message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M.
Constraints
* 1 β€ m β€ 100
* 1 β€ n β€ 109
Input
m n
Two integers m and n are given in a line.
Output
Print mn (mod 1,000,000,007) in a line.
Examples
Input
2 3
Output
8
Input
5 8
Output
390625
Submitted Solution:
```
def pow_mod(n, x, mod) :
if n == 1:
return n
ret = pow_mod(n*n, x//2, mod)
if x % 2 == 1:
ret *= n
return ret
n, x = list(map(int, input().split()))
print(pow_mod(n, x, 1000000007)
``` | instruction | 0 | 99,456 | 22 | 198,912 |
No | output | 1 | 99,456 | 22 | 198,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M.
Constraints
* 1 β€ m β€ 100
* 1 β€ n β€ 109
Input
m n
Two integers m and n are given in a line.
Output
Print mn (mod 1,000,000,007) in a line.
Examples
Input
2 3
Output
8
Input
5 8
Output
390625
Submitted Solution:
```
n,m = map(int, input().split())
a = n**m%1000000007
print(a)
``` | instruction | 0 | 99,457 | 22 | 198,914 |
No | output | 1 | 99,457 | 22 | 198,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M.
Constraints
* 1 β€ m β€ 100
* 1 β€ n β€ 109
Input
m n
Two integers m and n are given in a line.
Output
Print mn (mod 1,000,000,007) in a line.
Examples
Input
2 3
Output
8
Input
5 8
Output
390625
Submitted Solution:
```
m,n= map(int,input().split())
a=1
for i in range(n):
a*=m
if a>1000000007:
a%=1000000007
print(a)
``` | instruction | 0 | 99,458 | 22 | 198,916 |
No | output | 1 | 99,458 | 22 | 198,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M.
Constraints
* 1 β€ m β€ 100
* 1 β€ n β€ 109
Input
m n
Two integers m and n are given in a line.
Output
Print mn (mod 1,000,000,007) in a line.
Examples
Input
2 3
Output
8
Input
5 8
Output
390625
Submitted Solution:
```
def pow2(m, n):
M = 1000000007
if n == 0
return 1
res = pow2(m*m%M, n/2)
if n%2 == 1
res = res*m%M
return res
m,n = map(int,input().split())
print(pow2(m,n))
``` | instruction | 0 | 99,459 | 22 | 198,918 |
No | output | 1 | 99,459 | 22 | 198,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,551 | 22 | 199,102 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
def mat_dot(A, B, mod):
assert len(A[0]) == len(B), 'invalid_size'
L = len(A)
M = len(A[0])
N = len(B[0])
res = [[0]*N for _ in range(L)]
for i in range(L):
for j in range(N):
a = 0
for k in range(M):
a = (a+A[i][k]*B[k][j]) % mod
res[i][j] = a
return res
def mat_pow(A, x, mod):
N = len(A)
res = [[0]*N for _ in range(N)]
for i in range(N):
res[i][i] = 1
for i in range(x.bit_length()):
if 2**i & x:
res = mat_dot(res, A, mod)
A = mat_dot(A, A, mod)
return res
#if there exists K such that X**K %mod == Y % mod, return K % (mod-1)
#Otherwise, return -1
def bsgs(X, Y, mod):
Y %= mod
X %= mod
rm = int(mod**(0.5)) + 2
R = pow(pow(X, rm ,mod), mod-2, mod)
D = {Y: 0}
p = Y
for a in range(1, rm):
p = p*R % mod
D[p] = a
p = 1
b = 0
if p in D:
return (D[p]*rm + b)%(mod-1)
for b in range(1, rm):
p = (p*X) % mod
if p in D:
return (D[p]*rm + b)%(mod-1)
return -1
n, f1, f2, f3, c = map(int, input().split())
mod = 10**9+7
a1 = bsgs(5, f1, mod)
a2 = bsgs(5, f2, mod)
a3 = bsgs(5, f3, mod)
d = bsgs(5, c, mod)
A = [[1, 1, 1, 2*d, -6*d],
[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[0, 0, 0, 1, 1],
[0, 0, 0, 0, 1]]
B = mat_pow(A, n-3, mod - 1)
Ans = mat_dot(B, [[a3], [a2], [a1], [4], [1]], mod - 1)
print(pow(5, Ans[0][0], mod))
``` | output | 1 | 99,551 | 22 | 199,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,552 | 22 | 199,104 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
def mat_mult(A, B, MOD):
n, m, p = len(A), len(A[0]), len(B[0])
assert (len(B) == m)
C = [[0] * p for _ in range(n)]
for i in range(n):
for k in range(m):
Aik = A[i][k]
for j in range(p):
C[i][j] = (C[i][j] + Aik * B[k][j]) % MOD
return C
def ksm(A, n, MOD):
if (n == 0):
E = [[0 for i in range(len(A))] for j in range(len(A))]
for i in range(len(A)):
E[i][i] = 1
return E
if (n == 1):
