hdu 5446 lucas + 中国剩余定理 + 快速乘 （快速乘板子，中国剩余定理板子，lucas最新板子）

On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere deep in the cave, she found a treasure chest with a combination lock and some numbers on it. After quite a research, the mathematician found out that the correct combination to the lock would be obtained by calculating how many ways are there to pick m different apples among n of them and modulo it with M. M is the product of several different primes.
Input
On the first line there is an integer T(T≤20) representing the number of test cases.

Each test case starts with three integers n,m,k(1≤m≤n≤1018,1≤k≤10) on a line where k is the number of primes. Following on the next line are k different primes p1,...,pk. It is guaranteed that M=p1⋅p2⋅⋅⋅pk≤1018 and pi≤105 for every i∈{1,...,k}.
Output
For each test case output the correct combination on a line.
Sample Input
1
9 5 2
3 5
Sample Output
6

题意：M=p1*p2*…pk；求C（n，m）%M，pi小于10^5，n,m,M都是小于10^18。 pi为质数
题解：首先M=x*pi(1<=i<=k)
M不一定是质数 所以只能用Lucas定理求k次 C（n，m）%Pi得到 B[i];
最后会得到一个同余方程组
x≡B[0](mod p[0])
x≡B[1](mod p[1])
x≡B[2](mod p[2])
……
解这个同余方程组 用中国剩余定理
但ll*ll都会爆所以用快速乘

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5;
ll prim[N+100],B[N+100];
ll fac[N+100];


void init(ll p)
{
    fac[0]=1;
    for(int i=1;i<=p;i++)
        fac[i]=(fac[i-1]*i)%p;
}

ll qpow(ll a,ll b,ll p)
{
    a%=p;
    ll res=1;
    while(b)
    {
        if(b&1) res=res*a,res%=p;
        a*=a;
        a%=p;
        b>>=1;
    }
    return res;
}

ll C(ll n,ll m,ll p)
{
    if(m>n) return 0;
    return fac[n]*qpow(fac[m],p-2,p)%p*qpow(fac[n-m],p-2,p)%p;
}

ll lucas(ll n,ll m,ll p)
{
    if(m==0) return 1;
    else return (C(n%p,m%p,p)*lucas(n/p,m/p,p))%p;
}

ll exgcd(ll a,ll b,ll &x,ll &y)
{
   ll d;
   if(b==0) {x=1;y=0;return a;}
   d=exgcd(b,a%b,y,x);
   y-=a/b*x;
   return d;
}

ll fast_multi(ll m, ll n, ll mod)//快速乘法 
{
    ll ans = 0;//注意初始化是0，不是1 
    while (n)
    {
        if (n & 1)
            ans += m;
        m = (m + m) % mod;//和快速幂一样，只不过这里是加 
        m %= mod;//取模，不要超出范围 
        ans %= mod;
        n >>= 1;
    }
    return ans;
}


ll CTF(ll b[], ll w[], ll len)
{
   ll i,d,x,y,m,n,ret;
   ret=0;n=1;
   for(i=0;i<len;i++) n*=w[i];
   for(i=0;i<len;i++)
   {
       m=n/w[i];
       d=exgcd(w[i],m,x,y);
       ll re=fast_multi(y,m,n);
       re=fast_multi(re,b[i],n);
       // ret=(ret+(y*m%n)*b[i])%n;
       ret+=re;re%=n;
   }
   return (n+ret%n)%n;
}




int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll n,m;int k;
        scanf("%lld%lld%d",&n,&m,&k);
        for(int i=0;i<k;i++)
        {
           scanf("%lld",&prim[i]);
           init(prim[i]);
           B[i]=lucas(n,m,prim[i]);
        }
        printf("%lld\n",CTF(B,prim,k));
    }
}