text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
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
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)))
```
| 95,655 | [
0.428955078125,
-0.1871337890625,
-0.061492919921875,
0.2130126953125,
-0.389404296875,
-0.63720703125,
0.1624755859375,
-0.0667724609375,
0.16748046875,
0.95751953125,
0.430419921875,
0.175048828125,
0.13134765625,
-0.404052734375,
-0.312255859375,
-0.2205810546875,
-0.501953125,
... | 11 |
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
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))
```
| 95,656 | [
0.4267578125,
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0.293212890625,
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0.2225341796875,
0.83740234375,
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0.100830078125,
-0.2705078125,
-0.4091796875,
-0.09588623046875,
-0.482177734375,
-0.7... | 11 |
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)
```
No
| 95,657 | [
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0.1944580078125,
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-0.5478515625,
-0.... | 11 |
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:])
```
Yes
| 95,895 | [
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0.27294921875,
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0.0648193359375,
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-0.8046875,
-0.2042236328125,
0.0266265869140625,
-0.958984375,... | 11 |
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()
```
Yes
| 95,896 | [
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0.0296173095703125,
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-0.030059814453125,
-1.0078125,
-0.590... | 11 |
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])
```
Yes
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0.007228851318359375,
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-0.5219726562... | 11 |
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)))
```
No
| 95,899 | [
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0.22119140625,
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0.1085205078125,
0.003910064697265625,
-0.88427734375,
-0.2159423828125,
-0.0044708251953125,
-0.99560546875,
-... | 11 |
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])
```
No
| 95,900 | [
0.63818359375,
0.126220703125,
-0.2159423828125,
0.141845703125,
-0.39306640625,
-0.35888671875,
-0.27685546875,
0.262451171875,
-0.27587890625,
0.49072265625,
0.2071533203125,
0.10906982421875,
0.0227813720703125,
-0.87646484375,
-0.2413330078125,
-0.03179931640625,
-1.0029296875,
... | 11 |
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)
```
No
| 95,901 | [
0.6435546875,
0.1800537109375,
-0.2236328125,
0.162109375,
-0.405517578125,
-0.333984375,
-0.33447265625,
0.2452392578125,
-0.272216796875,
0.487548828125,
0.2156982421875,
0.09552001953125,
0.00020003318786621094,
-0.87255859375,
-0.1710205078125,
-0.024993896484375,
-1.0087890625,
... | 11 |
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()
```
No
| 95,902 | [
0.65576171875,
0.100830078125,
-0.1925048828125,
0.10418701171875,
-0.392822265625,
-0.327880859375,
-0.269775390625,
0.29296875,
-0.310302734375,
0.415771484375,
0.2264404296875,
0.149658203125,
0.054901123046875,
-0.84130859375,
-0.2235107421875,
-0.0162353515625,
-0.982421875,
-... | 11 |
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)
```
Yes
| 96,284 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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)
```
Yes
| 96,285 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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)
```
Yes
| 96,286 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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()
```
Yes
| 96,287 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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()
```
No
| 96,288 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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)
```
No
| 96,289 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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)
```
No
| 96,290 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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()
```
No
| 96,291 | [
0.486572265625,
0.10638427734375,
-0.297119140625,
0.396240234375,
-0.7568359375,
-0.45263671875,
-0.12249755859375,
0.286865234375,
-0.057525634765625,
1.01171875,
0.4365234375,
-0.034576416015625,
0.117431640625,
-0.7392578125,
-0.208251953125,
0.071044921875,
-0.7451171875,
-0.8... | 11 |
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()
```
Yes
| 96,300 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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()
```
Yes
| 96,301 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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))
```
Yes
| 96,302 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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()
```
Yes
| 96,303 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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)
```
No
| 96,304 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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)))
```
No
| 96,305 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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)
```
No
| 96,306 | [
0.466552734375,
-0.27001953125,
-0.0017576217651367188,
0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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)
```
No
| 96,307 | [
0.466552734375,
-0.27001953125,
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0.1588134765625,
-0.4599609375,
-0.6708984375,
-0.06756591796875,
-0.03875732421875,
0.081787109375,
1.1787109375,
0.701171875,
-0.057525634765625,
0.272216796875,
-0.662109375,
-0.343994140625,
0.049774169921875,
-0.49194335937... | 11 |
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])
```
Yes
| 96,446 | [
0.433349609375,
0.06536865234375,
-0.1724853515625,
-0.16455078125,
-0.68115234375,
-0.275634765625,
-0.29248046875,
0.18798828125,
0.308349609375,
0.90625,
0.54931640625,
-0.040740966796875,
-0.04742431640625,
-0.85888671875,
-0.40234375,
-0.055145263671875,
-0.62939453125,
-0.871... | 11 |
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')
```
Yes
| 96,623 | [
0.1864013671875,
-0.054718017578125,
-0.0011653900146484375,
-0.07952880859375,
-0.428466796875,
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-0.08428955078125,
0.32568359375,
0.227294921875,
0.79833984375,
0.182861328125,
0.007297515869140625,
0.15673828125,
-0.53759765625,
-0.2484130859375,
0.2568359375,
-0.3... | 11 |
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()
```
Yes
| 96,624 | [
0.1915283203125,
-0.043060302734375,
0.00875091552734375,
-0.09716796875,
-0.43408203125,
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-0.11309814453125,
0.323486328125,
0.2364501953125,
0.73291015625,
0.13720703125,
-0.0228271484375,
0.181396484375,
-0.47265625,
-0.297119140625,
0.294677734375,
-0.412841796875,... | 11 |
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')
```
Yes
| 96,625 | [
0.2113037109375,
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0.0225067138671875,
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0.313232421875,
0.237060546875,
0.79248046875,
0.22119140625,
0.0328369140625,
0.1348876953125,
-0.5439453125,
-0.2724609375,
0.293701171875,
-0.38916015625,
-0... | 11 |
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')
```
Yes
| 96,626 | [
0.2081298828125,
-0.054351806640625,
0.00012421607971191406,
-0.06939697265625,
-0.460693359375,
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-0.1046142578125,
0.328125,
0.239990234375,
0.7744140625,
0.1463623046875,
0.01605224609375,
0.1851806640625,
-0.5,
-0.25634765625,
0.278076171875,
-0.381103515625,
-0.84... | 11 |
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")
```
No
| 96,627 | [
0.1881103515625,
-0.046783447265625,
0.00122833251953125,
-0.04962158203125,
-0.439208984375,
-0.302978515625,
-0.07794189453125,
0.328125,
0.235595703125,
0.78857421875,
0.1783447265625,
0.026885986328125,
0.1602783203125,
-0.49609375,
-0.2298583984375,
0.268310546875,
-0.3586425781... | 11 |
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')
```
No
| 96,629 | [
0.2152099609375,
-0.045684814453125,
-0.0108642578125,
-0.0546875,
-0.46533203125,
-0.2900390625,
-0.109375,
0.32958984375,
0.2281494140625,
0.771484375,
0.173095703125,
-0.00408935546875,
0.1470947265625,
-0.495849609375,
-0.268798828125,
0.259521484375,
-0.369873046875,
-0.838867... | 11 |
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")
```
No
| 96,630 | [
0.1873779296875,
-0.019012451171875,
0.004055023193359375,
-0.051849365234375,
-0.425537109375,
-0.312744140625,
-0.09564208984375,
0.334716796875,
0.2275390625,
0.74951171875,
0.1671142578125,
0.0269317626953125,
0.1654052734375,
-0.49951171875,
-0.2724609375,
0.294677734375,
-0.378... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
try:
while input():
print('NO')
except EOFError:
pass
```
| 97,215 | [
0.1767578125,
0.1978759765625,
-0.04241943359375,
0.131591796875,
-0.3916015625,
-0.88916015625,
-0.11651611328125,
0.156982421875,
-0.006671905517578125,
0.60302734375,
0.1171875,
0.2188720703125,
0.175537109375,
-0.501953125,
-0.587890625,
-0.48095703125,
-0.49755859375,
-1.01074... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
q = input()
except EOFError:
break
print("no", flush=True)
```
| 97,216 | [
0.12744140625,
0.2215576171875,
-0.0120086669921875,
0.14794921875,
-0.39208984375,
-0.9150390625,
-0.142822265625,
0.154541015625,
0.00510406494140625,
0.61669921875,
0.055572509765625,
0.2174072265625,
0.1619873046875,
-0.499267578125,
-0.6494140625,
-0.481201171875,
-0.455078125,
... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
while(1):
try:
s=input()
except:
break
print("NO",flush=True)
```
| 97,217 | [
0.10986328125,
0.21875,
0.0178375244140625,
0.1297607421875,
-0.37890625,
-0.92529296875,
-0.138916015625,
0.10406494140625,
-0.049774169921875,
0.6201171875,
0.03350830078125,
0.1878662109375,
0.1361083984375,
-0.490234375,
-0.61376953125,
-0.513671875,
-0.44189453125,
-1.04296875... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
s=input()
print("NO")
except EOFError as e:
exit(0)
```
| 97,218 | [
0.126220703125,
0.2049560546875,
-0.0183258056640625,
0.1053466796875,
-0.38330078125,
-0.91064453125,
-0.1343994140625,
0.168212890625,
-0.0016012191772460938,
0.58740234375,
0.1287841796875,
0.212890625,
0.1385498046875,
-0.479736328125,
-0.57568359375,
-0.50537109375,
-0.467529296... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
x = input()
except:
break
print("NO")
```
| 97,219 | [
0.1519775390625,
0.2222900390625,
-0.0325927734375,
0.12451171875,
-0.40576171875,
-0.9072265625,
-0.1441650390625,
0.12255859375,
-0.047515869140625,
0.59423828125,
0.0906982421875,
0.1790771484375,
0.11669921875,
-0.495361328125,
-0.58056640625,
-0.471435546875,
-0.48681640625,
-... | 11 |
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
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
'''
```
| 97,220 | [
0.059600830078125,
0.254638671875,
0.01517486572265625,
0.229248046875,
-0.5224609375,
-0.642578125,
-0.09869384765625,
0.039154052734375,
-0.1917724609375,
0.7626953125,
0.2149658203125,
0.1689453125,
0.10302734375,
-0.81005859375,
-0.52734375,
-0.353515625,
-0.51220703125,
-1.106... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try :
line = input().strip()
print("No")
except EOFError:
break
```
| 97,221 | [
0.12646484375,
0.214111328125,
-0.0306854248046875,
0.1298828125,
-0.419189453125,
-0.90771484375,
-0.12176513671875,
0.133544921875,
0.013641357421875,
0.58642578125,
0.10760498046875,
0.175048828125,
0.12005615234375,
-0.484130859375,
-0.62890625,
-0.49072265625,
-0.486328125,
-1... | 11 |
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
Tags: *special, implementation, interactive
Correct Solution:
```
while True:
try:
input()
except EOFError:
break
print("NO")
```
| 97,222 | [
0.126220703125,
0.1861572265625,
-0.059722900390625,
0.1365966796875,
-0.378173828125,
-0.88916015625,
-0.139892578125,
0.1441650390625,
0.0445556640625,
0.58984375,
0.10003662109375,
0.25,
0.179443359375,
-0.4951171875,
-0.595703125,
-0.48876953125,
-0.5048828125,
-1.0146484375,
... | 11 |
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
```
Yes
| 97,223 | [
0.3779296875,
0.1983642578125,
-0.201416015625,
-0.058685302734375,
-0.58447265625,
-0.5654296875,
-0.202392578125,
0.38427734375,
-0.0858154296875,
0.7451171875,
-0.015289306640625,
0.1767578125,
0.06292724609375,
-0.53662109375,
-0.58349609375,
-0.57421875,
-0.387451171875,
-0.82... | 11 |
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)
```
Yes
| 97,224 | [
0.330322265625,
0.2093505859375,
-0.184326171875,
-0.0853271484375,
-0.599609375,
-0.58251953125,
-0.204833984375,
0.352294921875,
-0.1014404296875,
0.63037109375,
-0.0176849365234375,
0.112548828125,
0.112548828125,
-0.48193359375,
-0.56640625,
-0.5537109375,
-0.37548828125,
-0.88... | 11 |
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')
```
Yes
| 97,225 | [
0.326416015625,
0.1859130859375,
-0.1756591796875,
-0.06146240234375,
-0.61279296875,
-0.626953125,
-0.192138671875,
0.339111328125,
-0.0904541015625,
0.62646484375,
0.0297393798828125,
0.1134033203125,
0.163818359375,
-0.51318359375,
-0.5390625,
-0.57958984375,
-0.406005859375,
-0... | 11 |
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
```
Yes
| 97,226 | [
0.33056640625,
0.2060546875,
-0.2252197265625,
-0.059906005859375,
-0.58740234375,
-0.61767578125,
-0.0406494140625,
0.29541015625,
-0.09405517578125,
0.7080078125,
0.0130157470703125,
0.159912109375,
0.1783447265625,
-0.6181640625,
-0.44189453125,
-0.55322265625,
-0.45751953125,
-... | 11 |
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")
```
No
| 97,227 | [
0.29833984375,
0.1995849609375,
-0.173095703125,
-0.126708984375,
-0.634765625,
-0.6005859375,
-0.211669921875,
0.359130859375,
-0.0948486328125,
0.6337890625,
0.040557861328125,
0.1614990234375,
0.1019287109375,
-0.5390625,
-0.56689453125,
-0.638671875,
-0.406005859375,
-0.8232421... | 11 |
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")
```
No
| 97,228 | [
0.376708984375,
0.1846923828125,
-0.2392578125,
-0.07489013671875,
-0.59716796875,
-0.63818359375,
0.0044403076171875,
0.323974609375,
-0.108154296875,
0.6689453125,
0.028045654296875,
0.120361328125,
0.1617431640625,
-0.58984375,
-0.388427734375,
-0.490234375,
-0.48486328125,
-0.9... | 11 |
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)
```
No
| 97,229 | [
0.299072265625,
0.1756591796875,
-0.1580810546875,
-0.09979248046875,
-0.5615234375,
-0.62744140625,
-0.2435302734375,
0.3642578125,
-0.07061767578125,
0.6689453125,
0.042083740234375,
0.1619873046875,
0.07501220703125,
-0.53076171875,
-0.57763671875,
-0.625,
-0.391357421875,
-0.81... | 11 |
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")
```
No
| 97,230 | [
0.296630859375,
0.18994140625,
-0.1668701171875,
-0.114013671875,
-0.5927734375,
-0.6376953125,
-0.2215576171875,
0.378662109375,
-0.09710693359375,
0.619140625,
-0.0019178390502929688,
0.130859375,
0.119140625,
-0.50244140625,
-0.5556640625,
-0.62548828125,
-0.3955078125,
-0.85693... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence.