return A
k = ksm(A, n//2, MOD)
z = mat_mult(k, k, MOD)
if (n&1):
return (mat_mult(z, A, MOD))
else:
return z
def Fenjie(n):
k = {}
if (n==1):
return {}
a = 2
while (n>=2):
b = n%a
if (a*a > n):
k[n] = 1
break
if (b==0):
if (a in k):
k[a] += 1
else:
k[a] = 1
n = n//a
else:
a += 1
return k
def Euler(n):
if (n==1):
return 1
k = Fenjie(n)
m = n
for i in k:
m = m // i * (i-1)
return m
MOD = 10**9 + 7
n, b, c, d, e = list(map(int, input().split()))
l1 = [[0],[0],[1]]
l2 = [[0],[1],[0]]
l3 = [[1],[0],[0]]
l4 = [[6],[2],[0],[0],[0]]
a1 = [[1,1,1],[1,0,0],[0,1,0]]
a2 = [[3,-2,0,-1,1],[1,0,0,0,0],[0,1,0,0,0],[0,0,1,0,0],[0,0,0,1,0]]
if (n == 4):
print(e*e*b*c*d % MOD)
else:
a1 = ksm(a1, n-3, MOD-1)
a2 = ksm(a2, n-5, MOD-1)
b1 = mat_mult(a1, l1, MOD-1)
p1 = pow(b, b1[0][0], MOD)
c1 = mat_mult(a1, l2, MOD-1)
p2 = pow(c, c1[0][0], MOD)
d1 = mat_mult(a1, l3, MOD-1)
p3 = pow(d, d1[0][0], MOD)
n1 = mat_mult(a2, l4, MOD-1)[0][0]
p = pow(e, n1, MOD)
p = p1*p%MOD
p = p2*p%MOD
p = p3*p%MOD
print(p)
``` | output | 1 | 99,552 | 22 | 199,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,553 | 22 | 199,106 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
def baby_step_giant_step(g, y, p):
"""y = g^x (mod p)γζΊγγxγζ±γγ"""
m = int(p**0.5) + 1
# Baby-step
baby = {}
b = 1
for i in range(m):
baby[b] = i
b = (b * g) % p
# Giant-step
gm = pow(b, p-2, p)
giant = y
for i in range(m):
if giant in baby:
x = i*m + baby[giant]
return x
giant = (giant * gm) % p
return -1
def _mul(A, B, MOD):
C = [[0] * len(B[0]) for i in range(len(A))]
for i in range(len(A)):
for k in range(len(B)):
for j in range(len(B[0])):
C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % MOD
return C
def pow_matrix(A, n, MOD):
"""A**nγγγγͺγ³γ°γ«γγ£γ¦ζ±γγ"""
B = [[0] * len(A) for i in range(len(A))]
for i in range(len(A)):
B[i][i] = 1
while n > 0:
if n & 1:
B = _mul(A, B, MOD)
A = _mul(A, A, MOD)
n = n // 2
return B
n, f1, f2, f3, c = map(int, input().split())
MOD = 10**9 + 7
log_f1 = baby_step_giant_step(5, f1, MOD)
log_f2 = baby_step_giant_step(5, f2, MOD)
log_f3 = baby_step_giant_step(5, f3, MOD)
log_c = baby_step_giant_step(5, c, MOD)
matrix = [[0]*5 for i in range(5)]
matrix[0][0] = 1
matrix[0][1] = 1
matrix[0][2] = 1
matrix[0][3] = 2 * log_c
matrix[0][4] = -6 * log_c
matrix[1][0] = 1
matrix[2][1] = 1
matrix[3][3] = 1
matrix[3][4] = 1
matrix[4][4] = 1
matrix_n = pow_matrix(matrix, n - 3, MOD - 1)
ans = log_f3 * matrix_n[0][0] + log_f2 * matrix_n[0][1] + log_f1 * matrix_n[0][2] \
+ 4 * matrix_n[0][3] + matrix_n[0][4]
ans = ans % (MOD-1)
print(pow(5, ans, MOD))
``` | output | 1 | 99,553 | 22 | 199,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,554 | 22 | 199,108 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
n, f1, f2, f3, c = list(map(int,input().split()))
mat = [[1,1,1],[1,0,0],[0,1,0]]
final = [[1,0,0],[0,1,0],[0,0,1]]
nn = n - 3
N = 10**9 + 6
def prod(a, b):
m = [[0,0,0],[0,0,0],[0,0,0]]
for i in range(3):
for j in range(3):
m[i][j] = (a[i][0]*b[0][j] + a[i][1]*b[1][j]+a[i][2]*b[2][j]) % N
return m
while nn > 0:
if nn % 2 == 1:
final = prod(final, mat)
mat = prod(mat,mat)
nn //= 2
q = (final[0][0] * 3 + final[0][1] * 2 + final[0][2] * 1) % N
p = q - (n%N) + N
# p to potega c
ef3 = (final[0][0] * 1) % N
ef2 = (final[0][1] * 1) % N
ef1 = (final[0][2] * 1) % N
# print f1^ef1 *f2^ef2*f3^ef3 * c^p
def pot(a,w):
wyn = 1
while w > 0:
if w%2 == 1:
wyn = (wyn * a) % (N+1)
a = (a * a) % (N+1)
w //= 2
return wyn
l1 = pot(f1, ef1)
l2 = pot(f2, ef2)
l3 = pot(f3, ef3)
l4 = pot(c, p)
c = (l1*l2*l3*l4)%(N+1)
print(c)
``` | output | 1 | 99,554 | 22 | 199,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,555 | 22 | 199,110 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
MOD = int(1e9 + 7)
def matMOD(A):
f = lambda x: x%(MOD-1)
ret = [ list(map(f, i)) for i in A ]
return ret
def matmul(A, B):
a, b = len(A), len(A[0])
c = len(B[0])
