message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
Constraints
* 1\leq N,K\leq 100
* N and K are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
If we can choose K integers as above, print `YES`; otherwise, print `NO`.
Examples
Input
3 2
Output
YES
Input
5 5
Output
NO
Input
31 10
Output
YES
Input
10 90
Output
NO
Submitted Solution:
```
import sys
input_methods=['clipboard','file','key']
using_method=0
input_method=input_methods[using_method]
tin=lambda : map(int, input().split())
lin=lambda : list(tin())
mod=1000000007
#+++++
def main():
#a = int(input())
#b , c = tin()
#s = input()
b, c= tin()
return 'Yes' if (b + 1) // 2 >= c else 'No'
#+++++
isTest=False
def pa(v):
if isTest:
print(v)
def input_clipboard():
import clipboard
input_text=clipboard.get()
input_l=input_text.splitlines()
for l in input_l:
yield l
if __name__ == "__main__":
if sys.platform =='ios':
if input_method==input_methods[0]:
ic=input_clipboard()
input = lambda : ic.__next__()
elif input_method==input_methods[1]:
sys.stdin=open('inputFile.txt')
else:
pass
isTest=True
else:
pass
#input = sys.stdin.readline
ret = main()
if ret is not None:
print(ret)
``` | instruction | 0 | 56,989 | 5 | 113,978 |
No | output | 1 | 56,989 | 5 | 113,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
Constraints
* 1\leq N,K\leq 100
* N and K are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
If we can choose K integers as above, print `YES`; otherwise, print `NO`.
Examples
Input
3 2
Output
YES
Input
5 5
Output
NO
Input
31 10
Output
YES
Input
10 90
Output
NO
Submitted Solution:
```
n,k=map(int,input().split())
print('YES' if n=>2*(k-1) else 'NO')
``` | instruction | 0 | 56,990 | 5 | 113,980 |
No | output | 1 | 56,990 | 5 | 113,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
Constraints
* 1\leq N,K\leq 100
* N and K are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
If we can choose K integers as above, print `YES`; otherwise, print `NO`.
Examples
Input
3 2
Output
YES
Input
5 5
Output
NO
Input
31 10
Output
YES
Input
10 90
Output
NO
Submitted Solution:
```
import math
def solution(N, K):
if (math.ceil((N + 1) / 2) >= K):
return "YES"
else:
return "NO"
n, k = [int(s) for s in input().split(' ')]
solution(n, k)
``` | instruction | 0 | 56,991 | 5 | 113,982 |
No | output | 1 | 56,991 | 5 | 113,983 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,992 | 5 | 113,984 |
"Correct Solution:
```
N,A=list(map(int,input().split()))
print(N-A+1)
``` | output | 1 | 56,992 | 5 | 113,985 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,993 | 5 | 113,986 |
"Correct Solution:
```
[N,i] = list(map(int,input().split()))
print(N-i+1)
``` | output | 1 | 56,993 | 5 | 113,987 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,994 | 5 | 113,988 |
"Correct Solution:
```
a,b=map(int, input().split())
print((a-b)+1)
``` | output | 1 | 56,994 | 5 | 113,989 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,995 | 5 | 113,990 |
"Correct Solution:
```
N,I=[int(a) for a in input().split(" ")]
print(N-I+1)
``` | output | 1 | 56,995 | 5 | 113,991 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,996 | 5 | 113,992 |
"Correct Solution:
```
a,b=map(int,input().split())
print((a-b+1))
``` | output | 1 | 56,996 | 5 | 113,993 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,997 | 5 | 113,994 |
"Correct Solution:
```
n,k=map(int,input().split())
res=n-k+1
print(res)
``` | output | 1 | 56,997 | 5 | 113,995 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,998 | 5 | 113,996 |
"Correct Solution:
```
n,i = map(int,input().split())
print(1+n-i)
``` | output | 1 | 56,998 | 5 | 113,997 |
Provide a correct Python 3 solution for this coding contest problem.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5 | instruction | 0 | 56,999 | 5 | 113,998 |
"Correct Solution:
```
a,b=map(int,input().split(" "))
print(a-b+1)
``` | output | 1 | 56,999 | 5 | 113,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
N,i=map(int,input().split())
car=(N+1)-i
print(car)
``` | instruction | 0 | 57,000 | 5 | 114,000 |
Yes | output | 1 | 57,000 | 5 | 114,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
n, i = [int(_) for _ in input().split()]
print(n-i+1)
``` | instruction | 0 | 57,001 | 5 | 114,002 |
Yes | output | 1 | 57,001 | 5 | 114,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
x,y=input().split(" ")
x=int(x)
y=int(y)
print(x+1-y)
``` | instruction | 0 | 57,002 | 5 | 114,004 |
Yes | output | 1 | 57,002 | 5 | 114,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
n,i=map(int,input().split(" "))
print(n-i+1)
``` | instruction | 0 | 57,003 | 5 | 114,006 |
Yes | output | 1 | 57,003 | 5 | 114,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
N,i=int(input ("enter the number of cars and car ))
X=N-i+1
Print(x)
``` | instruction | 0 | 57,004 | 5 | 114,008 |
No | output | 1 | 57,004 | 5 | 114,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
n,i = map(int,input().spllit())
print(n-i+1)
``` | instruction | 0 | 57,005 | 5 | 114,010 |
No | output | 1 | 57,005 | 5 | 114,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
N , i = map( int input( ) )
x = (N - i) + 1
print ( x )
``` | instruction | 0 | 57,006 | 5 | 114,012 |
No | output | 1 | 57,006 | 5 | 114,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."
Constraints
* 1 \leq N \leq 100
* 1 \leq i \leq N
Input
Input is given from Standard Input in the following format:
N i
Output
Print the answer.
Examples
Input
4 2
Output
3
Input
1 1
Output
1
Input
15 11
Output
5
Submitted Solution:
```
print(sum([int(i) for i in input().split()])+1)
``` | instruction | 0 | 57,007 | 5 | 114,014 |
No | output | 1 | 57,007 | 5 | 114,015 |
Provide a correct Python 3 solution for this coding contest problem.
Problem H: Squid Multiplication
Squid Eiko loves mathematics. Especially she loves to think about integer. One day, Eiko found a math problem from a website.
"A sequence b ={ai + aj | i < j } is generated from a sequence a ={a0 , ... , an | ai is even if i is 0, otherwise ai is odd}. Given the sequence b , find the sequence a ."
This problem is easy for Eiko and she feels boring. So, she made a new problem by modifying this problem .
"A sequence b ={ai *aj | i < j } is generated from a sequence a ={ a0, ... , an | ai is even if i is 0, otherwise ai is odd}. Given the sequence b , find the sequence a ."
Your task is to solve the problem made by Eiko.
Input
Input consists of multiple datasets.
