range(len(s)), dif):
y = x
for k in j:
y ^= (2**k)
yield y
def calculate2(s, dif, arr):
y = int(s, 2)
for x in arr:
if(bin(y ^ x).count('1') == dif):
yield x
n, m = map(int, input().split())
result = []
(st, dif) = input().split()
total = calculate(st, int(dif))
for i in range(1, m):
st, dif = input().split()
total = calculate2(st, int(dif), total)
print(len(list(total)))
```
|
output
| 1
| 95,655
| 11
| 191,311
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya tries to break in a safe. He knows that a code consists of n numbers, and every number is a 0 or a 1. Vasya has made m attempts to enter the code. After each attempt the system told him in how many position stand the right numbers. It is not said in which positions the wrong numbers stand. Vasya has been so unlucky that he hasn’t entered the code where would be more than 5 correct numbers. Now Vasya is completely bewildered: he thinks there’s a mistake in the system and it is self-contradictory. Help Vasya — calculate how many possible code variants are left that do not contradict the previous system responses.
Input
The first input line contains two integers n and m (6 ≤ n ≤ 35, 1 ≤ m ≤ 10) which represent the number of numbers in the code and the number of attempts made by Vasya. Then follow m lines, each containing space-separated si and ci which correspondingly indicate Vasya’s attempt (a line containing n numbers which are 0 or 1) and the system’s response (an integer from 0 to 5 inclusively).
Output
Print the single number which indicates how many possible code variants that do not contradict the m system responses are left.
Examples
Input
6 2
000000 2
010100 4
Output
6
Input
6 3
000000 2
010100 4
111100 0
Output
0
Input
6 3
000000 2
010100 4
111100 2
Output
1
|
instruction
| 0
| 95,656
| 11
| 191,312
|
Tags: brute force
Correct Solution:
```
import os,io
from sys import stdout
# import collections
# import random
# import math
# from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache
# import sys
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(102400000)
# from functools import lru_cache
# @lru_cache(maxsize=None)
######################
# --- Maths Fns --- #
######################
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
def binomial_coefficient(n, k):
if 0 <= k <= n:
ntok = 1
ktok = 1
for t in range(1, min(k, n - k) + 1):
ntok *= n
ktok *= t
n -= 1
return ntok // ktok
else:
return 0
def powerOfK(k, max):
if k == 1:
return [1]
if k == -1:
return [-1, 1]
result = []
n = 1
while n <= max:
result.append(n)
n *= k
return result
def divisors(n):
i = 1
result = []
while i*i <= n:
if n%i == 0:
if n/i == i:
result.append(i)
else:
result.append(i)
result.append(n/i)
i+=1
return result
# @lru_cache(maxsize=None)
def digitsSum(n):
if n == 0:
return 0
r = 0
while n > 0:
r += n % 10
n //= 10
return r
######################
# ---- GRID Fns ---- #
######################
def isValid(i, j, n, m):
return i >= 0 and i < n and j >= 0 and j < m
def print_grid(grid):
for line in grid:
print(" ".join(map(str,line)))
######################
# ---- MISC Fns ---- #
######################
def kadane(a,size):
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def prefixSum(arr):
for i in range(1, len(arr)):
arr[i] = arr[i] + arr[i-1]
return arr
def ceil(n, d):
if n % d == 0:
return n // d
else:
return (n // d) + 1
# INPUTS --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
# for _ in range(t):
n, k = list(map(int, input().split()))
q = []
for _ in range(k):
a, b = list(map(lambda x: x.decode('utf-8').strip(), input().split()))
q.append((list(map(int, a)), int(b)))
code, correct = max(q, key=lambda x: x[1])
codeb = int("".join(map(str, code)), 2)
possibles = set()
def generate(n, correct, codeb, l, s):
if correct == 0:
while len(l) < n:
l.append(1)
p = int("".join(map(str, l)), 2)
s.add(p)
return
if n - len(l) < correct:
return
generate(n, correct-1, codeb, l+[0], s)
generate(n, correct, codeb, l+[1], s)
result = None
memo = {}
for code, correct in q:
codeb = int("".join(map(str, code)), 2)
newSetOfPossibles = set()
if correct in memo:
newSetOfPossibles = memo[correct]
else:
generate(n, correct, codeb, [], newSetOfPossibles)
memo[correct] = newSetOfPossibles
newSetOfPossibles = set(list(map(lambda x: x^codeb, list(newSetOfPossibles))))
if not result:
result = newSetOfPossibles
else:
result = result.intersection(newSetOfPossibles)
print(len(result))
```
|
output
| 1
| 95,656
| 11
| 191,313
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya tries to break in a safe. He knows that a code consists of n numbers, and every number is a 0 or a 1. Vasya has made m attempts to enter the code. After each attempt the system told him in how many position stand the right numbers. It is not said in which positions the wrong numbers stand. Vasya has been so unlucky that he hasn’t entered the code where would be more than 5 correct numbers. Now Vasya is completely bewildered: he thinks there’s a mistake in the system and it is self-contradictory. Help Vasya — calculate how many possible code variants are left that do not contradict the previous system responses.
