题解

#include <iostream>
using namespace std;
typedef long long ll;
long long n;
long long mi(ll a, ll b, ll p){
    ll ans = 1, cheng = a, t = b;
    while(t){
        if(t % 2 == 1) ans *= cheng;
        ans %= p;
        cheng = (cheng * cheng) % p;
        t /= 2;
    }
    return ans;
}
ll niyuan(ll i, ll p){
    ll b = p - 2;
    ll cheng = i, ans = 1;
    while(b){
        if(b % 2 == 1){
            ans = ans * cheng % p;
        }
        cheng = cheng * cheng % p;
        b /= 2;
    }
    return ans;
}
int main(){
    ll p;
    cin >> n >> p;
    for(int i=1; i<=n; i++){
        cout << niyuan(i, p) << " ";
    }
    return 0;
}