ret = [ [0] * c for i in range(a) ]
for i in range(a):
for j in range(c):
for k in range(b):
ret[i][j] += A[i][k] * B[k][j]
return ret
def matmul_log(A, m):
if m == 1:
return A
B = matmul_log(A, m//2)
if m % 2 == 0:
return matMOD(matmul(B, B))
else:
return matMOD(matmul(A, matmul(B, B)))
n, f1, f2, f3, c = map(int, input().split())
ini = [
[1, 0, 0, 1],
[0, 1, 0, 1],
[0, 0, 1, 1],
[4, 2, 0, 0, 0, 2]
]
m = [
[0, 0, 0, 0],
[1, 0, 0, 1],
[0, 1, 0, 1],
[0, 0, 1, 1]
]
m2 = [
[1, 0, 0, 0, 0, 1],
[1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 1],
[0, 0, 0, 1, 0, 1],
[0, 0, 0, 0, 1, 1]
]
M = matmul_log(m, n-1)
M2 = matmul_log(m2, n-1)
x, y, z = matmul([ini[0]], M)[0][0], matmul([ini[1]], M)[0][0], matmul([ini[2]], M)[0][0]
w = matmul([ini[3]], M2)[0][2]
#print(x, y, z)
#print(ini[3])
#print(w)
ans = (pow(c, w, MOD) * pow(f1, x, MOD) * pow(f2, y, MOD) * pow(f3, z, MOD)) % MOD
print(ans)
``` | output | 1 | 99,555 | 22 | 199,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,556 | 22 | 199,112 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
import os
from io import BytesIO, StringIO
#input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
DEBUG = False
debug_print = print if DEBUG else lambda *x,**y: None
def input_as_list():
return list(map(int, input().split()))
def array_of(f, *dim):
return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()
def main():
mod = 1000000007
def modmul(x, y):
return (x * y) % mod
def row(m, i, n=3):
return m[n * i:n * (i + 1)]
def col(m, i, n=3):
return m[i::n]
def vecmul(u, v):
s = 0
for i, j in zip(u, v):
s += (i * j) % (mod-1)
return s % (mod-1)
def matmul(a, b, n=3):
out = []
for i in range(n * n):
r, c = divmod(i, n)
out.append(vecmul(row(a, r), col(b, c)))
return out
def matadd(a, b, n=3):
out = []
for i in range(n * n):
out.append((a[i] + b[i]) % (mod-1))
return out
def matpow(m, p, n=3):
bs = str(bin(p)).lstrip('0b')
out = [0] * (n * n)
out[0::n + 1] = [1] * n
for b in reversed(bs):
if b == '1':
out = matmul(out, m)
m = matmul(m, m)
return out
def brute_force(n, f1, f2, f3, c):
i = 3
while i < n:
i += 1
f4 = pow(c, 2*i-6, mod) * f1 * f2 * f3 % mod
f1, f2, f3 = f2, f3, f4
return f3
def solve(n, f1, f2, f3, c):
f = [f1, f2, f3]
g = [modmul(c, f[0]), modmul(c * c, f[1]), modmul(c * c * c, f[2])]
mat = [1, 1, 1, 1, 0, 0, 0, 1, 0]
mat_n = matpow(mat, n - 3)
g_n = (pow(g[2], mat_n[0], mod)
* pow(g[1], mat_n[1], mod)
* pow(g[0], mat_n[2], mod)) % mod
c_n = pow(c, n, mod)
c_n_inv = pow(c_n, mod - 2, mod)
f_n = modmul(g_n, c_n_inv)
return f_n
def solve_from_stdin():
i = input_as_list()
print(solve(*i))
def test():
import random
sample = [random.randrange(4, 99) for _ in range(5)]
sample = [2, 2, 2, 2]
print(*sample)
for i in range(4,200):
print(i, *sample)
print(solve(i, *sample))
print(brute_force(i, *sample))
print()
solve_from_stdin()
main()
``` | output | 1 | 99,556 | 22 | 199,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,557 | 22 | 199,114 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
fib_matrix = [[1,1,1],
[1,0,0],
[0,1,0]]
def matrix_square(A, mod):
return mat_mult(A,A,mod)
def mat_mult(A,B, mod):
if mod is not None:
return [[(A[0][0]*B[0][0] + A[0][1]*B[1][0]+A[0][2]*B[2][0])%mod, (A[0][0]*B[0][1] + A[0][1]*B[1][1]+A[0][2]*B[2][1])%mod,(A[0][0]*B[0][2] + A[0][1]*B[1][2]+A[0][2]*B[2][2])%mod],
[(A[1][0]*B[0][0] + A[1][1]*B[1][0]+A[1][2]*B[2][0])%mod, (A[1][0]*B[0][1] + A[1][1]*B[1][1]+A[1][2]*B[2][1])%mod,(A[1][0]*B[0][2] + A[1][1]*B[1][2]+A[1][2]*B[2][2])%mod],
[(A[2][0]*B[0][0] + A[2][1]*B[1][0]+A[2][2]*B[2][0])%mod, (A[2][0]*B[0][1] + A[2][1]*B[1][1]+A[2][2]*B[2][1])%mod,(A[2][0]*B[0][2] + A[2][1]*B[1][2]+A[2][2]*B[2][2])%mod]]
def matrix_pow(M, power, mod):
#Special definition for power=0:
if power <= 0:
return M
powers = list(reversed([True if i=="1" else False for i in bin(power)[2:]])) #Order is 1,2,4,8,16,...