Submitted Solution:
```
data = []
i = int(input())
words = list(map(str, input().split(' ')))
j = int(input())
for t in range(j):
ok = list(map(str, input().split(' ')))
data.append([int(ok[0]), ok[1:]])
if i == 1:
inv = []
for x in range(len(data)):
if words[0] in data[x][1]:
print(x)
print('[:|:]')
quit()
print("Brand new problem!")
elif i == 2:
inv = []
a = words[0]
b = words[1]
for x in range(len(data)):
invx = 0
for i in range(data[x][0]):
for j in range(data[x][0]):
if i != j:
if data[x][1][i] == a and data[x][1][j] == b:
if i > j:
invx+=1
inv.append([x, invx])
inv.sort(key = lambda x : x[1])
if inv == []:
print("Brand new problem!")
else:
print(inv[0][0]+1)
print('[:' + (1+1-inv[0][1]) * '|' + ':]')
elif i == 3:
inv = []
a = words[0]
b = words[1]
c = words[2]
for x in range(len(data)):
invx = 0
for i in range(data[x][0]):
for j in range(data[x][0]):
for k in range(data[x][0]):
if i != j and j != k and i != k:
if data[x][1][i] == a and data[x][1][j] == b and data[x][1][k] == c:
if i > j:
invx+=1
if i > k:
invx += 1
if j > k:
invx += 1
inv.append([x, invx])
inv.sort(key = lambda x : x[1])
if inv == []:
print("Brand new problem!")
else:
print(inv[0][0]+1)
print('[:' + (3+1-inv[0][1]) * '|' + ':]')
elif i == 4:
inv = []
a = words[0]
b = words[1]
c = words[2]
d = words[3]
for x in range(len(data)):
invx = 0
for i in range(data[x][0]):
for j in range(data[x][0]):
for k in range(data[x][0]):
for l in range(data[x][0]):
if i != j and j != k and i != k and i != l and j != l and k != l:
if data[x][1][i] == a and data[x][1][j] == b and data[x][1][k] == c and data[x][1][l] == d:
if i > j:
invx+=1
if i > k:
invx += 1
if j > k:
invx += 1
if j > l:
invx += 1
if i > l:
invx += 1
if k > l:
invx += 1
inv.append([x, invx])
inv.sort(key = lambda x : x[1])
if inv == []:
print("Brand new problem!")
else:
print(inv[0][0]+1)
print('[:' + (6+1-inv[0][1]) * '|' + ':]')
```
No
| 97,265 | [
0.2259521484375,
0.294677734375,
0.0555419921875,
0.2052001953125,
-0.5361328125,
-0.59130859375,
-0.353271484375,
0.046356201171875,
-0.277099609375,
0.984375,
0.75634765625,
-0.026031494140625,
0.302001953125,
-0.68017578125,
-0.462646484375,
0.0672607421875,
-0.51513671875,
-0.6... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A widely known among some people Belarusian sport programmer Lesha decided to make some money to buy a one square meter larger flat. To do this, he wants to make and carry out a Super Rated Match (SRM) on the site Torcoder.com. But there's a problem — a severe torcoder coordinator Ivan does not accept any Lesha's problem, calling each of them an offensive word "duped" (that is, duplicated). And one day they nearely quarrelled over yet another problem Ivan wouldn't accept.
You are invited to act as a fair judge and determine whether the problem is indeed brand new, or Ivan is right and the problem bears some resemblance to those used in the previous SRMs.
You are given the descriptions of Lesha's problem and each of Torcoder.com archive problems. The description of each problem is a sequence of words. Besides, it is guaranteed that Lesha's problem has no repeated words, while the description of an archive problem may contain any number of repeated words.
The "similarity" between Lesha's problem and some archive problem can be found as follows. Among all permutations of words in Lesha's problem we choose the one that occurs in the archive problem as a subsequence. If there are multiple such permutations, we choose the one with the smallest number of inversions. Then the "similarity" of a problem can be written as <image>, where n is the number of words in Lesha's problem and x is the number of inversions in the chosen permutation. Note that the "similarity" p is always a positive integer.
The problem is called brand new if there is not a single problem in Ivan's archive which contains a permutation of words from Lesha's problem as a subsequence.
Help the boys and determine whether the proposed problem is new, or specify the problem from the archive which resembles Lesha's problem the most, otherwise.
Input
The first line contains a single integer n (1 ≤ n ≤ 15) — the number of words in Lesha's problem. The second line contains n space-separated words — the short description of the problem.
The third line contains a single integer m (1 ≤ m ≤ 10) — the number of problems in the Torcoder.com archive. Next m lines contain the descriptions of the problems as "k s1 s2 ... sk", where k (1 ≤ k ≤ 500000) is the number of words in the problem and si is a word of the problem description.
All words from all problem descriptions contain no more than 10 lowercase English letters. It is guaranteed that the total length of words in all problem descriptions does not exceed 500015.
Output
If Lesha's problem is brand new, print string "Brand new problem!" (without quotes).
Otherwise, on the first line print the index of the archive problem which resembles Lesha's problem most. If there are multiple such problems, print the one with the smallest index. On the second line print a string consisting of characters [:, character | repeated p times, and characters :], where p is the "similarity" between this problem and Lesha's one. The archive problems are numbered starting from one in the order in which they are given in the input.
Examples
Input
4
find the next palindrome
1
10 find the previous palindrome or print better luck next time
Output
1
[:||||||:]
Input
3
add two numbers
3
1 add
2 two two
3 numbers numbers numbers
Output
Brand new problem!
Input
4
these papers are formulas
3
6 what are these formulas and papers
5 papers are driving me crazy
4 crazy into the night
Output
1
[:||||:]
Input
3
add two decimals
5
4 please two decimals add
5 decimals want to be added
4 two add decimals add
4 add one two three
7 one plus two plus three equals six
Output
3
[:|||:]
Note
Let us remind you that the number of inversions is the number of pairs of words that follow in the permutation not in their original order. Thus, for example, if the original problem is "add two numbers", then permutation "numbers add two" contains two inversions — pairs of words "numbers" and "add", "numbers" and "two".
Sequence b1, b2, ..., bk is a subsequence of sequence a1, a2, ..., an if there exists such a set of indices 1 ≤ i1 < i2 < ... < ik ≤ n that aij = bj (in other words, if sequence b can be obtained from a by deleting some of its elements).
In the first test case the first problem contains the "find the palindrome next" permutation as a subsequence, in which the number of inversions equals 1 (words "palindrome" and "next").
In the second test case there is no problem that contains a permutation of words from Lesha's problem as a subsequence.
Submitted Solution:
```
import itertools
def check(curr_words, line):
if curr_words == []:
return True
for i in range(len(line)):
if line[i] == curr_words[0]:
return check(curr_words[1:], line[i+1:])
return False
n = int(input())
words = input().split()
m = int(input())
res, idx = 0, 0
for i in range(m):
line = input().split()[1:]
for p in itertools.permutations(range(n)):
curr_words = [words[j] for j in p]
cnt = 0
for j in range(n):
cnt += len([k for k in range(j+1, n) if p[k] < p[j]])
v = n * (n-1) // 2 - cnt + 1
if check(curr_words, line[:]) and v > res:
res, idx = v, i+1
if res > 0:
print(idx)
print('[:'+str('|'*res)+':]')
else:
print('Brand new wordslem!')
```
No
| 97,266 | [
0.2259521484375,
0.294677734375,
0.0555419921875,
0.2052001953125,
-0.5361328125,
-0.59130859375,
-0.353271484375,
0.046356201171875,
-0.277099609375,
0.984375,
0.75634765625,
-0.026031494140625,
0.302001953125,
-0.68017578125,
-0.462646484375,
0.0672607421875,
-0.51513671875,
-0.6... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integer arrays a_0, a_1, ..., a_{N - 1} and b_0, b_1, ..., b_{M - 1}. Calculate the array c_0, c_1, ..., c_{(N - 1) + (M - 1)}, defined by c_i = \sum_{j = 0}^i a_j b_{i - j} \bmod 998244353.
Constraints
* 1 \leq N, M \leq 524288
* 0 \leq a_i, b_i < 998244353
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N M
a_0 a_1 ... a_{N-1}
b_0 b_1 ... b_{M-1}
Output
Print the answer in the following format:
c_0 c_1 ... c_{(N - 1) + (M - 1)}
Output
Print the answer in the following format:
c_0 c_1 ... c_{(N - 1) + (M - 1)}
Examples
Input
4 5
1 2 3 4
5 6 7 8 9
Output
5 16 34 60 70 70 59 36
Input
1 1
10000000
10000000
Output
871938225
Submitted Solution:
```
import types
_atcoder_code = """