Each dataset is given by following formats.
n
b0 b1 ... bn*(n+1)/2-1
n is the number of odd integers in the sequence a. The range of n is 2 ≤ n ≤ 250. bi is separated by a space. Each bi is 1 ≤ bi ≤ 263-1. The end of the input consists of a single 0.
Output
For each dataset, you should output two lines. First line contains a0, an even number in the sequence a. The second line contains n odd elements separated by a space. The odd elements are sorted by increasing order. You can assume that the result is greater than or equal to 1 and less than or equal to 231-1.
Example
Input
3
6 10 14 15 21 35
2
30 42 35
0
Output
2
3 5 7
6
5 7 | instruction | 0 | 57,101 | 5 | 114,202 |
"Correct Solution:
```
# AOJ 1069: Squid Multiplication
# Python3 2018.7.10 bal4u
def gcd(a, b):
while b != 0:
r = a % b
a, b = b, r
return a
while True:
n = int(input())
if n == 0: break
even, odd = [], []
b = list(map(int, input().split()))
for i in b:
if i & 1: odd.append(i)
else: even.append(i)
even.sort()
odd.sort()
e1, e2, o1 = even[0], even[1], odd[0];
g = gcd(e1, o1)
e1, o1 = e1//g, o1//g
g = gcd(e2, o1)
e2, o1 = e2//g, o1//g
g = int((e1 * e2)**0.5)
print(g)
print(*[i//g for i in even])
``` | output | 1 | 57,101 | 5 | 114,203 |
Provide a correct Python 3 solution for this coding contest problem.
Problem H: Squid Multiplication
Squid Eiko loves mathematics. Especially she loves to think about integer. One day, Eiko found a math problem from a website.
"A sequence b ={ai + aj | i < j } is generated from a sequence a ={a0 , ... , an | ai is even if i is 0, otherwise ai is odd}. Given the sequence b , find the sequence a ."
This problem is easy for Eiko and she feels boring. So, she made a new problem by modifying this problem .
"A sequence b ={ai *aj | i < j } is generated from a sequence a ={ a0, ... , an | ai is even if i is 0, otherwise ai is odd}. Given the sequence b , find the sequence a ."
Your task is to solve the problem made by Eiko.
Input
Input consists of multiple datasets.
Each dataset is given by following formats.
n
b0 b1 ... bn*(n+1)/2-1
n is the number of odd integers in the sequence a. The range of n is 2 ≤ n ≤ 250. bi is separated by a space. Each bi is 1 ≤ bi ≤ 263-1. The end of the input consists of a single 0.
Output
For each dataset, you should output two lines. First line contains a0, an even number in the sequence a. The second line contains n odd elements separated by a space. The odd elements are sorted by increasing order. You can assume that the result is greater than or equal to 1 and less than or equal to 231-1.
Example
Input
3
6 10 14 15 21 35
2
30 42 35
0
Output
2
3 5 7
6
5 7 | instruction | 0 | 57,102 | 5 | 114,204 |
"Correct Solution:
```
while(True):
n = int(input())
if n == 0 : exit()
b = list(map(int, input().split()))
even, odd = b[:n], b[n:]
sq = even[0]*even[1]//odd[0]
l, r = 1, 2**32
while r-l > 1:
mid = (l+r)//2
if mid*mid <= sq : l = mid
else : r = mid
print(l)
print(" ".join(map(str, sorted([x//l for x in even]))))
``` | output | 1 | 57,102 | 5 | 114,205 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
6 3
((()))
4
3
1
Output
2
2
1 | instruction | 0 | 57,103 | 5 | 114,206 |
"Correct Solution:
```
from heapq import heappush, heappop, heapify
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N, Q = map(int, readline().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [0]*(2*N0)
lazy = [0]*(2*N0)
L0 = [0]*(2*N0)
L1 = [0]*(2*N0)
def init(A):
for i in range(N):
data[N0-1+i] = A[i]
for i in range(N0):
L0[N0-1+i] = L1[N0-1+i] = i
for i in range(N0-2, -1, -1):
data[i] = min(data[2*i+1], data[2*i+2])
L0[i] = L0[2*i+1]
L1[i] = L1[2*i+2]
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if not v:
continue
lazy[2*i-1] += v; lazy[2*i] += v
data[2*i-1] += v; data[2*i] += v
lazy[i-1] = 0
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] += x; data[R-1] += x
if L & 1:
lazy[L-1] += x; data[L-1] += x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
u = 1
def query(r):
propagates(*gindex(0, r))
R = N0 + r
R = N0 + r
while R:
if R & 1:
R -= 1
if data[R-1] < 2:
l0 = L0[R-1]
r0 = L1[R-1]+1
break
R >>= 1
else:
return 0
k = R-1
while k < N0-1:
v = lazy[k]
if v:
lazy[2*k+1] += v; lazy[2*k+2] += v
data[2*k+1] += v; data[2*k+2] += v
lazy[k] = 0
if data[2*k+2] < 2:
l0 = (l0 + r0) >> 1
k = 2*k+2
else:
r0 = (l0 + r0) >> 1
k = 2*k+1
return r0
que = []
*s, = map("()".index, readline().strip())
A = [0]*N
C = [0]*N
cur = 0
for i in range(N):
if s[i]:
que.append(i)
C[i] = 1
cur -= 1
else:
cur += 1
A[i] = cur
heapify(que)
init(A)
for i in range(Q):
q = int(readline())
if s[q-1] == 0:
while que and s[que[0]] == 0:
v = heappop(que)
C[v] = 0
if not que or q-1 <= que[0]:
write("%d\n" % q)
else:
k = heappop(que)
C[k] = 0
s[k] = 0
s[q-1] = 1
heappush(que, q-1)
write("%d\n" % (k+1))
update(k, q-1, 2)
else:
v = query(q-1)
if v == q-1:
write("%d\n" % q)
else:
s[v] = 1
s[q-1] = 0
if C[v] == 0:
heappush(que, v)
C[v] = 1
write("%d\n" % (v + 1))
update(v, q-1, -2)
solve()
``` | output | 1 | 57,103 | 5 | 114,207 |
Provide a correct Python 3 solution for this coding contest problem.
Tunnel formula
One day while exploring an abandoned mine, you found a long formula S written in the mine. If you like large numbers, you decide to take out the choke and add `(` or `)` so that the result of the formula calculation is as large as possible. If it has to be a mathematical formula even after adding it, how many can it be at the maximum?
There is enough space between the letters, and you can add as many `(` or `)` as you like. If the final formula is a formula, you may write `(` or `)` so that the correspondence of the first parenthesis is broken (see Sample 2). Also, here, <expr> defined by the following BNF is called a mathematical formula. All numbers in the formula are single digits.
