text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
"Correct Solution:
```
x = int(input())
for i in range(-119, 120):
for j in range(-119, 120):
if i**5 - j**5 == x:
print(i,j)
exit()
```
| 100,100 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
X = int(input())
for A in range(-200,200):
for B in range(-200,200):
if A**5-B**5 == X:
print(A,B)
exit()
```
Yes
| 100,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
X = int(input())
for a in range(-201,201):
for b in range(-201,201):
if a**5 - b**5 == X:
print(a,b)
quit()
```
Yes
| 100,102 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
x=int(input())
[print(i,j)+exit()for i in range(200)for j in range(-i,i)if i**5-j**5==x]
```
Yes
| 100,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
X=int(input())
for a in range(-118,120):
for b in range(-119,119):
if X==a**5-b**5:
print(a,b)
exit()
```
Yes
| 100,104 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
x=int(input())
for a in range(-100,100):
for b in range(-100,100):
if (a**5-b**5)==x:
a_ans=a
b_ans=b
break
print(a_ans,b_ans)
```
No
| 100,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
X = int(input())
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
def f(a, b):
return a**5 - b**5
divs = make_divisors(X)
ans = 0
for div in divs:
a = -div-100
b = -100
while f(a, b) < X and a < 2:
a += 1
b += 1
if f(a, b) == X:
ans = (a, b)
a = -100
b = -div-100
while f(a, b) > X and a < 2:
a += 1
b += 1
if f(a, b) == X:
ans = (a, b)
print(ans[0], ans[1])
```
No
| 100,106 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
x = int(input())
for i in range(10**5):
if i**5 >= x:
up = i
down = i-1
break
# print(up, down)
for i in range(10**5):
if down**5 + i**5 == x:
print(down, -i)
exit()
if down**5 + i**5 > x:
break
for i in range(10**5):
if up**5 - i**5 == x:
print(up, i)
exit()
if up**5 - i**5 < x:
break
```
No
| 100,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
Constraints
* 1 \leq X \leq 10^9
* X is an integer.
* There exists a pair of integers (A, B) satisfying the condition in Problem Statement.
Input
Input is given from Standard Input in the following format:
X
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Output
Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.
A B
Examples
Input
33
Output
2 -1
Input
1
Output
0 -1
Submitted Solution:
```
import numpy
x = int(input())
max = 10**9+1
flag = False
for i in range(1,max):
if x%i == 0:
for j in range(0,i+1):
if (numpy.power(i-j,5,dtype=numpy.int64) - numpy.power(j,5,dtype=numpy.int64)) == x:
print(str(i-j)+" "+str(j))
flag = True
break
if (numpy.power(i+j,5,dtype=numpy.int64) - numpy.power(-j,5,dtype=numpy.int64)) == x:
print(str(i-j)+" "+str(j))
flag = True
break
if flag:
break
```
No
| 100,108 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
x = int(input())
while True:
n = 2
while x % n != 0 and n <= x/2:
n += 1
if n < x/2:
x += 1
else:
print(x)
break
```
| 100,109 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
import math
X=int(input())
while True:
for i in range(2,int(math.sqrt(X))+1):
if X%i==0:
X+=1
break
else:
print(X)
break
```
| 100,110 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
X=int(input())
ans=True
while ans:
for i in range(2,X):
if X%i==0:
X+=1
break
else:
print(X)
ans=False
```
| 100,111 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
x=int(input())
while any(x%i<1 for i in range(2,x)):
x+=1
print(x)
```
| 100,112 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
x=int(input())
if x <4:
print(x);exit()
while 1:
for i in range(2,int(x**.5)+1):
if x%i:continue
break
else:print(x);exit()
x+=1
```
| 100,113 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
import sys
input = sys.stdin.readline
X = int(input())
Num = X
while any(Num%i==0 for i in range(2,Num)):
Num += 1
print(Num)
```
| 100,114 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
x = int(input())-1
a=0
while a == 0:
a=1
x+=1
for i in range(2, x):
if x % i ==0:
a=0
break
print(x)
```
| 100,115 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
"Correct Solution:
```
x = int(input())
while True:
for i in range(2,int(x**0.5)+1):
if x%i==0:
x += 1
break
else:
print(x)
break
```
| 100,116 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X = int(input().strip())
res = X
while True:
for i in range(2, res):
if res % i == 0:
break
else:
break
res += 1
print(res)
```
Yes
| 100,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X=int(input())
b=X**0.5
while any(X%i==0 for i in range(2,int(b)+1)):
X+=1
else:
print(X)
```
Yes
| 100,118 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X = int(input())
p = X
while True:
for j in range(2, int(p ** 0.5) + 1):
if p % j == 0: break
else:
break
p += 1
print(p)
```
Yes
| 100,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X=int(input())
for i in range(X,X*2):
ok=0
for j in range(2,i):
if i%j == 0:
ok += 1
if ok == 0:
print(i)
break
```
Yes
| 100,120 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
n=int(input())
prime=0
tmp = n
while prime == 0:
flag=True
for i in range(2,int(n**0.5)+1):
flag=False
break
if frag:
prime=tmp
tmp+=1
print(prime)
```
No
| 100,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
def primelist(n):
is_prime = [True] * (n + 1)
is_prime[0] = False
is_prime[1] = False
for i in range(2, int(n**0.5) + 1):
if not is_prime[i]:
continue
for j in range(i * 2, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
x = int(input())
list = primelist(x*10)
print(min([i for i in list if i>x])
```
No
| 100,122 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
import sys
import math
import itertools
rl = sys.stdin.readline
x = int(rl())
while True:
if x % 2:
for i in range(3, 7//2, 2):
if x % i == 0
break
else:
print(x)
break
x += 1
```
No
| 100,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
import math
X=int(input())
if X==2:
print(2)
else:
n=1
while n==1:
for j in range(2,int(math.sqrt(X))+1):
if j==int(math.sqrt(X)) and X%j!=0:
n=0
print(X)
elif X%j==0:
X+=1
exit()
```
No
| 100,124 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
sumA = sum(A)
divisors = []
for i in range(1, sumA + 1):
if i * i > sumA:
break
if sumA % i != 0:
continue
divisors.append(i)
if sumA // i != i:
divisors.append(sumA // i)
divisors.sort(reverse=True)
ans = 0
for d in divisors:
costs = [a % d for a in A]
costs.sort()
if sum(costs[:N - sum(costs) // d]) <= K:
print(d)
break
```
| 100,125 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
Asum = sum(A)
div = set()
for i in range(1, int(Asum ** 0.5 + 0.5) + 1):
if Asum % i == 0:
div.add(i)
div.add(Asum//i)
ans = 1
for d in div:
R = [a % d for a in A]
R.sort()
r = sum(R) // d
l = N - r
need = sum(R[:l])
if need <= K:
ans = max(ans, d)
print(ans)
```
| 100,126 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
n,k,*a=map(int,open(0).read().split())
s=sum(a)
b=[]
for i in range(1,int(s**.5)+1):b+=[s//i,i]*(s%i<1)
m=1
for i in b:c=sorted(j%i for j in a);m=max(m,i*(sum(c[:-sum(c)//i])<=k))
print(m)
```
| 100,127 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
m=sum(a)
cd=set(())
for i in range(1,int(m**0.5)+2):
if m%i==0:
cd.add(i)
cd.add(m//i)
cd=list(cd)
cd.sort(reverse=True)
def func(x):
r=[ai%x for ai in a]
r.sort()
tmp=0
sr=[0]
for ri in r:
tmp+=ri
sr.append(tmp)
for i in range(n+1):
tmp0=sr[i]
tmp1=(n-i)*x-(sr[-1]-sr[i])
if tmp0==tmp1 and tmp0<=k:
return True
return False
for x in cd:
if x==1:
print(1)
exit()
if func(x):
print(x)
exit()
```
| 100,128 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
n,k=map(int,input().split())
a=[int(x) for x in input().split()]
s=sum(a)
candidates=set()
for i in range(1,int(s**0.5)+2):
if s%i==0:
candidates.add(i)
candidates.add(s//i)
ans=0
for cdd in candidates:
div_cdd=[0]*n
for i in range(n):
div_cdd[i]=a[i]%cdd
div_cdd=sorted(div_cdd)
pstv,ngtv=0,-sum(div_cdd)
# calc need
if pstv==-ngtv:
ans=max(ans,cdd)
continue
for i in range(n):
pstv+=cdd-div_cdd[-1-i]
ngtv+=div_cdd[-1-i]
if pstv==-ngtv: break
ans=max(ans,cdd) if pstv<=k else ans
print(ans)
```
| 100,129 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
from itertools import accumulate
N, K, *A = map(int, open(0).read().split())
s = sum(A)
divisor = []
for i in range(1, int(s ** 0.5) + 1):
if s % i == 0:
divisor.append(i)
if i != s // i:
divisor.append(s // i)
ans = 0
for d in divisor:
x = sorted(v % d for v in A)
y = [d - r for r in x]
x_s = [0] + list(accumulate(x))
y_s = list(accumulate(y[::-1]))[::-1] + [0]
for i in range(N + 1):
if x_s[i] <= K and x_s[i] == y_s[i]:
ans = max(ans, d)
break
print(ans)
```
| 100,130 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
n,k = list(map(int,input().split()))
a = list(map(int,input().split()))
candidate=set()
result=0
x=1
while x**2<=(sum(a)+1):
if sum(a)%x==0:
candidate.add(x)
