具体解析见：https://blog.csdn.net/qq_17807067/article/details/127231568

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
 
int n,p;
ll inv[20000528];
 
int main(){
    scanf("%d %d",&n,&p);
    inv[1]=1;//1的逆元是1
    printf("%ld\n",inv[1]);
    for(int i=2;i<=n;i++){
        inv[i]=(p-p/i)*inv[p%i]%p;
        printf("%ld\n",inv[i]);
    }
    return 0;
}