0\le a,b <2^{31},2\le p < 2^{31} (快速幂/取模公式) P1226

1\le a\le10^9,1\le b\le10^{2\times10^7},1\le m\le 10^8 (欧拉函数/拓展欧拉定理) P5091

20分代码：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
signed main()
{
    ios::sync_with_stdio(false), cin.tie(0);
    ll a, b = 0, r = 1, p = 1e9 + 7;
    cin >> a;
    ll t = p, phi = p;
    for (ll i = 2; i * i <= p; ++i)
    {
        if (t % i == 0)
        {
            phi = phi - phi / i;
            while (t % i == 0)
            {
                t /= i;
            }
        }
    }
    if (t > 1)
    {
        phi = phi - phi / t;
    }
    char c;
    bool flag = false;
    while (cin >> c)
    {
        b = b * 10 + (c - '0');
        if (b >= phi)
        {
            flag = true;
            b %= phi;
        }
    }
    if (flag)
    {
        b += phi;
    }
    for (; b; b >>= 1, a = a * a % p)
    {
        if (b & 1)
        {
            r = r * a % p;
        }
    }
    cout << r;
    return 0;
}

12分代码：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
signed main()
{
    ios::sync_with_stdio(false), cin.tie(0);
    ll a, b, r = 1, p = 1e9 + 7;
    cin >> a >> b;
    for (; b; b >>= 1, a = a * a % p)
    {
        if (b & 1)
        {
            r = r * a % p;
        }
    }
    cout << r;
    return 0;
}

4分代码：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
signed main()
{
    ios::sync_with_stdio(false), cin.tie(0);
    ll a, b, r = 1, p = 1e9 + 7;
    cin >> a >> b;
    while (b--)
    {
        r *= a;
    }
    cout << r;
    return 0;
}