求M的N次方的最后三位

#include <stdlib.h>
#include <iostream>
#include "oj.h"
using namespace std;
//思路：每次相乘，仅保留积的最后三位
unsigned int  GetLastThreeNumOfResult (unsigned int  m, unsigned int  n)
{
	unsigned int result = m %1000;
	
	for (int i =1; i < n ;++i)
	{
		result = (result*m)%1000;
	}
	return result;
}

int main()
{
	int m,n;
	while (cin>>m>>n)
	{
		cout<<GetLastThreeNumOfResult(m,n)<<endl;
	}
	return 0;
}