FZU 2020 :组合 【lucas】

Problem Description

给出组合数C(n,m), 表示从n个元素中选出m个元素的方案数。例如C(5,2) = 10, C(4,2) = 6.可是当n,m比较大的时候，C(n,m)很大！于是xiaobo希望你输出 C(n,m) mod p的值！

思路：水题，练一下ｌｕｃａｓ

#include<iostream>
#include<cstdio>
#include <math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#define MOD 1000003
#define maxn 2009
#define LL long long
using namespace std;
LL mpow(LL a,LL n,LL p)
{
        if(n==0)return 1;
        if(n==1)return a%p;
        if(n&1)return (a*mpow(a,n-1,p))%p;
        else
        {
                LL u=mpow(a,n>>1,p)%p;
                return (u*u)%p;
        }
}
LL C(LL n,LL m,LL p)
{
        if(m==0)return 1;
        if(m>n-m)m=n-m;
        LL up=1,down=1;
        for(int i=1;i<=m;i++){
                up=(up*(n-i+1))%p;
                down=(down*i)%p;
        }
        return up*mpow(down,p-2,p)%p;
}
long long lucas(long long n,long long m,long long p)
{
        if(m==0)return 1;
        return C(n%p,m%p,p)*lucas(n/p,m/p,p);
}
int main()
{
        long long m,n,p;
        int t;
        scanf("%d",&t);
        while(t--)
        {
                scanf("%I64d%I64d%I64d",&n,&m,&p);
                printf("%I64d ",lucas(n,m,p));
        }
        return 0;
}