<expr> :: = "(" <expr> ")"
| <term> "+" <term>
| <term> "-" <term>
<term> :: = <digit> | <expr>
<digit> :: = "0" | "1" | "2" | "3" | "4"
| "5" | "6" | "7" | "8" | "9"
Constraints
* 3 ≤ | S | ≤ 200
S represents a mathematical formula.
Input Format
Input is given from standard input in the following format.
S
Output Format
Output the answer as an integer.
Sample Input 1
1- (2 + 3-4 + 5)
Sample Output 1
Five
1- (2 + 3- (4 + 5)) is the maximum.
Sample Input 2
1- (2 + 3 + 4)
Sample Output 2
0
(1- (2 + 3) + 4) is the maximum.
Sample Input 3
1- (2 + 3)
Sample Output 3
-Four
Note that 1- (2) + (3) is not the formula here.
Example
Input
1-(2+3-4+5)
Output
5 | instruction | 0 | 57,118 | 5 | 114,236 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 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 pf(s): return print(s, flush=True)
def main():
s = S()
fm = {}
a = []
for c in s:
if '0' <= c <= '9':
ci = int(c)
if len(a) > 0 and isinstance(a[-1], int):
a[-1] = a[-1] * 10 + ci
else:
a.append(ci)
else:
a.append(c)
def f(a):
key = tuple(a)
if key in fm:
return fm[key]
if len(a) == 2:
fm[key] = [inf,-inf]
return [inf,-inf]
for i in range(len(a)):
if a[i] != '(':
if i > 0:
a = a[i:]
break
for i in range(len(a)-1,-1,-1):
if a[i] != ')':
a = a[:i+1]
break
if len(a) == 1:
r = [a[0],a[0]]
fm[key] = r
return r
ri = [inf]
ra = [-inf]
l = len(a)
for i in range(1,len(a)-1):
if not a[i] in ['+','-'] or (i > 1 and a[i-2] == '(') or (i+2 < l and a[i+2] == ')'):
continue
fl = f(a[:i])
fr = f(a[i+1:])
if a[i] == '+':
ri.append(fl[0]+fr[0])
ra.append(fl[1]+fr[1])
else:
ri.append(fl[0]-fr[1])
ra.append(fl[1]-fr[0])
r = [min(ri), max(ra)]
fm[key] = r
return r
r = f(a)
# print(r)
# for k in sorted(fm.keys(), key=lambda x: [len(str(x)), str(x)]):
# print('k,fm[k]', ''.join(map(str,k)),fm[k])
return r[1]
print(main())
``` | output | 1 | 57,118 | 5 | 114,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tunnel formula
One day while exploring an abandoned mine, you found a long formula S written in the mine. If you like large numbers, you decide to take out the choke and add `(` or `)` so that the result of the formula calculation is as large as possible. If it has to be a mathematical formula even after adding it, how many can it be at the maximum?
There is enough space between the letters, and you can add as many `(` or `)` as you like. If the final formula is a formula, you may write `(` or `)` so that the correspondence of the first parenthesis is broken (see Sample 2). Also, here, <expr> defined by the following BNF is called a mathematical formula. All numbers in the formula are single digits.
<expr> :: = "(" <expr> ")"
| <term> "+" <term>
| <term> "-" <term>
<term> :: = <digit> | <expr>
<digit> :: = "0" | "1" | "2" | "3" | "4"
| "5" | "6" | "7" | "8" | "9"
Constraints
* 3 ≤ | S | ≤ 200
S represents a mathematical formula.
Input Format
Input is given from standard input in the following format.
S
Output Format
Output the answer as an integer.
Sample Input 1
1- (2 + 3-4 + 5)
Sample Output 1
Five
1- (2 + 3- (4 + 5)) is the maximum.
Sample Input 2
1- (2 + 3 + 4)
Sample Output 2
0
(1- (2 + 3) + 4) is the maximum.
Sample Input 3
1- (2 + 3)
Sample Output 3
-Four
Note that 1- (2) + (3) is not the formula here.
Example
Input
1-(2+3-4+5)
Output
5
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 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 pf(s): return print(s, flush=True)
def main():
s = S()
fm = {}
def sa(s):
a = []
for c in s:
if '0' <= c <= '9':
ci = int(c)
if len(a) > 0 and isinstance(a[-1], int):
a[-1] = a[-1] * 10 + ci
else:
a.append(ci)
else:
a.append(c)
return a
def f(s, sf):
key = (s, sf)
if key in fm:
return fm[key]
# print(s,sf)
a = sa(s)
l = len(a)
if l == 1:
if isinstance(a[0], int):
return a[0]
return None
r = None
for i in range(0,l-2):
if not isinstance(a[i], int):
continue
if a[i-1] == '(' or a[i+1] == ')':
continue
tl = f(''.join(map(str, a[:i+1])), sf)
if tl is None:
continue
if a[i+1] == '+':
tr = f(''.join(map(str, a[i+2:])), sf)
if not tr is None:
tl += tr
else:
tl = None
else:
tr = f(''.join(map(str, a[i+2:])), not sf)
if not tr is None:
tl -= tr
else:
tl = None
if tl is None:
continue
if sf:
if r is None or r < tl:
r = tl
else:
if r is None or r > tl:
r = tl
kc = 0
for c in a:
if c == '(':
kc += 1
elif c == ')':
kc -= 1
if kc < 0:
kc = 0
if kc > 0:
a += [')'] * kc
kc = 0
for c in a[::-1]:
if c == '(':
kc -= 1
if kc < 0:
kc = 0
elif c == ')':
kc += 1
if kc > 0:
a = ['('] * kc + a
while '(' in a:
# print('while (', a)
i = len(a) - 1
while '(' != a[i]:
i -= 1
j = a.index(')', i)
if j-i < 2:
break
tr = a[i+1]
for k in range(i+2,j,2):
if a[k] == '+':
tr += a[k+1]
else:
tr -= a[k+1]
a[i:j+1] = [tr]
if not '(' in a:
# print('not (', a)
tr = a[0]
for i in range(1,len(a),2):
if a[i] == '+':
tr += a[i+1]
else:
tr -= a[i+1]
if sf:
if r is None or r < tr:
r = tr
else:
if r is None or r > tr:
r = tr
fm[key] = r
return r
r = f(s, True)
return r
print(main())
``` | instruction | 0 | 57,119 | 5 | 114,238 |
No | output | 1 | 57,119 | 5 | 114,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tunnel formula
One day while exploring an abandoned mine, you found a long formula S written in the mine. If you like large numbers, you decide to take out the choke and add `(` or `)` so that the result of the formula calculation is as large as possible. If it has to be a mathematical formula even after adding it, how many can it be at the maximum?
There is enough space between the letters, and you can add as many `(` or `)` as you like. If the final formula is a formula, you may write `(` or `)` so that the correspondence of the first parenthesis is broken (see Sample 2). Also, here, <expr> defined by the following BNF is called a mathematical formula. All numbers in the formula are single digits.