# Python port of AtCoder Library.
__version__ = '0.0.1'
"""
atcoder = types.ModuleType('atcoder')
exec(_atcoder_code, atcoder.__dict__)
_atcoder__bit_code = """
def _ceil_pow2(n: int) -> int:
x = 0
while (1 << x) < n:
x += 1
return x
def _bsf(n: int) -> int:
x = 0
while n % 2 == 0:
x += 1
n //= 2
return x
"""
atcoder._bit = types.ModuleType('atcoder._bit')
exec(_atcoder__bit_code, atcoder._bit.__dict__)
_atcoder__math_code = """
import typing
def _is_prime(n: int) -> bool:
'''
Reference:
M. Forisek and J. Jancina,
Fast Primality Testing for Integers That Fit into a Machine Word
'''
if n <= 1:
return False
if n == 2 or n == 7 or n == 61:
return True
if n % 2 == 0:
return False
d = n - 1
while d % 2 == 0:
d //= 2
for a in (2, 7, 61):
t = d
y = pow(a, t, n)
while t != n - 1 and y != 1 and y != n - 1:
y = y * y % n
t <<= 1
if y != n - 1 and t % 2 == 0:
return False
return True
def _inv_gcd(a: int, b: int) -> typing.Tuple[int, int]:
a %= b
if a == 0:
return (b, 0)
# Contracts:
# [1] s - m0 * a = 0 (mod b)
# [2] t - m1 * a = 0 (mod b)
# [3] s * |m1| + t * |m0| <= b
s = b
t = a
m0 = 0
m1 = 1
while t:
u = s // t
s -= t * u
m0 -= m1 * u # |m1 * u| <= |m1| * s <= b
# [3]:
# (s - t * u) * |m1| + t * |m0 - m1 * u|
# <= s * |m1| - t * u * |m1| + t * (|m0| + |m1| * u)
# = s * |m1| + t * |m0| <= b
s, t = t, s
m0, m1 = m1, m0
# by [3]: |m0| <= b/g
# by g != b: |m0| < b/g
if m0 < 0:
m0 += b // s
return (s, m0)
def _primitive_root(m: int) -> int:
if m == 2:
return 1
if m == 167772161:
return 3
if m == 469762049:
return 3
if m == 754974721:
return 11
if m == 998244353:
return 3
divs = [2] + [0] * 19
cnt = 1
x = (m - 1) // 2
while x % 2 == 0:
x //= 2
i = 3
while i * i <= x:
if x % i == 0:
divs[cnt] = i
cnt += 1
while x % i == 0:
x //= i
i += 2
if x > 1:
divs[cnt] = x
cnt += 1
g = 2
while True:
for i in range(cnt):
if pow(g, (m - 1) // divs[i], m) == 1:
break
else:
return g
g += 1
"""
atcoder._math = types.ModuleType('atcoder._math')
exec(_atcoder__math_code, atcoder._math.__dict__)
_atcoder_modint_code = """
from __future__ import annotations
import typing
# import atcoder._math
class ModContext:
context = []
def __init__(self, mod: int) -> None:
assert 1 <= mod
self.mod = mod
def __enter__(self) -> None:
self.context.append(self.mod)
def __exit__(self, exc_type: typing.Any, exc_value: typing.Any,
traceback: typing.Any) -> None:
self.context.pop()
@classmethod
def get_mod(cls) -> int:
return cls.context[-1]
class Modint:
def __init__(self, v: int = 0) -> None:
self._mod = ModContext.get_mod()
if v == 0:
self._v = 0
else:
self._v = v % self._mod
def val(self) -> int:
return self._v
def __iadd__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
self._v += rhs._v
else:
self._v += rhs
if self._v >= self._mod:
self._v -= self._mod
return self
def __isub__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
self._v -= rhs._v
else:
self._v -= rhs
if self._v < 0:
self._v += self._mod
return self
def __imul__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
self._v = self._v * rhs._v % self._mod
else:
self._v = self._v * rhs % self._mod
return self
def __ifloordiv__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
inv = rhs.inv()._v
else:
inv = atcoder._math._inv_gcd(rhs, self._mod)[1]
self._v = self._v * inv % self._mod
return self
def __pos__(self) -> Modint:
return self
def __neg__(self) -> Modint:
return Modint() - self
def __pow__(self, n: int) -> Modint:
assert 0 <= n
return Modint(pow(self._v, n, self._mod))
def inv(self) -> Modint:
eg = atcoder._math._inv_gcd(self._v, self._mod)
assert eg[0] == 1
return Modint(eg[1])
def __add__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
result = self._v + rhs._v
if result >= self._mod:
result -= self._mod
return raw(result)
else:
return Modint(self._v + rhs)
def __sub__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
result = self._v - rhs._v
if result < 0:
result += self._mod
return raw(result)
else:
return Modint(self._v - rhs)
def __mul__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
return Modint(self._v * rhs._v)
else:
return Modint(self._v * rhs)
def __floordiv__(self, rhs: typing.Union[Modint, int]) -> Modint:
if isinstance(rhs, Modint):
inv = rhs.inv()._v
else:
inv = atcoder._math._inv_gcd(rhs, self._mod)[1]
return Modint(self._v * inv)
def __eq__(self, rhs: typing.Union[Modint, int]) -> bool:
if isinstance(rhs, Modint):
return self._v == rhs._v
else:
return self._v == rhs
def __ne__(self, rhs: typing.Union[Modint, int]) -> bool:
if isinstance(rhs, Modint):
return self._v != rhs._v
else:
return self._v != rhs
def raw(v: int) -> Modint:
x = Modint()
x._v = v
return x
"""
atcoder.modint = types.ModuleType('atcoder.modint')
atcoder.modint.__dict__['atcoder'] = atcoder
atcoder.modint.__dict__['atcoder._math'] = atcoder._math
exec(_atcoder_modint_code, atcoder.modint.__dict__)
ModContext = atcoder.modint.ModContext
Modint = atcoder.modint.Modint
_atcoder_convolution_code = """
import typing
# import atcoder._bit
# import atcoder._math
# from atcoder.modint import ModContext, Modint
_sum_e = {} # _sum_e[i] = ies[0] * ... * ies[i - 1] * es[i]
def _butterfly(a: typing.List[Modint]) -> None:
g = atcoder._math._primitive_root(a[0].mod())
n = len(a)
h = atcoder._bit._ceil_pow2(n)
if a[0].mod() not in _sum_e:
es = [0] * 30 # es[i]^(2^(2+i)) == 1
ies = [0] * 30
cnt2 = atcoder._bit._bsf(a[0].mod() - 1)
e = Modint(g).pow((a[0].mod() - 1) >> cnt2)
ie = e.inv()
for i in range(cnt2, 1, -1):
# e^(2^i) == 1
es[i - 2] = e
ies[i - 2] = ie
e *= e
ie *= ie
sum_e = [0] * 30
now = Modint(1)
for i in range(cnt2 - 2):
sum_e[i] = es[i] * now
now *= ies[i]
_sum_e[a[0].mod()] = sum_e
else:
sum_e = _sum_e[a[0].mod()]
for ph in (1, h + 1):
w = 1 << (ph - 1)
p = 1 << (h - ph)
now = Modint(1)
for s in range(w):
offset = s << (h - ph + 1)
for i in range(p):
left = a[i + offset]
right = a[i + offset + p] * now
a[i + offset] = left + right
a[i + offset + p] = left - right
now *= sum_e[atcoder._bit._bsf(~s)]
_sum_ie = {} # _sum_ie[i] = es[0] * ... * es[i - 1] * ies[i]
def _butterfly_inv(a: typing.List[Modint]) -> None:
g = atcoder._math._primitive_root(a[0].mod())
n = len(a)
h = atcoder._bit.ceil_pow2(n)
if a[0].mod() not in _sum_ie:
es = [0] * 30 # es[i]^(2^(2+i)) == 1
ies = [0] * 30
cnt2 = atcoder._bit._bsf(a[0].mod() - 1)
e = Modint(g).pow((a[0].mod() - 1) >> cnt2)
ie = e.inv()
for i in range(cnt2, 1, -1):
# e^(2^i) == 1
es[i - 2] = e
ies[i - 2] = ie
e *= e
ie *= ie
sum_ie = [0] * 30
now = Modint(1)
for i in range(cnt2 - 2):
sum_ie[i] = ies[i] * now
now *= es[i]
_sum_ie[a[0].mod()] = sum_ie
else:
sum_ie = _sum_ie[a[0].mod()]
for ph in range(h, 0, -1):
w = 1 << (ph - 1)
p = 1 << (h - ph)
inow = Modint(1)
for s in range(w):
offset = s << (h - ph + 1)
for i in range(p):
left = a[i + offset]
right = a[i + offset + p]
a[i + offset] = left + right
a[i + offset + p] = (a[0].mod() +
left.val() - right.val()) * inow.val()
inow *= sum_ie[atcoder._bit._bsf(~s)]
def convolution_mod(a: typing.List[Modint],
b: typing.List[Modint]) -> typing.List[Modint]:
n = len(a)
m = len(b)
if n == 0 or m == 0:
return []
if min(n, m) <= 60:
if n < m:
n, m = m, n
a, b = b, a
ans = [Modint(0) for _ in range(n + m - 1)]
for i in range(n):
for j in range(m):
ans[i + j] += a[i] * b[j]
return ans
z = 1 << atcoder._bit._ceil_pow2(n + m - 1)
while len(a) < z:
a.append(Modint(0))
_butterfly(a)
while len(b) < z:
b.append(Modint(0))
_butterfly(b)
for i in range(z):
a[i] *= b[i]
_butterfly_inv(a)
a = a[:n + m - 1]
iz = Modint(z).inv()
for i in range(n + m - 1):
a[i] *= iz
return a
def convolution(mod: int, a: typing.List[typing.Any],
b: typing.List[typing.Any]) -> typing.List[typing.Any]:
n = len(a)
m = len(b)
if n == 0 or m == 0:
return []
with ModContext(mod):
a2 = list(map(Modint, a))
b2 = list(map(Modint, b))
return list(map(lambda c: c.val(), convolution_mod(a2, b2)))
def convolution_int(
a: typing.List[int], b: typing.List[int]) -> typing.List[int]:
n = len(a)
m = len(b)
if n == 0 or m == 0:
return []
mod1 = 754974721 # 2^24
mod2 = 167772161 # 2^25
mod3 = 469762049 # 2^26
m2m3 = mod2 * mod3
m1m3 = mod1 * mod3
m1m2 = mod1 * mod2
m1m2m3 = mod1 * mod2 * mod3
i1 = atcoder._math._inv_gcd(mod2 * mod3, mod1)[1]
i2 = atcoder._math._inv_gcd(mod1 * mod3, mod2)[1]
i3 = atcoder._math._inv_gcd(mod1 * mod2, mod3)[1]
c1 = convolution(mod1, a, b)
c2 = convolution(mod2, a, b)
c3 = convolution(mod3, a, b)
c = [0] * (n + m - 1)
for i in range(n + m - 1):
c[i] += (c1[i] * i1) % mod1 * m2m3
c[i] += (c2[i] * i2) % mod2 * m1m3
c[i] += (c3[i] * i3) % mod3 * m1m2
c[i] %= m1m2m3
return c
"""
atcoder.convolution = types.ModuleType('atcoder.convolution')
atcoder.convolution.__dict__['atcoder'] = atcoder
atcoder.convolution.__dict__['atcoder._bit'] = atcoder._bit
atcoder.convolution.__dict__['atcoder._math'] = atcoder._math
atcoder.convolution.__dict__['ModContext'] = atcoder.modint.ModContext
atcoder.convolution.__dict__['Modint'] = atcoder.modint.Modint
exec(_atcoder_convolution_code, atcoder.convolution.__dict__)
convolution_int = atcoder.convolution.convolution_int
# https://atcoder.jp/contests/practice2/tasks/practice2_f
import sys
# from atcoder.convolution import convolution_int
def main() -> None:
n, m = map(int, sys.stdin.readline().split())
a = list(map(int, sys.stdin.readline().split()))
b = list(map(int, sys.stdin.readline().split()))
c = convolution_int(a, b)
print(' '.join([str(ci % 998244353) for ci in c]))
if __name__ == '__main__':
main()
```
No
| 97,604 | [
0.482421875,
0.05938720703125,
-0.471435546875,
-0.00643157958984375,
-0.51708984375,
-0.45068359375,
-0.275634765625,
0.188232421875,
0.2071533203125,
0.47607421875,
0.402099609375,
-0.340087890625,
0.154296875,
-0.8583984375,
-0.2314453125,
-0.1326904296875,
-0.391845703125,
-1.1... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
n = int(input())
R = [int(i) for i in input().split()]
L = [abs(R[i]-R[i+1]) for i in range(n-1)]
ans = [0 for _ in range(n)]
ans[0] = L[0]
for i in range(1, n-1):
ans[i] = max(L[i], L[i]-ans[i-1])
if i - 2 >= 0:
ans[i] = max(ans[i], L[i]-L[i-1]+ans[i-2])
print(max(ans))
```
Yes
| 98,320 | [
0.422607421875,
0.373779296875,
0.01216888427734375,
0.443603515625,
-0.66552734375,
-0.42822265625,
0.2115478515625,
0.10223388671875,
-0.1839599609375,
0.93115234375,
0.8193359375,
0.05224609375,
0.10211181640625,
-0.412109375,
-0.537109375,
0.2109375,
-0.96435546875,
-0.93164062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=998244353
