HDOJ 1097（阶乘尾数，水题）

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16669    Accepted Submission(s): 5936

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

 

AC code：

#include <iostream>
using namespace std;
int main()
{
	int a,b;
	while(scanf("%d%d",&a,&b)!=EOF)
	{
		a%=10;
		if(a==0||a==5||a==6||a==1) printf("%d\n",a);
		else if(a==4||a==9)
		{
			printf("%d\n",b%2==0?a*a%10:a);	
		}
		else
		{
			if(b%4==1) printf("%d\n",a);
			else if(b%4==2) printf("%d\n",a*a%10);
			else if(b%4==3) printf("%d\n",a*a*a%10);
			else printf("%d\n",a*a*a*a%10);	
		}	
	}
	return 0;	
}