标签:
题意: 大组合数取余 (素数连乘)
思路:
对于答案 X
X % pi = ai === C(m,n) % pi;
然后就是用孙子定理求出X, ai 用 卢卡斯定理求得
中间 LL * LL 会爆, 运用按位乘法
对于 m * n % K, 把 m 看成 二进制形式的多项式, 拆开和 n 相乘, 再取余
#include<bits/stdc++.h> using namespace std; const int maxn = 1e5 + 131; typedef long long LL; LL Pow_Mod(LL a, LL b, LL p) { LL ret = 1; while(b) { if(b & 1) ret = (ret * a) % p; a = a * a % p; b >>= 1; } return ret; } void Exgcd(LL a, LL b, LL& d, LL& x, LL& y) { if(b == 0) { d = a, x = 1, y = 0; } else { Exgcd(b,a%b,d,y,x); y -= x * (a / b); } } ///////////////////////////// Lucas LL Fac[maxn], Inv[maxn]; void Init(LL n) { Fac[0] = 1; for(LL i = 1; i < n; ++i) Fac[i] = Fac[i-1] * i % n; Inv[n-1] = Pow_Mod(Fac[n-1], n-2, n); for(LL i = n-2; i >= 0; --i) Inv[i] = Inv[i+1] * (i + 1) % n; } LL C(LL m, LL n, LL p) { if(n > m || m < 0 || n < 0) return 0; return (Fac[m] * Inv[n]) % p * Inv[m-n] % p; } LL Lucas(LL m, LL n, LL p) { if(n == 0) return 1; return Lucas(m/p, n/p, p) * C(m%p, n%p, p) % p; } ////////////////////////////////// LL Ai[maxn], Pi[maxn]; LL mul(LL a, LL b, LL p) { a = (a % p + p) % p; b = (b % p + p) % p; LL ret = 0; while(b) { if(b & 1) ret = (ret + a) % p; b >>= 1; a <<= 1; a %= p; } return ret; } LL China(int n, LL *a, LL *m) { LL x, y, d, M = 1; LL ret = 0; for(int i = 1; i <= n; ++i) M = M * m[i]; for(int i = 1; i <= n; ++i) { LL w = M / m[i]; //y = Pow_Mod(w, m[i]-2, m[i]); WA Exgcd(m[i],w,d,d,y) ; ret = (ret + mul(a[i], mul(y, w, M), M)) % M; } return ret; } int main() { int t; scanf("%d",&t); while(t--) { LL m, n; int k; scanf("%lld%lld%d", &m, &n, &k); for(int i = 1; i <= k; ++i) { scanf("%llu",&Pi[i]); Init(Pi[i]); Ai[i] = Lucas(m,n,Pi[i]); } LL ans = China(k,Ai,Pi); printf("%lld\n",ans); } }
标签:
原文地址:http://www.cnblogs.com/aoxuets/p/4808180.html