matrices = [None for _ in powers]
matrices[0] = M
for i in range(1,len(powers)):
matrices[i] = matrix_square(matrices[i-1], mod)
result = None
for matrix, power in zip(matrices, powers):
if power:
if result is None:
result = matrix
else:
result = mat_mult(result, matrix, mod)
return result
fib_matrix2 = [[1,1,1,1,0],
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,0,1,1],
[0,0,0,0,1]]
def matrix_square2(A, mod):
return mat_mult2(A,A,mod)
def mat_mult2(A,B, mod):
if mod is not None:
m=[]
for i in range(5):
m.append([])
for j in range(5):
sums=0
for k in range(5):
sums+=A[i][k]*B[k][j]
m[-1].append(sums%mod)
return m
def matrix_pow2(M, power, mod):
#Special definition for power=0:
if power <= 0:
return M
powers = list(reversed([True if i=="1" else False for i in bin(power)[2:]])) #Order is 1,2,4,8,16,...
matrices = [None for _ in powers]
matrices[0] = M
for i in range(1,len(powers)):
matrices[i] = matrix_square2(matrices[i-1], mod)
result = None
for matrix, power in zip(matrices, powers):
if power:
if result is None:
result = matrix
else:
result = mat_mult2(result, matrix, mod)
return result
n,f1,f2,f3,c=map(int,input().split())
f3pow=matrix_pow(fib_matrix, n-2, 1000000006)[0][2]
f1pow=matrix_pow(fib_matrix, n-2, 1000000006)[1][2]
f2pow=matrix_pow(fib_matrix, n-2, 1000000006)[1][1]
cpow=2*matrix_pow2(fib_matrix2, n, 1000000006)[2][4]
ans=pow(c,cpow,1000000007)
ans=(ans*pow(f1,f1pow,1000000007))%1000000007
ans=(ans*pow(f2,f2pow,1000000007))%1000000007
ans=(ans*pow(f3,f3pow,1000000007))%1000000007
print(ans)
``` | output | 1 | 99,557 | 22 | 199,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}. | instruction | 0 | 99,558 | 22 | 199,116 |
Tags: dp, math, matrices, number theory
Correct Solution:
```
import math
md = 10**9+7
def multiply(M,N):
md2 = 10**9+6
R = [[0 for i in range(3)] for j in range(3)]
for i in range(0, 3):
for j in range(0, 3):
for k in range(0, 3):
R[i][j] += (M[i][k] * N[k][j])%md2
R[i][j] %= md2
return R
def power(mat, n):
res = [[1,0,0],[0,1,0],[0,0,1]]
while n:
if n&1: res = multiply(res, mat)
mat = multiply(mat, mat)
n//=2
return res
n, f1, f2, f3, c = map(int, input().split())
f1 = (f1*c)%md
f2 = (f2*c**2)%md
f3 = (f3*c**3)%md
#print(f1, f2, f3)
mat = [[1,1,1],[1,0,0],[0,1,0]]
res = power(mat, n-3)
#print(res)
pw1, pw2, pw3 = res[0][2], res[0][1], res[0][0]
f1 = pow(f1, pw1, md)
f2 = pow(f2, pw2, md)
f3 = pow(f3, pw3, md)
ans = ((f1 * f2)%md * f3)%md
c = pow(c, md-2, md)
ans *= pow(c, n%(md-1), md)
ans %= md
print(ans)
``` | output | 1 | 99,558 | 22 | 199,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let f_{x} = c^{2x-6} β
f_{x-1} β
f_{x-2} β
f_{x-3} for x β₯ 4.
You have given integers n, f_{1}, f_{2}, f_{3}, and c. Find f_{n} mod (10^{9}+7).
Input
The only line contains five integers n, f_{1}, f_{2}, f_{3}, and c (4 β€ n β€ 10^{18}, 1 β€ f_{1}, f_{2}, f_{3}, c β€ 10^{9}).
Output
Print f_{n} mod (10^{9} + 7).
Examples
Input
5 1 2 5 3
Output
72900
Input
17 97 41 37 11
Output
317451037
Note
In the first example, f_{4} = 90, f_{5} = 72900.
In the second example, f_{17} β 2.28 Γ 10^{29587}.
