hdu-3037-组合数取模-Lucas定理

http://acm.hdu.edu.cn/showproblem.php?pid=3037

题意很简单 求C（n+m，m）%p，P是小于1e5的素数

n,m《1e18

那么得到

 

     

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iostream>
using namespace std;
  long long mod =1000000007;

long long fast_m(long long a,long long b)
{
    long long x,ans=1;
    x=a;
    while(b)
    {
        if (b&1)
            ans=(ans*x)%mod;
        x=(x*x)%mod;
        b/=2;
    }
    return ans%mod;
}
long long fac[100005];
long long Lucas(long long n,long long m,long long p)
{
    long long ret=1;
    while(n&&m)
    {
        long long a=n%p,b=m%p;
        if(a<b) return 0;
        ret=(ret*fac[a]*fast_m(fac[b]*fac[a-b]%p,p-2))%p;
        n/=p;
        m/=p;
    }
    return ret;
}
int main()
{
    long long i;

    int t;
    cin>>t;
    while(t--)
    {
        long long n,m;
        cin>>n>>m>>mod;
        fac[0]=1;
    for(  i=1; i<=mod; i++)
        fac[i]=fac[i-1]*i%mod;
        printf("%lld\n",Lucas(n+m,m,mod));
    }
    return 0;
}