HDOJ 1097（阶乘尾数，水题）

最新推荐文章于 2021-04-01 22:58:00 发布

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文章标签：

#output #numbers #each #input #bt

本文介绍了一个经典的算法问题——快速计算a的b次方结果的最后一位数字。通过分析不同基数的幂次方尾数规律，给出了一种高效算法实现方案。此算法特别针对大整数情况进行了优化。

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;	
}