nyoj473 A^B Problem （快速幂）

• 题目473
• 题目信息
• 运行结果
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描述

Give you two numbers a and b,how to know the a^b‘s the last digit number.It looks so easy,but everybody is too lazy to slove this problem,so they remit to you who is wise.

输入

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

输出

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

样例输入

7 66
8 800

样例输出

9
6

提示

There is no such case in which a = 0 && b = 0。

来源

hdu

上传者

ACM_丁国强

 
#include <stdio.h>
int main()
{
	int a,b,_a,s;
	while(scanf("%d %d",&a,&b)!=EOF)
	{
		if(a==0&&b==0)
		break;
		s=1;
		while(b)
		{
			if(s>=10)
			s=s%10;
			if(a>=10)
			a=a%10;
			if(b%2==1)
			s=s*a;
			a=a*a;
			b=b/2;	
		}
		if(s>=10)
		s=s%10;
		printf("%d\n",s);
	}
	return 0;
}                

nyoj473 A^B Problem （快速幂）

原文：http://blog.csdn.net/su20145104009/article/details/46770453