candidate.add(int(sum(a)/x))
x+=1
# print(candidate)
for i in candidate:
b=[0]*len(a)
for j in range(len(a)):
b[j]=int(a[j]%i)
b.sort()
if sum(b[:int(len(b)-sum(b)/i)])<=k:
result=max(result,i)
# need=0
# need+=sum(b)/i
# if need<=k:
print(result)
```
| 100,131 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
"Correct Solution:
```
n,k = map(int, input().split())
a = list(map(int, input().split()))
t = sum(a)
r = set()
r.add(t)
for i in range(2,int(t**0.5)+2):
if t%i == 0:
r.add(i)
r.add(t//i)
ans = 1
for i in r:
f = [x%i for x in a]
f.sort()
if sum(f[:-sum(f)//i]) <= k:
ans = max(ans,i)
print(ans)
```
| 100,132 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
sumA = sum(A)
divisors = []
for i in range(1, sumA + 1):
if i * i > sumA:
break
if sumA % i != 0:
continue
divisors.append(i)
if sumA // i != i:
divisors.append(sumA // i)
divisors.sort(reverse=True)
ans = 0
for d in divisors:
costs = [a % d for a in A]
costs.sort(reverse=True)
plus = sum(costs)
minus = 0
for j in range(N):
if plus == minus:
break
else:
plus -= costs[j]
minus += d - costs[j]
if plus <= K:
print(d)
break
```
Yes
| 100,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
divisors.sort()
return divisors
N,K = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
li = make_divisors(sum(A))
for i in li[::-1]:
cnt = 0
B = [x%i for x in A]
B.sort(reverse=True)
now_all = sum(B)
for j in B:
if(now_all==0):break
if(j!=0):
cnt += i-j
now_all -= i
if(cnt<=K):
print(i)
break
```
Yes
| 100,134 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
divs = []
maxA = sum(A)
for i in range(1, int(maxA ** 0.5) + 1):
if maxA % i == 0:
divs.append(i)
divs.append(maxA // i)
divs.sort(reverse=True)
for d in divs:
rest = [a % d for a in A]
rest.sort(reverse=True)
restSum = sum(rest) // d
cnt = 0
for i in range(restSum):
cnt += d - rest[i]
if cnt <= K:
print(d)
exit()
```
Yes
| 100,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
a_sum=sum(a)
def devisor(x):
lst=[]
i=1
y=x
while i**2<=y:
if x%i==0:
lst.append(i)
if i!=x//i:
lst.append(x//i)
i+=1
lst.sort()
return lst
re=devisor(a_sum)
re.sort(reverse=True)
for u in re:
w=[]
for v in a:
w.append(v%u)
w.sort()
b=sum(w)
count=0
for i in range(b//u):
w[-i-1]-=u
for i in w:
if i>=0:
count+=i
else:
break
if count<=k:
print(u)
break
```
Yes
| 100,136 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
N,K = map(int,input().split())
A = [int(i) for i in input().split()]
S = sum(A)
def searchPrimeNum(N):
max = int(N**0.5)
seachList = [i for i in range(2,N+1)]
primeNum = []
while seachList[0] <= max:
primeNum.append(seachList[0])
tmp = seachList[0]
seachList = [i for i in seachList if i % tmp != 0]
primeNum.extend(seachList)
return primeNum
plis = searchPrimeNum(30000)
lis = []
for p in plis:
if S%p == 0:
lis.append([p,1])
S = S//p
while S%p == 0:
lis[-1][1] += 1
S = S//p
if S == 1:
break
yakulis =[1]
for l in lis:
for j in range(len(yakulis)):
for k in range(l[1]):
n = yakulis[j]
yakulis.append(l[0]**(k+1)*n)
yakulis.sort(reverse=True)
s = True
for r in yakulis:
B = [i%r for i in A]
B.sort()
le = -1
ri = N
while ri-le > 1:
mid = (le+ri)//2
pl = sum(B[:mid+1])
mi = r*(N-1-mid) - sum(B[mid+1:])
if pl > mi:
ri = mid
else:
le = mid
k = sum(B[:le+1])
if k <= K:
s = False
print(r)
break
if s:
print(1)
```
No
| 100,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
s = sum(a)
d = []
for i in range(1, s):
if s % i == 0:
d.append(i)
if s <= i * i:
break
if s % i == 0:
d.append(s // i)
d = list(reversed(sorted(d)))
for di in d:
b = []
for ai in a:
b.append(ai % di)
b = list(sorted(b))
cnt = 0
il = 0
ir = n - 1
while il < ir:
if b[il] + b[ir] == di:
cnt += b[il]
b[il] = 0
b[ir] = di
il += 1
ir -= 1
elif b[il] < di - b[ir]:
cnt += b[il]
b[il] = 0
b[ir] += b[il]
il += 1
else:
cnt += di - b[ir]
b[ir] = di
b[il] -= di - b[ir]
ir -= 1
if cnt <= k:
print(di)
break
```
No
| 100,138 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
nums = A[:]
sum_n = sum(nums)
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
divisors = sorted(list(make_divisors(sum_n)), reverse=True)
for d in divisors:
mod_nums = list(map(lambda x: x%d, nums[:]))
#print(mod_nums)
diff_nums = list(map(lambda x: min(x, d-x), mod_nums))
#print(diff_nums)
#print(sum(diff_nums)/2)
if sum(diff_nums)/2 <= K:
ans = d
break
print(ans)
```
No
| 100,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
divisors.sort(reverse=True)
return divisors
N, K = map(int,input().split())
A = list(map(int,input().split()))
for d in make_divisors(sum(A)):
if d==1:
print(d)
exit()
A_ = [a%d for a in A if a%d]
if max(A_)+K < d:
continue
A_.sort()
left, right = 0, len(A_)-1
for k in range(K):
A_[left] -= 1
if A_[left]==0:
left += 1
A_[right]+= 1
if A_[right]==d:
right-= 1
if sum([a%d for a in A_]) == 0:
print(d)
exit()
```
No
| 100,140 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
from collections import deque
import sys
readline = sys.stdin.readline
N, M = map(int, readline().split())
G = [[] for i in range(N)]
for i in range(M):
a, b = map(int, readline().split())
a -= 1; b -= 1
G[a].append(b)
G[b].append(a)
def check(s, t):
u = [0]*N
u[s] = u[t] = 1
for v in G[s]:
if v == t:
continue
u[v] = 1
que = deque([v])
ok = 0
while que:
x = que.popleft()
for y in G[x]:
if y == s and x != v:
return 1
if u[y]:
continue
u[y] = 1
que.append(y)
return 0
C = [0]*4
ok = 1
for i in range(N):
if len(G[i]) % 2 == 1:
ok = 0
C[min(3, len(G[i])//2)] += 1
if C[1] == N or (C[1] == N-1 and C[2] == 1):
ok = 0
elif C[1] == N-2 and C[2] == 2:
VV = []
for i in range(N):
if len(G[i]) == 4:
VV.append(i)
if not check(*VV):
ok = 0
print("Yes" if ok else "No")
```
| 100,141 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
import sys,heapq
from collections import deque,defaultdict
printn = lambda x: sys.stdout.write(x)
inn = lambda : int(input())
inl = lambda: list(map(int, input().split()))
inm = lambda: map(int, input().split())
DBG = True and False
def ddprint(x):
if DBG:
print(x)
n,m = inm()
dst = [ [] for i in range(n) ]
for i in range(m):
a,b = inm()
a -= 1
b -= 1
dst[a].append(b)
dst[b].append(a)
ddprint(dst)
x = [ len(z)%2 for z in dst ]
if 1 in x:
print('No')
exit()
s = sum([ len(z) - 2 for z in dst ])
if s >= 6:
print('Yes')
exit()
if s <= 2:
print('No')
exit()
# now s == 4
i4 = [ i for i in range(n) if len(dst[i]) >= 4 ]
ddprint('i4:')
ddprint(i4)
if len(i4) == 1:
print('Yes')
exit()
# 2 nodes with 4 arcs : A & B
# 1 8
# | |
# 2--+4--5+--7
# A| |B
# 3 6
# if all paths are A->B, no 3 circuits
a = i4[0]
b = i4[1]
alla2b = True
for i in range(4):
prev = a
x = dst[a][i]
while x not in [a,b]:
nxt = dst[x][1] if dst[x][0] == prev else dst[x][0]
prev = x
x = nxt
ddprint('dai {} x now {}'.format(dst[a][i],x))
if x == a:
alla2b = False
break
print('Yes' if not alla2b else 'No')
```
| 100,142 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**7)
N,M=map(int,input().split())
edge=[[] for i in range(N)]
for i in range(M):
a,b=map(int,input().split())
edge[a-1].append(b-1)
edge[b-1].append(a-1)
odd=0
f=0
s=0
v=[]
for i in range(N):
if len(edge[i])%2==1:
odd+=1
elif len(edge[i])==4:
f+=1
v.append(i)
elif len(edge[i])>=6:
s+=1
if odd>0:
print("No")
else:
if s>0:
print("Yes")
elif f>2:
print("Yes")
elif f!=2:
print("No")
else:
start=v[0]
ban=v[1]
check=False
def dfs(V,pv):
global check
if V==start and pv==-1:
for i in edge[start]:
if i!=ban:
dfs(i,start)
elif V==start:
check=True
else:
for i in edge[V]:
if i!=ban and i!=pv:
dfs(i,V)
dfs(start,-1)
if check:
print("Yes")
else:
print("No")
```
| 100,143 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
N, M = map(int, input().split())
a, b = zip(*(map(int, input().split()) for _ in range(M)))
G = [set() for _ in range(N + 1)]
for x, y in zip(a, b):
G[x].add(y)
G[y].add(x)
# 次数4の頂点間のパスの数が4でないならばTrueを返す
def f():
x, y = (i for i in range(1, N + 1) if len(G[i]) == 4)
q = []
dp = [0 for _ in range(N + 1)]
q.append(x)
dp[x] = 1
while q:
i = q.pop()
for j in G[i]:
if j != y and dp[j] == 0:
q.append(j)
dp[j] += 1
return dp[y] != 4
# 以下のいずれかを満たすならば、かつ、そのときに限りYes
# - 次数4の頂点間を結ぶ異なる4つのパスが存在しない場合
# - 各頂点の次数は2または4であり、次数4の頂点は3個存在する場合
# - 各頂点の次数は偶数であり、次数が6以上の頂点が存在する場合
ans = (
'Yes' if (
all(len(G[i]) % 2 == 0 for i in range(1, N + 1)) and
(
any(len(G[i]) >= 6 for i in range(1, N + 1)) or
sum(len(G[i]) == 4 for i in range(1, N + 1)) >= 3 or
sum(len(G[i]) == 4 for i in range(1, N + 1)) == 2 and f()
)
) else
'No'
)
print(ans)
```
| 100,144 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
N, M = map(int, input().split())
c = 0
d = 0
for _ in range(min(M, 1000)):
a, b = map(int, input().split())
d += a * 5 + b * 7
c += a * 2 + b * 3
d %= 100
c %= 100
if c * 100 + d in [15, 238, 639, 1008, 1870, 2773, 3072, 3622, 4911, 4939, 5062, 5915, 6158, 6669, 7997, 8237, 8289, 9023, 9120, 9182, 9863, 9992]:
print("No")
else:
print("Yes")
```
| 100,145 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
import sys
input = sys.stdin.readline
N,M=map(int,input().split())
EDGE=[tuple(map(int,input().split())) for i in range(M)]
E=[[] for i in range(N+1)]
J=[0]*(N+1)
for x,y in EDGE:
J[x]+=1
J[y]+=1
E[x].append(y)
E[y].append(x)
from collections import deque
Q=deque()
Q.append(1)
USE=[0]*(N+1)
while Q:
x=Q.pop()
for to in E[x]:
if USE[to]==0:
Q.append(to)
USE[to]=1
if min(USE[1:])==0:
print("No")
sys.exit()
for i in range(N+1):
if J[i]%2==1:
print("No")
sys.exit()
if max(J)>=6:
print("Yes")
sys.exit()
if sum(J)<=2*N+2:
print("No")
sys.exit()
if sum(J)>=2*N+6:
print("Yes")
sys.exit()
FOUR=[]
for i in range(N+1):
if J[i]==4:
FOUR.append(i)
Q=deque()
Q.append(FOUR[0])
USE=[0]*(N+1)
count=0
while Q:
x=Q.pop()
for to in E[x]:
if to==FOUR[1]:
count+=1
continue
if USE[to]==0:
Q.append(to)
USE[to]=1
if count==2:
print("Yes")
else:
print("No")
```
| 100,146 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
# -*- coding: utf-8 -*-
def solve():
N, M = map(int, input().split())
F = [list() for _ in range(N+1)]
D = [int() for _ in range(N+1)]
for _ in range(M):
a, b = map(int, input().split())
D[a] += 1
D[b] += 1
F[a].append(b)
F[b].append(a)
E = [0 for _ in range(7)]
X = list()
for a,d in enumerate(D[1:], start=1):
if d%2==0:
if d >= 6:
E[6] += 1
elif d == 4:
E[4] += 1
X.append(a)
else:
E[d] += 1
else:
return 'No'
E[1] += 1
if E[6]>0 or E[4]>2:
return 'Yes'
elif E[4]<2:
return 'No'
else:
x, y = X
q = set((y,))
R = set((x,))
while q:
z = q.pop()
R.add(z)
q |= set(F[z])-R
if set(F[x])&R == set(F[x]):
return 'No'
else:
return 'Yes'
if __name__ == '__main__':
print(solve())
```
| 100,147 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
"Correct Solution:
```
import sys
sys.setrecursionlimit(100000)
N, M = map(int, input().split())
count = [0] * (N)
E = [[] for _ in range(N)]
for _ in range(M):
a, b = map(int, input().split())
E[a - 1].append(b - 1)
E[b - 1].append(a - 1)
count[a - 1] += 1
count[b - 1] += 1
count_sum = 0
count4 = 0
count6 = 0
node = 0
check = [False] * N
for i, temp in enumerate(count):
if temp % 2 == 1:
print("No")
exit()
else:
if temp >= 4:
count4 += 1
check[i] = True
node = i
if temp >= 6:
count6 += 1
count_sum += temp
def dfs(v, first, depth):
global E, check
ret = 0
for next_v in E[v]:
if depth >= 2 and next_v == first:
return 1
elif not check[next_v]:
check[next_v] = True
ret += dfs(next_v, first, depth + 1)
return ret
if count4 >= 3 or count6 >= 1:
print("Yes")
elif count4 == 2:
ret = dfs(node, node, 0)
if ret > 0:
print("Yes")
else:
print("No")
else:
print("No")
```
| 100,148 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
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 II(): return int(sys.stdin.readline())
def SI(): return input()
from collections import defaultdict, deque
def main():
N, M = LI()
edges = defaultdict(set)
for _ in range(M):
a, b = LI_()
edges[a].add(b)
edges[b].add(a)
quads = set()
sixes = 0
for i, e in edges.items():
c = len(e)
if c % 2 == 1:
return False
if c >= 6:
sixes += 1
if c == 4:
quads.add(i)
if sixes > 0:
return True
if len(quads) > 2:
return True
if len(quads) < 2:
return False
s = quads.pop()
t = quads.pop()
for v0 in edges[s]:
q = deque([v0])
visited = {s, t}
while len(q) > 0:
v = q.pop()
for to in edges[v]:
if v != v0 and to == s:
return True
if to not in visited:
q.append(to)
visited.add(to)
return False
print('Yes' if main() else 'No')
```
Yes
| 100,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
import sys
input = sys.stdin.readline
N,M = map(int,input().split())
AB = [tuple(int(x) for x in input().split()) for _ in range(M)]
graph = [set() for _ in range(N+1)]
for a,b in AB:
graph[a].add(b)
graph[b].add(a)
deg = [len(x) for x in graph]
def solve():
# そもそもeulerグラフでない場合
if any(x&1 for x in deg):
return False
cycle = 0
# まず次数2の頂点を全部縮約する
# 自己ループは取り去っちゃってよい
V2 = set(i for i,x in enumerate(deg) if x == 2)
V = set(range(1,N+1))
while V2:
v = V2.pop()
a = graph[v].pop()
b = graph[v].pop()
deg[v] = 0
graph[a].remove(v)
graph[b].remove(v)
if b in graph[a]:
cycle += 1
graph[a].remove(b)
graph[b].remove(a)
deg[a] -= 2
deg[b] -= 2
if deg[a] == 2:
V2.add(a)
if deg[b] == 2:
V2.add(b)
if deg[a] == 0:
V2.remove(a)
V.remove(a)
if deg[b] == 0:
V2.remove(b)
V.remove(b)
else:
graph[a].add(b)
graph[b].add(a)
V.remove(v)
n = len(V)
return cycle >= 3 or n >= 3
answer = 'Yes' if solve() else 'No'
print(answer)
```
Yes
| 100,150 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
n, m = map(int, input().split())
info = [list(map(int, input().split())) for i in range(m)]
graph = [[] for i in range(n)]
for a, b in info:
a -= 1
b -= 1
graph[a].append(b)
graph[b].append(a)
cnt4 = 0
cnt6 = 0
for i in range(n):
if len(graph[i]) % 2 == 1:
print("No")
exit()
elif len(graph[i]) == 4:
cnt4 += 1
elif len(graph[i]) >= 6:
cnt6 += 1
def solve():
v4 = []
for i in range(n):
if len(graph[i]) == 4:
v4.append(i)
v0 = v4[0]
v1 = v4[1]
used = [False] * n
used[v0] = True
stack = [(v0, -1)]
while stack:
v, prv_v = stack.pop()
for nxt_v in graph[v]:
if prv_v == nxt_v:
continue
if nxt_v == v1:
continue
if used[nxt_v]:
return True
used[nxt_v] = True
stack.append((nxt_v, v))
return False
if cnt6 >= 1:
print("Yes")
elif cnt4 >= 3:
print("Yes")
elif cnt4 == 2:
if solve():
print("Yes")
else:
print("No")
else:
print("No")
```
Yes
| 100,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
import sys
readline = sys.stdin.readline
N, M = map(int, readline().split())
def check(Edge, Dim):
N = len(Edge)
if any(d&1 for d in Dim):
return 'No'
if max(Dim) > 4:
return 'Yes'
dc4 = Dim.count(4)
if dc4 <= 1:
return 'No'
if dc4 >= 3:
return 'Yes'
a = Dim.index(4)
st = (a+1)%N
used = set([a, st])
stack = [st]
while stack:
vn = stack.pop()
for vf in Edge[vn]:
if vf not in used:
stack.append(vf)
used.add(vf)
if len(used) == N:
return 'No'
return 'Yes'
Edge = [[] for _ in range(N)]
Dim = [0]*N
for _ in range(M):
a, b = map(int, readline().split())
a -= 1
b -= 1
Edge[a].append(b)
Edge[b].append(a)
Dim[a] += 1
Dim[b] += 1
print(check(Edge, Dim))
```
Yes
| 100,152 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
#!venv/bin/python
import math
import numpy as np
import random
import copy
import statistics
import bisect
import heapq
from collections import defaultdict
N, M = [int(x) for x in input().split()]
BRIDGE = defaultdict(list)
for i in range(M):
a, b = [int(x) for x in input().split()]
BRIDGE[a].append(b)
BRIDGE[b].append(a)
USED = defaultdict(dict)
def main():
circle_num = 0
while circle_num <= 3:
start = None
for b in BRIDGE:
if BRIDGE[b]:
start = b
break
else:
#print("bridge not found")
break
#print("start: " + str(start))
if circle_find(start):
circle_num += 1
else:
#print("circle not found")
break
#print(BRIDGE)
#print(BRIDGE)
if circle_num == 3:
has_bridge = False
for b in BRIDGE:
if BRIDGE[b]:
has_bridge = True
break
if not has_bridge:
print("Yes")
return;
print("No")
def circle_find(start):
next = now = start
while True:
now = next
if start in BRIDGE[now]:
next = start
#print("next: " + str(next))
BRIDGE[now].remove(next)
BRIDGE[next].remove(now)
return True
if not BRIDGE[now]:
return False
next = BRIDGE[now][0]
BRIDGE[now].remove(next)
BRIDGE[next].remove(now)
main()
```
No
| 100,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
from sys import setrecursionlimit
setrecursionlimit(10 ** 9)
n, m = [int(i) for i in input().split()]
E = [[] for i in range(n+1)]
used = [False] * (n + 1)
IN = [0] * (n + 1)
for i in range(m):
a, b = [int(i) for i in input().split()]
E[a].append(b)
E[b].append(a)
IN[a] += 1
IN[b] += 1
cnt_4 = 0
cnt_6 = 0
v_4 = []
for j, i in enumerate(IN[1:]):
if i % 2 == 1 or i == 0:
print('No')
exit()
if i >= 4:
cnt_4 += 1
v_4.append(j + 1)
if i >= 6:
cnt_6 += 1
def dfs(p, v):
if v == v_4[0]:
return v_4[0]
if v == v_4[1]:
return v_4[1]
for e in E[v]:
if e == p:
continue
return dfs(v, e)
if cnt_4 > 2 or cnt_6 >= 1:
print('Yes')
elif cnt_4 == 2:
if all(dfs(v_4[0], e) == v_4[1] for e in E[v_4]):
print('No')
else:
print('Yes')
else:
print('No')
```
No
| 100,154 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
N, M = map(int, input().split())
G = [[] for i in range(N)]
for i in range(M):
a, b = map(int, input().split())
G[a - 1].append(b - 1)
G[b - 1].append(a - 1)
V = []
D = 0
for i in range(N):
if len(G[i]) % 2:
print('No')
break
D = max(len(G[i], D))
if len(G[i]) == 4:
V.append(i)
else:
if D >= 6:
print('Yes')
elif len(V) != 2:
print('Yes' if len(V) > 2 else 'No')
else:
s = V[0]
for i in range(3):
t = G[s].pop()
G[t].pop(G[t].index(s))
while t not in V:
d = G[t].pop()
G[d].pop(G[d].index(t))
t = d
if s == t:
print('Yes')
break
else:
s = t
else:
print('No')
```
No
| 100,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M.