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def Answer(a):
till=0
ans=-inf
for i in a:
till+=i
ans=max(till,ans)
till=max(0,till)
return ans
n=Int()
a=array()
a=[abs(a[i]-a[i+1]) for i in range(n-1)]
# print(*a)
n-=1
a=[a[i]*(-1)**i for i in range(n)]
b=[-i for i in a]
print(max(Answer(a),Answer(b)))
```
Yes
| 98,321 | [
0.332275390625,
0.329345703125,
-0.06658935546875,
0.5712890625,
-0.65576171875,
-0.379150390625,
0.1551513671875,
0.126953125,
-0.0562744140625,
0.9287109375,
0.78125,
-0.00429534912109375,
0.1484375,
-0.44775390625,
-0.4921875,
0.25341796875,
-0.9296875,
-1.0537109375,
-0.18725... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
def f(l, r, p):
if l > r: return 0
return p[r] - p[l - 1] if l % 2 == 1 else -f(l - 1, r, p) + p[l - 1]
def main():
read = lambda: tuple(map(int, input().split()))
n = read()[0]
v = read()
p = [0]
pv = 0
for i in range(n - 1):
cp = abs(v[i] - v[i + 1]) * (-1) ** i
pv += cp
p += [pv]
mxc, mxn = 0, 0
mnc, mnn = 0, 0
for i in range(n):
cc, cn = f(1, i, p), f(2, i, p)
mxc, mxn = max(mxc, cc - mnc), max(mxn, cn - mnn)
mnc, mnn = min(mnc, cc), min(mnn, cn)
return max(mxc, mxn)
print(main())
```
Yes
| 98,322 | [
0.415771484375,
0.36083984375,
0.041015625,
0.41796875,
-0.66015625,
-0.404541015625,
0.19189453125,
0.0823974609375,
-0.2215576171875,
0.9228515625,
0.83984375,
0.01776123046875,
0.09228515625,
-0.391357421875,
-0.53369140625,
0.2205810546875,
-1.01171875,
-0.96240234375,
-0.160... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
from sys import stdin, stdout
n = int(stdin.readline())
values = list(map(int, stdin.readline().split()))
first = []
second = []
ans = float('-inf')
for i in range(n - 1):
first.append(abs(values[i] - values[i + 1]) * (-1) ** i)
second.append(abs(values[i] - values[i + 1]) * (-1) ** (i + 1))
cnt = 0
for i in range(n - 1):
cnt += first[i]
ans = max(ans, cnt)
if cnt < 0:
cnt = 0
cnt = 0
for i in range(n - 1):
cnt += second[i]
ans = max(ans, cnt)
if cnt < 0:
cnt = 0
stdout.write(str(ans))
```
Yes
| 98,323 | [
0.3525390625,
0.370849609375,
0.0792236328125,
0.449462890625,
-0.6884765625,
-0.424560546875,
0.152099609375,
0.11358642578125,
-0.1805419921875,
0.9970703125,
0.74169921875,
0.0472412109375,
0.158935546875,
-0.3779296875,
-0.5009765625,
0.203369140625,
-0.9619140625,
-0.991210937... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().strip().split()))
b = [0]*(n-1)
c = [0]*(n-1)
for i in range(n-1):
if (i+1)%2 == 0:
b[i] = (abs(a[i] - a[i+1]))
else:
b[i] = (abs(a[i] - a[i+1]))*(-1)
for i in range(1,n-1):
if i%2 == 0:
c[i] = (abs(a[i] - a[i+1]))
else:
c[i] = (abs(a[i] - a[i+1]))*(-1)
d1 = [-1000000001]*(n-1)
d2 = [-1000000001]*(n-1)
for i in range(n-1):
if i > 0:
d1[i] = max(b[i]+d1[i-1],max(0,b[i]))
else:
d1[i] = max(b[i],0)
for i in range(n-1):
if i > 0:
d2[i] = max(c[i]+d2[i-1],max(0,c[i]))
else:
d2[i] = max(c[i],0)
print(max(max(d1),max(d2)))
## print(b)
## print(d1)
## print(c)
## print(d2)
```
No
| 98,324 | [
0.39111328125,
0.40234375,
0.03192138671875,
0.4365234375,
-0.69140625,
-0.39404296875,
0.17138671875,
0.1026611328125,
-0.22021484375,
0.95166015625,
0.8232421875,
0.051971435546875,
0.17041015625,
-0.381103515625,
-0.4990234375,
0.231689453125,
-0.95947265625,
-0.93701171875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
from collections import defaultdict as dd
def main():
n = int(input())
A = list(map(int, input().split()))
# print(A)
B = []
for i in range(1, len(A)):
B.append(abs(A[i]-A[i-1]))
# print(B)
Dp = dd(int)
Dm = dd(int)
Dp[0]=0
for i in range(n-1):
Dm[i] = Dp[i-1] + B[i]
Dp[i] = max(Dm[i-1] - B[i], 0)
print(max(Dm[n-2], Dp[n-2]))
if __name__ == "__main__":
main()
```
No
| 98,325 | [
0.345458984375,
0.310546875,
0.020782470703125,
0.463134765625,
-0.6796875,
-0.350341796875,
0.192138671875,
0.0909423828125,
-0.17333984375,
0.96923828125,
0.728515625,
-0.007465362548828125,
0.1378173828125,
-0.37158203125,
-0.52294921875,
0.218017578125,
-1.01171875,
-0.93408203... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
a = LI()
r = t = abs(a[0]-a[1])
for i in range(3,n,2):
t += abs(a[i]-a[i-1]) - abs(a[i-1]-a[i-2])
if r < t:
r = t
elif t < 0:
t = 0
if n < 3:
return r
t = abs(a[1]-a[2])
for i in range(4,n,2):
t += abs(a[i]-a[i-1]) - abs(a[i-1]-a[i-2])
if r < t:
r = t
elif t < 0:
t = 0
return r
print(main())
```
No
| 98,326 | [
0.3837890625,
0.37158203125,
0.0523681640625,
0.45068359375,
-0.72900390625,
-0.352294921875,
0.19921875,
0.055908203125,
-0.1463623046875,
0.9755859375,
0.76220703125,
-0.00685882568359375,
0.118896484375,
-0.3837890625,
-0.51513671875,
0.248046875,
-0.99853515625,
-1.0263671875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:
<image>
In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.
Output
Print the only integer — the maximum value of f.
Examples
Input
5
1 4 2 3 1
Output
3
Input
4
1 5 4 7
Output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
a = LI()
r = t = abs(a[0]-a[1])
for i in range(3,n,2):
t += abs(a[i]-a[i-1]) - abs(a[i-1]-a[i-2])
if r < t:
r = t
elif t < 0:
t = 0
t = 0
for i in range(2,n,2):
t += abs(a[i]-a[i-1]) - abs(a[i-1]-a[i-2])
if r < t:
r = t
elif t < 0:
t = 0
return r
print(main())
```
No
| 98,327 | [
0.3837890625,
0.37158203125,
0.0523681640625,
0.45068359375,
-0.72900390625,
-0.352294921875,
0.19921875,
0.055908203125,
-0.1463623046875,
0.9755859375,
0.76220703125,
-0.00685882568359375,
0.118896484375,
-0.3837890625,
-0.51513671875,
0.248046875,
-0.99853515625,
-1.0263671875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
while True:
N = int(input())
if N == 0:
break
table = [[0 for i in range(N+1)] for j in range(N+1)]
for i in range(N):
n = [int(i) for i in input().split()]
for j in range(N):
table[i][j] = n[j]
for i in range(N):
for j in range(N):
table[i][N] += table[i][j]
table[N][j] += table[i][j]
for i in range(N):
table[N][N] += table[i][N]
for i in range(N+1):
for j in range(N+1):
print(str(table[i][j]).rjust(5), end="")
print("")
```
Yes
| 98,521 | [
0.48779296875,
-0.028289794921875,
-0.10205078125,
-0.00909423828125,
-0.58642578125,
-0.318115234375,
-0.045928955078125,
0.1048583984375,
0.0599365234375,
0.54931640625,
0.45654296875,
0.0653076171875,
-0.004058837890625,
-0.3759765625,
-0.57666015625,
0.11163330078125,
-0.39208984... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
# AOJ 0102: Matrix-like Computation
# Python3 2018.6.17 bal4u
while True:
n = int(input())
if n == 0: break
arr = [[0 for r in range(n+2)] for c in range(n+2)]
for r in range(n):
arr[r] = list(map(int, input().split()))
arr[r].append(sum(arr[r]))
for c in range(n+1):
s = 0
for r in range(n): s += arr[r][c]
arr[n][c] = s
for r in range(n+1):
for c in range(n+1): print(format(arr[r][c], '5d'), end='')
print()
```
Yes
| 98,522 | [
0.467041015625,
-0.0172576904296875,
-0.131591796875,
-0.0148773193359375,
-0.351806640625,
-0.27685546875,
-0.06591796875,
0.0662841796875,
0.050994873046875,
0.51513671875,
0.61474609375,
0.04644775390625,
-0.0904541015625,
-0.472900390625,
-0.77197265625,
0.07696533203125,
-0.2152... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
while True:
n = int(input())
if n==0:
break
A=[]
for a in range(n):
x =list(map(int, input().split()))
x.append(sum(x))
A.append(x)
Y=[]
for j in range(n+1):
y=0
for i in range(n):
y+=A[i][j]
Y.append(y)
A.append(Y)
for i in range(n+1):
for j in range(n+1):
a=A[i][j]
a=str(a)
a=a.rjust(5)
print(a,end="")
print()
```
Yes
| 98,523 | [
0.497314453125,
0.0040740966796875,
-0.0014448165893554688,
0.02178955078125,
-0.57861328125,
-0.294677734375,
-0.152587890625,
0.100341796875,
0.1619873046875,
0.556640625,
0.427978515625,
0.08319091796875,
-0.033416748046875,
-0.4287109375,
-0.57666015625,
0.0733642578125,
-0.34570... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
while True:
inputCount = int(input())
if inputCount == 0:
break
table = []
for lp in range(inputCount):
content = [int(item) for item in input().split(" ")]
content.append(sum(content))
table.append(content)
table.append([])
for col in range(inputCount + 1):
total = 0
for row in range(inputCount):
total += table[row][col]
table[inputCount].append(total)
for array in table:
print("".join("{:>5}".format(item) for item in array))
```
Yes
| 98,524 | [
0.4794921875,
0.0164031982421875,
-0.05419921875,
0.01177215576171875,
-0.55810546875,
-0.272705078125,
-0.118408203125,
0.247802734375,
0.263916015625,
0.5419921875,
0.47802734375,
0.01739501953125,
-0.099609375,
-0.315673828125,
-0.6455078125,
0.1343994140625,
-0.2098388671875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
while True:
n = int(input())
if n != 0:
sum_col = [0 for i in range(n+1)]
for i in range(n):
list_row = list(map(int,input().split(" ")))
sum_row = sum(list_row)
list_row.append(sum_row)
for j in range(len(list_row)):
sum_col[j] += list_row[j]
print((" ").join(list(map(str,list_row))))
print((" ").join(list(map(str,sum_col))))
else:
break
```
No
| 98,525 | [
0.52783203125,
-0.01422882080078125,
-0.06854248046875,
-0.044769287109375,
-0.52197265625,
-0.25,
-0.089599609375,
0.207763671875,
0.16455078125,
0.56103515625,
0.45751953125,
0.039642333984375,
0.0721435546875,
-0.35400390625,
-0.5712890625,
0.0972900390625,
-0.270263671875,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
while True:
n=int(input())
if not n: break
a=[list(map(int,input().split())) for _ in range(n)]
t=[[0 for _ in range(n+1)] for _ in range(n+1)]
for i in range(n):
for j in range(n):
t[i][j]=a[i][j]
t[i][n]+=a[i][j]
t[n][j]+=a[i][j]
t[n][n]+=a[i][j]
print(t)
```
No
| 98,526 | [
0.52685546875,
0.044189453125,
-0.10723876953125,
-0.00162506103515625,
-0.517578125,
-0.277099609375,
-0.058807373046875,
0.1561279296875,
0.143798828125,
0.603515625,
0.396484375,
0.08221435546875,
-0.055633544921875,
-0.380615234375,
-0.6416015625,
0.07318115234375,
-0.2783203125,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
total = [0] * (n + 1)
for i in range(n):
a = [int(i) for i in input().split()]
a.append(sum(a))
print(" ".join(map(str, a)))
for j in range(n + 1):
total[j] += a[j]
print(" ".join(map(str, total)))
```
No
| 98,527 | [
0.5751953125,
0.007366180419921875,
-0.05462646484375,
0.0008273124694824219,
-0.5244140625,
-0.298095703125,
-0.032440185546875,
0.1448974609375,
0.17724609375,
0.61328125,
0.395263671875,
0.055908203125,
-0.035064697265625,
-0.2861328125,
-0.5673828125,
0.05364990234375,
-0.2810058... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to develop a tiny little part of spreadsheet software.