Input
The first input line contains two integers n and m (6 ≤ n ≤ 35, 1 ≤ m ≤ 10) which represent the number of numbers in the code and the number of attempts made by Vasya. Then follow m lines, each containing space-separated si and ci which correspondingly indicate Vasya’s attempt (a line containing n numbers which are 0 or 1) and the system’s response (an integer from 0 to 5 inclusively).
Output
Print the single number which indicates how many possible code variants that do not contradict the m system responses are left.
Examples
Input
6 2
000000 2
010100 4
Output
6
Input
6 3
000000 2
010100 4
111100 0
Output
0
Input
6 3
000000 2
010100 4
111100 2
Output
1
Submitted Solution:
```
import os,io
from sys import stdout
# import collections
# import random
# import math
# from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache
# import sys
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(102400000)
# from functools import lru_cache
# @lru_cache(maxsize=None)
######################
# --- Maths Fns --- #
######################
def primes(n):
sieve = [True] * n
for i in range(3,int(n**0.5)+1,2):
if sieve[i]:
sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
return [2] + [i for i in range(3,n,2) if sieve[i]]
def binomial_coefficient(n, k):
if 0 <= k <= n:
ntok = 1
ktok = 1
for t in range(1, min(k, n - k) + 1):
ntok *= n
ktok *= t
n -= 1
return ntok // ktok
else:
return 0
def powerOfK(k, max):
if k == 1:
return [1]
if k == -1:
return [-1, 1]
result = []
n = 1
while n <= max:
result.append(n)
n *= k
return result
def divisors(n):
i = 1
result = []
while i*i <= n:
if n%i == 0:
if n/i == i:
result.append(i)
else:
result.append(i)
result.append(n/i)
i+=1
return result
# @lru_cache(maxsize=None)
def digitsSum(n):
if n == 0:
return 0
r = 0
while n > 0:
r += n % 10
n //= 10
return r
######################
# ---- GRID Fns ---- #
######################
def isValid(i, j, n, m):
return i >= 0 and i < n and j >= 0 and j < m
def print_grid(grid):
for line in grid:
print(" ".join(map(str,line)))
######################
# ---- MISC Fns ---- #
######################
def kadane(a,size):
max_so_far = 0
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def prefixSum(arr):
for i in range(1, len(arr)):
arr[i] = arr[i] + arr[i-1]
return arr
def ceil(n, d):
if n % d == 0:
return n // d
else:
return (n // d) + 1
# INPUTS --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
# for _ in range(t):
n, k = list(map(int, input().split()))
q = []
for _ in range(k):
a, b = list(map(lambda x: x.decode('utf-8').strip(), input().split()))
q.append((list(map(int, a)), int(b)))
code, correct = max(q, key=lambda x: x[1])
codeb = int("".join(map(str, code)), 2)
possibles = set()
def generate(n, correct, l):
if correct == 0:
while len(l) < n:
l.append(1)
p = int("".join(map(str, l)), 2)
possibles.add(p^codeb)
return
if n - len(l) < correct:
return
generate(n, correct-1, l+[0])
generate(n, correct, l+[1])
generate(n, correct, [])
impossible = set()
total = 0
for possibleCode in possibles:
for attempt, match in q:
attempt = "".join(list(map(str, attempt)))
attempt = int(attempt, base=2)
r = (possibleCode^attempt)
r = "{0:{f}{w}b}".format(r, w=n, f='0')
t = r.count('0')
if t != match:
impossible.add(possibleCode)
break
total += 1
print(total)
```
|
instruction
| 0
| 95,657
| 11
| 191,314
|
No
|
output
| 1
| 95,657
| 11
| 191,315
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
ans = 0
ans = [[0] * (N+1) for _ in range(N+1)]
depth = 1
def dfs(l, r, d):
global depth
if l == r:
return
depth = max(depth, d)
m = (l + r) // 2
for i in range(l, m+1):
for j in range(m, r+1):
ans[i][j] = ans[i][j] = d
dfs(l, m, d+1)
dfs(m+1, r, d+1)
dfs(1, N, 1)
for i in range(1, N):
print(*ans[i][i+1:])
```
|
instruction
| 0
| 95,895
| 11
| 191,790
|
Yes
|
output
| 1
| 95,895
| 11
| 191,791
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
#import sys
#input = sys.stdin.readline
def main():
N = int( input())
ANS = [ [0]*N for _ in range(N)]
for i in range(11):
if N <= pow(2,i):
M = i
break
# for s in range(1, N, 2):
# for i in range(N-s):
# ANS[i][i+s] = 1
for t in range(1,M+1):
w = pow(2,t-1)
for s in range(1,N+1, 2):
if w*s >= N:
break
for i in range(N-s*w):
if ANS[i][i+w*s] == 0:
ANS[i][i+w*s] = t
for i in range(N-1):
print( " ".join( map( str, ANS[i][i+1:])))
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 95,896
| 11
| 191,792
|
Yes
|
output
| 1
| 95,896
| 11
| 191,793
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
def num(i,j):
if i==j:
return -1
