求m^n的后三位数（时间复杂度O(logn)）

Problem description:

Given two integers m and n, find the integer represented by the last three digits of $m^n$.

For example, if $m=3$ and $n=9$, the returned number should be $683(3^9=19683)$.

Please give an algorithm with $O(logn)$ complexity.

Note:

In a computer program, when $m$ or $n$ is large, exponential calculations may cause data overflow. Your algorithm should be able to avoid this problem.

Input:

Line 1: two integers $m$ $n$, split by space

Output:

Line 1: one integer

Sample Input 1:

3 9

Sample Output 1:

683

Sample Input 2:

2 10

Sample Output 2:

24

Solution:

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

int power(int m,int n)
{
    
    
	int lastthree;
	if (n==0)
		return 1;
    else if(n==1) 
        return m%1000;
    else if(n%2==1){
    
    
     	lastthree = (m%1000)*power(m,(n-1)/2)*power(m,(n-1)/2);
        return lastthree%1000;
    }
	else if(n%2==0){
    
    
	 	lastthree = power(m,n/2)*power(m,n/2);
        return lastthree%1000;
    }
}

int main() 
{
    
    
	int m, n;
	
	scanf("%d%d",&m,&n);
	
    cout << abs(power(m, n)) << endl;
	
	return 0;
}

注意：

1. 要求输出$m^n$的后三位，每次只需要计算$m$的后三位即可，因此使用$\%1000$求每次运算的后三位数。
2. 采用“分治”（divide and conquer）思想，采用递归，来满足$O(logn)$的时间复杂度。
3. 输出注意取绝对值。
4. 注意$n=0$的情况。