<expr> :: = "(" <expr> ")"
| <term> "+" <term>
| <term> "-" <term>
<term> :: = <digit> | <expr>
<digit> :: = "0" | "1" | "2" | "3" | "4"
| "5" | "6" | "7" | "8" | "9"
Constraints
* 3 ≤ | S | ≤ 200
S represents a mathematical formula.
Input Format
Input is given from standard input in the following format.
S
Output Format
Output the answer as an integer.
Sample Input 1
1- (2 + 3-4 + 5)
Sample Output 1
Five
1- (2 + 3- (4 + 5)) is the maximum.
Sample Input 2
1- (2 + 3 + 4)
Sample Output 2
0
(1- (2 + 3) + 4) is the maximum.
Sample Input 3
1- (2 + 3)
Sample Output 3
-Four
Note that 1- (2) + (3) is not the formula here.
Example
Input
1-(2+3-4+5)
Output
5
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 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 pf(s): return print(s, flush=True)
def main():
s = S()
fm = {}
def sa(s):
a = []
for c in s:
if '0' <= c <= '9':
ci = int(c)
if len(a) > 0 and isinstance(a[-1], int):
a[-1] = a[-1] * 10 + ci
else:
a.append(ci)
else:
a.append(c)
return a
def f(s, sf):
key = (s, sf)
if key in fm:
return fm[key]
# print(s,sf)
if s[0] == '(' and s[-1] == ')':
s = s[1:-1]
a = sa(s)
l = len(a)
if l == 1:
if isinstance(a[0], int):
return a[0]
return None
r = None
for i in range(0,l-2):
if not isinstance(a[i], int):
continue
if a[i-1] == '(' or a[i+1] == ')':
continue
tl = f(''.join(map(str, a[:i+1])), sf)
if tl is None:
continue
if a[i+1] == '+':
tr = f(''.join(map(str, a[i+2:])), sf)
if not tr is None:
tl += tr
else:
tl = None
else:
tr = f(''.join(map(str, a[i+2:])), not sf)
if not tr is None:
tl -= tr
else:
tl = None
if tl is None:
continue
if sf:
if r is None or r < tl:
r = tl
else:
if r is None or r > tl:
r = tl
kc = 0
for c in a:
if c == '(':
kc += 1
elif c == ')':
kc -= 1
if kc < 0:
kc = 0
if kc > 0:
a += [')'] * kc
kc = 0
for c in a[::-1]:
if c == '(':
kc -= 1
if kc < 0:
kc = 0
elif c == ')':
kc += 1
if kc > 0:
a = ['('] * kc + a
while '(' in a:
# print('while (', a)
i = len(a) - 1
while '(' != a[i]:
i -= 1
j = a.index(')', i)
if j-i < 2:
break
tr = a[i+1]
for k in range(i+2,j,2):
if a[k] == '+':
tr += a[k+1]
else:
tr -= a[k+1]
a[i:j+1] = [tr]
if not '(' in a:
# print('not (', a)
tr = a[0]
for i in range(1,len(a),2):
if a[i] == '+':
tr += a[i+1]
else:
tr -= a[i+1]
if sf:
if r is None or r < tr:
r = tr
else:
if r is None or r > tr:
r = tr
fm[key] = r
return r
r = f(s, True)
return r
print(main())
``` | instruction | 0 | 57,120 | 5 | 114,240 |
No | output | 1 | 57,120 | 5 | 114,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tunnel formula
One day while exploring an abandoned mine, you found a long formula S written in the mine. If you like large numbers, you decide to take out the choke and add `(` or `)` so that the result of the formula calculation is as large as possible. If it has to be a mathematical formula even after adding it, how many can it be at the maximum?
There is enough space between the letters, and you can add as many `(` or `)` as you like. If the final formula is a formula, you may write `(` or `)` so that the correspondence of the first parenthesis is broken (see Sample 2). Also, here, <expr> defined by the following BNF is called a mathematical formula. All numbers in the formula are single digits.
<expr> :: = "(" <expr> ")"
| <term> "+" <term>
| <term> "-" <term>
<term> :: = <digit> | <expr>
<digit> :: = "0" | "1" | "2" | "3" | "4"
| "5" | "6" | "7" | "8" | "9"
Constraints
* 3 ≤ | S | ≤ 200
S represents a mathematical formula.
Input Format
Input is given from standard input in the following format.
S
Output Format
Output the answer as an integer.
Sample Input 1
1- (2 + 3-4 + 5)
Sample Output 1
Five
1- (2 + 3- (4 + 5)) is the maximum.
Sample Input 2
1- (2 + 3 + 4)
Sample Output 2
0
(1- (2 + 3) + 4) is the maximum.
Sample Input 3
1- (2 + 3)
Sample Output 3
-Four
Note that 1- (2) + (3) is not the formula here.
Example
Input
1-(2+3-4+5)
Output
5
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 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 pf(s): return print(s, flush=True)
def main():
s = S()
fm = {}
a = []
for c in s:
if '0' <= c <= '9':
ci = int(c)
if len(a) > 0 and isinstance(a[-1], int):
a[-1] = a[-1] * 10 + ci
else:
a.append(ci)
else:
a.append(c)
def f(a):
key = tuple(a)
if key in fm:
return fm[key]
if len(a) == 2:
fm[key] = [inf,-inf]
return [inf,-inf]
for i in range(len(a)):
if a[i] != '(':
if i > 0:
a = a[i:]
break
for i in range(len(a)-1,-1,-1):
if a[i] != ')':
a = a[:i+1]
break
if len(a) == 1:
r = [a[0],a[0]]
fm[key] = r
return r
ri = [inf]
ra = [-inf]
for i in range(1,len(a)-1):
if not a[i] in ['+','-']:
continue
fl = f(a[:i])
fr = f(a[i+1:])
if a[i] == '+':
ri.append(fl[0]+fr[0])
ra.append(fl[1]+fr[1])
else:
ri.append(fl[0]-fr[1])
ra.append(fl[1]-fr[0])
r = [min(ri), max(ra)]
fm[key] = r
return r
r = f(a)
return r[1]
print(main())
``` | instruction | 0 | 57,121 | 5 | 114,242 |
No | output | 1 | 57,121 | 5 | 114,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tunnel formula
One day while exploring an abandoned mine, you found a long formula S written in the mine. If you like large numbers, you decide to take out the choke and add `(` or `)` so that the result of the formula calculation is as large as possible. If it has to be a mathematical formula even after adding it, how many can it be at the maximum?
There is enough space between the letters, and you can add as many `(` or `)` as you like. If the final formula is a formula, you may write `(` or `)` so that the correspondence of the first parenthesis is broken (see Sample 2). Also, here, <expr> defined by the following BNF is called a mathematical formula. All numbers in the formula are single digits.
