HDU 5446 Unknown Treasure

Problem Description 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  

lucas定理+中国剩余定理

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100005;
LL T,n,m,k,a[maxn],b[maxn],f[maxn];

LL PowMod(LL a,LL b,LL MOD){  
    LL ret=1;  
    a=a%MOD;
    while(b){  
        if(b&1) ret=(ret*a)%MOD;  
        a=(a*a)%MOD;  
        b>>=1;  
    }  
    return ret;  
}  

LL C(LL n,LL m,LL p)
{
    if (n<m) return 0;
    if (n==m) return 1;
    LL ans=1;
    ans=f[n]*PowMod(f[n-m]*f[m],p-2,p)%p;
    return ans;
}

LL lucas(LL n,LL m,LL p){  
    LL ret=1;  
    while(n&&m&&ret){   
        ret=ret*C(n%p,m%p,p)%p;  
        n/=p;  
        m/=p;  
    }  
    return ret;  
}  

void egcd(LL a,LL b,LL&d,LL&x,LL&y)
{
    if (!b){d=a,x=1,y=0;}
    else 
    {
        egcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}

LL lmes() {
    LL M=a[0],R=b[0],x,y,d;
    for (int i=1; i<k; i++) {
       egcd(M,a[i],d,x,y);
       if ((b[i]-R)%d) return -1;
       x=(b[i]-R)/d*x%(a[i]/d);
       R+=x*M;
       M=M/d*a[i];
       R%=M;
    }
    return (R+M)%M;
}

int main()
{
    cin>>T;
    while (T--)
    {
        cin>>n>>m>>k;
        for (int i=0;i<k;i++)
        {
            cin>>a[i]; 
            for (LL j=f[0]=1;j<=a[i];j++) f[j]=(f[j-1]*j)%a[i];
            b[i]=(a[i]+lucas(n,m,a[i]))%a[i];
        }
        cout<<lmes()<<endl;
    }
    return 0;
}