S=bin(i)[2:][::-1]+"0"*20
T=bin(j)[2:][::-1]+"0"*20
for index in range(min(len(S),len(T))):
if S[index]!=T[index]:
return index+1
return -1
N=int(input())
a=[[0 for j in range(N-1-i)] for i in range(N-1)]
for i in range(N-1):
for j in range(i+1,N):
a[i][j-i-1]=num(i,j)
for i in range(N-1):
print(*a[i])
```
|
instruction
| 0
| 95,897
| 11
| 191,794
|
Yes
|
output
| 1
| 95,897
| 11
| 191,795
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
n = int(input())
ans = [[-1] * (n - i - 1) for i in range(n - 1)]
s = [(0, n)]
t = 1
while s:
p, q = s.pop()
if p + 1 == q:
continue
m = (q - p) // 2 + p
for i in range(p, m):
for j in range(m, q):
ans[i][j - i - 1] = t
t += 1
s.append((p, m))
s.append((m, q))
for row in ans:
print(' '.join(map(str, row)))
```
|
instruction
| 0
| 95,899
| 11
| 191,798
|
No
|
output
| 1
| 95,899
| 11
| 191,799
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
N=int(input())
ans=[]
for i in range(N-1):
tmp=[]
for j in range(i+1,N):
if i%2 != j%2:
tmp.append(1)
else:
tmp.append((i+2)//2+1)
ans.append(tmp)
for i in range(N-1):
print(*ans[i])
```
|
instruction
| 0
| 95,900
| 11
| 191,800
|
No
|
output
| 1
| 95,900
| 11
| 191,801
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
res = [[0] * x for x in range(N - 1, 0, -1)]
for x in range(N - 1):
for y in range(N - 1 - x):
b = 0
for k in range(10):
if y + 1 < pow(2, k): break
b = k + 1
res[x][y] = b
for r in res: print(*r)
```
|
instruction
| 0
| 95,901
| 11
| 191,802
|
No
|
output
| 1
| 95,901
| 11
| 191,803
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
def f_d():
n = int(input())
for i in range(n-1):
print(" ".join([str(i+1)]*(n-i-1)))
if __name__ == "__main__":
f_d()
```
|
instruction
| 0
| 95,902
| 11
| 191,804
|
No
|
output
| 1
| 95,902
| 11
| 191,805
|
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 = lambda: sys.stdin.readline().rstrip()
T = int(input())
P = 10 ** 9 + 7
for _ in range(T):
N, b = map(int, input().split())
A = sorted([int(a) for a in input().split()])
if b == 1:
print(N % 2)
continue
a = A.pop()
pre = a
s = 1
ans = pow(b, a, P)
while A:
a = A.pop()
s *= b ** min(pre - a, 30)
if s >= len(A) + 5:
ans -= pow(b, a, P)
if ans < 0: ans += P
while A:
a = A.pop()
ans -= pow(b, a, P)
if ans < 0: ans += P
print(ans)
break
if s:
s -= 1
ans -= pow(b, a, P)
if ans < 0: ans += P
pre = a
else:
s = 1
ans = -ans
if ans < 0: ans += P
ans += pow(b, a, P)
if ans >= P: ans -= P
pre = a
else:
print(ans)
```
|
instruction
| 0
| 96,284
| 11
| 192,568
|
Yes
|
output
| 1
| 96,284
| 11
| 192,569
|
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 os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# sys.setrecursionlimit(300000)
# from heapq import *
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
M = 1000000007
m = 1000000003
for _ in range(N()):
n,p = RL()
a = RLL()
a.sort(reverse=True)
s = 0
ss = 0
for i in a:
t = pow(p,i,M)
tt = pow(p,i,m)
if s==0 and ss==0:
s+=t
ss+=tt
else:
s-=t
ss-=tt
s%=M
ss%=m
print(s)
```
|
instruction
| 0
| 96,285
| 11
| 192,570
|
Yes
|
output
| 1
| 96,285
| 11
| 192,571
|
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:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=-10**6, 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)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(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)
MOD=10**9+7
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
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(50001)]
pp=[]
def SieveOfEratosthenes(n=50000):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for i in range(50001):
if prime[i]:
pp.append(i)
#---------------------------------running code------------------------------------------
t=int(input())
for i in range(t):
n,p=map(int,input().split())
power=[int(i) for i in input().split()]
if p==1:
print(n%2)
else:
power.sort(reverse=True)
ans=[0,0]
ok=True
for i in range(n):
k=power[i]
if ans[0]==0:
ans=[1,k]
elif ans[1]==k:
ans[0]-=1
else:
while ans[1]>k:
ans[1]-=1
ans[0]*=p
if ans[0]>=(n-i+1):
#print(ans)
ok=False
ind=i
break
if ok==False:
output=((ans[0]*pow(p,ans[1],mod))%mod)
for j in range(ind,n):
output=((output-pow(p,power[j],mod))%mod)
print(output)
break
else:
ans[0]-=1
if ok:
print((ans[0]*pow(p,ans[1],mod))%mod)
```
|
instruction
| 0
| 96,286
| 11
| 192,572
|
Yes
|
output
| 1
| 96,286
| 11
| 192,573
|
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
def print(val):
sys.stdout.write(str(val) + '\n')
def prog():
for _ in range(int(input())):
n,p = map(int,input().split())
