結果
| 問題 |
No.1973 Divisor Sequence
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-06-10 21:49:09 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 97 ms / 2,000 ms |
| コード長 | 2,951 bytes |
| コンパイル時間 | 198 ms |
| コンパイル使用メモリ | 82,984 KB |
| 実行使用メモリ | 77,764 KB |
| 最終ジャッジ日時 | 2024-09-21 07:28:57 |
| 合計ジャッジ時間 | 2,479 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 22 |
ソースコード
def isPrimeMR(n):
if n==1:
return 0
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
if n in L:
return 1
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
import sys,random,bisect
from collections import deque,defaultdict,Counter
from heapq import heapify,heappop,heappush
from itertools import cycle, permutations
from math import log,gcd
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
mod = 10**9 + 7
def mat_mul(X,Y):
n,m = len(X),len(Y[0])
res = [[0 for j in range(m)] for i in range(n)]
for i in range(n):
for j in range(m):
for k in range(len(Y)):
res[i][j] += X[i][k] * Y[k][j]
res[i][j] %= mod
return res
N,M = mi()
pf = primeFactor(M)
res = 1
for p in pf:
e = pf[p]
A = [[0 for j in range(e+1)] for i in range(e+1)]
for i in range(e+1):
for j in range(e+1):
if i+j <= e:
A[j][i] = 1
E = [[1] for i in range(e+1)]
tmp = N-1
while tmp:
if tmp&1:
E = mat_mul(A,E)
A = mat_mul(A,A)
tmp >>= 1
s = 0
for i in range(e+1):
s += E[i][0]
s %= mod
res *= s
res %= mod
print(res)