結果
問題 | No.391 CODING WAR |
ユーザー |
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提出日時 | 2020-03-09 00:42:21 |
言語 | Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1) |
結果 |
AC
|
実行時間 | 827 ms / 2,000 ms |
コード長 | 2,297 bytes |
コンパイル時間 | 98 ms |
コンパイル使用メモリ | 12,800 KB |
実行使用メモリ | 49,920 KB |
最終ジャッジ日時 | 2024-11-07 20:21:21 |
合計ジャッジ時間 | 7,320 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 16 |
ソースコード
from collections import defaultdictclass Combinatorics:def __init__(self, N, mod):'''Preprocess for calculating binomial coefficients nCr (0 <= r <= n, 0 <= n <= N)over the finite field Z/(mod)Z.Input:N (int): maximum nmod (int): a prime number. The order of the field Z/(mod)Z over which nCr is calculated.'''self.mod = modself.fact = {i: None for i in range(N+1)} # n!self.inverse = {i: None for i in range(1, N+1)} # inverse of n in the field Z/(MOD)Zself.fact_inverse = {i: None for i in range(N+1)} # inverse of n! in the field Z/(MOD)Z# preprocessself.fact[0] = self.fact[1] = 1self.fact_inverse[0] = self.fact_inverse[1] = 1self.inverse[1] = 1for i in range(2, N+1):self.fact[i] = i * self.fact[i-1] % self.modq, r = divmod(self.mod, i)self.inverse[i] = (- (q % self.mod) * self.inverse[r]) % self.modself.fact_inverse[i] = self.inverse[i] * self.fact_inverse[i-1] % self.moddef perm(self, n, r):'''Calculate nPr = n! / (n-r)! % mod'''if n < r or n < 0 or r < 0:return 0else:return (self.fact[n] * self.fact_inverse[n-r]) % self.moddef binom(self, n, r):'''Calculate nCr = n! /(r! (n-r)!) % mod'''if n < r or n < 0 or r < 0:return 0else:return self.fact[n] * (self.fact_inverse[r] * self.fact_inverse[n-r] % self.mod) % self.moddef hom(self, n, r):'''Calculate nHr = {n+r-1}Cr % mod.Assign r objects to one of n classes.Arrangement of r circles and n-1 partitions:o o o | o o | | | o | | | o o | | o'''if n == 0 and r > 0:return 0if n >= 0 and r == 0:return 1return self.binom(n + r - 1, r)N, M = map(int, input().split())MOD = 10**9 + 7com = Combinatorics(M, MOD)ans = 0for i in range(M):if i % 2 == 0:ans = (ans + com.binom(M, i) * pow(M - i, N, MOD)) % MODelse:ans = (ans - com.binom(M, i) * pow(M - i, N, MOD)) % MODprint(ans)