线性求逆元

大佬链接

[https://www.cnblogs.com/qdscwyy/p/7795368.html]

代码

A[i]=-(p/i)*A[p%i];

#include<iostream>
#include<cstdio>
#define N 3000005
using namespace std;

typedef long long ll;

ll _inv[N];  
void pre(int MOD){  
    _inv[0]=_inv[1]=1;  
    for(int i=2;i<N;i++)
        _inv[i]=((MOD-MOD/i)*_inv[MOD%i])%MOD;  
}    
int main(){  
    int n,p;
    cin>>n>>p;
    pre(p);
    for(int i=1;i<=n;++i)
        printf("%d\n",_inv[i]);
    return 0;  
}