Submitted Solution:
```
import sys
# from bisect import bisect_right
# gcd
# from fractions import gcd
# from math import ceil, floor
# from copy import deepcopy
# from itertools import accumulate
# l = ['a', 'b', 'b', 'c', 'b', 'a', 'c', 'c', 'b', 'c', 'b', 'a']
# print(S.most_common(2)) # [('b', 5), ('c', 4)]
# print(S.keys()) # dict_keys(['a', 'b', 'c'])
# print(S.values()) # dict_values([3, 5, 4])
# print(S.items()) # dict_items([('a', 3), ('b', 5), ('c', 4)])
# from collections import Counter
# import math
# from functools import reduce
#
# fin = open('in_1.txt', 'r')
# sys.stdin = fin
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
# template
# def solve(L, A, B, M):
# ret, digits = 0, 0
# for d in range(18, 0, -1):
# lo = 10 ** (d - 1)
# hi = 10 ** d - 1
# if hi < A or A + B * (L - 1) < lo:
# continue
# imin = 0 if lo <= A else (lo - A + B - 1) // B
# init = A + B * imin
# if init > hi:
# continue
# imax = L - 1 if A + B * (L - 1) <= hi else imin + (hi - init) // B
# n = imax - imin + 1
# p = 10 ** d % M
# a = matpow([[0,1,0], [0,0,1], [1,1,1]], n, M)
# # a[0][3] = sum p^i for i in [0, n-1]
# # a[1][3] = sum i * p^i for i in [0, n-1]
# # sum (A + B * (imax - i)) * p^i for i in [0, n-1]
# sub = (A + B * imax) % M * a[1][3] % M + M - (B * a[0][3] % M)
# ret += sub % M * pow10(digits, M) % M
# digits += d * (imax - imin + 1)
# return ret % M
def pow10(p, mod):
if p % 2 == 1:
return 10 * pow10(p - 1, mod) % mod
elif p > 0:
sub = pow10(p // 2, mod)
return sub * sub % mod
else:
return 1
def matpow(a, p, mod):
if p % 2 == 1:
return matmul(a, matpow(a, p - 1, mod), mod)
elif p > 0:
b = matpow(a, p // 2, mod)
return matmul(b, b, mod)
else:
n = len(a)
return [[1 if i == j else 0 for j in range(n)] for i in range(n)]
def matmul(a, b, mod):
n = len(a)
ret = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
for k in range(n):
for j in range(n):
ret[i][j] += a[i][k] * b[k][j]
for j in range(n):
ret[i][j] %= mod
return ret
if __name__ == '__main__':
# write code
mod = 10**9+7
n, a1, a2, a3, c = mi()
a = matpow([[0, 1, 0], [0, 0, 1], [1, 1, 1]], n - 2, mod)
# print(a)
ans = pow(a1, a[1][0], mod) * pow(a2, a[1][1], mod) * pow(a3, a[1][2], mod) % mod
# print(ans)
cmat = matpow([[0, 1, 0, 0, 0], [0, 0, 1, 0, 0], [1, 1, 1, 2, 0], [0, 0, 0, 1, 1], [0, 0, 0, 0, 1]], n - 3, mod)
# print(c)
ans *= pow(c, (cmat[2][3] + cmat[2][4]), mod)
ans %= mod
print(ans)
``` | instruction | 0 | 99,564 | 22 | 199,128 |
No | output | 1 | 99,564 | 22 | 199,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n distinct integers x_1,x_2,β¦,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
The first line of each test case contains two integers n,k (2 β€ n β€ 2 β
10^5, -10^{18} β€ k β€ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,β¦,x_n (-10^{18} β€ x_i β€ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board. | instruction | 0 | 99,662 | 22 | 199,324 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import math
def gcd(a,b):
return math.gcd(a,b)
for _ in range(int(input())):
n,k=map(int,input().split())
l=list(map(int,input().split()))
g=0
l.sort()
for i in range(1,n):
g=gcd(g,l[i]-l[0])
if (k-l[0])%g==0:
print('YES')
else:
print('NO')
``` | output | 1 | 99,662 | 22 | 199,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n distinct integers x_1,x_2,β¦,x_n are written on the board. Nezzar can perform the following operation multiple times.
* Select two integers x,y (not necessarily distinct) on the board, and write down 2x-y. Note that you don't remove selected numbers.
Now, Nezzar wonders if it is possible to have his favorite number k on the board after applying above operation multiple times.
Input
The first line contains a single integer t (1 β€ t β€ 10^5) β the number of test cases.
The first line of each test case contains two integers n,k (2 β€ n β€ 2 β
10^5, -10^{18} β€ k β€ 10^{18}).
The second line of each test case contains n distinct integers x_1,x_2,β¦,x_n (-10^{18} β€ x_i β€ 10^{18}).
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, print "YES" on a single line if it is possible to have k on the board. Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Example
Input
6
2 1
1 2
3 0
2 3 7
2 -1
31415926 27182818
2 1000000000000000000
1 1000000000000000000
2 -1000000000000000000
-1000000000000000000 123
6 80
-5 -20 13 -14 -2 -11
Output
YES
YES
NO
YES
YES
NO
Note
In the first test case, the number 1 is already on the board.
In the second test case, Nezzar could perform the following operations to write down k=0 on the board:
* Select x=3 and y=2 and write down 4 on the board.
* Select x=4 and y=7 and write down 1 on the board.
* Select x=1 and y=2 and write down 0 on the board.
In the third test case, it is impossible to have the number k = -1 on the board. | instruction | 0 | 99,664 | 22 | 199,328 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def gcd(x, y):
while y != 0:
(x, y) = (y, x % y)
return x
def main():
t=int(input())
allAns=[]
for _ in range(t):
n,k=readIntArr()
a=readIntArr()
cumugcd=0
for i in range(1,n):
cumugcd=gcd(cumugcd,abs(a[i]-a[0]))
if abs(k-a[0])%cumugcd==0:
ans='YES'
else:
ans='NO'
allAns.append(ans)
multiLineArrayPrint(allAns)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main()
``` | output | 1 | 99,664 | 22 | 199,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x.
