快速冪 & 求乘法逆元

快速冪求a^b%c

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int n, a, b, c;

ll quick_mul(int a, int b, int c)
{
    ll ans = 1 % c;
    while (b)
    {
        if (b & 1)
            ans = (1ll * ans * a) % c;
        b >>= 1;
        a = (1ll * a * a) % c;
    }
    return ans;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n;
    while (n--)
    {
        cin >> a >> b >> c;
        cout << quick_mul(a, b, c) << endl;
    }
    return 0;
}

求乘法逆元

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int n, a, p;

ll quick_mul(int a, int b, int c)
{
    ll ans = 1 % c;
    while (b)
    {
        if (b & 1)
            ans = (1ll * ans * a) % c;
        b >>= 1;
        a = (1ll * a * a) % c;
    }
    return ans;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n;
    while (n--)
    {
        cin >> a >> p;
        int ans = quick_mul(a, p - 2, p);
        if (a % p == 0)
            cout << "impossible" << endl;
        else
            cout << ans << endl;
    }
    return 0;
}