Write a program which adds up columns and rows of given table as shown in the following figure:
<image>
Input
The input consists of several datasets. Each dataset consists of:
n (the size of row and column of the given table)
1st row of the table
2nd row of the table
:
:
nth row of the table
The input ends with a line consisting of a single 0.
Output
For each dataset, print the table with sums of rows and columns. Each item of the table should be aligned to the right with a margin for five digits. Please see the sample output for details.
Example
Input
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
4
52 96 15 20
86 22 35 45
45 78 54 36
16 86 74 55
0
Output
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
52 96 15 20 183
86 22 35 45 188
45 78 54 36 213
16 86 74 55 231
199 282 178 156 815
Submitted Solution:
```
def main():
while True:
n = int(input())
if n == 0:
break
for i in range(n):
num = 0
m = map(int, input().split())
for j in m:
print("{:5}".format(j), end='')
num += j
print("{:5}".format(num))
if __name__ == '__main__':
main()
```
No
| 98,528 | [
0.59619140625,
0.01593017578125,
-0.07666015625,
-0.04986572265625,
-0.5068359375,
-0.239013671875,
-0.055419921875,
0.1822509765625,
0.082275390625,
0.50927734375,
0.427978515625,
0.0296630859375,
0.0011243820190429688,
-0.271240234375,
-0.650390625,
0.0784912109375,
-0.309814453125... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
You are a student looking for a job. Today you had an employment examination for an IT company. They asked you to write an efficient program to perform several operations. First, they showed you an N \times N square matrix and a list of operations. All operations but one modify the matrix, and the last operation outputs the character in a specified cell. Please remember that you need to output the final matrix after you finish all the operations.
Followings are the detail of the operations:
WR r c v
(Write operation) write a integer v into the cell (r,c) (1 \leq v \leq 1,000,000)
CP r1 c1 r2 c2
(Copy operation) copy a character in the cell (r1,c1) into the cell (r2,c2)
SR r1 r2
(Swap Row operation) swap the r1-th row and r2-th row
SC c1 c2
(Swap Column operation) swap the c1-th column and c2-th column
RL
(Rotate Left operation) rotate the whole matrix in counter-clockwise direction by 90 degrees
RR
(Rotate Right operation) rotate the whole matrix in clockwise direction by 90 degrees
RH
(Reflect Horizontal operation) reverse the order of the rows
RV
(Reflect Vertical operation) reverse the order of the columns
Input
First line of each testcase contains nine integers. First two integers in the line, N and Q, indicate the size of matrix and the number of queries, respectively (1 \leq N,Q \leq 40,000). Next three integers, A B, and C, are coefficients to calculate values in initial matrix (1 \leq A,B,C \leq 1,000,000), and they are used as follows: A_{r,c} = (r * A + c * B) mod C where r and c are row and column indices, respectively (1\leq r,c\leq N). Last four integers, D, E, F, and G, are coefficients to compute the final hash value mentioned in the next section (1 \leq D \leq E \leq N, 1 \leq F \leq G \leq N, E - D \leq 1,000, G - F \leq 1,000). Each of next Q lines contains one operation in the format as described above.
Output
Output a hash value h computed from the final matrix B by using following pseudo source code.
h <- 314159265
for r = D...E
for c = F...G
h <- (31 * h + B_{r,c}) mod 1,000,000,007
where "<-" is a destructive assignment operator, "for i = S...T" indicates a loop for i from S to T (both inclusive), and "mod" is a remainder operation.
Examples
Input
2 1 3 6 12 1 2 1 2
WR 1 1 1
Output
676573821
Input
2 1 3 6 12 1 2 1 2
RL
Output
676636559
Input
2 1 3 6 12 1 2 1 2
RH
Output
676547189
Input
39989 6 999983 999979 999961 1 1000 1 1000
SR 1 39989
SC 1 39989
RL
RH
RR
RV
Output
458797120
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N, Q, A, B, C, D, E, F, G = map(int, readline().split())
d = 0; rx = 0; ry = 0
*X, = range(N)
*Y, = range(N)
def fc(d, x, y):
if d == 0:
return x, y
if d == 1:
return y, N-1-x
if d == 2:
return N-1-x, N-1-y
return N-1-y, x
mp = {}
for i in range(Q):
c, *g = readline().strip().split()
c0, c1 = c
if c0 == "R":
if c1 == "L":
d = (d - 1) % 4
elif c1 == "R":
d = (d + 1) % 4
elif c1 == "H":
if d & 1:
rx ^= 1
else:
ry ^= 1
else: #c1 == "V":
if d & 1:
ry ^= 1
else:
rx ^= 1
elif c0 == "S":
a, b = map(int, g); a -= 1; b -= 1
if c1 == "R":
if d & 1:
if rx != ((d & 2) > 0):
a = N-1-a; b = N-1-b
X[a], X[b] = X[b], X[a]
else:
if ry != ((d & 2) > 0):
a = N-1-a; b = N-1-b
Y[a], Y[b] = Y[b], Y[a]
else: #c1 == "C":
if d & 1:
if ((d & 2) == 0) != ry:
a = N-1-a; b = N-1-b
Y[a], Y[b] = Y[b], Y[a]
else:
if ((d & 2) > 0) != rx:
a = N-1-a; b = N-1-b
X[a], X[b] = X[b], X[a]
elif c0 == "C": #c == "CP":
y1, x1, y2, x2 = map(int, g); x1 -= 1; y1 -= 1; x2 -= 1; y2 -= 1
x1, y1 = fc(d, x1, y1)
x2, y2 = fc(d, x2, y2)
if rx:
x1 = N-1-x1; x2 = N-1-x2
if ry:
y1 = N-1-y1; y2 = N-1-y2
key1 = (X[x1], Y[y1]); key2 = (X[x2], Y[y2])
if key1 not in mp:
xa, ya = key1
mp[key2] = (ya*A + xa*B + A + B) % C
else:
mp[key2] = mp[key1]
else: #c == "WR":
y, x, v = map(int, g); x -= 1; y -= 1
x, y = fc(d, x, y)
if rx:
x = N-1-x
if ry:
y = N-1-y
key = (X[x], Y[y])
mp[key] = v
MOD = 10**9 + 7
h = 314159265
for y in range(D-1, E):
for x in range(F-1, G):
x0, y0 = fc(d, x, y)
if rx:
x0 = N-1-x0
if ry:
y0 = N-1-y0
x0 = X[x0]; y0 = Y[y0]
key = (x0, y0)
if key in mp:
v = mp[key]
else:
v = ((y0+1)*A + (x0+1)*B) % C
h = (31 * h + v) % MOD
write("%d\n" % h)
solve()
```
| 98,549 | [
0.270263671875,
-0.03875732421875,
-0.176513671875,
-0.222900390625,
-0.60888671875,
-0.468505859375,
-0.082763671875,
-0.103271484375,
0.00478363037109375,
0.85546875,
0.9326171875,
0.237548828125,
0.1497802734375,
-0.93115234375,
-0.42919921875,
-0.17626953125,
-0.34619140625,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
import sys
#sys.stdin = open("input.txt")
#T = int(input())
def trova(a,b,x):
if x>=a:
return a
if x <= b:
while x==C[b]:
b = b-1
return b
m = (a+b)//2
if x == C[m]:
while x == C[m]:
m = m-1
return m
if a == b:
while C[a] == x:
a = a-1
return a
if x < C[m]:
trova(m,b,x)
return
if x > C[m]:
trova(a,m,x)
return
T = 1
t = 0
while t<T:
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
n = 0
C = []
while n<N:
C.append(A[n]-B[n])
n +=1
C.sort(reverse = True)
n = 0
out = 0
#print (C)
inizio = 0
fine = N-1
while inizio<fine:
if C[inizio]+C[fine] > 0:
out += fine - inizio
inizio +=1
else:
fine -=1
print (out)
t +=1
```
Yes
| 98,766 | [
0.6767578125,
0.3271484375,
-0.06939697265625,
0.064453125,
-0.77978515625,
-0.39208984375,
-0.3203125,
0.194580078125,
0.2359619140625,
0.927734375,
0.68115234375,
0.257080078125,
-0.0210723876953125,
-0.6162109375,
-0.34375,
-0.01479339599609375,
-0.60302734375,
-1.1083984375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
from bisect import bisect_right as left
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
m=[]
ans=0
for i in range(n):
m.append(a[i]-b[i])
m.sort()
p,ne=[],[]
ans=0
for i in range(n):
if m[i]>0:
ans-=1
ans+=n-left(m,-m[i])
print(ans//2)
```
Yes
| 98,767 | [
0.6376953125,
0.2841796875,
-0.22607421875,
0.06317138671875,
-0.90673828125,
-0.43701171875,
-0.2413330078125,
0.09112548828125,
0.309326171875,
1.080078125,
0.6630859375,
0.1488037109375,
0.1578369140625,
-0.796875,
-0.436767578125,
-0.086669921875,
-0.6826171875,
-1.078125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
import sys
# from math import ceil,floor,tan
import bisect
RI = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
n = int(ri())
a = RI()
b= RI()
c = [a[i]-b[i] for i in range(n)]
c.sort()
i=0
while i < len(c) and c[i] < 0 :
i+=1
ans = 0
for i in range(i,n):
pos = bisect.bisect_left(c,1-c[i])
if pos < i:
ans+=(i-pos)
print(ans)
```
Yes
| 98,768 | [
0.59814453125,
0.388671875,
-0.146240234375,
0.1019287109375,
-0.95458984375,
-0.396484375,
-0.301513671875,
0.116455078125,
0.275390625,
1.0634765625,
0.63134765625,
0.2154541015625,
-0.030487060546875,
-0.7255859375,
-0.475341796875,
-0.1331787109375,
-0.67431640625,
-1.026367187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
m = list(map(int, input().split()))
for i in range(n):
l[i] -= m[i]
l.sort()
i = 0
j = n - 1
count = 0
while j > i:
if l[j] + l[i] > 0:
count += j - i
j -= 1
else:
i += 1
print(count)
```
Yes
| 98,769 | [
0.72119140625,
0.300537109375,
-0.273681640625,
0.06488037109375,
-0.8525390625,
-0.410888671875,
-0.31494140625,
0.2117919921875,
0.345458984375,
0.95751953125,
0.607421875,
0.2042236328125,
0.10968017578125,
-0.72412109375,
-0.348388671875,
-0.06829833984375,
-0.64208984375,
-1.0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
from bisect import *
qtd = int(input())
dif = sorted([int(a) - int(b) for a, b in zip([int(x) for x in input().split()],[int(x) for x in input().split()])])
#todas as combinacoes dos positivos:
totalpos = qtd
for el in dif:
if el <= 0:
totalpos -= 1
else:
break
output = int(totalpos*(totalpos-1)/2)
i = 0
while(dif[i] < 0):
maisDir = bisect_left(dif, -dif[i], i, qtd)-1
output += qtd - maisDir
i += 1
print(output)
```
No
| 98,770 | [
0.498291015625,
0.231201171875,
-0.1297607421875,
0.17529296875,
-0.8466796875,
-0.3779296875,
-0.29345703125,
0.11285400390625,
0.31201171875,
1.111328125,
0.70556640625,
0.1922607421875,
0.07135009765625,
-0.65576171875,
-0.323486328125,
-0.006381988525390625,
-0.58984375,
-1.051... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
from bisect import bisect_left as bisect
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=[a[i]-b[i] for i in range(n)]
c.sort()
cu=0
for i in c:
x=1-i
z=bisect(c,x)
if x<0:
cu+=(n-z)-1
else:
cu+=(n-z)
print(cu//2)
```
No
| 98,771 | [
0.5712890625,
0.251220703125,
-0.1617431640625,
0.087158203125,
-0.939453125,
-0.404296875,
-0.2103271484375,
0.058929443359375,
0.346923828125,
1.109375,
0.70654296875,
0.240478515625,
0.1822509765625,
-0.83349609375,
-0.380126953125,
-0.10235595703125,
-0.6845703125,
-1.142578125... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
all_data = input()
n = int(all_data)
print(n)
x1 = input()
x2 = input()
l1 = [int(x) for x in x1.split(' ')]
l2 = [int(x) for x in x2.split(' ')]
print(l1, l2)
cnt = 0
for i in range(n):
for j in range(i + 1, n):
if l1[i] + l1[j] > l2[i] + l2[j]:
cnt += 1
print(cnt)
```
No
| 98,772 | [
0.6982421875,
0.253173828125,
-0.1717529296875,
0.08563232421875,
-0.72119140625,
-0.39013671875,
-0.1915283203125,
0.09588623046875,
0.329345703125,
0.94873046875,
0.564453125,
0.2333984375,
0.0087738037109375,
-0.78173828125,
-0.4326171875,
-0.0821533203125,
-0.66650390625,
-0.98... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The next lecture in a high school requires two topics to be discussed. The i-th topic is interesting by a_i units for the teacher and by b_i units for the students.
