P3811 【模板】乘法逆元

P3811 【模板】乘法逆元

本题只能用线性筛逆元，其他方法的话超时，不过还是写出来

 

线性筛逆元

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3e6+5;
int inv[maxn];
int main() {
    int n, p; scanf("%d%d",&n,&p);
    inv[1] = 1; puts("1");
    for (int i = 2; i <= n; ++i) {
        inv[i] = (ll)(p-p/i)*inv[p%i]%p;
        printf("%d\n",inv[i]);
    }
    return 0;
}

 

费马小定理+快速幂

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll qpow(ll a, ll b, ll mod) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res*a%mod;
        a = a*a%mod;
        b >>= 1;
    }
    return res;
}
int main() {
    int n, p; scanf("%d%d",&n,&p);
    puts("1");
    for (int i = 2; i <= n; ++i) {
        printf("%lld\n",qpow(i,p-2,p));
    }
    return 0;
}

 

扩展欧几里得

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
void exgcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) x = 1, y = 0;
    else exgcd(b, a%b, y, x), y -= a/b * x;
}
int main() {
    int n, p; scanf("%d%d",&n,&p);
    puts("1");
    for (int i = 2; i <= n; ++i) {
        ll x, y;
        exgcd(i,p,x,y);
        if (x < 0) x += p;
        printf("%lld\n",x);
    }
    return 0;
}