vals = list(map(int,input().split()))
if p == 1:
print(n%2)
else:
vals.sort(reverse = True)
curr_power = 0
last = vals[0]
too_large = False
for a in range(n):
k = vals[a]
if k < last and curr_power > 0:
for i in range(1,last - k+1):
curr_power *= p
if curr_power > n-a:
too_large = True
break
if too_large:
curr_power %= 1000000007
curr_power = curr_power * pow(p, last - i ,1000000007)
mod_diff = curr_power % 1000000007
for b in range(a,n):
k = vals[b]
mod_diff = (mod_diff - pow(p, k ,1000000007)) % 1000000007
print(mod_diff)
break
else:
if curr_power > 0:
curr_power -= 1
else:
curr_power += 1
else:
if curr_power > 0:
curr_power -= 1
else:
curr_power += 1
last = k
if not too_large:
print((curr_power*pow(p,last,1000000007))%1000000007)
prog()
```
|
instruction
| 0
| 96,287
| 11
| 192,574
|
Yes
|
output
| 1
| 96,287
| 11
| 192,575
|
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 = 0
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 == 0:
left = x
else:
slot[x] += 1
if x == left:
left = 0
slot.pop(x)
elif slot[x] % p == 0:
slot[x+1] += 1
slot.pop(x)
if x+1 == left:
left = 0
slot.pop(x+1)
i+=1
if left == 0:
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
if res == 1:
print(left,n,p)
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
| 96,288
| 11
| 192,576
|
No
|
output
| 1
| 96,288
| 11
| 192,577
|
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:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
[1,1,1,1,4,4]
We want the split to be as even as possible each time
"""
T = getInt()
for _ in range(1,T+1):
N, P = getInts()
A = getInts()
if T == 9999 and _ < 9999: continue
A.sort()
diff = 0
prev = A[-1]
i = N-1
while i >= 0:
diff *= pow(P,prev-A[i],MOD)
j = i
while j > 0 and A[j-1] == A[j]: j -= 1
num = i-j+1
if num <= diff:
diff -= num
else:
num -= diff
if num % 2: diff = 1
else: diff = 0
prev = A[i]
i = j-1
if T == 9999 and _ == 9999:
print(i,j)
print((diff * pow(P,prev,MOD)) % MOD)
```
|
instruction
| 0
| 96,289
| 11
| 192,578
|
No
|
output
| 1
| 96,289
| 11
| 192,579
|
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:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os, sys, heapq as h, time
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def isInt(s): return '0' <= s[0] <= '9'
MOD = 10**9 + 7
"""
[1,1,1,1,4,4]
We want the split to be as even as possible each time
"""
def solve():
N, P = getInts()
A = getInts()
A.sort()
diff = 0
prev = A[-1]
for i in range(N-1,-1,-1):
diff *= pow(P,prev-A[i],MOD)
diff %= MOD
if diff:
diff -= 1
else:
diff = 1
prev = A[i]
return diff * pow(P,prev,MOD)
for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
```
|
instruction
| 0
| 96,290
| 11
| 192,580
|
No
|
output
| 1
| 96,290
| 11
| 192,581
|
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 = 0
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 == 0:
left = x
else:
slot[x] += 1
if x == left:
left = 0
slot.pop(x)
elif slot[x] % p == 0:
slot[x+1] += 1
slot.pop(x)
if x+1 == left:
left = 0
slot.pop(x+1)
i+=1
if left == 0:
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
| 96,291
| 11
| 192,582
|
No
|
output
| 1
| 96,291
| 11
| 192,583
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
t = int(input())
for q in range(t):
n, x, y = map(int, input().split())
Bob = list(map(int, input().split()))
Cnt = [ [0, i] for i in range(n+1) ]
Ans = [ -1] * n
Occ = [ [] for i in range(n+1) ]
for i in range(n):
Bob[i]-=1
Cnt[Bob[i]][0] +=1
Occ[Bob[i]].append(i)
Cnt.sort(reverse = True)
#print("\n\nNew test case\n", n, x, y)
#print("Cnt ", Cnt)
lvl = Cnt[0][0]
i=0
xcpy = x
while x > 0:
#print("Deleting from ", i)
while x>0 and Cnt[i][0] >= lvl:
#print("Now: ", Cnt[i])
Cnt[i][0]-=1
col = Cnt[i][1]
Ans[Occ[col].pop()] = col
x-=1
i+=1
if i==n or Cnt[i][0] < lvl:
lvl = Cnt[0][0]
i = 0
Cnt.sort(reverse = True)
#print("Cnt = ", Cnt)
x = xcpy
if Cnt[0][0]*2 - (n-x) > n-y:
print("NO")
continue
Pos = []
Clr = []
for i in range(n):
if Ans[i]==-1:
Pos.append( [Bob[i], i])
Clr.append( Bob[i])
m = len(Pos)
Pos.sort()
Clr.sort()
offset = m//2
nocnt = n-y
nocolor = Cnt[-1][1]
for i in range(m):
pos = Pos[i][1]
c = Clr[(offset+i)%m]
if i+nocnt==m:
Ans[pos] = nocolor
nocnt-=1
continue
if Pos[i][0]==c:
assert(nocnt > 0)
Ans[pos] = nocolor
nocnt -=1
else:
Ans[pos] = c
assert(nocnt==0)
print("YES")
for c in Ans:
print(c+1, end = ' ')
print()
```
|
instruction
| 0
| 96,300
| 11
| 192,600
|
Yes
|
output
| 1
| 96,300
| 11
| 192,601
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
from sys import stdin, stdout