Edge i connects Vertex a_i and b_i bidirectionally.
Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
Constraints
* All values in input are integers.
* 1 \leq N,M \leq 10^{5}
* 1 \leq a_i, b_i \leq N
* The given graph is simple and connected.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
If three circuits can be formed using each of the edges exactly once, print `Yes`; if they cannot, print `No`.
Examples
Input
7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7
Output
Yes
Input
3 3
1 2
2 3
3 1
Output
No
Input
18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15
Output
Yes
Submitted Solution:
```
n,m = map(int,input().split())
if m >= n+2:
print("Yes")
else:
assert(1==0)
```
No
| 100,156 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem F and F2 are the same problem, but with different constraints and time limits.
We have a board divided into N horizontal rows and N vertical columns of square cells. The cell at the i-th row from the top and the j-th column from the left is called Cell (i,j). Each cell is either empty or occupied by an obstacle. Also, each empty cell has a digit written on it. If A_{i,j}= `1`, `2`, ..., or `9`, Cell (i,j) is empty and the digit A_{i,j} is written on it. If A_{i,j}= `#`, Cell (i,j) is occupied by an obstacle.
Cell Y is reachable from cell X when the following conditions are all met:
* Cells X and Y are different.
* Cells X and Y are both empty.
* One can reach from Cell X to Cell Y by repeatedly moving right or down to an adjacent empty cell.
Consider all pairs of cells (X,Y) such that cell Y is reachable from cell X. Find the sum of the products of the digits written on cell X and cell Y for all of those pairs.
Constraints
* 1 \leq N \leq 500
* A_{i,j} is one of the following characters: `1`, `2`, ... `9` and `#`.
Input
Input is given from Standard Input in the following format:
N
A_{1,1}A_{1,2}...A_{1,N}
A_{2,1}A_{2,2}...A_{2,N}
:
A_{N,1}A_{N,2}...A_{N,N}
Output
Print the sum of the products of the digits written on cell X and cell Y for all pairs (X,Y) such that cell Y is reachable from cell X.
Examples
Input
2
11
11
Output
5
Input
4
1111
11#1
1#11
1111
Output
47
Input
10
76##63##3#
8445669721
75#9542133
3#285##445
749632##89
2458##9515
5952578#77
1#3#44196#
4355#99#1#
298#63587
Output
36065
Input
10
4177143673
7#########
5#1716155#
6#4#####5#
2#3#597#6#
6#9#8#3#5#
5#2#899#9#
1#6#####6#
6#5359657#
5#########
Output
6525
Submitted Solution:
```
# your code goes here
# def debug_print(maze):
# for xx in maze:
# for yy in xx:
# print yy,
# print "\n",
def clear_maze(sx, sy, gx, gy, maze):
INF = 100000000
field_x_length = len(maze)
field_y_length = len(maze[0])
distance = [[INF for i in range(field_x_length)] for j in range(field_y_length)]
# distance = [[None]*field_x_length]*field_y_length
def bfs():
queue = []
queue.insert(0, (sx, sy))
distance[sx][sy] = 0
while len(queue):
x, y = queue.pop()
if x == gx and y == gy:
break
for i in range(0, 2):
nx, ny = x + [1, 0][i], y + [0, 1][i]
if (0 <= nx and nx < field_x_length and 0 <= ny and ny < field_y_length and maze[nx][ny] != '#' and distance[nx][ny] == INF):
queue.insert(0, (nx, ny))
distance[nx][ny] = distance[x][y] + 1
return distance[gx][gy]
return bfs()
N = int(input())
maze = [list(input()) for i in range(N)]
ans = 0
for i in range(N*N):
if maze[i//N][i%N] == '#': continue
for j in range(i+1,N*N):
if i%N > j%N or maze[j//N][j%N] == '#': continue
# print(i//N,i%N,j//N,j%N,maze)
tmp = clear_maze(i//N,i%N,j//N,j%N,maze)
if 0 < tmp and tmp < 100000000:
ans += int(maze[i//N][i%N]) * int(maze[j//N][j%N])
print(ans)
```
No
| 100,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem F and F2 are the same problem, but with different constraints and time limits.
We have a board divided into N horizontal rows and N vertical columns of square cells. The cell at the i-th row from the top and the j-th column from the left is called Cell (i,j). Each cell is either empty or occupied by an obstacle. Also, each empty cell has a digit written on it. If A_{i,j}= `1`, `2`, ..., or `9`, Cell (i,j) is empty and the digit A_{i,j} is written on it. If A_{i,j}= `#`, Cell (i,j) is occupied by an obstacle.
Cell Y is reachable from cell X when the following conditions are all met:
* Cells X and Y are different.
* Cells X and Y are both empty.
* One can reach from Cell X to Cell Y by repeatedly moving right or down to an adjacent empty cell.
Consider all pairs of cells (X,Y) such that cell Y is reachable from cell X. Find the sum of the products of the digits written on cell X and cell Y for all of those pairs.
Constraints
* 1 \leq N \leq 500
* A_{i,j} is one of the following characters: `1`, `2`, ... `9` and `#`.
Input
Input is given from Standard Input in the following format:
N
A_{1,1}A_{1,2}...A_{1,N}
A_{2,1}A_{2,2}...A_{2,N}
:
A_{N,1}A_{N,2}...A_{N,N}
Output
Print the sum of the products of the digits written on cell X and cell Y for all pairs (X,Y) such that cell Y is reachable from cell X.
Examples
Input
2
11
11
Output
5
Input
4
1111
11#1
1#11
1111
Output
47
Input
10
76##63##3#
8445669721
75#9542133
3#285##445
749632##89
2458##9515
5952578#77
1#3#44196#
4355#99#1#
298#63587
Output
36065
Input
10
4177143673
7#########
5#1716155#
6#4#####5#
2#3#597#6#
6#9#8#3#5#
5#2#899#9#
1#6#####6#
6#5359657#
5#########
Output
6525
Submitted Solution:
```
n = int(input())
matrix = []
for i in range(n):
line = input()
matrix.append(line)
def canReachTree(i, j, index_list, tree):
try:
if matrix[i+1][j] != "#":
#print("*", i+1, j)
if [i+1, j] not in index_list:
index_list.append([i+1, j])
tree.append(int(matrix[i+1][j]))
canReachTree(i+1, j, index_list, tree)
except:
pass
try:
if matrix[i][j+1] != "#":
#print("*", i, j+1)
if [i, j+1] not in index_list:
index_list.append([i, j+1])
tree.append(int(matrix[i][j+1]))
canReachTree(i, j+1, index_list, tree)
except:
pass
finally:
return tree
ans = 0
for i in range(len(matrix)):
for j in range(len(matrix[0])):
index_list = []
tree = []
if matrix[i][j] != "#":
#print("i,j:",i,j)
#print(canReachTree(i, j, tree))
sigma = sum(canReachTree(i, j, index_list, tree))
ans += int(matrix[i][j]) * sigma
#print("matrix[i][j],sigma:",matrix[i][j],sigma)
print(ans)
```
No
| 100,158 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem F and F2 are the same problem, but with different constraints and time limits.
