快速幂取余

#include<iostream>
#include<cmath>
using namespace std;

long long f(long long a,long long b,long long n)
{
    long long t,y;
    t=1;
    y=a;
    while (b!=0)
    {
        if (b&1==1)
            t=t*y%n;
        y=y*y%n;
        b=b>>1;
    }
    return t;
}

int PowerMod(int a, int b, int c)
{
    int ans = 1;
    a = a % c;
    while(b>0)
    {
        if(b % 2 == 1)
            ans = (ans * a) % c;
        b = b/2;
        a = (a * a) % c;
    }
    return ans;
}

int main()
{
    long long a,res;
    cin >> a;
    cout << f(a,a,10);
    return 0;
}

借鉴：https://www.cnblogs.com/PegasusWang/archive/2013/03/13/2958150.html