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, stdin.readline().split())
a = stdin.readline().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:
break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans = [0] * n
for i in range(s):
l, x = heappop(q)
ans[d[x].pop()] = x
l += 1
if l:
heappush(q, (l, x))
p = []
while q:
l, x = heappop(q)
p.extend(d[x])
if p:
h = (n - s) // 2
y = n - y
q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:
ans[x] = e
y -= 1
else:
stdout.write("NO\n")
return
else:
ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:
ans[p[i]] = e
y -= 1
print("YES")
print(' '.join(ans))
T = int(stdin.readline())
for t in range(T):
solve()
```
|
instruction
| 0
| 96,301
| 11
| 192,602
|
Yes
|
output
| 1
| 96,301
| 11
| 192,603
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
from collections import defaultdict
import heapq
T = int(input())
for _ in range(T):
N, A, B = [int(x) for x in input().split(' ')]
b = [int(x) for x in input().split(' ')]
a = [0 for _ in range(N)]
split = defaultdict(list)
for i, x in enumerate(b):
split[x].append((i, x))
heap = []
for x in split.values():
heapq.heappush(heap, (-len(x), x))
for _ in range(A):
_, cur = heapq.heappop(heap)
i, x = cur.pop()
a[i] = x
if len(cur):
heapq.heappush(heap, (-len(cur), cur))
if heap:
rot = -heap[0][0]
rem = [x for cur in heap for x in cur[1]]
d = N - B
if 2*rot-d > len(rem):
print('NO')
continue
heap[0] = (heap[0][0] + d, heap[0][1])
rot = -min(x[0] for x in heap)
unused = list(set(range(1, N+2))-set(b))[0]
#print(rem, rot)
for i in range(d):
rem[i] = (rem[i][0], unused)
#print(rem)
for i in range(len(rem)):
a[rem[i][0]] = rem[(i-rot+len(rem))%len(rem)][1]
print('YES')
print(' '.join(str(x) for x in a))
```
|
instruction
| 0
| 96,302
| 11
| 192,604
|
Yes
|
output
| 1
| 96,302
| 11
| 192,605
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
from collections import defaultdict
from heapq import heapify, heappop, heappush
def solve():
n, s, y = map(int, input().split())
a = input().split()
d = defaultdict(list)
for i, x in enumerate(a):
d[x].append(i)
for i in range(1, n + 2):
e = str(i)
if e not in d:break
q = [(-len(d[x]), x) for x in d.keys()]
heapify(q)
ans,p = [0] * n,[]
for i in range(s):
l, x = heappop(q);ans[d[x].pop()] = x;l += 1
if l:heappush(q, (l, x))
while q:l, x = heappop(q);p.extend(d[x])
if p:
h = (n - s) // 2;y = n - y;q = p[h:] + p[:h]
for x, z in zip(p, q):
if a[x] == a[z]:
if y:ans[x] = e;y -= 1
else:print("NO");return
else:ans[x] = a[z]
for i in range(n - s):
if y and ans[p[i]] != e:ans[p[i]] = e;y -= 1
print("YES");print(' '.join(ans))
for t in range(int(input())):solve()
```
|
instruction
| 0
| 96,303
| 11
| 192,606
|
Yes
|
output
| 1
| 96,303
| 11
| 192,607
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
from itertools import cycle
t=int(input())
for _ in range(t):
n,x,y=list(map(lambda x:int(x),input().split()))
a=list(map(lambda x:int(x),input().split()))
hashmap={}
elements_in=set()
elements_all=set()
if n==x:
print("YES")
print(*a)
continue
for i in range(1,n+2):
elements_all.add(i)
for ind,i in enumerate(a):
elements_in.add(i)
if i not in hashmap:
hashmap[i]=[]
hashmap[i].append(ind)
residue=y-x
elements_not_in=elements_all.difference(elements_in)
original=[0 for i in range(n)]
items=[[a,list(b)]for a,b in list(hashmap.items())]
indices=cycle([i for i in range(len(items))])
index=next(indices)
counter=1
remains={}
while counter<=n-x:
if len(items[index][1])==0:
index = next(indices)
continue
if items[index][0] not in remains:
remains[items[index][0]]=[]
remains[items[index][0]].append(items[index][1][0])
hashmap[items[index][0]] =items[index][1][1:]
items[index][1]=items[index][1][1:]
index=next(indices)
counter+=1
items = [[a, list(b)] for a, b in list(hashmap.items())]
indices = cycle([i for i in range(len(items))])
index = next(indices)
counter=1
while counter <= x:
if len(items[index][1]) == 0:
index = next(indices)
continue
original[items[index][1][0]] = items[index][0]
hashmap[items[index][0]] = items[index][1][1:]
items[index][1] = items[index][1][1:]
index = next(indices)
counter += 1
hashmap={k:len(v) for k,v in remains.items()}
is_possible = False
counter=1
while counter<=residue:
is_possible=False
flag=0
for j in hashmap:
if hashmap[j] == 0:
continue
for i in range(n-1,-1,-1):
if j!=a[i] and original[i]==0:
is_possible=True
hashmap[j]-=1
original[i]=j
flag=1
break
if flag==1:
break
if is_possible==False:
break
counter+=1
if is_possible==False and residue!=0:
print("NO")
else:
elements_not_in=list(elements_not_in)
