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=;//1e5越界
 const ll M=1e9+;
 using namespace std;
 ll fac[N]={,},inv[N]={,},fi[N]={,};//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=;i<N;i++)
     {
         fac[i]=fac[i-]*i%M;
         inv[i]=(M-M/i)*inv[M%i]%M;
         fi[i]=inv[i]*fi[i-]%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-,m-)<<endl;//求组合数
     }
     return ;
 }

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 + ;
 const LL mod = 1e9 + ;

 void exgcd(LL a, LL b, LL &x, LL &y)    //拓展欧几里得算法
 {
     if(!b) x = , y = ;
     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=; i<=n; i++){
         printf("%lld\n", niYuan(i,p));
     }

 }