HDU 1097 a hard puzzle

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

簡單來說，就是求前一個數的最後一個數的b次方的問題，很容易就看出來有規律，四個一循環，所以很容易就解決了。

#include<stdio.h>
#include<math.h>
int main()
{
	int a,b,a1,b1,sum;
	while(scanf("%d%d",&a,&b)!=EOF)
	{
		a1=a%10;
		b1=b%4;
		if(b1==0){
			sum=((int)pow(a1,4))%10;
	     	printf("%d\n",sum);
		}
		else{
			sum=((int)pow(a1,b1))%10;
	     	printf("%d\n",sum);
		} 
	}
 }