for i in range(0,len(original)):
if original[i]==0:
original[i]=elements_not_in[0]
print("YES")
print(*original)
```
|
instruction
| 0
| 96,304
| 11
| 192,608
|
No
|
output
| 1
| 96,304
| 11
| 192,609
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter
import heapq
t = int(stdin.readline())
for _ in range(t):
n, x, y = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
unused = set(list(range(1, n+2))) - set(a)
r = list(unused)[0]
ans = [r]*n
c = Counter(a)
h = []
for k in c:
v = c[k]
heapq.heappush(h, (-v, k))
for _ in range(x):
v, k = heapq.heappop(h)
for i, elem in enumerate(a):
if elem == k and ans[i] == r:
ans[i] = k
break
if v+1< 0:
heapq.heappush(h, (v+1, k))
cnt = 0
while h:
v, k = heapq.heappop(h)
for i in range(n):
if ans[i] == r and a[i] != k:
ans[i] = k
cnt += 1
v += 1
if v == 0:
break
if cnt == y-x:
break
if cnt < y-x:
print("NO")
continue
print("YES")
print(" ".join(map(str, ans)))
```
|
instruction
| 0
| 96,305
| 11
| 192,610
|
No
|
output
| 1
| 96,305
| 11
| 192,611
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
from itertools import cycle
t=int(input())
for _ in range(t):
n,x,y=list(map(lambda x:int(x),input().split()))
a=list(map(lambda x:int(x),input().split()))
hashmap={}
elements_in=set()
elements_all=set()
for i in range(1,n+2):
elements_all.add(i)
for ind,i in enumerate(a):
elements_in.add(i)
if i not in hashmap:
hashmap[i]=[]
hashmap[i].append(ind)
residue=y-x
elements_not_in=elements_all.difference(elements_in)
original=[0 for i in range(n)]
items=[[a,list(b)]for a,b in list(hashmap.items())]
indices=cycle([i for i in range(len(items))])
index=next(indices)
counter=1
# original[items[index][1][0]] = items[index][0]
remains={}
while counter<=n-x:
if len(items[index][1])==0:
index = next(indices)
continue
if items[index][0] not in remains:
remains[items[index][0]]=[]
remains[items[index][0]].append(items[index][1][0])
hashmap[items[index][0]] =items[index][1][1:]
items[index][1]=items[index][1][1:]
index=next(indices)
counter+=1
items = [[a, list(b)] for a, b in list(hashmap.items())]
indices = cycle([i for i in range(len(items))])
index = next(indices)
counter=1
while counter <= x:
if len(items[index][1]) == 0:
index = next(indices)
continue
original[items[index][1][0]] = items[index][0]
hashmap[items[index][0]] = items[index][1][1:]
items[index][1] = items[index][1][1:]
index = next(indices)
counter += 1
hashmap={k:len(v) for k,v in remains.items()}
is_possible = False
counter=1
while counter<=residue:
is_possible=False
flag=0
for j in hashmap:
if hashmap[j] == 0:
continue
for i in range(0,n):
if j!=a[i] and original[i]==0:
is_possible=True
hashmap[j]-=1
original[i]=j
flag=1
break
if flag==1:
break
if is_possible==False:
break
counter+=1
if is_possible==False:
print("NO")
else:
elements_not_in=list(elements_not_in)
for i in range(0,len(original)):
if original[i]==0:
original[i]=elements_not_in[0]
print("YES")
print(*original)
```
|
instruction
| 0
| 96,306
| 11
| 192,612
|
No
|
output
| 1
| 96,306
| 11
| 192,613
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the game of Mastermind, there are two players — Alice and Bob. Alice has a secret code, which Bob tries to guess. Here, a code is defined as a sequence of n colors. There are exactly n+1 colors in the entire universe, numbered from 1 to n+1 inclusive.
When Bob guesses a code, Alice tells him some information about how good of a guess it is, in the form of two integers x and y.
The first integer x is the number of indices where Bob's guess correctly matches Alice's code. The second integer y is the size of the intersection of the two codes as multisets. That is, if Bob were to change the order of the colors in his guess, y is the maximum number of indices he could get correct.
For example, suppose n=5, Alice's code is [3,1,6,1,2], and Bob's guess is [3,1,1,2,5]. At indices 1 and 2 colors are equal, while in the other indices they are not equal. So x=2. And the two codes have the four colors 1,1,2,3 in common, so y=4.
<image> Solid lines denote a matched color for the same index. Dashed lines denote a matched color at a different index. x is the number of solid lines, and y is the total number of lines.
You are given Bob's guess and two values x and y. Can you find one possibility of Alice's code so that the values of x and y are correct?
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.
The first line of each test case contains three integers n,x,y (1≤ n≤ 10^5, 0≤ x≤ y≤ n) — the length of the codes, and two values Alice responds with.
