O(N）求组合数

$思路：在数学表达式上n的阶乘的逆元=（n+1)的阶乘的逆元乘以(n+1)$

模板CODE

#include
#include
#define endl '\n'
const int mod = 1e9 + 7, N = 1e5;
using namespace std;
int fact[N+10], invf[N+10];
// C(n,m) = n! /  m!*(n-m)!
//716327852
int qmi(int m, int n = mod-2)
{
	int res = 1;
	while(n)
	{
		if(n & 1) res = 1ll*res*m%mod;
		m = 1ll*m*m%mod;
		n >>= 1;
	}
	return res;
}

int comb(int n, int m)
{
	return 1ll*fact[n]*invf[m]%mod*invf[n-m]%mod;
}

void solve()
{
	int n, m; cin >> n >> m;
	cout << comb(n, m) << endl;
}

int main()
{
	int t; cin >> t;
	fact[0] = invf[0] = 1;
	for(int i = 1; i <= N; i ++)
	{
		fact[i] = 1ll*fact[i-1]*i%mod;
	}
	invf[N] = qmi(fact[N]);
	cout << invf[N] << endl;
	for(int i = N-1; i >= 1; i --)
	{
		invf[i] = 1ll*invf[i+1]*(i+1)%mod;
	}

	while(t --)
	{
		solve();
	}
	return 0;
}