<expr> :: = "(" <expr> ")"
| <term> "+" <term>
| <term> "-" <term>
<term> :: = <digit> | <expr>
<digit> :: = "0" | "1" | "2" | "3" | "4"
| "5" | "6" | "7" | "8" | "9"
Constraints
* 3 ≤ | S | ≤ 200
S represents a mathematical formula.
Input Format
Input is given from standard input in the following format.
S
Output Format
Output the answer as an integer.
Sample Input 1
1- (2 + 3-4 + 5)
Sample Output 1
Five
1- (2 + 3- (4 + 5)) is the maximum.
Sample Input 2
1- (2 + 3 + 4)
Sample Output 2
0
(1- (2 + 3) + 4) is the maximum.
Sample Input 3
1- (2 + 3)
Sample Output 3
-Four
Note that 1- (2) + (3) is not the formula here.
Example
Input
1-(2+3-4+5)
Output
5
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 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 pf(s): return print(s, flush=True)
def main():
s = S()
fm = {}
def sa(s):
a = []
for c in s:
if '0' <= c <= '9':
ci = int(c)
if len(a) > 0 and isinstance(a[-1], int):
a[-1] = a[-1] * 10 + ci
else:
a.append(ci)
else:
a.append(c)
return a
def f(s, sf):
key = (s, sf)
if key in fm:
return fm[key]
# print(s,sf)
a = sa(s)
l = len(a)
r = None
for i in range(2,l-2):
if not isinstance(a[i], int):
continue
if a[i-1] == '(' or a[i+1] == ')':
continue
tl = f(''.join(map(str, a[:i+1])), sf)
if tl is None:
continue
if a[i+1] == '+':
tr = f(''.join(map(str, a[i+2:])), sf)
if not tr is None:
tl += tr
else:
tl = None
else:
tr = f(''.join(map(str, a[i+2:])), not sf)
if not tr is None:
tl -= tr
else:
tl = None
if tl is None:
continue
if sf:
if r is None or r < tl:
r = tl
else:
if r is None or r > tl:
r = tl
kc = 0
for c in a:
if c == '(':
kc += 1
elif c == ')':
kc -= 1
if kc < 0:
kc = 0
if kc > 0:
a += [')'] * kc
kc = 0
for c in a[::-1]:
if c == '(':
kc -= 1
if kc < 0:
kc = 0
elif c == ')':
kc += 1
if kc > 0:
a = ['('] * kc + a
while '(' in a:
# print('while (', a)
i = len(a) - 1
while '(' != a[i]:
i -= 1
j = a.index(')', i)
if j-i < 2:
break
tr = a[i+1]
for k in range(i+2,j,2):
if a[k] == '+':
tr += a[k+1]
else:
tr -= a[k+1]
a[i:j+1] = [tr]
if not '(' in a:
# print('not (', a)
tr = a[0]
for i in range(1,len(a),2):
if a[i] == '+':
tr += a[i+1]
else:
tr -= a[i+1]
if sf:
if r is None or r < tr:
r = tr
else:
if r is None or r > tr:
r = tr
fm[key] = r
return r
r = f(s, True)
return r
print(main())
``` | instruction | 0 | 57,122 | 5 | 114,244 |
No | output | 1 | 57,122 | 5 | 114,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1. | instruction | 0 | 57,180 | 5 | 114,360 |
Tags: dp, implementation, math, number theory
Correct Solution:
```
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.
Copyright 2019 Cheran Senthilkumar <hello@cheran.io>
"""
import os
from atexit import register
from io import BytesIO
stdout = BytesIO()
register(lambda: os.write(1, stdout.getvalue()))
readline = BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = lambda: readline().decode().rstrip('\r\n')
def print(*args, sep=' ', end='\n'):
stdout.write(sep.join(map(str, args)).encode())
stdout.write(end.encode())
def gcd(a, b):
while b:
a, b = b, a % b
return a
def main():
n = int(input())
mat = [format(int(input(), 16), '0%db' % n) for _ in range(n)]
t = [0] * (n + 1)
for i in range(n):
cnt1, cnt2 = 1, 1
prev1, prev2 = mat[i][0], mat[0][i]
for j in range(1, n):
if mat[i][j] == prev1:
cnt1 += 1
else:
t[cnt1], cnt1 = 1, 1
if mat[j][i] == prev2:
cnt2 += 1
else:
t[cnt2], cnt2 = 1, 1
prev1, prev2 = mat[i][j], mat[j][i]
t[cnt1], t[cnt2] = 1, 1
res = n
for i in range(n):
if t[i] == 1:
res = gcd(i, res)
print(res)
if __name__ == '__main__':
main()
``` | output | 1 | 57,180 | 5 | 114,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1. | instruction | 0 | 57,184 | 5 | 114,368 |
Tags: dp, implementation, math, number theory
Correct Solution:
```
from sys import stdin,stdout
def hexa(c):
if c=='0':
return "0000"
if c=='1':
return "0001"
if c=='2':
return "0010"
if c=='3':
return "0011"
if c=='4':
return "0100"
if c=='5':
return "0101"
if c=='6':
return "0110"
if c=='7':
return "0111"
if c=='8':
return "1000"
if c=='9':
return "1001"
if c=='A':
return "1010"
if c=='B':
return "1011"
if c=='C':
return "1100"
if c=='D':
return "1101"
if c=='E':
return "1110"
if c=='F':
return "1111"
#hexa = ['0000','0001','0010','0011','0100','0101','0110','0111','1000','1001','1010','1011','1100','1101','1110','1111']
def gcd(a,b):
if b==0:
return a
return gcd(b,a%b)
n=int(stdin.readline())
a=[0]*n
for _ in range(n):
s = stdin.readline().strip()
t=""
for i in s:
t+=hexa(i)
a[_] = t
r=[]
t= 1
for i in range(1,n):
if a[i]==a[i-1]:
t+=1
else:
r.append(t)
t=1
r.append(t)
c=[]
t=1
for i in range(1,n):
f=1
for j in range(n):
if a[j][i-1]!=a[j][i]:
f=0
break
if f==1:
t+=1
else:
c.append(t)
t=1
c.append(t)
ans = r[0]
for i in r+c:
ans = gcd(ans,i)
stdout.write("%d" %ans)
``` | output | 1 | 57,184 | 5 | 114,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1. | instruction | 0 | 57,186 | 5 | 114,372 |
Tags: dp, implementation, math, number theory
Correct Solution:
```
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.
