結果
| 問題 |
No.1973 Divisor Sequence
|
| コンテスト | |
| ユーザー |
terasa
|
| 提出日時 | 2022-06-11 00:22:07 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 2,276 bytes |
| コンパイル時間 | 148 ms |
| コンパイル使用メモリ | 82,824 KB |
| 実行使用メモリ | 86,884 KB |
| 最終ジャッジ日時 | 2024-09-21 07:30:23 |
| 合計ジャッジ時間 | 4,987 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 11 TLE * 1 -- * 10 |
ソースコード
import sys
import pypyjit
import itertools
import heapq
import math
from collections import deque, defaultdict
import bisect
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 6)
pypyjit.set_param('max_unroll_recursion=-1')
def index_lt(a, x):
'return largest index s.t. A[i] < x or -1 if it does not exist'
return bisect.bisect_left(a, x) - 1
def index_le(a, x):
'return largest index s.t. A[i] <= x or -1 if it does not exist'
return bisect.bisect_right(a, x) - 1
def index_gt(a, x):
'return smallest index s.t. A[i] > x or len(a) if it does not exist'
return bisect.bisect_right(a, x)
def index_ge(a, x):
'return smallest index s.t. A[i] >= x or len(a) if it does not exist'
return bisect.bisect_left(a, x)
class Matpow:
def __init__(self, N, A, p):
self.N = N
self.A = A
self.p = p
self.digit = 60
self.doubling = [None] * self.digit
self.doubling[0] = A
for i in range(1, self.digit):
self.doubling[i] = self.mul(self.doubling[i - 1], self.doubling[i - 1])
def pow(self, n):
E = [[1 if i == j else 0 for j in range(self.N)] for i in range(self.N)]
acc = E
for k in range(self.digit):
if n & (1 << k):
acc = self.mul(acc, self.doubling[k])
return acc
def mul(self, A, B):
C = [[0 for _ in range(self.N)] for _ in range(self.N)]
for i in range(self.N):
for j in range(self.N):
for k in range(self.N):
C[i][j] += A[i][k] * B[k][j]
C[i][j] %= self.p
return C
N, M = map(int, input().split())
mod = 10 ** 9 + 7
def divs(n):
d = []
for i in range(1, n + 1):
if i * i > n:
break
if i * i == n:
d.append(i)
break
if n % i == 0:
d.append(i)
d.append(n // i)
return sorted(d)
D = divs(M)
idx = {}
for i, n in enumerate(D):
idx[n] = i
A = []
for d in D:
a = [0] * len(D)
for n in divs(M // d):
a[idx[n]] = 1
A.append(a)
An = Matpow(len(D), A, mod).pow(N - 1)
ans = 0
for i in range(len(D)):
for j in range(len(D)):
ans += An[i][j]
ans %= mod
print(ans)
terasa