求组合数、求逆元、求阶乘 O(n)

在O(n)的时间内求组合数、求逆元、求阶乘。·。·

#include <iostream>
#include <cstdio>
#define ll long long
const int N=200005;//1e5越界
const ll M=1e9+7;
using namespace std;
ll fac[N]={1,1},inv[N]={1,1},fi[N]={1,1};//fac[i]是i的阶乘，inv[i]是i的逆元,fi[i]是i之前的很多逆元求得阶乘，（将除i取模变为乘i的逆元取模
 
void init()
{
    for(int i=2;i<N;i++)
    {
        fac[i]=fac[i-1]*i%M;
        inv[i]=(M-M/i)*inv[M%i]%M;
        fi[i]=inv[i]*fi[i-1]%M;
    }//递推保存fac阶乘，和fi各个逆元取模相乘
}
ll C(ll a,ll b)
{
    return fac[a]*fi[b]%M*fi[a-b]%M;//C(a,b)=a!/(b!*(a-b)!)
}
int main()
{
    init();
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        cout<<C(m+n-4,m-2)<<endl;//求组合数
    }
    return 0;
}

拓展欧几里得求逆元：

#include<bits/stdc++.h>
#define pi acos(-1)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const LL INF = 0x3f3f3f3f;
const int maxn = 3e6 + 10;
const LL mod = 1e9 + 7;


void exgcd(LL a, LL b, LL &x, LL &y)    //拓展欧几里得算法
{
    if(!b) x = 1, y = 0;
    else
    {
        exgcd(b, a % b, y, x);
        y -= x * (a / b);
    }
}

LL niYuan(LL a, LL b)   //求a对b取模的逆元
{
    LL x, y;
    exgcd(a, b, x, y);
    return (x + b) % b;
}


int main()
{
    LL n, p;
    cin >> n >> p;
    for(int i=1; i<=n; i++){
        printf("%lld
", niYuan(i,p));
    }


}