You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l β€ x β€ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them.
Input
The first line contains integer n β the number of queries (1 β€ n β€ 10000).
Each of the following n lines contain two integers li, ri β the arguments for the corresponding query (0 β€ li β€ ri β€ 1018).
Output
For each query print the answer in a separate line.
Examples
Input
3
1 2
2 4
1 10
Output
1
3
7
Note
The binary representations of numbers from 1 to 10 are listed below:
110 = 12
210 = 102
310 = 112
410 = 1002
510 = 1012
610 = 1102
710 = 1112
810 = 10002
910 = 10012
1010 = 10102 | instruction | 0 | 99,779 | 22 | 199,558 |
Tags: bitmasks, constructive algorithms
Correct Solution:
```
for i in range(int(input())):
l,r=map(int,input().split())
while(l|(l+1)<=r): l|=l+1
print(l)
# Made By Mostafa_Khaled
``` | output | 1 | 99,779 | 22 | 199,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x.
You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l β€ x β€ r, and <image> is maximum possible. If there are multiple such numbers find the smallest of them.
Input
The first line contains integer n β the number of queries (1 β€ n β€ 10000).
Each of the following n lines contain two integers li, ri β the arguments for the corresponding query (0 β€ li β€ ri β€ 1018).
Output
For each query print the answer in a separate line.
Examples
Input
3
1 2
2 4
1 10
Output
1
3
7
Note
The binary representations of numbers from 1 to 10 are listed below:
110 = 12
210 = 102
310 = 112
410 = 1002
510 = 1012
610 = 1102
710 = 1112
810 = 10002
910 = 10012
1010 = 10102 | instruction | 0 | 99,783 | 22 | 199,566 |
Tags: bitmasks, constructive algorithms
Correct Solution:
```
"""
pppppppppppppppppppp
ppppp ppppppppppppppppppp
ppppppp ppppppppppppppppppppp
pppppppp pppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppp pppppppp
ppppppppppppppppppppp ppppppp
ppppppppppppppppppp ppppp
pppppppppppppppppppp
"""
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
for _ in range(int(data())):
left, right = sp()
answer = left
for i in range(65):
if not(left & (1 << i)) and (answer + (1 << i) <= right):
answer += (1 << i)
out(answer)
``` | output | 1 | 99,783 | 22 | 199,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,893 | 22 | 199,786 |
Tags: implementation, math
Correct Solution:
```
import sys
import math
def main():
str_list = sys.stdin.readline().strip("\n").split(" ")
number_list = [int(str(i)) for i in str_list]
t = number_list[0]
x = number_list[1]
res = math.ceil(t / x)
if(t % x == 0):
res += 1
res *= x
print(res)
return res
main()
``` | output | 1 | 99,893 | 22 | 199,787 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,894 | 22 | 199,788 |
Tags: implementation, math
Correct Solution:
```
n, k = map(int, input().split())
print((int(n / k) + 1) * k)
``` | output | 1 | 99,894 | 22 | 199,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,895 | 22 | 199,790 |
Tags: implementation, math
Correct Solution:
```
import sys
str_list = sys.stdin.readline().strip("\n").split(" ")
number_list = [int(str(i)) for i in str_list]
t = number_list[0]
x = number_list[1]
y = (t + x)%x
res = t+x - y
print(res)
``` | output | 1 | 99,895 | 22 | 199,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,896 | 22 | 199,792 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin
def parse():
line = stdin.readline().split()
N, K = map( int, line)
quo, rem = int(N/K), N%K
ans = (quo+1)*K
print( ans)
parse()
``` | output | 1 | 99,896 | 22 | 199,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,897 | 22 | 199,794 |
Tags: implementation, math
Correct Solution:
```
n,k=map(int,input().split())
print(n+k-n%k)
``` | output | 1 | 99,897 | 22 | 199,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,898 | 22 | 199,796 |
Tags: implementation, math
Correct Solution:
```
n, k = map(int, input().split())
ans = (n // k + 1) * k
print(ans)
``` | output | 1 | 99,898 | 22 | 199,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,899 | 22 | 199,798 |
Tags: implementation, math
Correct Solution:
```
#A
cin=lambda:map(int,input().split())
n,k=cin()
print((n//k)*k+k)
``` | output | 1 | 99,899 | 22 | 199,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39 | instruction | 0 | 99,900 | 22 | 199,800 |
Tags: implementation, math
Correct Solution:
```
s= input("").split(' ')
n= int(s[0])
k= int(s[1])
if k>n:
print(str(k))
else:
a=int(n/k)
counter= k*a
while True:
if(counter>n):
print(str(counter))
break
counter=counter+k
``` | output | 1 | 99,900 | 22 | 199,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
[n, k] = map(int, input() .split())
a = int(n//k + 1)
b = int(k*a)
print(b)
``` | instruction | 0 | 99,901 | 22 | 199,802 |
Yes | output | 1 | 99,901 | 22 | 199,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
from math import ceil
c, d=map(int, input().split())
if c%d==0:
g=ceil(c/d)+1
print(d*g)
else:
g=ceil(c/d)
print(d*g)
``` | instruction | 0 | 99,902 | 22 | 199,804 |
Yes | output | 1 | 99,902 | 22 | 199,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
t,x = list(map(int,input().strip().split(' ')))
tmp = (t//x)*x
res = tmp if tmp > t else tmp+x
print(res)
``` | instruction | 0 | 99,903 | 22 | 199,806 |
Yes | output | 1 | 99,903 | 22 | 199,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
#import sys
#sys.stdin = open('in', 'r')
#n = int(input())
#a = [int(x) for x in input().split()]
n,k = map(int, input().split())
print(k*(n//k) + k)
``` | instruction | 0 | 99,904 | 22 | 199,808 |
Yes | output | 1 | 99,904 | 22 | 199,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
from math import sqrt
n , k = map(int,input().split())
i = n
while i <= sqrt(10**9):
if i > n and i % k == 0 :
print(i)
break
i +=1
``` | instruction | 0 | 99,905 | 22 | 199,810 |
No | output | 1 | 99,905 | 22 | 199,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
from math import ceil
c, d=map(int, input().split())
if c%d==0:
g=ceil(c/d)+1
print(d*g)
else:
g=ceil(c/d)+1
print(d*g)
``` | instruction | 0 | 99,906 | 22 | 199,812 |
No | output | 1 | 99,906 | 22 | 199,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
n, k = map(int, input().split())
print((n + 1) // k * k)
``` | instruction | 0 | 99,907 | 22 | 199,814 |
No | output | 1 | 99,907 | 22 | 199,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.
