eg:

n=10，10%2==0,

10/2=5, 5%2==1,4*

5/2==2, 2%2==0,4*256

0/2=0,  1024

递归算法

#include<iostream>
using namespace std;
long long quick_pow(int b,int e)
{
    if(b==0)
        return 0;
    if(e==0)
        return 1;
    if(e%2==0)
    {
        int temp=quick_pow(b,e/2);
        return temp*temp;
    }
    if(e%2!=0)
        return b*quick_pow(b,e-1);
}
int main()
{
    long long base,exponet;//底数，指数
    cin>>base>>exponet;
    cout<<quick_pow(base,exponet);
    return 0;
}

非递归算法

#include<iostream>
using namespace std;
long long quick_pow(int base,int exponent)
{
    long long result=1;
    while(exponent>0)
    {
        if(exponent%2!=0)
            result*=base;
        base*=base;
        exponent/=2;
    }
    return result;
}
int main()
{
    long long base,exponent;
    cin>>base>>exponent;
    cout<<quick_pow(base,exponent);
    return 0;
}