The pair of topics i and j (i < j) is called good if a_i + a_j > b_i + b_j (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of topics.
The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the interestingness of the i-th topic for the teacher.
The third line of the input contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the interestingness of the i-th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5
4 8 2 6 2
4 5 4 1 3
Output
7
Input
4
1 3 2 4
1 3 2 4
Output
0
Submitted Solution:
```
# limit for array size
N = 100001;
# Max size of tree
tree = [0] * (2 * N);
# function to build the tree
def build(arr) :
# insert leaf nodes in tree
for i in range(n) :
tree[n + i] = arr[i];
# build the tree by calculating parents
for i in range(n - 1, 0, -1) :
tree[i] = tree[i << 1] + tree[i << 1 | 1];
# function to update a tree node
def updateTreeNode(p, value) :
# set value at position p
tree[p + n] = value;
p = p + n;
# move upward and update parents
i = p;
while i > 1 :
tree[i >> 1] = tree[i] + tree[i ^ 1];
i >>= 1;
# function to get sum on interval [l, r)
def query(l, r) :
res = 0;
# loop to find the sum in the range
l += n;
r += n;
while l < r :
if (l & 1) :
res += tree[l];
l += 1
if (r & 1) :
r -= 1;
res += tree[r];
l >>= 1;
r >>= 1
return res;
from collections import defaultdict
import bisect
#a,b = map(int,input().strip().split())
n = int(input().strip())
a = [int(i) for i in input().strip().split()]
b = [int(i) for i in input().strip().split()]
minus = [(a[i] - b[i],i) for i in range(n)]
minus.sort()
order = [i[1] for i in minus]
minus = [i[0] for i in minus]
total = 0
temp = [0 for i in range(n)]
build(temp)
#print(minus)
for i in range(n):
result = a[i] - b[i]
ans = bisect.bisect_right(minus,result)
#print(i,result,ans)
total += n - ans
if ans < n:
total -= query(ans + 1,n + 1)
updateTreeNode(order[i], 1)
#print(tree[:10])
print(total)
```
No
| 98,773 | [
0.72265625,
0.36181640625,
-0.146484375,
0.0909423828125,
-0.6630859375,
-0.5546875,
-0.37158203125,
0.242919921875,
0.515625,
0.76025390625,
0.830078125,
-0.020477294921875,
0.117919921875,
-0.61572265625,
-0.23095703125,
0.2626953125,
-0.64111328125,
-0.84765625,
-0.40087890625... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enter a positive integer n of 4,000 or less, with a pair of integers a, b, c, d in the range 0-1000.
a + b + c + d = n
Create a program that outputs the number of combinations that satisfy the conditions.
Input
Given multiple datasets. Each dataset is given n on one row. Please process until the end of the input.
The number of datasets does not exceed 50.
Output
For each data set, output the number of combinations of a, b, c, and d on one line.
Example
Input
2
3
35
Output
10
20
8436
Submitted Solution:
```
def Kosuu(n,sak):
if n <= sak:
kazu = n + 1
sta = 0
end = n
else:
kazu = sak * 2 + 1 - n
sta = n - sak
end = sak
return [kazu,sta,end]
while True:
try:
Sum = 0
n = int(input())
ABl = Kosuu(n,2000)
for i in range(ABl[1],ABl[2] + 1):
Sum += Kosuu(i,1000)[0] * Kosuu(n -i,1000)[0]
print(Sum)
except EOFError:
break
```
Yes
| 99,353 | [
0.5947265625,
0.257568359375,
-0.08087158203125,
0.2369384765625,
-0.54931640625,
-0.63671875,
-0.16552734375,
0.380126953125,
0.283203125,
0.66015625,
0.63232421875,
-0.0533447265625,
-0.0313720703125,
-0.4794921875,
-0.54443359375,
0.003932952880859375,
-0.580078125,
-0.821289062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enter a positive integer n of 4,000 or less, with a pair of integers a, b, c, d in the range 0-1000.
a + b + c + d = n
Create a program that outputs the number of combinations that satisfy the conditions.
Input
Given multiple datasets. Each dataset is given n on one row. Please process until the end of the input.
The number of datasets does not exceed 50.
Output
For each data set, output the number of combinations of a, b, c, and d on one line.
Example
Input
2
3
35
Output
10
20
8436
Submitted Solution:
```
ab = [0 for _ in range(2001)]
for a in range(1001):
for b in range(1001):
ab[a+b] += 1
while True:
try:
n = int(input())
except:
break
ans = sum(ab[i]*ab[n-i] for i in range(2001) if n-i >= 0 and n-i <= 2000)
print(ans)
```
Yes
| 99,354 | [
0.57373046875,
-0.0576171875,
0.268310546875,
0.166748046875,
-0.475830078125,
-0.412353515625,
-0.023193359375,
0.21142578125,
0.261474609375,
0.6640625,
0.7001953125,
-0.26025390625,
0.052215576171875,
-0.56005859375,
-0.56591796875,
-0.11444091796875,
-0.51025390625,
-0.6953125,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enter a positive integer n of 4,000 or less, with a pair of integers a, b, c, d in the range 0-1000.
a + b + c + d = n
Create a program that outputs the number of combinations that satisfy the conditions.
Input
Given multiple datasets. Each dataset is given n on one row. Please process until the end of the input.
The number of datasets does not exceed 50.
Output
For each data set, output the number of combinations of a, b, c, and d on one line.
Example
Input
2
3
35
Output
10
20
8436
Submitted Solution:
```
import sys
hist = [0 for i in range(4001)]
for i in range(1001):
for j in range(1001):
hist[i + j] += 1
for line in sys.stdin:
ans = 0
n = int(line)
for i in range(min(n, 2000) + 1):
ans += (hist[i] * hist[n-i])
print(ans)
```
Yes
| 99,356 | [
0.63671875,
-0.07916259765625,
0.3876953125,
0.119384765625,
-0.615234375,
-0.2464599609375,
-0.225830078125,
0.289794921875,
0.29296875,
0.6865234375,
0.57861328125,
-0.287353515625,
0.100830078125,
-0.495361328125,
-0.68359375,
-0.00424957275390625,
-0.60791015625,
-0.818359375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enter a positive integer n of 4,000 or less, with a pair of integers a, b, c, d in the range 0-1000.
a + b + c + d = n
Create a program that outputs the number of combinations that satisfy the conditions.
Input
Given multiple datasets. Each dataset is given n on one row. Please process until the end of the input.
The number of datasets does not exceed 50.
Output
For each data set, output the number of combinations of a, b, c, and d on one line.
Example
Input
2
3
35
Output
10
20
8436
Submitted Solution:
```
while True:
m = 0
try:
n = int(input().strip())
for a in range(10001):
if n - a <0:
break
for b in range(10001):
if n - (a+b) <0:
break
for c in range(10001):
if n - (a+b+c) <0:
break
for d in range(10001):
if n - (a+b+c+d) <0:
break
if a+b+c+d == n:
m += 1
print(m)
except EOFError:
break
```
No
| 99,357 | [
0.51171875,
0.006206512451171875,
0.21484375,
0.20947265625,
-0.4619140625,
-0.5048828125,
-0.0263519287109375,
0.2108154296875,
0.216552734375,
0.79052734375,
0.70361328125,
-0.1141357421875,
0.1180419921875,
-0.6162109375,
-0.54833984375,
-0.10101318359375,
-0.52490234375,
-0.827... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enter a positive integer n of 4,000 or less, with a pair of integers a, b, c, d in the range 0-1000.
a + b + c + d = n
Create a program that outputs the number of combinations that satisfy the conditions.
Input
Given multiple datasets. Each dataset is given n on one row. Please process until the end of the input.
The number of datasets does not exceed 50.
Output
For each data set, output the number of combinations of a, b, c, and d on one line.
Example
Input
2
3
35
Output
10
20
8436
Submitted Solution:
```
import sys
for line in sys.stdin.readlines():
n = int(line.rstrip())
count = 0
for i in range(min(2001,n+1)):
if n - i >= 0:
count += (i+1)*(n-i+1)
print(count)
```
No
| 99,358 | [
0.48681640625,
-0.0379638671875,
0.267822265625,
0.2374267578125,
-0.58447265625,
-0.4384765625,
-0.130615234375,
0.2205810546875,
0.1473388671875,
0.75732421875,
0.689453125,
-0.240478515625,
0.00016808509826660156,
-0.494140625,
-0.66650390625,
-0.1297607421875,
-0.471923828125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Enter a positive integer n of 4,000 or less, with a pair of integers a, b, c, d in the range 0-1000.
a + b + c + d = n
Create a program that outputs the number of combinations that satisfy the conditions.
Input
Given multiple datasets. Each dataset is given n on one row. Please process until the end of the input.
The number of datasets does not exceed 50.
Output
For each data set, output the number of combinations of a, b, c, and d on one line.
