【快速幂/费马定理/逆元】AcWing876.快速幂求逆元——逆元证明

AcWing876.快速幂求逆元

题解

#include <iostream>

using namespace std;

typedef long long LL;

int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}

int qmi(int a, int b, int p)
{
    int res = 1;
    while(b)
    {
        if(b & 1) res = res * (LL)a % p;
        b >>= 1;
        a = a * (LL)a % p;
    }
    return res;
}

int main()
{
    int n, a, p;
    cin >> n;
    while(n -- )
    {
        cin >> a >> p;
        if(gcd(a, p) != 1) cout << "impossible" << endl;
        else    cout << qmi(a, p - 2, p) << endl;
    }
    return 0;
}