线性求组合数C(n,m)n固定（使用逆元）

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1, mod = 1e9 + 7;
unsigned int C[N], g[N];
int main()
{
    int n, ans = 1;
    scanf("%d", &n);
    C[0] = g[0] = g[1] = 1;
    //逆元
    for (int i = 2; i < n / 2 + 1; ++i)
        g[i] = (1ULL * g[mod % i] * (mod - mod / i)) % mod;
    //(n-m+1)*(n-m+2)*...*(n-m+m)/1*2*...*m
    //上面乘多一个，下面乘多一个，转化为乘逆元
    for (int i = 1; i <= n / 2 + 1; ++i)
        C[i] = (1ULL * ((1ULL * C[i - 1] * (n - i)) % mod) * g[i]) % mod;
    return 0;
}