Example
Input
2
3
35
Output
10
20
8436
Submitted Solution:
```
def solve(n):
ans=0
for a in range(n+1):
for b in range(n+1):
if a+b>n:
break
for c in range(n+1):
if n-(a+b+c)>=0:
ans+=1
else:
break
return ans
while True:
try:
n=int(input())
print(solve(n))
except EOFError:
break
```
No
| 99,359 | [
0.5048828125,
-0.058013916015625,
0.11260986328125,
0.2296142578125,
-0.443603515625,
-0.37744140625,
-0.060760498046875,
0.290283203125,
0.2333984375,
0.73388671875,
0.748046875,
-0.12359619140625,
0.033905029296875,
-0.56005859375,
-0.564453125,
-0.1365966796875,
-0.53076171875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
input()
d={(0,0):1}
r=s=i=0
for x in map(int,input().split()):s^=x;i^=1;c=d.get((s,i),0);r+=c;d[s,i]=c+1
print(r)
```
Yes
| 99,527 | [
0.556640625,
0.040374755859375,
-0.306396484375,
-0.43603515625,
-0.6494140625,
-0.82763671875,
-0.1011962890625,
0.170166015625,
0.10748291015625,
1.033203125,
0.697265625,
-0.1771240234375,
0.165283203125,
-0.94873046875,
-0.55078125,
0.038299560546875,
-0.402099609375,
-0.820800... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
xor = [a[0]]
for i in range(1, n):
xor.append(xor[-1]^a[i])
d = {0: [1, 0]}
for i in range(n):
if xor[i] not in d:
d[xor[i]] = [0, 0]
d[xor[i]][(i+1)%2] += 1
#for k in d: print(d[k])
funny = 0
for k in d:
odd = d[k][1]
even = d[k][0]
funny += (odd*(odd-1))//2
funny += (even*(even-1))//2
print(funny)
```
Yes
| 99,528 | [
0.56298828125,
0.0236663818359375,
-0.298828125,
-0.433349609375,
-0.63134765625,
-0.833984375,
-0.1063232421875,
0.1514892578125,
0.1051025390625,
1.0224609375,
0.69482421875,
-0.1829833984375,
0.1590576171875,
-0.92822265625,
-0.54248046875,
0.048309326171875,
-0.408935546875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys, re
from collections import deque, defaultdict, Counter
from math import sqrt, hypot, factorial, pi, sin, cos, radians, log10
if sys.version_info.minor >= 5: from math import gcd
else: from fractions import gcd
from heapq import heappop, heappush, heapify, heappushpop
from bisect import bisect_left, bisect_right
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul, xor
from copy import copy, deepcopy
from functools import reduce, partial
from fractions import Fraction
from string import ascii_lowercase, ascii_uppercase, digits
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def round(x): return int((x*2+1) // 2)
def fermat(x, y, MOD): return x * pow(y, MOD-2, MOD) % MOD
def lcm(x, y): return (x * y) // gcd(x, y)
def lcm_list(nums): return reduce(lcm, nums, 1)
def gcd_list(nums): return reduce(gcd, nums, nums[0])
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
N = INT()
A = LIST()
acc = [0] + list(accumulate(A, xor))
C1 = Counter()
for i in range(0, N+1, 2):
C1[acc[i]] += 1
C2 = Counter()
for i in range(1, N+1, 2):
C2[acc[i]] += 1
ans = 0
for k, v in C1.items():
if v >= 2:
ans += v * (v-1) // 2
for k, v in C2.items():
if v >= 2:
ans += v * (v-1) // 2
print(ans)
```
Yes
| 99,529 | [
0.5595703125,
0.030303955078125,
-0.306884765625,
-0.423095703125,
-0.6416015625,
-0.83935546875,
-0.107421875,
0.16455078125,
0.10809326171875,
1.0283203125,
0.697265625,
-0.1763916015625,
0.1650390625,
-0.93896484375,
-0.54833984375,
0.043212890625,
-0.40185546875,
-0.822265625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
import sys
from collections import defaultdict
input = sys.stdin.readline
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().strip().split()))
xr = 0
ev = defaultdict(int)
od = defaultdict(int)
od[0] += 1
ans = 0
for i in range(n):
xr ^= arr[i]
if i & 1:
ans += od[xr]
od[xr] += 1
else:
ans += ev[xr]
ev[xr] += 1
print(ans)
```
Yes
| 99,530 | [
0.568359375,
0.027252197265625,
-0.303466796875,
-0.440185546875,
-0.64892578125,
-0.81689453125,
-0.1044921875,
0.1495361328125,
0.10650634765625,
1.037109375,
0.68017578125,
-0.197509765625,
0.1522216796875,
-0.9267578125,
-0.56494140625,
0.05096435546875,
-0.3828125,
-0.79931640... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
count=0;cc={}
def work(a,i,j):
global cc,count
if j==i:return a[i]
if (i,j) in cc:return cc[(i,j)]
m=(j+i-1)//2
a1=work(a,i,m)
a2=work(a,m+1,j)
#print((i,j),a1,a2)
if a1==a2:
if (i,j) not in cc:
count+=1
cc[(i,j)]=a1^a2
return a1^a2
n=int(input())
a=list(map(int,input().split()))
if n%2==0:
work(a,0,n-1)
if n>4:work(a,0,n-3);work(a,2,n-1)
else:
work(a,1,n-1)
work(a,0,n-2)
print(count)
```
No
| 99,531 | [
0.5615234375,
0.02783203125,
-0.301513671875,
-0.423095703125,
-0.63037109375,
-0.8408203125,
-0.10894775390625,
0.1600341796875,
0.11553955078125,
1.0302734375,
0.6953125,
-0.1839599609375,
0.163818359375,
-0.923828125,
-0.54345703125,
0.047882080078125,
-0.40185546875,
-0.8198242... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
# Bismillahirahmanirahim
# Soru 1
#
# nv = list(map(int, input().split()))
#
# n = nv[0] - 1
# v = nv[1]
# if v >= n:
# print(n)
# quit()
# k = 2
# money = v
# while n - v > 0:
# n = n - 1
# money += k
# k += 1
# print(money)
#Soru2
# def carpan(n):
# lst = []
# sq = n**0.5 + 1
# for i in range(2,int(sq)+1):
# if n%i == 0:
# lst.append(i)
# return lst
#
#
# n = int(input())
# lst = list(map(int, input().split()))
# lst = sorted(lst, reverse=True)
# mn = lst[-1]
# total = sum(lst)
# large = 0
# for i in lst:
# mx = i
# lst1 = carpan(mx)
# for j in lst1:
# if mx + mn - (mx/j + mn*j) > large:
# large = mx + mn - (mx/j + mn*j)
# print(int(total-large))
# Soru 3
n = int(input())
lst = list(map(int, input().split()))
total = 0
for i in range(1,n):
lst[i] = lst[i] ^ lst[i - 1]
def ikili(n):
return n *(n-1)/2
dct = {0:[0, 0]}
for i in lst:
dct[i] = [0, 0]
for i in range(n):
if i %2 == 0:
dct[lst[i]][0] += 1
else:
dct[lst[i]][1] += 1
total += dct[0][1]
dct.pop(0)
for i in dct:
total += ikili(dct[i][0]) + ikili(dct[i][1])
print(int(total))
```
No
| 99,532 | [
0.51708984375,
0.0189361572265625,
-0.2666015625,
-0.40673828125,
-0.6640625,
-0.810546875,
-0.0902099609375,
0.09466552734375,
0.08380126953125,
1.009765625,
0.70458984375,
-0.1396484375,
0.1650390625,
-1.021484375,
-0.54443359375,
0.04840087890625,
-0.42724609375,
-0.8408203125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
x,cnt=0,0
for i in range(n-1):
x^=a[i]
y=0
for j in range(n):
y^=a[j]
if((j-i+1)%2==0) and (x==y):
cnt+=1
x=0
print(cnt)
```
No
| 99,533 | [
0.56298828125,
0.0236663818359375,
-0.298828125,
-0.433349609375,
-0.63134765625,
-0.833984375,
-0.1063232421875,
0.1514892578125,
0.1051025390625,
1.0224609375,
0.69482421875,
-0.1829833984375,
0.1590576171875,
-0.92822265625,
-0.54248046875,
0.048309326171875,
-0.408935546875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sasha likes programming. Once, during a very long contest, Sasha decided that he was a bit tired and needed to relax. So he did. But since Sasha isn't an ordinary guy, he prefers to relax unusually. During leisure time Sasha likes to upsolve unsolved problems because upsolving is very useful.
Therefore, Sasha decided to upsolve the following problem:
You have an array a with n integers. You need to count the number of funny pairs (l, r) (l ≤ r). To check if a pair (l, r) is a funny pair, take mid = (l + r - 1)/(2), then if r - l + 1 is an even number and a_l ⊕ a_{l+1} ⊕ … ⊕ a_{mid} = a_{mid + 1} ⊕ a_{mid + 2} ⊕ … ⊕ a_r, then the pair is funny. In other words, ⊕ of elements of the left half of the subarray from l to r should be equal to ⊕ of elements of the right half. Note that ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
It is time to continue solving the contest, so Sasha asked you to solve this task.
Input
The first line contains one integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the size of the array.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^{20}) — array itself.
Output
Print one integer — the number of funny pairs. You should consider only pairs where r - l + 1 is even number.
Examples
Input
5
1 2 3 4 5
Output
1
Input
6
3 2 2 3 7 6
Output
3
Input
3
42 4 2
Output
0
Note
Be as cool as Sasha, upsolve problems!
In the first example, the only funny pair is (2, 5), as 2 ⊕ 3 = 4 ⊕ 5 = 1.
In the second example, funny pairs are (2, 3), (1, 4), and (3, 6).
In the third example, there are no funny pairs.
Submitted Solution:
```
n = int(input())
arr_str = input().split()
arr = [int(x) for x in arr_str]
cum_sum_arr = []
cum_sum_arr.append(0)
cum_sum = 0
for i in range(0, len(arr)):
cum_sum = cum_sum^arr[i]
cum_sum_arr.append(cum_sum)
print(cum_sum_arr)
count_pair = 0
sum2ind = dict()
sum2ind[0] = [0]
for r in range(1, len(cum_sum_arr)):
if cum_sum_arr[r] in sum2ind:
for l_minus_1 in sum2ind[cum_sum_arr[r]]:
if (r - l_minus_1) % 2 == 0:
count_pair += 1
sum2ind[cum_sum_arr[r]].append(r)
else:
sum2ind[cum_sum_arr[r]] = [r]
print(count_pair)
```
No
| 99,534 | [
0.56298828125,
0.0236663818359375,
-0.298828125,
-0.433349609375,
-0.63134765625,
-0.833984375,
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0.1514892578125,
0.1051025390625,
1.0224609375,
0.69482421875,
-0.1829833984375,
0.1590576171875,
-0.92822265625,
-0.54248046875,
0.048309326171875,
-0.408935546875,
-0... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Uh oh! Applications to tech companies are due soon, and you've been procrastinating by doing contests instead! (Let's pretend for now that it is actually possible to get a job in these uncertain times.)
You have completed many programming projects. In fact, there are exactly n types of programming projects, and you have completed a_i projects of type i. Your résumé has limited space, but you want to carefully choose them in such a way that maximizes your chances of getting hired.
You want to include several projects of the same type to emphasize your expertise, but you also don't want to include so many that the low-quality projects start slipping in. Specifically, you determine the following quantity to be a good indicator of your chances of getting hired:
$$$ f(b_1,…,b_n)=∑_{i=1}^n b_i(a_i-b_i^2). $$$
Here, b_i denotes the number of projects of type i you include in your résumé. Of course, you cannot include more projects than you have completed, so you require 0≤ b_i ≤ a_i for all i.
Your résumé only has enough room for k projects, and you will absolutely not be hired if your résumé has empty space, so you require ∑_{i=1}^n b_i=k.
Find values for b_1,…, b_n that maximize the value of f(b_1,…,b_n) while satisfying the above two constraints.
Input
The first line contains two integers n and k (1≤ n≤ 10^5, 1≤ k≤ ∑_{i=1}^n a_i) — the number of types of programming projects and the résumé size, respectively.
The next line contains n integers a_1,…,a_n (1≤ a_i≤ 10^9) — a_i is equal to the number of completed projects of type i.
Output
In a single line, output n integers b_1,…, b_n that achieve the maximum value of f(b_1,…,b_n), while satisfying the requirements 0≤ b_i≤ a_i and ∑_{i=1}^n b_i=k. If there are multiple solutions, output any.
Note that you do not have to output the value f(b_1,…,b_n).
Examples
Input
10 32
1 2 3 4 5 5 5 5 5 5
Output
1 2 3 3 3 4 4 4 4 4
Input
5 8
4 4 8 2 1
Output
2 2 2 1 1
Note
For the first test, the optimal answer is f=-269. Note that a larger f value is possible if we ignored the constraint ∑_{i=1}^n b_i=k.
For the second test, the optimal answer is f=9.
Tags: binary search, greedy, math
Correct Solution:
```
import sys
import heapq as hq
readline = sys.stdin.readline
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
eps = 10**-7
def solve():
n, k = nm()
a = nl()
ans = [0]*n
ok = 10**9; ng = -4*10**18
while ok - ng > 1:
mid = (ok + ng) // 2
ck = 0
for i in range(n):
d = 9 - 12 * (mid + 1 - a[i])
if d < 0:
continue
ck += min(a[i], int((3 + d**.5) / 6 + eps))
# print(mid, ck)
if ck > k:
ng = mid
else:
ok = mid
for i in range(n):
d = 9 - 12 * (ok + 1 - a[i])
if d < 0:
continue
ans[i] = min(a[i], int((3 + d**.5) / 6 + eps))
# print(ans)
rk = k - sum(ans)
l = list()
for i in range(n):
if ans[i] < a[i]:
hq.heappush(l, (-a[i] + 3 * ans[i]**2 - 3 * ans[i] + 1, i))
for _ in range(rk):
v, i = hq.heappop(l)
ans[i] += 1
if ans[i] < a[i]:
hq.heappush(l, (-a[i] + 3 * ans[i]**2 - 3 * ans[i] + 1, i))
print(*ans)
return
solve()
```
| 99,593 | [
0.437255859375,
-0.184814453125,
-0.01319122314453125,
0.2286376953125,
-0.5517578125,
-0.5244140625,
-0.0428466796875,
0.046142578125,
0.1318359375,
0.7177734375,
0.9384765625,
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0.52685546875,
-0.73876953125,
-0.270263671875,
0.2471923828125,
-0.75146484375,
-0.868... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.
* `0 p x`: a_p \gets a_p + x
* `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}.
Constraints
* 1 \leq N, Q \leq 500,000
* 0 \leq a_i, x \leq 10^9
* 0 \leq p < N
* 0 \leq l_i < r_i \leq N
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N Q
a_0 a_1 ... a_{N - 1}
\textrm{Query}_0
\textrm{Query}_1
:
\textrm{Query}_{Q - 1}
Output
For each query of the latter type, print the answer.