Copyright 2019 Cheran Senthilkumar <hello@cheran.io>
"""
import os
from atexit import register
from io import BytesIO
stdout = BytesIO()
register(lambda: os.write(1, stdout.getvalue()))
readline = BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = lambda: readline().decode().rstrip('\r\n')
def print(*args, sep=' ', end='\n'):
stdout.write(sep.join(map(str, args)).encode())
stdout.write(end.encode())
def gcd(a, b):
while b:
a, b = b, a % b
return a
def main():
n = int(readline())
mat = [format(int(readline(), 16), '0%db' % n) for _ in range(n)]
t = [0] * (n + 1)
for i in range(n):
cnt1, cnt2 = 1, 1
prev1, prev2 = mat[i][0], mat[0][i]
for j in range(1, n):
if mat[i][j] == prev1:
cnt1 += 1
else:
t[cnt1], cnt1 = 1, 1
if mat[j][i] == prev2:
cnt2 += 1
else:
t[cnt2], cnt2 = 1, 1
prev1, prev2 = mat[i][j], mat[j][i]
t[cnt1], t[cnt2] = 1, 1
res = n
for i in range(n):
if t[i] == 1:
res = gcd(i, res)
print(res)
if __name__ == '__main__':
main()
``` | output | 1 | 57,186 | 5 | 114,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
import sys
n = int(sys.stdin.readline())
a = [[] for i in range(n)]
for i in range(n):
a[i] = bin((1 << n) | int(sys.stdin.readline(), 16))[3:]
from math import gcd
ans = 0
i = 0
while i < n:
j = i + 1
while j < n and a[i] == a[j]:
j += 1
ans = gcd(ans, j - i)
k = 0
while k < n:
l = k + 1
while l < n and a[i][k] == a[i][l]:
l += 1
ans = gcd(ans, k - l)
k = l
i = j
print(ans)
``` | instruction | 0 | 57,188 | 5 | 114,376 |
Yes | output | 1 | 57,188 | 5 | 114,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
from sys import stdin,stdout
def gcd(a,b):
if b==0:
return a
else:
return gcd(b,a%b)
n=int(stdin.readline())
temp=""
ans=n
p=0
for i in range(n):
s=bin((1<<n)+int(stdin.readline(),16))[3:]
if i==0 or temp!=s:
temp=s
ans=gcd(ans,p)
p=1
pp=1
for j in range(1,n):
if s[j]==s[j-1]:
pp+=1
else:
ans=gcd(ans,pp)
pp=1
else:
p+=1
ans=gcd(ans,p)
print(ans)
``` | instruction | 0 | 57,189 | 5 | 114,378 |
Yes | output | 1 | 57,189 | 5 | 114,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
from math import gcd
n = int(input())
a = [[2]*(n+1) for _ in range(n+1)]
for i in range(n):
j = 0
for x in map(lambda c: int(c, 16), input()):
for d in range(3, -1, -1):
a[i][j] = 1 if (1 << d) & x else 0
j += 1
a[i][-1] = 2
a[-1] = [2] * (n+1)
ans = 0
for i in range(n):
prev, cont = a[i][0], 1
for j in range(1, n+1):
if prev != a[i][j]:
if ans == 0:
ans = cont
else:
ans = gcd(ans, cont)
prev, cont = a[i][j], 1
else:
cont += 1
for j in range(n):
prev, cont = a[0][j], 1
for i in range(1, n+1):
if prev != a[i][j]:
ans = gcd(ans, cont)
prev, cont = a[i][j], 1
else:
cont += 1
print(ans)
``` | instruction | 0 | 57,190 | 5 | 114,380 |
Yes | output | 1 | 57,190 | 5 | 114,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
from math import gcd
from functools import reduce
import sys
import os
d=sys.stdin.readlines()
if d[-1][-1] != "\n":
d[-1]+="\n"
n = int(d[0])
cur_tab = [0 for _ in range(n+1)]
def group(l):
cur = 1
for x in range(1,n):
if l[x] == l[x-1]:
cur+=1
else:
cur_tab[cur]=1
cur=1
cur_tab[cur]=1
if cur_tab[1]==1:
print(1)
exit(0)
mat=[None for _ in range(n)]
euh=0
for x in d[1:]:
line = x[:-1]
s=str(bin(int(line, 16)))[2:].zfill(n)
mat[euh]=s
euh+=1
group(s)
for i in range(n):
cur=1
for x in range(1,n):
if mat[x][i] == mat[x-1][i]:
cur+=1
else:
cur_tab[cur]=1
cur=1
cur_tab[cur]=1
if cur_tab[1]==1:
print(1)
exit(0)
print(reduce(gcd, [index for index in range(n+1) if cur_tab[index] == 1]))
``` | instruction | 0 | 57,191 | 5 | 114,382 |
Yes | output | 1 | 57,191 | 5 | 114,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
import math
import sys
import bisect
#input=sys.stdin.readline
#t=int(input())
t=1
def d2b(x):
s1=''
for i in range(3,-1,-1):
k=x>>i
if k&1:
s1+='1'
else:
s1+='0'
return s1
def divisor(x):
i=2
l1=[]
while i*i<=x:
if x%i==0:
if x//i==i:
l1.append(i)
else:
l1.append(x//i)
l1.append(i)
i+=1
return l1
for _ in range(t):
#mod=10**9+7
n=int(input())
#n,k=map(int,input().split())
#l=list(map(int,input().split()))
#s=input()
l=[]
for i in range(n):
s=input()
s2=''
for j in range(n//4):
if s[j].isdigit():
s2+=d2b(int(s[j]))
else:
s2+=d2b(ord(s[j])-55)
l.append(s2)
l2=divisor(n)
l2.sort(reverse=True)
val=[]
flag=True
ans=1
#print(l2)
for i1 in range(len(l2)):
flag=True
for i in range(l2[i1]):
for j in range(l2[i1]):
#print()
for k1 in range(n//l2[i1]):
for k2 in range(n//l2[i1]):
#print(i+k1*(l2[i1])+1,j+k2*(l2[i1])+1)
if l[i][j]!=l[i+k1*(l2[i1])][j+k2*(l2[i1])]:
flag=False
if flag==False:
break
if flag==False:
break
if flag==False:
break
if flag==True:
ans=l2[i]
break
print(ans)
``` | instruction | 0 | 57,192 | 5 | 114,384 |
No | output | 1 | 57,192 | 5 | 114,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
ans=0
def check22(w):
global ans
for i in range(0,n,w):
for j in range(0,n,w):
for p in range(w):
for q in range(w):
if mat[p+i][q+j]!=mat[i][j]:
return
ans=max(ans,w)
n=int(input())
mat=[]
ss=n
for i in range(n):
s=int(input(),16)
k=list(bin(s))
x=[0 for _ in range(n)]
for r in range(2,len(k)):
x[r-2]=int(k[r])
mat.append(x)
for w in range(2,n//2+1):
check22(w)
check22(n)
print(ans)
``` | instruction | 0 | 57,193 | 5 | 114,386 |
No | output | 1 | 57,193 | 5 | 114,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
n = int(input())
A = []
C = []
x = input()
A.append(x)
for j in x:
d = list(map(int,list(bin(int(j, 16))[2:])))
for _ in range(len(d), 4):
C.append(0)
C.extend(d)
ok = []
for i in range(1, n + 1):
if (n % i == 0):
c = 1
for j in range(n//i):
if (not all(x == C[j * i] for x in C[j * i : (j + 1) * i])):
c = 0
break
if c == 1:
ok.append(i)
for i in range(n - 1):
x = input()
A.append(x)
ok.reverse()
for i in ok:
if (n % i == 0):
c = 1
for j in range(n//i):
if (not all(x == C[j * i] for x in C[j * i : (j + 1) * i])):
c = 0
break
if c == 1:
ans = i
break
print(i)
``` | instruction | 0 | 57,194 | 5 | 114,388 |
No | output | 1 | 57,194 | 5 | 114,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a binary matrix A of size n × n. Let's denote an x-compression of the given matrix as a matrix B of size n/x × n/x such that for every i ∈ [1, n], j ∈ [1, n] the condition A[i][j] = B[⌈ i/x ⌉][⌈ j/x ⌉] is met.