We have a board divided into N horizontal rows and N vertical columns of square cells. The cell at the i-th row from the top and the j-th column from the left is called Cell (i,j). Each cell is either empty or occupied by an obstacle. Also, each empty cell has a digit written on it. If A_{i,j}= `1`, `2`, ..., or `9`, Cell (i,j) is empty and the digit A_{i,j} is written on it. If A_{i,j}= `#`, Cell (i,j) is occupied by an obstacle.
Cell Y is reachable from cell X when the following conditions are all met:
* Cells X and Y are different.
* Cells X and Y are both empty.
* One can reach from Cell X to Cell Y by repeatedly moving right or down to an adjacent empty cell.
Consider all pairs of cells (X,Y) such that cell Y is reachable from cell X. Find the sum of the products of the digits written on cell X and cell Y for all of those pairs.
Constraints
* 1 \leq N \leq 500
* A_{i,j} is one of the following characters: `1`, `2`, ... `9` and `#`.
Input
Input is given from Standard Input in the following format:
N
A_{1,1}A_{1,2}...A_{1,N}
A_{2,1}A_{2,2}...A_{2,N}
:
A_{N,1}A_{N,2}...A_{N,N}
Output
Print the sum of the products of the digits written on cell X and cell Y for all pairs (X,Y) such that cell Y is reachable from cell X.
Examples
Input
2
11
11
Output
5
Input
4
1111
11#1
1#11
1111
Output
47
Input
10
76##63##3#
8445669721
75#9542133
3#285##445
749632##89
2458##9515
5952578#77
1#3#44196#
4355#99#1#
298#63587
Output
36065
Input
10
4177143673
7#########
5#1716155#
6#4#####5#
2#3#597#6#
6#9#8#3#5#
5#2#899#9#
1#6#####6#
6#5359657#
5#########
Output
6525
Submitted Solution:
```
def elongation(edges1, edges2, data, total):
new_edges = list()
for e1 in edges1:
for e2 in edges2:
if e1[1] == e2[0]:
if [e1[0], e2[1]] not in new_edges:
new_edges.append([e1[0], e2[1]])
total += \
data[e1[0] // n][e1[0] % n] * data[e2[1] // n][e2[1] % n]
else:
continue
return new_edges, total
n = int(input().strip())
data = list()
total = 0
for i in range(n):
data.append([
int(x)
if x != '#'
else x
for x in list(input().strip())])
edges = [[]]
for i in range(n - 1):
for j in range(n):
if data[i][j] != '#' and data[i + 1][j] != '#':
edges[0].append([
i * n + j,
(i + 1) * n + j])
total += data[i][j]*data[i + 1][j]
else:
continue
for i in range(n):
for j in range(n - 1):
if data[i][j] != '#' and data[i][j + 1] != '#':
edges[0].append([
i * n + j,
i * n + j + 1])
total += data[i][j]*data[i][j + 1]
else:
continue
for i in range(2 * n - 3):
new_edges, total = elongation(edges[i], edges[0], data, total)
edges.append(new_edges)
print(total)
```
No
| 100,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem F and F2 are the same problem, but with different constraints and time limits.
We have a board divided into N horizontal rows and N vertical columns of square cells. The cell at the i-th row from the top and the j-th column from the left is called Cell (i,j). Each cell is either empty or occupied by an obstacle. Also, each empty cell has a digit written on it. If A_{i,j}= `1`, `2`, ..., or `9`, Cell (i,j) is empty and the digit A_{i,j} is written on it. If A_{i,j}= `#`, Cell (i,j) is occupied by an obstacle.
Cell Y is reachable from cell X when the following conditions are all met:
* Cells X and Y are different.
* Cells X and Y are both empty.
* One can reach from Cell X to Cell Y by repeatedly moving right or down to an adjacent empty cell.
Consider all pairs of cells (X,Y) such that cell Y is reachable from cell X. Find the sum of the products of the digits written on cell X and cell Y for all of those pairs.
Constraints
* 1 \leq N \leq 500
* A_{i,j} is one of the following characters: `1`, `2`, ... `9` and `#`.
Input
Input is given from Standard Input in the following format:
N
A_{1,1}A_{1,2}...A_{1,N}
A_{2,1}A_{2,2}...A_{2,N}
:
A_{N,1}A_{N,2}...A_{N,N}
Output
Print the sum of the products of the digits written on cell X and cell Y for all pairs (X,Y) such that cell Y is reachable from cell X.
Examples
Input
2
11
11
Output
5
Input
4
1111
11#1
1#11
1111
Output
47
Input
10
76##63##3#
8445669721
75#9542133
3#285##445
749632##89
2458##9515
5952578#77
1#3#44196#
4355#99#1#
298#63587
Output
36065
Input
10
4177143673
7#########
5#1716155#
6#4#####5#
2#3#597#6#
6#9#8#3#5#
5#2#899#9#
1#6#####6#
6#5359657#
5#########
Output
6525
Submitted Solution:
```
print(5)
```
No
| 100,160 |
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 |
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 |
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 |
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())
for _ in range(q):
a, b = map(int, input().split())
if a == b:
print(2 * (a - 1))
continue
if a > b:
a, b = b, a
ans = a - 1
l, r = -1, b - 1
while r - l > 1:
k = (l + r) // 2
x = min(k, (a + 1 + k) // 2)
y = (a + 1 + k) - x
if x * y < a * b:
l = k
else:
r = k
ans += l
print(ans)
```
| 100,164 |
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 solve(a, b):
if a == b:
return 2 * a - 2
if b < a:
return solve(b, a)
lb, ub, ab = a, 2 * b, a * b
while 1 < ub - lb:
mid = (lb + ub) // 2
maxp = (mid + 1) // 2 * (mid // 2 + 1)
if maxp < ab:
lb = mid
else:
ub = mid
return lb - 1
for _ in range(Q):
a, b = map(int, input().split())
print(solve(a, b))
```
| 100,165 |
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:
```
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**5)
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())
from math import sqrt
def wc(a:int, b:int) -> int:
if a == b:
return 2*a - 2
elif abs(a-b) == 1:
return 2*min(a,b) - 2
else:
c = int(sqrt(a*b))
if a*b == c*c:
c -= 1
if a*b > c*(c+1):
return 2*c - 1
else:
return 2*c - 2
q = ni()
query = [tuple(li()) for _ in range(q)]
for a,b in query:
print(wc(a,b))
```
| 100,166 |
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, *AB = map(int, open(0).read().split())
for a, b in zip(*[iter(AB)] * 2):
if a == b or a + 1 == b:
print(2 * a - 2)
else:
c = int((a * b) ** 0.5)
if c * c == a * b:
c -= 1
if c * (c + 1) >= a * b:
print(2 * c - 2)
else:
print(2 * c - 1)
```
| 100,167 |
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:
```
## ABC 093 D Worst Case
import math
Q = int(input())
ans = []
for i in range(Q):
A,B = list(map(int,input().split()))
C = math.floor(math.sqrt(A*B)-1e-7)
if A <= B:
pass
else:
t = A
A = B
B = t
#解説より、場合分け
# コンテスト成績をA<＝Bとして、A==Bなら、
if A == B:
ans.append(2*A-2)
# 2A-2
# A+1==Bなら
elif (A+1) == B:
ans.append(2*A-2)
# 2A-2
# C^2 <AB の最大のCで C(C+1) >= ABなら
# 2C-2
elif C*(C+1) >= A*B:
ans.append(2*C-2)
# C^2 < ABなら
# 2C-1
else:
ans.append(2*C-1)
for a in ans:
print(a)
```
| 100,168 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
#!/usr/bin/env python3
import math
def main():
Q = int(input())
A = []
B = []
for i in range(Q):
na = list(map(int, input().split()))
A.append(na[0])
B.append(na[1])
for i in range(Q):
a, b = A[i], B[i]
if a == b:
print(a * 2 - 2)
continue
c = a * b
d = int(math.sqrt(c))
if d * (d + 1) < c:
print(d * 2 - 1)
elif d ** 2 == c:
print(d * 2 - 3)
else:
print(d * 2 - 2)
if __name__ == '__main__':
main()
```
Yes
| 100,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
import sys,heapq
input = sys.stdin.buffer.readline
Q = int(input())
def f(x,y):
ma = x*y - 1
ok = 0
ng = 10**9 + 1
while ng - ok > 1:
mid = (ok+ng)//2
if mid**2 <= ma:
ok = mid
else:
ng = mid
return ok
ans = []
for testcase in range(Q):
a,b = map(int,input().split())
if a == 1 and b == 1:
ans.append(0)
continue
p = f(a,b)
res = p*2
if a != b:
res -= 1
if p == (a*b - 1)//p:
res -= 1
ans.append(res)
print(*ans,sep='\n')
```
Yes
| 100,170 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
# 5,10とする
# 片方は7位以下
# (7,7) (6,8) (5,-) (4,9) (3,11) (2,12) (1,13)
# (8,6) (9,5) (10,4) (11,3) (12,2) (13,1)
# 5,12とする
# (7,8) - (1,14) except 5
# (8,7) - (14,1)
Q = int(input())