The second line of each test case contains n integers b_1,…,b_n (1≤ b_i≤ n+1) — Bob's guess, where b_i is the i-th color of the guess.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
Output
For each test case, on the first line, output "YES" if there is a solution, or "NO" if there is no possible secret code consistent with the described situation. You can print each character in any case (upper or lower).
If the answer is "YES", on the next line output n integers a_1,…,a_n (1≤ a_i≤ n+1) — Alice's secret code, where a_i is the i-th color of the code.
If there are multiple solutions, output any.
Example
Input
7
5 2 4
3 1 1 2 5
5 3 4
1 1 2 1 2
4 0 4
5 5 3 3
4 1 4
2 3 2 3
6 1 2
3 2 1 1 1 1
6 2 4
3 3 2 1 1 1
6 2 6
1 1 3 2 1 1
Output
YES
3 1 6 1 2
YES
3 1 1 1 2
YES
3 3 5 5
NO
YES
4 4 4 4 3 1
YES
3 1 3 1 7 7
YES
2 3 1 1 1 1
Note
The first test case is described in the statement.
In the second test case, x=3 because the colors are equal at indices 2,4,5. And y=4 because they share the colors 1,1,1,2.
In the third test case, x=0 because there is no index where the colors are the same. But y=4 because they share the colors 3,3,5,5.
In the fourth test case, it can be proved that no solution exists.
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
from heapq import heappush, heappop, heapify
DEBUG = False
def solve(N, X, Y, B):
# Want X matching, and Y - X in derangement, and pad rest (pad possibly mixed with the derangements)
match = X
derange = Y - X
pad = N - match - derange
if DEBUG:
print()
print("test", t + 1)
print("derange", derange, "match", match, "pad", pad)
print("B")
print(B)
padVal = next(iter(set(range(1, N + 1)) - set(B)))
A = [padVal for i in range(N)]
if DEBUG:
print("after pad")
print(A)
pairs = []
unpaired = defaultdict(list)
for i, x in enumerate(B):
assert len(unpaired) <= 1
if not unpaired or x in unpaired:
unpaired[x].append(i)
else:
y, = unpaired.keys()
pairs.append((i, unpaired[y].pop()))
if not unpaired[y]:
del unpaired[y]
mustSkip = None
while derange:
if not pairs:
return 'NO'
i, j = pairs.pop()
assert B[i] != B[j]
if derange == 1:
A[i] = B[j]
mustSkip = j
derange -= 1
else:
A[i], A[j] = B[j], B[i]
derange -= 2
for i, x in enumerate(A):
if match:
if x == padVal and i != mustSkip:
A[i] = B[i]
match -= 1
else:
break
if match:
return 'NO'
if DEBUG:
print('after match')
print(A)
if DEBUG:
aFreq = Counter(A)
bFreq = Counter(B)
xCheck = sum(1 for a, b in zip(A, B) if a == b)
yCheck = 0
for k in aFreq.keys() & bFreq.keys():
yCheck += min(aFreq[k], bFreq[k])
assert xCheck == X
assert yCheck == Y
return "YES\n" + " ".join(str(x) for x in A)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
T = int(input())
for t in range(T):
N, X, Y = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]
ans = solve(N, X, Y, B)
print(ans)
```
|
instruction
| 0
| 96,307
| 11
| 192,614
|
No
|
output
| 1
| 96,307
| 11
| 192,615
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 40).
Output
Output a single string.
Examples
Input
2
Output
Adams
Input
8
Output
Van Buren
Input
29
Output
Harding
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem F
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
x = int(input())
a = [1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267]
print(a[x-1])
```
|
instruction
| 0
| 96,446
| 11
| 192,892
|
Yes
|
output
| 1
| 96,446
| 11
| 192,893
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
n=int(input())
s=input().split()
t=False
if n==1:
if s[0]=='0':
print('NO')
exit()
else:
print('YES')
exit()
else:
for i in range(n):
if s[i]=='0':
if t:
print('NO')
exit()
else:
t=True
if not t:
print('NO')
else:
print('YES')
```
|
instruction
| 0
| 96,623
| 11
| 193,246
|
Yes
|
output
| 1
| 96,623
| 11
| 193,247
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
def main():
n = int(input())
arr = list(map(int, input().split()))
total = sum(arr)
if n == 1 and total == 1:
print("YES")
elif n >= 2 and total == n - 1:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 96,624
| 11
| 193,248
|
Yes
|
output
| 1
| 96,624
| 11
| 193,249
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
n = int(input())
x = input().count('0')
print('YES' if x==(n!=1) else 'NO')
```
|
instruction
| 0
| 96,625
| 11
| 193,250
|
Yes
|
output
| 1
| 96,625
| 11
| 193,251
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
if (n == 1 and a != [1]) or (n != 1 and a.count(1) != n - 1):
print('NO')
else:
print('YES')
```
|
instruction
| 0
| 96,626
| 11
| 193,252
|
Yes
|
output
| 1
| 96,626
| 11
| 193,253
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
n = input()
x = input()
x = x.split()
if 1000>=int(n)>=1:
if int(n)== len(x):
if int(n)== 1 and x[0]=="0":
print("NO")
elif x.count("0") <=1:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
```
|
instruction
| 0
| 96,627
| 11
| 193,254
|
No
|
output
| 1
| 96,627
| 11
| 193,255
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
n=int(input())
s=input()
if n==1 and s=='1':
print('YES')
elif s.count('0')<=1:
print('YES')
elif s.count('0')==0 and s.count('1')>1:
print('NO')
else:
print('NO')
```
|
instruction
| 0
| 96,629
| 11
| 193,258
|
No
|
output
| 1
| 96,629
| 11
| 193,259
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with n buttons. Determine if it is fastened in a right way.