Input
The only line contains two integers n and k (1 β€ n, k β€ 109).
Output
Print the smallest integer x > n, so it is divisible by the number k.
Examples
Input
5 3
Output
6
Input
25 13
Output
26
Input
26 13
Output
39
Submitted Solution:
```
a,b=map(int,input().split())
m=list(range(b,a+b,b))
if a==b:
print(a*2)
elif a<b:
print(b)
else:
for i in m:
if i>a :
print(i)
exit(0)
``` | instruction | 0 | 99,908 | 22 | 199,816 |
No | output | 1 | 99,908 | 22 | 199,817 |
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])
``` | instruction | 0 | 100,106 | 22 | 200,212 |
No | output | 1 | 100,106 | 22 | 200,213 |
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 | instruction | 0 | 100,125 | 22 | 200,250 |
"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
``` | output | 1 | 100,125 | 22 | 200,251 |
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 | instruction | 0 | 100,129 | 22 | 200,258 |
"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)
``` | output | 1 | 100,129 | 22 | 200,259 |
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 | instruction | 0 | 100,130 | 22 | 200,260 |
"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)
``` | output | 1 | 100,130 | 22 | 200,261 |
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
``` | instruction | 0 | 100,133 | 22 | 200,266 |
Yes | output | 1 | 100,133 | 22 | 200,267 |
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
``` | instruction | 0 | 100,134 | 22 | 200,268 |
Yes | output | 1 | 100,134 | 22 | 200,269 |
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)
``` | instruction | 0 | 100,137 | 22 | 200,274 |
No | output | 1 | 100,137 | 22 | 200,275 |
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)
``` | instruction | 0 | 100,139 | 22 | 200,278 |
No | output | 1 | 100,139 | 22 | 200,279 |
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()
``` | instruction | 0 | 100,140 | 22 | 200,280 |
No | output | 1 | 100,140 | 22 | 200,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,493 | 22 | 200,986 |
Tags: combinatorics, math, number theory
Correct Solution:
```
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now----------------------------------------------------
n, k = map(int, input().split())
M = 998244353
ft = [1]*500001
for i in range(1, 500001):
ft[i] = (ft[i-1]*i) % M
ans = 0
for i in range(1, n+1):
a = (n//i) - 1
b = k-1
if (a < b): break
ans = ans + (ft[a])*pow(ft[b]*ft[a-b], M-2, M)
ans = ans % M
print(ans)
``` | output | 1 | 100,493 | 22 | 200,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,494 | 22 | 200,988 |
Tags: combinatorics, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
p=998244353
pri=p
fac=[1 for i in range((5*(10**5))+1)]
for i in range(2,len(fac)):
fac[i]=(fac[i-1]*(i%pri))%pri
def modi(x):
return (pow(x,p-2,p))%p;
def ncr(n,r):
x=(fac[n]*((modi(fac[r])%p)*(modi(fac[n-r])%p))%p)%p
return x;
n,k=map(int,input().split())
total=0
for i in range(1,n//k+1):
total+=ncr(n//i-1,k-1)
total%=pri
print(total)
``` | output | 1 | 100,494 | 22 | 200,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,495 | 22 | 200,990 |
Tags: combinatorics, math, number theory
Correct Solution:
```
n,k=map(int,input().split())
ans=0
fact=[0]*(n+1)
invfact=[0]*(n+1)
fact[0]=1
invfact[0]=1
m=998244353
ans=0
for j in range(1,n+1):
fact[j]=(fact[j-1]*j)%m
invfact[j]=pow(fact[j],m-2,m)
def comb( n, k):
if n<k or n<0 :
return 0
return (fact[n]*invfact[n-k]*invfact[k])%m
for j in range(1,n+1):
ans+=comb((n//j)-1,k-1)
ans=ans%m
print(ans)
``` | output | 1 | 100,495 | 22 | 200,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,497 | 22 | 200,994 |
Tags: combinatorics, math, number theory
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(x) for x in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def bo(i):
return ord(i)-ord('a')
def mo(a, m) :
m0 = m
y = 0
x = 1
if (m == 1) :
return 0
while (a > 1) :
q = a // m
t = m
m = a % m
a = t
t = y
y = x - q * y
x = t
if (x < 0) :
x = x + m0
return x
mod=998244353
n,k=mi()
ft=[1 for i in range(n+1)]
mt=[1 for i in range(n+1)]
for i in range(1,n+1):
ft[i]=(ft[i-1]*i)%mod