Obviously, x-compression is possible only if x divides n, but this condition is not enough. For example, the following matrix of size 2 × 2 does not have any 2-compression:
01 10
For the given matrix A, find maximum x such that an x-compression of this matrix is possible.
Note that the input is given in compressed form. But even though it is compressed, you'd better use fast input.
Input
The first line contains one number n (4 ≤ n ≤ 5200) — the number of rows and columns in the matrix A. It is guaranteed that n is divisible by 4.
Then the representation of matrix follows. Each of n next lines contains n/4 one-digit hexadecimal numbers (that is, these numbers can be represented either as digits from 0 to 9 or as uppercase Latin letters from A to F). Binary representation of each of these numbers denotes next 4 elements of the matrix in the corresponding row. For example, if the number B is given, then the corresponding elements are 1011, and if the number is 5, then the corresponding elements are 0101.
Elements are not separated by whitespaces.
Output
Print one number: maximum x such that an x-compression of the given matrix is possible.
Examples
Input
8
E7
E7
E7
00
00
E7
E7
E7
Output
1
Input
4
7
F
F
F
Output
1
Note
The first example corresponds to the matrix:
11100111 11100111 11100111 00000000 00000000 11100111 11100111 11100111
It is easy to see that the answer on this example is 1.
Submitted Solution:
```
from math import gcd
n = int(input())
a = []
for i in range(n):
row = input()
a.append(bin(int(row,16))[2:].zfill(n))
# print(*a, sep='\n')
vis = [[False]*n]*n
xs = set()
for i in range(n):
br = False
for j in range(n):
if vis[i][j]:
continue
currx = 1
d = a[i][j]
expand = True
while expand and currx+i < n and currx+j < n:
for ki in range(i, i+currx):
if a[ki][j+currx] != d:
expand = False
break
if expand:
for kj in range(j, j+currx):
if a[i+currx][kj] != d:
expand = False
break
if expand:
for ki in range(i, i+currx):
vis[ki][j+currx] = True
for kj in range(j, j+currx):
vis[i+currx][kj] = True
currx += 1
# print(currx)
kj = j + currx
xs.add(currx)
if currx == 1:
br = True
break
j += currx
if br:
break
print(xs)
# print(min(xs))
ans = n
for x in xs:
ans = gcd(ans, x)
print(ans)
``` | instruction | 0 | 57,195 | 5 | 114,390 |
No | output | 1 | 57,195 | 5 | 114,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input
The first line contains a non-empty string s — the sum Xenia needs to count. String s contains no spaces. It only contains digits and characters "+". Besides, string s is a correct sum of numbers 1, 2 and 3. String s is at most 100 characters long.
Output
Print the new sum that Xenia can count.
Examples
Input
3+2+1
Output
1+2+3
Input
1+1+3+1+3
Output
1+1+1+3+3
Input
2
Output
2
Submitted Solution:
```
s=input()
ar=[]
for i in s:
if i!='+':
ar.append(i)
ar.sort()
t=""
for i in ar:
t=t+i+'+'
t=t[:-1]
print(t)
``` | instruction | 0 | 57,444 | 5 | 114,888 |
Yes | output | 1 | 57,444 | 5 | 114,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
def f(x):
bi = 0
ci = 0
cnt1 = -10 ** 10
minsum1 = 10 ** 10
cnt2 = -10 ** 10
minsum2 = 10 ** 10
for i in range(len(A)):
if bi - minsum1 > cnt1:
cnt1 = bi - minsum1
if bi <= minsum1:
minsum1 = bi
if ci - minsum2 > cnt2:
cnt2 = ci - minsum2
if ci <= minsum2:
minsum2 = ci
bi += A[i] - x
ci += x - A[i]
if bi - minsum1 > cnt1:
cnt1 = bi - minsum1
if bi <= minsum1:
minsum1 = bi
if ci - minsum2 > cnt2:
cnt2 = ci - minsum2
if ci <= minsum2:
minsum2 = ci
return max(cnt1, cnt2, 0)
n = int(input())
A = list(map(int, input().split()))
l = min(A)
r = max(A)
for _ in range(70):
p = l + (r - l) / 3
q = r - (r - l) / 3
x = f(p)
y = f(q)
if x > y:
l = p
elif y > x:
r = q
else:
l = p
r = q
print(f(l))
``` | instruction | 0 | 57,564 | 5 | 115,128 |
Yes | output | 1 | 57,564 | 5 | 115,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
import sys
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().split()]
eps = 1e-12
def f(x):
mx = a[0] - x
tsmx = 0.0
mn = a[0] - x
tsmn = 0.0
for ai in a:
tsmx = max(tsmx + ai - x, ai - x)
mx = max(tsmx, mx)
tsmn = min(tsmn + ai - x, ai - x)
mn = min(tsmn, mn)
return abs(mx), abs(mn)
l = min(a)
r = max(a)
f1, f2 = f(l)
for i in range(0, 90):
m = (l + r) / 2
f1, f2 = f(m)
if f1 > f2:
l = m
else:
r = m
A, B = f(l)
print(min(A,B))
``` | instruction | 0 | 57,566 | 5 | 115,132 |
Yes | output | 1 | 57,566 | 5 | 115,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
n=int(input())
A=list(map(int,input().split()))
B=[]
x=float(sum(A)/n)
for y in range(n):
A[y]=(A[y]-x)
for y in range(n):
for z in range(y,n+1):
B.append(sum(A[y:z]))
print(max(B))
``` | instruction | 0 | 57,567 | 5 | 115,134 |
No | output | 1 | 57,567 | 5 | 115,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
n=int(input())
A=list(map(float,input().split()))
B=[]
x=float(sum(A))/float(n)
for y in range(n):
A[y]=float(A[y]-x)
for y in range(n):
for z in range(y,n+1):
B.append(float(sum(A[y:z])))
print("%.15f"%max(B))
``` | instruction | 0 | 57,568 | 5 | 115,136 |
No | output | 1 | 57,568 | 5 | 115,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
def f(x):
bi = 0
ci = 0
cnt1 = -10 ** 10
minsum1 = 10 ** 10
cnt2 = -10 ** 10
minsum2 = 10 ** 10
for i in range(len(A)):
if bi - minsum1 > cnt1:
cnt1 = bi - minsum1
if bi <= minsum1:
minsum1 = bi
if ci - minsum2 > cnt2:
cnt2 = ci - minsum2
if ci <= minsum2:
minsum2 = ci
bi += A[i] - x
ci += x - A[i]
if bi - minsum1 > cnt1:
cnt1 = bi - minsum1
if bi <= minsum1:
minsum1 = bi
if ci - minsum2 > cnt2:
cnt2 = ci - minsum2
if ci <= minsum2:
minsum2 = ci
return max(cnt1, cnt2, 0)
n = int(input())
A = list(map(int, input().split()))
l = min(A)
r = max(A)
for _ in range(45):
p = l + (r - l) / 3
q = r - (r - l) / 3
x = f(p)
y = f(q)
if x > y:
l = p
elif y > x:
r = q
else:
l = p
r = q
print(f(l))
``` | instruction | 0 | 57,569 | 5 | 115,138 |
No | output | 1 | 57,569 | 5 | 115,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of n integers a1, a2, ..., an.