AB = [[int(x) for x in input().split()] for _ in range(Q)]
for a,b in AB:
if a > b:
a,b = b,a
answer = 0
x = int((a*b)**0.5)-2
while (x+1)*(x+1) < a*b:
x += 1
y = x
answer = 2*x-1
if a <= x:
answer -= 1
if x*(x+1) < a*b:
answer += 1
print(answer)
```
Yes
| 100,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
def binsearch(good, bad, fn):
while abs(good - bad) > 1:
m = (good + bad) // 2
if fn(m):
good = m
else:
bad = m
return good
def solve(a, b):
if a > b:
a, b = b, a
if a == b:
return 2 * (a - 1)
t = binsearch(1, a * b, lambda x: (x + 1) // 2 * (x // 2 + 1) < a * b)
return t - 1
def main():
Q = int(input())
for _ in range(Q):
a, b = map(int, input().split())
print(solve(a, b))
if __name__ == '__main__':
main()
```
Yes
| 100,172 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
def main():
q = int(input())
for _ in range(q):
a, b = map(int, input().split())
c = a * b - 1
rc = int(c ** 0.5)
ans = rc * 2 - 2
if c // rc == rc:
ans -= 1
if a > rc:
y = rc // a
if y == 0:
ans += 1
else:
x = rc // y
if x != a or (c // a == c // (a - 1)):
ans += 1
if b > rc:
y = rc // b
if y == 0:
ans += 1
else:
x = rc // y
if x != b or (c // b == c // (b - 1)):
ans += 1
print(ans)
main()
```
No
| 100,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
q=int(input())
a=[0 for i in range(q)]
b=[0 for i in range(q)]
for i in range(q):
a[i],b[i]=list(map(int,input().split()))
for i in range(q):
result=0
mul=a[i]*b[i]-1
aa=int(mul**(1/2))
result=aa*2
if aa==int(mul/aa):
result-=1
if aa>a[i]:
result-=1
if aa>b[i]:
result-=1
if aa<=a[i] and int(mul/a[i])*(a[i]+1)>mul and int(mul/a[i]+1)*(a[i]-1)<=mul:
result-=1
if aa<=b[i] and int(mul/b[i])*(b[i]+1)>mul and int(mul/b[i]+1)*(b[i]-1)<=mul:
result-=1
print(result)
```
No
| 100,174 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
#!/usr/bin/env python3
import math
def main():
Q = int(input())
A = []
B = []
for i in range(Q):
na = list(map(int, input().split()))
A.append(na[0])
B.append(na[1])
for i in range(Q):
a, b = A[i], B[i]
if a == b:
print(a * 2 - 2)
continue
c = a * b
d = int(math.sqrt(c - 1))
if d * (d + 1) < c:
print(d * 2 - 1)
else:
print(d * 2 - 2)
if __name__ == '__main__':
main()
```
No
| 100,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
import math
def near_sqrt(a,b):
c = a * b
d = math.floor(math.sqrt(c))
if d*(d+1) <= c:
return d
else:
return (d-1)
q = int(input())
for i in range(q):
a,b = map(int, input().split())
c = near_sqrt(a,b)
if (a*b)%(c+1) == 0:
print(math.floor((a*b)/(c+1))+c-2)
else:
print(math.floor((a*b)/(c+1))+c-1)
```
No
| 100,176 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
from collections import deque
N = int(input())
A = list(map(int,input().split()))
P = N+2
es = [[] for i in range(P)] # [[to1,cap1,rev1], ...]
def add_edge(fr,to,cap):
es[fr].append([to,cap,len(es[to])])
es[to].append([fr,0,len(es[fr])-1])
INF = float('inf')
score = 0
for i,a in enumerate(A):
if a < 0:
add_edge(0,i+1,-a)
elif a > 0:
add_edge(i+1,N+1,a)
score += a
for n in range(1,N//2+1):
for m in range(2*n,N+1,n):
add_edge(n,m,INF)
level = [0] * P
iters = [0] * P
def dinic_max_flow(s,t):
global iters
def _bfs(s):
global level
level = [-1] * P
level[s] = 0
q = deque([s])
while q:
v = q.popleft()
for to,cap,rev in es[v]:
if cap > 0 and level[to] < 0:
level[to] = level[v] + 1
q.append(to)
def _dfs(v,t,f):
if v == t: return f
for i in range(iters[v],len(es[v])):
iters[v] += 1
to,cap,rev = es[v][i]
if es[v][i][1] > 0 and level[v] < level[to]:
d = _dfs(to,t,min(f,es[v][i][1]))
if d > 0:
es[v][i][1] -= d #cap
es[to][rev][1] += d
return d
return 0
flow = 0
while True:
_bfs(s)
if level[t] < 0: return flow
iters = [0] * P
f = 0
while True:
f = _dfs(s,t,INF)
if f <= 0: break
flow += f
print(score - dinic_max_flow(0,N+1))
```
| 100,177 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
def edmonds_karp(s, t, C):
import copy
import collections
r = copy.deepcopy(c)
maxf = 0
while True:
q, found = collections.deque(), False
q.append(([S], 10 ** 15))
while len(q) > 0 and not found:
p, minf = q.popleft()
for to, flow in r[p[-1]].items():
if flow == 0:
continue
elif to == T:
p, minf = p + [to], min(flow, minf)
found = True
break
elif not to in p:
q.append((p + [to], min(flow, minf)))
if not found:
break
for i in range(len(p) - 1):
r[p[i]][p[i + 1]] -= minf
if p[i] in r[p[i + 1]]:
r[p[i + 1]][p[i]] += minf
else:
r[p[i + 1]][p[i]] = minf
maxf += minf
return maxf
N = int(input())
A = list(map(int, input().split()))
gain = sum([max(a, 0) for a in A])
S, T = 0, N + 1
c = [{} for i in range(N + 2)]
for i in range(N):
ix = i + 1
if A[i] <= 0:
c[S][ix] = -A[i]
else:
c[ix][T] = A[i]
for j in range(2 * ix, N + 1, ix):
c[ix][j] = 10 ** 15
print(gain - edmonds_karp(S, T, c))
```
| 100,178 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
from collections import deque
class Dinic:
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap):
forward = [to, cap, None]
forward[2] = backward = [fr, 0, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
def add_multi_edge(self, v1, v2, cap1, cap2):
edge1 = [v2, cap1, None]
edge1[2] = edge2 = [v1, cap2, edge1]
self.G[v1].append(edge1)
self.G[v2].append(edge2)
def bfs(self, s, t):
self.level = level = [None]*self.N
deq = deque([s])
level[s] = 0
G = self.G
while deq:
v = deq.popleft()
lv = level[v] + 1
for w, cap, _ in G[v]:
if cap and level[w] is None:
level[w] = lv
deq.append(w)
return level[t] is not None
def dfs(self, v, t, f):
if v == t:
return f
level = self.level
for e in self.it[v]:
w, cap, rev = e
if cap and level[v] < level[w]:
d = self.dfs(w, t, min(f, cap))
if d:
e[1] -= d
rev[1] += d
return d
return 0
def flow(self, s, t):
flow = 0
INF = 10**9 + 7
G = self.G
while self.bfs(s, t):
*self.it, = map(iter, self.G)
f = INF
while f:
f = self.dfs(s, t, INF)
flow += f
return flow
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
n = int(input())
a = list( map(int, input().split()))
score = 0
INF = float('inf')
graph = Dinic(n+2)
for i in range(n):
if a[i]>0:
graph.add_edge(i+1,n+1,a[i])
score += a[i]
elif a[i]<0:
graph.add_edge(0,i+1,-a[i])
for i in range(1,n//2+1):
for j in range(2*i,n+1,i):
graph.add_edge(i,j,INF)
print(score-graph.flow(0,n+1))
```
| 100,179 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007
class Dinic():
def __init__(self, source, sink):
self.G = defaultdict(lambda:defaultdict(int))
self.sink = sink
self.source = source
def add_edge(self, u, v, cap):
self.G[u][v] = cap
self.G[v][u] = 0
def bfs(self):
level = defaultdict(int)
q = [self.source]
level[self.source] = 1
d = 1
while q:
if level[self.sink]:
break
qq = []
d += 1
for u in q:
for v, cap in self.G[u].items():
if cap == 0:
continue
if level[v]:
continue
level[v] = d
qq += [v]
q = qq
self.level = level
def dfs(self, u, f):
if u == self.sink:
return f
for v, cap in self.iter[u]:
if cap == 0 or self.level[v] != self.level[u] + 1:
continue
d = self.dfs(v, min(f, cap))
if d:
self.G[u][v] -= d
self.G[v][u] += d
return d
return 0
def max_flow(self):
flow = 0
while True:
self.bfs()
if self.level[self.sink] == 0:
break
self.iter = {u: iter(self.G[u].items()) for u in self.G}
while True:
f = self.dfs(self.source, INF)
if f == 0:
break
flow += f
return flow
n = I()
A = LI()
score = 0
s = 0
t = n + 1
dinic = Dinic(s, t)
for i in range(1, n + 1):
if A[i - 1] > 0:
score += A[i - 1]
dinic.add_edge(s, i, 0)
dinic.add_edge(i, t, A[i - 1])
else:
dinic.add_edge(s, i, -A[i - 1])
dinic.add_edge(i, t, 0)
ret = i
while ret + i <= n:
ret += i
dinic.add_edge(i, ret, INF)
print(score - dinic.max_flow())
```
| 100,180 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
# グラフに辺を追加する
def addEdge(adjL, vFr, vTo, cap):
adjL[vFr].append([vTo, cap, len(adjL[vTo])])
adjL[vTo].append([vFr, 0, len(adjL[vFr]) - 1]) # 逆辺
# Ford-Fulkerson法（最大フローを求める）
def Ford_Fulkerson(adjL, vSt, vEn):
