题目描述

利用快速幂 费马小定理求组合数

C++ 代码

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 100010, mod = 1e9 + 7;
//利用快速幂与费马小定理



int fact[N], infact[N];


int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}


int main()
{
    fact[0] = infact[0] = 1;
    for (int i = 1; i < N; i ++ )
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;//利用快速幂求逆元
    }


    int n;
    scanf("%d", &n);
    while (n -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%d\n", (LL)fact[a] * infact[b] % mod * infact[a - b] % mod);//随时取模，防止溢出
    }

    return 0;
}