Problem Description：

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise. 

Input： 

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30) 

Output： 

For each test case, you should output the a^b's last digit number. 

Sample Input： 

7 66

8 800 

Sample Output：

9

6 

 解题思路：

这道题就是求a的b次方的最后一位数字，典型的快速幂，因为是要求最后一位，所以在进行快速幂之前，我们可以对原数进行取个位数字之后再运算，因为决定最后一位数字的始终就是个位数字，也就是%10。 

程序代码： 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int quickmul(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%10;
		a=(a*a)%10;
		b>>=1;
	}
	return ans%10;
}
int main()
{
	int n,m;
	while(cin>>n>>m)
	{
		cout<<quickmul(n%10,m)<<endl;//关键n%10
	}
	return 0;
}