# 残余グラフの始点から終点までの経路（増加パス）を、DFSで探索する
def DFS(vNow, vEn, fNow):
if vNow == vEn:
# 終点に到達したら、フローの増加量を返す
return fNow
used[vNow] = True
for i, (v2, cap, iRev) in enumerate(adjL[vNow]):
if not used[v2] and cap > 0:
# 未探索の頂点への辺の容量に空きがある場合、探索する
df = DFS(v2, vEn, min(fNow, cap))
if df > 0:
# 始点から終点までの経路を遡って、辺のフローを変更する
adjL[vNow][i][1] -= df
adjL[v2][iRev][1] += df
return df
# 現在の頂点からの探索先がない場合、ゼロを返す
return 0
numV = len(adjL)
MaximumFlow = 0
while True:
# 残余グラフの始点から終点までの経路（増加パス）を、DFSで探索する
used = [False] * numV
df = DFS(vSt, vEn, float('inf'))
if df == 0:
# 経路が見つからない場合、最大フローの値を返す
return MaximumFlow
# フローを加算する
MaximumFlow += df
N = int(input())
As = list(map(int, input().split()))
adjList = [[] for v in range(N + 2)]
for i, A in enumerate(As, 1):
if A <= 0:
addEdge(adjList, 0, i, -A)
else:
addEdge(adjList, i, N + 1, A)
for i in range(1, N + 1):
for j in range(2 * i, N + 1, i):
addEdge(adjList, i, j, float('inf'))
# Ford-Fulkerson法（最大フローを求める）
mf = Ford_Fulkerson(adjList, 0, N + 1)
print(sum([A for A in As if A > 0]) - mf)
```
| 100,181 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
from collections import deque
import sys
sys.setrecursionlimit(200000)
n = int(input())
houseki = list(map(int,input().split()))
g = [[] for i in range(n+2)]
INF = float("inf")
MAX = 0
for i in range(n):
if houseki[i] <= 0:
g[0].append([i+1,-houseki[i],len(g[i+1])])
g[i+1].append([0,0,len(g[0])-1])
else:
g[n+1].append([i+1,0,len(g[i+1])])
g[i+1].append([n+1,houseki[i],len(g[n+1])-1])
MAX += houseki[i]
j = (i+1)*2
while j <= n:
g[i+1].append([j,INF,len(g[j])])
g[j].append([i+1,0,len(g[i+1])-1])
j += i+1
def bfs(s,t):
global level
que = deque([s])
level[s] = 0
while que:
v = que.popleft()
lv = level[v] +1
for y, cap, rev in g[v]:
if cap and level[y] is None:
level[y] = lv
que.append(y)
return level[t] if level[t] else 0
def dfs(x,t,f):
if x == t:
return f
for j in range(it[x],len(g[x])):
it[x] = j
y, cap, rev = g[x][j]
if cap and level[x] < level[y]:
d = dfs(y,t,min(f,cap))
if d:
g[x][j][1] -= d
g[y][rev][1] += d
return d
return 0
flow = 0
f = INF = float("inf")
level = [None]*(n+2)
while bfs(0,n+1):
it = [0]* (n+2)
f = INF
while f:
f = dfs(0,n+1,INF)
flow += f
level = [None]*(n+2)
print(MAX-flow)
```
| 100,182 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
"""
https://atcoder.jp/contests/arc085/tasks/arc085_c
"""
from collections import defaultdict
from collections import deque
def Ford_Fulkerson_Func(s,g,lines,cost):
N = len(cost)
ans = 0
queue = deque([ [s,float("inf")] ])
ed = [True] * N
ed[s] = False
route = [0] * N
route[s] = -1
while queue:
now,flow = queue.pop()
for nex in lines[now]:
if ed[nex]:
flow = min(cost[now][nex],flow)
route[nex] = now
queue.append([nex,flow])
ed[nex] = False
if nex == g:
ans += flow
break
else:
continue
break
else:
return False,ans
t = g
s = route[t]
while s != -1:
cost[s][t] -= flow
if cost[s][t] == 0:
lines[s].remove(t)
if cost[t][s] == 0:
lines[t].add(s)
cost[t][s] += flow
t = s
s = route[t]
return True,ans
def Ford_Fulkerson(s,g,lines,cost):
ans = 0
while True:
fl,nans = Ford_Fulkerson_Func(s,g,lines,cost)
if fl:
ans += nans
continue
else:
break
return ans
N = int(input())
a = list(map(int,input().split()))
S = N+1
T = N+2
lis = defaultdict(set)
cost = [[0] * (N+3) for i in range(N+3)]
for i in range(N):
if a[i] > 0:
lis[i+1].add(T)
cost[i+1][T] += a[i]
elif a[i] < 0:
lis[S].add(i+1)
cost[S][i+1] -= a[i]
for i in range(1,N+1):
for j in range(i*2,N+1,i):
lis[i].add(j)
cost[i][j] += float("inf")
m = Ford_Fulkerson(S,T,lis,cost)
ss = 0
for i in range(N):
if a[i] > 0:
ss += a[i]
print (ss-m)
```
| 100,183 |
Provide a correct Python 3 solution for this coding contest problem.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import time,random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
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 list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
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 pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
class Flow():
def __init__(self, e, N):
self.E = e
self.N = N
def max_flow(self, s, t):
r = 0
e = self.E
def f(c, cap):
v = self.v
v[c] = 1
if c == t:
return cap
for i in range(self.N):
if v[i] or e[c][i] <= 0:
continue
cp = min(cap, e[c][i])
k = f(i, cp)
if k > 0:
e[c][i] -= k
e[i][c] += k
return k
return 0
while True:
self.v = [None] * self.N
fs = f(s, inf)
if fs == 0:
break
r += fs
return r
def main():
n = I()
a = LI()
s = n
t = n + 1
e = [[0] * (n+2) for _ in range(n+2)]
for i in range(n):
c = a[i]
if c < 0:
e[s][i] = -c
ii = i + 1
for j in range(ii*2, n+1, ii):
e[i][j-1] = inf
else:
e[i][t] = c
fl = Flow(e, n+2)
r = fl.max_flow(s,t)
return sum(map(lambda x: max(0,x), a)) - r
# start = time.time()
print(main())
# pe(time.time() - start)
```
| 100,184 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
# Dinic's algorithm
from collections import deque
class Dinic:
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap):
forward = [to, cap, None]
forward[2] = backward = [fr, 0, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
def add_multi_edge(self, v1, v2, cap1, cap2):
edge1 = [v2, cap1, None]
edge1[2] = edge2 = [v1, cap2, edge1]
self.G[v1].append(edge1)
self.G[v2].append(edge2)
def bfs(self, s, t):
self.level = level = [None]*self.N
deq = deque([s])
level[s] = 0
G = self.G
while deq:
v = deq.popleft()
lv = level[v] + 1
for w, cap, _ in G[v]:
if cap and level[w] is None:
level[w] = lv
deq.append(w)
return level[t] is not None
def dfs(self, v, t, f):
if v == t:
return f
level = self.level
for e in self.it[v]:
w, cap, rev = e
if cap and level[v] < level[w]:
d = self.dfs(w, t, min(f, cap))
if d:
e[1] -= d
rev[1] += d
return d
return 0
def flow(self, s, t):
flow = 0
INF = 10**9 + 7
G = self.G
while self.bfs(s, t):
*self.it, = map(iter, self.G)
f = INF
while f:
f = self.dfs(s, t, INF)
flow += f
return flow
N=int(input())
a=list(map(int,input().split()))
jew=Dinic(N+2)
ans=0
for i in range(N):
if a[i]>=0:
ans+=a[i]
jew.add_edge(0,i+1,a[i])
else:
jew.add_edge(i+1,N+1,-a[i])
inf=10**15
for i in range(1,N+1):
for j in range(1,N//i+1):
jew.add_edge(i*j,i,inf)
f=jew.flow(0,N+1)
print(ans-f)
```
Yes
| 100,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
N=int(input())
P=[int(i) for i in input().split()]
inf = 10**20
table=[[0]*(N+2) for i in range(N+2)]
for i in range(1,N+1):
if P[i-1]>0:
table[i][N+1]=P[i-1]
else:
table[0][i]=-P[i-1]
for j in range(2*i,N+1,i):
table[i][j]=inf
#print(table)
def fk(x,t,f):
#print(x)
visit[x]=True
if x==t:
return f
for i in range(N+2):
if (not visit[i]) and table[x][i]>0:
df=fk(i,t,min(f,table[x][i]))
if df>0:
table[x][i]-=df
table[i][x]+=df
return df
return 0
ans=0
while True:
visit=[False]*(N+2)
df=fk(0,N+1,inf)
if df>0:
ans+=df
else:
break
num=sum([p for p in P if p>0])
print(num-ans)
```
Yes
| 100,186 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
from copy import deepcopy
N = int(input())
a0 = list(map(int, input().split()))
for i in range(N, 0, -1):
if a0[i - 1] < 0:
tmp = 0
b = []
for j in range(1, 101):
if i * j - 1 >= N:
break
tmp += a0[i * j - 1]
b.append(i * j - 1)
if tmp < 0:
for j in b:
a0[j] = 0
for k1 in range(N, 0, -1):
for k2 in range(k1-1, 0, -1):
a = deepcopy(a0)
for j in range(1, 101):
if k1*j-1 >= N:
break
a[k1*j-1] = 0
for j in range(1, 101):
if k2*j-1 >= N:
break
a[k2*j-1] = 0
for i in range(N, 0, -1):
if a[i - 1] < 0:
tmp = 0
b = []
for j in range(1, 101):
if i * j - 1 >= N:
break
tmp += a[i * j - 1]
b.append(i * j - 1)
if tmp < 0:
for j in b:
a[j] = 0
for i in range(N, 0, -1):
if a0[i - 1] < 0:
tmp = 0
b = []
for j in range(1, 101):
if i * j - 1 >= N:
break
tmp += a0[i * j - 1]
b.append(i * j - 1)
if tmp < 0:
for j in b:
a0[j] = 0
if sum(a) > sum(a0):
a0 = a
print(sum(a0))
```
Yes
| 100,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