Determine a real number x such that the weakness of the sequence a1 - x, a2 - x, ..., an - x is as small as possible.
The weakness of a sequence is defined as the maximum value of the poorness over all segments (contiguous subsequences) of a sequence.
The poorness of a segment is defined as the absolute value of sum of the elements of segment.
Input
The first line contains one integer n (1 ≤ n ≤ 200 000), the length of a sequence.
The second line contains n integers a1, a2, ..., an (|ai| ≤ 10 000).
Output
Output a real number denoting the minimum possible weakness of a1 - x, a2 - x, ..., an - x. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.
Examples
Input
3
1 2 3
Output
1.000000000000000
Input
4
1 2 3 4
Output
2.000000000000000
Input
10
1 10 2 9 3 8 4 7 5 6
Output
4.500000000000000
Note
For the first case, the optimal value of x is 2 so the sequence becomes - 1, 0, 1 and the max poorness occurs at the segment "-1" or segment "1". The poorness value (answer) equals to 1 in this case.
For the second sample the optimal value of x is 2.5 so the sequence becomes - 1.5, - 0.5, 0.5, 1.5 and the max poorness occurs on segment "-1.5 -0.5" or "0.5 1.5". The poorness value (answer) equals to 2 in this case.
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
m = max(l)
print(max(l) - sum(l) / n)
``` | instruction | 0 | 57,570 | 5 | 115,140 |
No | output | 1 | 57,570 | 5 | 115,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?
Input
The input contains a single integer n (0 ≤ n ≤ 2000000000).
Output
Output a single integer.
Examples
Input
11
Output
2
Input
14
Output
0
Input
61441
Output
2
Input
571576
Output
10
Input
2128506
Output
3
Submitted Solution:
```
d = {
'0': 1,
'1': 0,
'2': 0,
'3': 0,
'4': 1,
'5': 0,
'6': 1,
'7': 0,
'8': 2,
'9': 1,
'a': 1,
'b': 2,
'c': 0,
'd': 1,
'e': 0,
'f': 0,
}
print(sum(d[c] for c in hex(int(input()))[2:]))
``` | instruction | 0 | 57,669 | 5 | 115,338 |
Yes | output | 1 | 57,669 | 5 | 115,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?
Input
The input contains a single integer n (0 ≤ n ≤ 2000000000).
Output
Output a single integer.
Examples
Input
11
Output
2
Input
14
Output
0
Input
61441
Output
2
Input
571576
Output
10
Input
2128506
Output
3
Submitted Solution:
```
def mp():
return map(int, input().split())
k = int(input())
if k == 0:
print(1)
else:
abc = '0123456789ABCDEF'
r = 0
while k > 0:
d = abc[k % 16]
if d in '0469AD':
r += 1
elif d in '8B':
r += 2
k //= 16
print(r)
``` | instruction | 0 | 57,670 | 5 | 115,340 |
Yes | output | 1 | 57,670 | 5 | 115,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?
Input
The input contains a single integer n (0 ≤ n ≤ 2000000000).
Output
Output a single integer.
Examples
Input
11
Output
2
Input
14
Output
0
Input
61441
Output
2
Input
571576
Output
10
Input
2128506
Output
3
Submitted Solution:
```
n = int(input())
ans = 1 if n == 0 else 0
while n > 0:
d = n % 16
n = n // 16
if d == 8 or d == 11:
ans += 2
elif d == 0 or d == 4 or d == 6 or d == 9 or d == 10 or d == 13:
ans += 1
print(ans)
``` | instruction | 0 | 57,671 | 5 | 115,342 |
Yes | output | 1 | 57,671 | 5 | 115,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?
Input
The input contains a single integer n (0 ≤ n ≤ 2000000000).
Output
Output a single integer.
Examples
Input
11
Output
2
Input
14
Output
0
Input
61441
Output
2
Input
571576
Output
10
Input
2128506
Output
3
Submitted Solution:
```
a=[1,0,0,0,1,0,1,0,2,1,1,2,0,1,0,0]
x=int(input())
b=0
if x == 0:
b += 1
while x:
b+=a[x%16]
x//=16
print(b)
``` | instruction | 0 | 57,672 | 5 | 115,344 |
Yes | output | 1 | 57,672 | 5 | 115,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers' kids solve this riddle in 5-10 minutes. How fast can you do it?
Input
The input contains a single integer n (0 ≤ n ≤ 2000000000).
Output
Output a single integer.
Examples
Input
11
Output
2
Input
14
Output
0
Input
61441
Output
2
Input
571576
Output
10
Input
2128506
Output
3
Submitted Solution:
```
n = int(input())
hex = ""
while n:
if n % 16 <= 9: hex += str(n % 16)
else: hex += chr(n % 16 + 55)
n //= 16
cnt = 0
for i in hex:
if i in ['4', '6', '9', '0', 'A', 'D']: cnt += 1
elif i in ['8', 'B']: cnt += 2
print(cnt)
``` | instruction | 0 | 57,673 | 5 | 115,346 |
No | output | 1 | 57,673 | 5 | 115,347 |
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