mt[i]= mo(ft[i],mod)
if n<k:
print(0)
exit(0)
if k==1:
print(n)
exit(0)
c=0
for i in range(1,n) :
if n//i<k:
break
c=(c%mod+ft[n//i-1]*mt[k-1]*mt[n//i-k])%mod
print(c)
``` | output | 1 | 100,497 | 22 | 200,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,498 | 22 | 200,996 |
Tags: combinatorics, math, number theory
Correct Solution:
```
from math import sqrt
n,k = [int(x) for x in input().split()]
if n//k > 500:
modNum = 998244353
def choose(n,k):
if k >= n:
return 1
base = 1
div = min(k,n-k)
for x in range(1,div+1):
base *= n
base //= x
n -= 1
return base%998244353
out = 0
for a in range(n//k):
start = a+1
amt = n//start
out += choose(amt-1, k-1)
print(out%modNum)
else:
primes = [2]
for num in range(3,n+1,2):
sq = sqrt(num)
valid = True
for a in primes:
if num%a == 0:
valid = False
break
if a > sq:
break
if valid:
primes.append(num)
def factored(n,primes):
counts = {x:0 for x in primes}
for a in primes:
base = a
while base <= n:
counts[a] += n//base
base *= a
return counts
modNum = 998244353
out = 0
for a in range(n//k):
start = a+1
amt = n//start
first = factored(amt-1, primes)
second = factored(k-1, primes)
third = factored(amt-k, primes)
thing = {x:first[x]-second[x]-third[x] for x in primes}
base = 1
for x in thing:
for y in range(thing[x]):
base *= x
base = base%998244353
out += base
print(out%modNum)
``` | output | 1 | 100,498 | 22 | 200,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,499 | 22 | 200,998 |
Tags: combinatorics, math, number theory
Correct Solution:
```
"""
NTC here
"""
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
profile = 0
def iin(): return int(input())
def lin(): return list(map(int, input().split()))
N = 5*10**5
factorialInv = [None] * (N + 1)
naturalInv = [None] * (N + 1)
fact = [None] * (N + 1)
def InverseofNumber(p):
naturalInv[0] = naturalInv[1] = 1
for i in range(2, N + 1, 1):
naturalInv[i] = pow(i, p-2, p)
def InverseofFactorial(p):
factorialInv[0] = factorialInv[1] = 1
for i in range(2, N + 1, 1):
factorialInv[i] = (naturalInv[i] *
factorialInv[i - 1]) % p
def factorial(p):
fact[0] = 1
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
def ncr(N, R, p):
# n C r = n!*inverse(r!)*inverse((n-r)!)
ans = ((fact[N] * factorialInv[R])% p *
factorialInv[N - R])% p
# print("INNER", N, 'c', R, ans)
return ans
def main():
n, k = lin()
if k==1:
print(n)
return
if k==n:
print(1)
elif k>n:
print(0)
else:
md = 998244353
InverseofNumber(md)
InverseofFactorial(md)
factorial(md)
ans = ncr(n-1, k-1, md)
ans %= md
for i in range(2, n+1):
x1 = n//i
if x1>0 and x1>=k and k-1:
# print("INNER2", i)
ans += ncr(x1-1, k-1, md)
ans %= md
print(ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
if profile:
import cProfile
cProfile.run('main()')
else:
main()
``` | output | 1 | 100,499 | 22 | 200,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,500 | 22 | 201,000 |
Tags: combinatorics, math, number theory
Correct Solution:
```
n,k=map(int,input().split())
mod = 998244353
f = [1]
for i in range(n):
f += [f[-1]*(i+1) % mod]
def comb(a, b):
return f[a]*pow(f[b], mod-2, mod)*pow(f[a-b], mod-2, mod) % mod
ans=0
for i in range(1,n+1):
m=(n//i)-1
if m<(k-1):break
ans+=comb(m,k-1)
ans%=mod
print(ans)
``` | output | 1 | 100,500 | 22 | 201,001 |
Provide tags and a correct Python 2 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n.
Input
The only line contains two integers n and k (1 β€ n, k β€ 5 β
10^5).
Output
Print one integer β the number of stable arrays [a_1, a_2, ..., a_k] such that 1 β€ a_1 < a_2 < ... < a_k β€ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,501 | 22 | 201,002 |
Tags: combinatorics, math, number theory
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=998244353
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
def inv(x):
return pow(x,mod-2,mod)
n,k=li()
fac=[1]*(n+1)
for i in range(2,n+1):
fac[i]=(fac[i-1]*i)%mod
ans=0
for i in range(1,n+1):
tp=n/i
if tp<k:
break
temp=(fac[tp-1]*inv(fac[k-1]))%mod
temp=(temp*inv(fac[tp-k]))%mod
ans=(ans+temp)%mod
pn(ans)
``` | output | 1 | 100,501 | 22 | 201,003 |
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