Input
The first line contains integer n (1 ≤ n ≤ 1000) — the number of buttons on the jacket.
The second line contains n integers ai (0 ≤ ai ≤ 1). The number ai = 0 if the i-th button is not fastened. Otherwise ai = 1.
Output
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Examples
Input
3
1 0 1
Output
YES
Input
3
1 0 0
Output
NO
Submitted Solution:
```
n=int(input())
l=sum(map(int,input().split()))
if n==1 or n==l+1: print("YES")
else: print("NO")
```
|
instruction
| 0
| 96,630
| 11
| 193,260
|
No
|
output
| 1
| 96,630
| 11
| 193,261
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,215
| 11
| 194,430
|
Tags: *special, implementation, interactive
Correct Solution:
```
try:
while input():
print('NO')
except EOFError:
pass
```
|
output
| 1
| 97,215
| 11
| 194,431
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,216
| 11
| 194,432
|
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
q = input()
except EOFError:
break
print("no", flush=True)
```
|
output
| 1
| 97,216
| 11
| 194,433
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,217
| 11
| 194,434
|
Tags: *special, implementation, interactive
Correct Solution:
```
while(1):
try:
s=input()
except:
break
print("NO",flush=True)
```
|
output
| 1
| 97,217
| 11
| 194,435
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,218
| 11
| 194,436
|
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
s=input()
print("NO")
except EOFError as e:
exit(0)
```
|
output
| 1
| 97,218
| 11
| 194,437
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,219
| 11
| 194,438
|
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
x = input()
except:
break
print("NO")
```
|
output
| 1
| 97,219
| 11
| 194,439
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,220
| 11
| 194,440
|
Tags: *special, implementation, interactive
Correct Solution:
```
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
lx=lambda x:map(int,input().split(x))
#from math import log10 ,log2,ceil,factorial as fac,gcd
#from itertools import combinations_with_replacement as cs
#from functools import reduce
#from bisect import bisect_right as br,bisect_left as bl
#from collections import Counter
import sys
#for _ in range(t()):
while 1:
try:
n=ss()
if n=="Is it rated?":
print("NO")
else:
print("YES")
except:
break
sys.stdout.flush()
'''
1+3+9
(3**n-1)//2
'''
```
|
output
| 1
| 97,220
| 11
| 194,441
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,221
| 11
| 194,442
|
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try :
line = input().strip()
print("No")
except EOFError:
break
```
|
output
| 1
| 97,221
| 11
| 194,443
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
|
instruction
| 0
| 97,222
| 11
| 194,444
|
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
input()
except EOFError:
break
print("NO")
```
|
output
| 1
| 97,222
| 11
| 194,445
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
try:
while True:
q=input()
print('NO')
except:
pass
```
|
instruction
| 0
| 97,223
| 11
| 194,446
|
Yes
|
output
| 1
| 97,223
| 11
| 194,447
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
print("NO\n"*99)
```
|
instruction
| 0
| 97,224
| 11
| 194,448
|
Yes
|
output
| 1
| 97,224
| 11
| 194,449
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
print(100*'NO''\n')
```
|
instruction
| 0
| 97,225
| 11
| 194,450
|
Yes
|
output
| 1
| 97,225
| 11
| 194,451
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
while True:
try:
s = input()
if s == "Is it rated?":
print('NO')
else:
break
except:
break
```
|
instruction
| 0
| 97,226
| 11
| 194,452
|
Yes
|
output
| 1
| 97,226
| 11
| 194,453
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
s = input()
print("NO")
```
|
instruction
| 0
| 97,227
| 11
| 194,454
|
No
|
output
| 1
| 97,227
| 11
| 194,455
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
n=input()
if(n=="Is it rated?"):
print("NO")
```
|
instruction
| 0
| 97,228
| 11
| 194,456
|
No
|
output
| 1
| 97,228
| 11
| 194,457
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
try:
s = input()
print("NO")
except:
exit(0)
```
|
instruction
| 0
| 97,229
| 11
| 194,458
|
No
|
output
| 1
| 97,229
| 11
| 194,459
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Interaction
This is an interactive problem. You need to read participants' queries from standard input and print your responses to standard output. You don't know the number of queries upfront, so you'll need to process them as you get them; you'll know you're done once you reach the end of the file.
In each query, you will be asked the question, written in one line. You have to answer it correctly, patiently and without any display of emotions. Your response is case-insensitive.
Please make sure to use the stream flushing operation after each response in order not to leave part of your output in some buffer.
Example
Input
Is it rated?
Is it rated?
Is it rated?
Output
NO
NO
NO
Submitted Solution:
```
print("No")
```
|
instruction
| 0
| 97,230
| 11
| 194,460
|
No
|
output
| 1
| 97,230
| 11
| 194,461
|
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