HDU 5446

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);
	}
}