[转]组合数取模 Lucas定理
对于C(n, m) mod p。这里的n,m,p(p为素数)都很大的情况。就不能再用C(n, m) = C(n - 1,m) + C(n - 1, m - 1)的公式递推了。
这里用到Lusac定理
For non-negative integers m and n and a prime p, the following congruence relation holds:
where
and
are the base p expansions of m and n respectively.
对于单独的C(ni, mi) mod p,已知C(n, m) mod p = n!/(m!(n - m)!) mod p。显然除法取模,这里要用到m!(n-m)!的逆元。
根据费马小定理:
已知(a, p) = 1,则 ap-1 ≡ 1 (mod p), 所以 a*ap-2 ≡ 1 (mod p)。
也就是 (m!(n-m)!)的逆元为 (m!(n-m)!)p-2 ;
代码:
#include <iostream>
using namespace std;
typedef long long LL;
LL exp_mod(LL a, LL b, LL p) { LL res = 1; while(b != 0) { if(b&1) res = (res * a) % p; a = (a*a) % p; b >>= 1; } return res; } LL Comb(LL a, LL b, LL p) { if(a < b) return 0; if(a == b) return 1; if(b > a - b) b = a - b; LL ans = 1, ca = 1, cb = 1; for(LL i = 0; i < b; ++i) { ca = (ca * (a - i))%p; cb = (cb * (b - i))%p; } ans = (ca*exp_mod(cb, p - 2, p)) % p; return ans; } LL Lucas(int n, int m, int p) { LL ans = 1; while(n&&m&&ans) { ans = (ans*Comb(n%p, m%p, p)) % p; n /= p; m /= p; } return ans; } int main() { Read(); int n, m, p; while(~scanf("%d%d%d", &n, &m, &p)) { printf("%lld\n", Lucas(n, m, p)); } return 0; }
如果这是你所爱的,就不要让自己后悔~~~