HDU 5446

来源：互联网发布：mysql从excel导入数据编辑：程序博客网时间：2024/06/18 05:58

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); }}///////////////////////////// LucasLL 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]);  WAExgcd(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);}}

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