Example
Input
5 5
1 2 3 4 5
1 0 5
1 2 4
0 3 10
1 0 5
1 0 3
Output
15
7
25
6
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
class FenwickTree:
"""FenwickTree (Binary Indexed Tree, 0-index)
Queries:
1. add(i, val): add val to i-th value
2. sum(n): sum(bit[0] + ... + bit[n-1])
complexity: O(log n)
See: http://hos.ac/slides/20140319_bit.pdf
"""
def __init__(self, a_list):
self.N = len(a_list)
self.bit = a_list[:]
for _ in range(self.N, 1 << (self.N - 1).bit_length()):
self.bit.append(0)
for i in range(self.N-1):
self.bit[i | (i+1)] += self.bit[i]
def add(self, i, val):
while i < self.N:
self.bit[i] += val
i |= i + 1
def sum(self, n):
ret = 0
while n >= 0:
ret += self.bit[n]
n = (n & (n + 1)) - 1
return ret
def query(self, low, high):
return self.sum(high) - self.sum(low)
def yosupo():
# https://judge.yosupo.jp/problem/point_add_range_sum
_, Q = map(int, input().split())
fwt = FenwickTree([int(x) for x in input().split()])
ans = []
for _ in range(Q):
type_, l, r = map(int, input().split())
if type_ == 0:
fwt.add(l, r)
else:
ans.append(fwt.query(l-1, r-1))
print(*ans, sep="\n")
def aoj():
# https://onlinejudge.u-aizu.ac.jp/courses/library/3/DSL/2/DSL_2_B
N, Q = map(int, input().split())
fwt = FenwickTree([0] * N)
for _ in range(Q):
type_, l, r = map(int, input().split())
if type_ == 0:
fwt.add(l-1, r)
else:
print(fwt.query(l-2, r-1))
if __name__ == "__main__":
yosupo()
# aoj()
```
Yes
| 100,085 | [
0.371826171875,
-0.140380859375,
-0.060760498046875,
0.2333984375,
-0.29443359375,
-0.13232421875,
-0.0124664306640625,
0.227294921875,
0.280029296875,
0.84765625,
0.380126953125,
-0.256103515625,
-0.059173583984375,
-0.74560546875,
-0.44775390625,
0.25244140625,
-0.407958984375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.
* `0 p x`: a_p \gets a_p + x
* `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}.
Constraints
* 1 \leq N, Q \leq 500,000
* 0 \leq a_i, x \leq 10^9
* 0 \leq p < N
* 0 \leq l_i < r_i \leq N
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N Q
a_0 a_1 ... a_{N - 1}
\textrm{Query}_0
\textrm{Query}_1
:
\textrm{Query}_{Q - 1}
Output
For each query of the latter type, print the answer.
Example
Input
5 5
1 2 3 4 5
1 0 5
1 2 4
0 3 10
1 0 5
1 0 3
Output
15
7
25
6
Submitted Solution:
```
import sys
class Fenwick_Tree:
def __init__(self, n):
self._n = n
self.data = [0] * n
def add(self, p, x):
assert 0 <= p < self._n
p += 1
while p <= self._n:
self.data[p - 1] += x
p += p & -p
def sum(self, l, r):
assert 0 <= l <= r <= self._n
return self._sum(r) - self._sum(l)
def _sum(self, r):
s = 0
while r > 0:
s += self.data[r - 1]
r -= r & -r
return s
def main():
input = sys.stdin.readline
n, q = map(int, input().split())
fw = Fenwick_Tree(n)
for i, a in enumerate(map(int, input().split())): fw.add(i, a)
for _ in range(q):
t, a, b = map(int, input().split())
if t == 0:
fw.add(a, b)
else:
print(fw.sum(a, b))
if __name__ == "__main__":
main()
```
Yes
| 100,086 | [
0.372314453125,
-0.1087646484375,
-0.10870361328125,
0.216552734375,
-0.41357421875,
-0.1571044921875,
-0.053009033203125,
0.2587890625,
0.434326171875,
0.80712890625,
0.56640625,
-0.077392578125,
0.05487060546875,
-0.6064453125,
-0.44677734375,
0.2117919921875,
-0.4814453125,
-0.6... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.
* `0 p x`: a_p \gets a_p + x
* `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}.
Constraints
* 1 \leq N, Q \leq 500,000
* 0 \leq a_i, x \leq 10^9
* 0 \leq p < N
* 0 \leq l_i < r_i \leq N
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N Q
a_0 a_1 ... a_{N - 1}
\textrm{Query}_0
\textrm{Query}_1
:
\textrm{Query}_{Q - 1}
Output
For each query of the latter type, print the answer.
Example
Input
5 5
1 2 3 4 5
1 0 5
1 2 4
0 3 10
1 0 5
1 0 3
Output
15
7
25
6
Submitted Solution:
```
class SegmentTree:
def __init__(self, a):
self.padding = 0
self.n = len(a)
self.N = 2 ** (self.n-1).bit_length()
self.seg_data = [self.padding]*(self.N-1) + a + [self.padding]*(self.N-self.n)
for i in range(2*self.N-2, 0, -2):
self.seg_data[(i-1)//2] = self.seg_data[i] + self.seg_data[i-1]
def __len__(self):
return self.n
def update(self, i, x):
idx = self.N - 1 + i
self.seg_data[idx] += x
while idx:
idx = (idx-1) // 2
self.seg_data[idx] = self.seg_data[2*idx+1] + self.seg_data[2*idx+2]
def query(self, i, j):
# [i, j)
if i == j:
return self.seg_data[self.N - 1 + i]
else:
idx1 = self.N - 1 + i
idx2 = self.N - 2 + j # 閉区間にする
result = self.padding
while idx1 < idx2 + 1:
if idx1&1 == 0: # idx1が偶数
result = result + self.seg_data[idx1]
if idx2&1 == 1: # idx2が奇数
result = result + self.seg_data[idx2]
idx2 -= 1
idx1 //= 2
idx2 = (idx2 - 1)//2
return result
@property
def data(self):
return self.seg_data[self.N-1:self.N-1+self.n]
N, Q = map(int, input().split())
A = list(map(int, input().split()))
st = SegmentTree(A)
ans = []
for _ in range(Q):
t, i, j = map(int, input().split())
if t:
ans.append(st.query(i, j))
else:
st.update(i, j)
print(*ans, sep='\n')
```
Yes
| 100,087 | [
0.422119140625,
0.08673095703125,
-0.0745849609375,
0.453857421875,
-0.45556640625,
-0.265869140625,
-0.0623779296875,
0.22119140625,
0.2578125,
0.98046875,
0.2119140625,
0.07476806640625,
-0.10113525390625,
-0.77392578125,
-0.53564453125,
0.3271484375,
-0.392578125,
-0.76123046875... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.
* `0 p x`: a_p \gets a_p + x
* `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}.
Constraints
* 1 \leq N, Q \leq 500,000
* 0 \leq a_i, x \leq 10^9
* 0 \leq p < N
* 0 \leq l_i < r_i \leq N
* All values in Input are integer.
Input
Input is given from Standard Input in the following format:
N Q
a_0 a_1 ... a_{N - 1}
\textrm{Query}_0
\textrm{Query}_1
:
\textrm{Query}_{Q - 1}
Output
For each query of the latter type, print the answer.
Example
Input
5 5
1 2 3 4 5
1 0 5
1 2 4
0 3 10
1 0 5
1 0 3
Output
15
7
25
6
Submitted Solution:
```
class FenwickTree():
def __init__(self, n):
self.n = n
self.data = [0] * n
def build(self, arr):
#assert len(arr) <= n
for i, a in enumerate(arr):
self.data[i] = a
for i in range(1, self.n + 1):
if i + (i & -i) < self.n:
self.data[i + (i & -i) - 1] += self.data[i - 1]
def add(self, p, x):
#assert 0 <= p < self.n
p += 1
while p <= self.n:
self.data[p - 1] += x
p += p & -p
def sum(self, r):
s = 0
while r:
s += self.data[r - 1]
r -= r & -r
return s
def range_sum(self, l, r):
#assert 0 <= l <= r <= self.n
return self.sum(r) - self.sum(l)
import sys
input = sys.stdin.buffer.readline
N, Q = map(int, input().split())
A = tuple(map(int, input().split()))
ft = FenwickTree(N)
ft.build(A)
res = []
for _ in range(Q):
q, x, y = map(int, input().split())
if q:
res.append(str(ft.range_sum(x, y)))
else:
ft.add(x, y)
print('\n'.join(res))
```
No
| 100,092 | [
0.39892578125,
-0.08184814453125,
-0.10400390625,
0.27783203125,
-0.354248046875,
-0.11932373046875,
0.0196380615234375,
0.30078125,
0.41943359375,
0.8271484375,
0.51708984375,
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0.019256591796875,
-0.6396484375,
-0.462890625,
0.2027587890625,
-0.56396484375,
-0.645... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th.
The score of a participant is the product of his/her ranks in the two contests.
Process the following Q queries:
* In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
Constraints
* 1 \leq Q \leq 100
* 1\leq A_i,B_i\leq 10^9(1\leq i\leq Q)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
Q
A_1 B_1
:
A_Q B_Q
Output
For each query, print the maximum possible number of participants whose scores are smaller than Takahashi's.
Example
Input
8
1 4
10 5
3 3
4 11
8 9
22 40
8 36
314159265 358979323
Output
1
12
4
11
14
57
31
671644785
"Correct Solution:
```
q=int(input())
def check(t,a,b):
k=(2*a+t)//2
return k*(2*a+t-k)<a*b
for i in range(q):
a,b=sorted(map(int,input().split()))
if a==b or a==b-1:
print(2*a-2)
continue
l,r=1,b-a
while l+1<r:
t=(l+r)//2
if check(t,a,b):
l=t
else:
r=t
if check(r,a,b):
print(2*a-2+r)
elif check(l,a,b):
print(2*a-2+l)
else:
print(2*a-1)
```
| 100,161 | [
0.490966796875,
0.047882080078125,
-0.3173828125,
0.291748046875,
-0.406982421875,
-0.302734375,
-0.10760498046875,
0.19580078125,
-0.1424560546875,
0.923828125,
0.413330078125,
0.0113677978515625,
0.125732421875,
-0.96240234375,
-0.72900390625,
0.1851806640625,
-0.68359375,
-0.919... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th.
The score of a participant is the product of his/her ranks in the two contests.
Process the following Q queries:
* In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
Constraints
* 1 \leq Q \leq 100
* 1\leq A_i,B_i\leq 10^9(1\leq i\leq Q)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
Q
A_1 B_1
:
A_Q B_Q
Output
For each query, print the maximum possible number of participants whose scores are smaller than Takahashi's.
Example
Input
8
1 4
10 5
3 3
4 11
8 9
22 40
8 36
314159265 358979323
Output
1
12
4
11
14
57
31
671644785
"Correct Solution:
```
from math import sqrt
import sys
sdin = sys.stdin.readline
q = int(sdin())
ab = []
for i in range(q):
ab.append(tuple(map(int, sdin().split())))
for a, b in ab:
if b-a >= 0 and b-a <= 1:
print(2*a - 2)
else:
if not (sqrt(a*b) - int(sqrt(a*b))):
c = int(sqrt(a*b) - 1)
else:
c = int(sqrt(a*b))
if c*(c+1) >= a*b:
print(2*c - 2)
else:
print(2*c - 1)
```
| 100,162 | [
0.44189453125,
0.16943359375,
-0.340087890625,
0.251220703125,
-0.391357421875,
-0.2861328125,
-0.1256103515625,
0.090576171875,
-0.15380859375,
0.9580078125,
0.395263671875,
0.013458251953125,
0.1568603515625,
-0.85986328125,
-0.66943359375,
0.2178955078125,
-0.7421875,
-0.9052734... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th.
The score of a participant is the product of his/her ranks in the two contests.
Process the following Q queries:
* In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
Constraints
* 1 \leq Q \leq 100
* 1\leq A_i,B_i\leq 10^9(1\leq i\leq Q)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
Q
A_1 B_1
:
A_Q B_Q
Output
For each query, print the maximum possible number of participants whose scores are smaller than Takahashi's.
Example
Input
8
1 4
10 5
3 3
4 11
8 9
22 40
8 36
314159265 358979323
Output
1
12
4
11
14
57
31
671644785
"Correct Solution:
```
from math import sqrt
Q = int(input())
for i in range(Q):
A, B = list(map(int, input().split()))
sq = sqrt(A*B)
sq_int = int(sq)
ans = sq_int*2 - 2
if sq_int**2==A*B and A!=B:
ans -= 1
if sq_int*(sq_int+1)<A*B:
ans += 1
#print(A*B, sq_int)
print(ans)
```
| 100,163 | [
0.477294921875,
0.1690673828125,
-0.379150390625,
0.288818359375,
-0.46337890625,
-0.2149658203125,
-0.159912109375,
0.1435546875,
-0.06719970703125,
0.91748046875,
0.450439453125,
0.088134765625,
0.08782958984375,
-0.8876953125,
-0.6787109375,
0.307373046875,
-0.68212890625,
-0.90... | 11 |
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