from collections import deque
class Dinic:
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap):
forward = [to, cap, None]#(行き先、容量、逆辺の参照)
forward[2] = backward = [fr, 0, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
def add_multi_edge(self, v1, v2, cap1, cap2):
edge1 = [v2, cap1, None]
edge1[2] = edge2 = [v1, cap2, edge1]
self.G[v1].append(edge1)
self.G[v2].append(edge2)
def bfs(self, s, t):
self.level = level = [None]*self.N
deq = deque([s])
level[s] = 0
G = self.G
while deq:
v = deq.popleft()
lv = level[v] + 1
for w, cap, _ in G[v]:
if cap and level[w] is None:
level[w] = lv
deq.append(w)
return level[t] is not None
def dfs(self, v, t, f):
if v == t:
return f
level = self.level
for e in self.it[v]:
w, cap, rev = e
if cap and level[v] < level[w]:
d = self.dfs(w, t, min(f, cap))
if d:
e[1] -= d
rev[1] += d
return d
return 0
def flow(self, s, t):
flow = 0
INF = 10**9 + 7
G = self.G
while self.bfs(s, t):
*self.it, = map(iter, self.G)
f = INF
while f:
f = self.dfs(s, t, INF)
flow += f
return flow
N = int(input())
a = list(map(int, input().split()))
D = Dinic(N+2)
inf = 10**18
profit = 0
cost = 0
for i in range(N):
if a[i]>=0:
D.add_edge(i, N+1, a[i])
else:
D.add_edge(N, i, -a[i])
cost+=a[i]
profit+=abs(a[i])
for i in range(1, N+1):
for k in range(2, N//i+1):
D.add_edge(i-1, i*k-1, inf)
f = D.flow(N, N+1)
print(profit-f+cost)
```
Yes
| 100,188 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
N = int(input())
As = list(map(int, input().split()))
while True:
vs = [sum(As[x::x + 1]) for x in range(N)]
vMin = min(vs)
if vMin >= 0:
break
xMin = N - 1 - vs[::-1].index(vMin)
for x in range(xMin, N, xMin + 1):
As[x] = 0
print(vs[0])
```
No
| 100,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
c,ans = [1]*n,sum(a)
def f(x):
num = 0
for i in range(x,n+1,x):
if c[i-1]: num+=a[i-1]
if num<0:
for i in range(x,n+1,x): c[i-1] = 0
return num
else: return 0
for i in range(n,0,-1): ans-=f(i)
print(ans)
```
No
| 100,190 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
flag = 1
for i in a:
if i > 0:
flag = 0
if flag:
print(0)
else:
flag_ = 1
while flag_:
ans = sum(a)
prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
b = [[] for i in range(100)]
for i in prime:
j = i
while j <= n:
b[i].append(a[j - 1])
j += i
c = []
flag_ = 0
for i in b:
if sum(i) < 0:
flag_ = 1
c.append(sum(i))
dis = 0
mi = 0
for i in range(100):
if c[i] < mi:
mi = c[i]
dis = i
if mi >= 0:
break
for i in range(1, n + 1):
if i % dis == 0:
a[i - 1] = 0
ans += -mi
print(ans)
```
No
| 100,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N gemstones labeled 1 through N.
You can perform the following operation any number of times (possibly zero).
* Select a positive integer x, and smash all the gems labeled with multiples of x.
Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan). However, a_i may be negative, in which case you will be charged money.
By optimally performing the operation, how much yen can you earn?
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the maximum amount of money that can be earned.
Examples
Input
6
1 2 -6 4 5 3
Output
12
Input
6
100 -100 -100 -100 100 -100
Output
200
Input
5
-1 -2 -3 -4 -5
Output
0
Input
2
-1000 100000
Output
99000
Submitted Solution:
```
class Ford_Fulkerson:
def __init__(self, v, inf=float("inf")):
self.V = v
self.inf = inf
self.G = [[] for _ in range(v)]
self.used = [False for _ in range(v)]
def addEdge(self, fm, to, cap):
self.G[fm].append({'to':to, 'cap':cap, 'rev':len(self.G[to])})
self.G[to].append({'to':fm, 'cap':0, 'rev':len(self.G[fm])-1})
def dfs(self, v, t, f):
if v == t: return f
self.used[v] = True
for i in range(len(self.G[v])):
e = self.G[v][i]
if self.used[e["to"]] != True and e['cap'] > 0:
d = self.dfs(e['to'], t ,min(f, e['cap']))
if d > 0:
e['cap'] -= d
self.G[e['to']][e['rev']]['cap'] += d
return d
return 0
def max_flow(self,s,t):
flow = 0
while True:
self.used = [False for i in range(self.V)]
f = self.dfs(s,t,self.inf)
if f == 0: return flow
flow += f
from sys import stdin, setrecursionlimit
def IL():return list(map(int, stdin.readline().split()))
setrecursionlimit(1000000)
def main():
N = int(input())
a = IL()
d = Ford_Fulkerson(102)
s=N
t=N+1
for i in range(N):
d.addEdge(s,i,max(-a[i],0))
d.addEdge(i,t,max(a[i],0))
x = i+1
y=x+x
for j in range(y,N+1):
d.addEdge(x-1,y-1,float("inf"))
res = sum([max(i,0) for i in a])
print(res - d.max_flow(s,t))
if __name__ == "__main__": main()
```
No
| 100,192 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
import sys
def solve():
a = int(input())
b = int(input())
if a == b:
print(1)
return
t = a ^ b
N = bitlen(t)
a = a & (2**N - 1)
b = b & (2**N - 1)
blen = bitlen(b)
sb = b & (2**(blen - 1) - 1)
sblen = bitlen(sb)
s = 2**sblen - 1
ymax = b | s
T = 2**(N - 1)
ans = T - a
if ymax < T + a:
ans += ymax + 1 - a
else:
ans += T
print(ans)
def bitlen(x):
return 1 + bitlen(x >> 1) if x else 0
if __name__ == '__main__':
solve()
```
| 100,193 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
import sys
readline = sys.stdin.readline
def calc(A, B):
if A == B:
return 1
for i in range(61, -1, -1):
if A&(1<<i) and B&(1<<i):
A ^= 1<<i
B ^= 1<<i
if not A&(1<<i) and B&(1<<i):
break
Bd = B ^ (1<<(B.bit_length())-1)
BM = ((1<<Bd.bit_length())-1)
B = B | BM
C = (1<<(B.bit_length())-1)
res = (B-A+1) + (C-max(BM+1, A))
return res
A = int(readline())
B = int(readline())
print(calc(A, B))
```
| 100,194 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
import math
A = int(input())
B = int(input())
if B < A: A,B = B,A
if A == B:
print(1)
else:
t = int(math.floor(math.log2(B)))
while (A & 2**t) == (B & 2**t):
if A & 2**t:
A -= 2**t
B -= 2**t
t-=1
ans = 2**t - A # [A,2^t-1]
r = t-1
while 2**t + 2**r > B and r>=0:
r-=1
# [2^t,2^t+2^{r+1}-1] \cup [2^t+A,2^{t+1}-1]
C = 2**(r+1)
if C < A:
ans += C + 2**t - A
else:
ans += 2**t
print(ans)
```
| 100,195 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
import sys
def solve():
a = int(input())
b = int(input())
if a == b:
print(1)
return
t = a ^ b
N = len(bin(t)) - 2
t = 1 << N
a = a & (t - 1)
b = b & (t - 1)
blen = len(bin(b)) - 2
sb = b & (2**(blen - 1) - 1)
if sb == 0:
sblen = 0
else:
sblen = len(bin(sb)) - 2
s = (1<<sblen) - 1
ymax = b | s
T = 1<<(N - 1)
ans = T - a
if ymax < T + a:
ans += ymax - T + 1 + T - a
else:
ans += T
print(ans)
if __name__ == '__main__':
solve()
```
| 100,196 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
a=int(input())
b=int(input())
if a==b:
print(1)
exit()
abin=bin(a)[2:]
bbin=bin(b)[2:]
la=len(abin)
lb=len(bbin)
if la==lb:
while abin[0]==bbin[0]:
abin=abin[1:]
bbin=bbin[1:]
while len(abin)>1 and abin[0]=="0":
abin=abin[1:]
cbin=bbin[1:]
while cbin and cbin[0]=="0":
cbin=cbin[1:]
c=2**(len(cbin))-1
a=int(abin,2)
b=int(bbin,2)
lb=len(bbin)
lbnd=2**(lb-1)+c
hbnd=2**(lb-1)+a
if lbnd>=hbnd:
print(2**lb-a)
else:
print(2**(lb-1)-a+lbnd-a+1)
```
| 100,197 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
# D
res = 0
A = int(input())
B = int(input())
def pow2_floor_log2(x):
res = 1
while res*2 <= x:
res *= 2
return res
if A == B:
res += 1
while A < B:
# find log2A and log2B
C = pow2_floor_log2(B)
if A < C:
# A to C
res += C - A
if B > C:
B_ = C + pow2_floor_log2(B - C)*2 - 1
else:
B_ = C
if C + A <= B_:
res += C
break
else:
res += (C - A) + (B_ - C + 1)
break
else:
A -= C
B -= C
print(res)
```
| 100,198 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
"Correct Solution:
```
def solve(a, b):
if a == b:
return 1
f = 1 << 60
r = 0
while f:
if r == 0 and b & f and a & f == 0:
r = f
elif r > 0 and b & f:
k = f << 1
break
f >>= 1
else:
k = 1
a &= r - 1
if k > a:
return (r << 1) - a
return 2 * (r - a) + k
a = int(input())
b = int(input())
print(solve(a, b))
```
| 100,199 |
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