NYOJA^B Problem

标签：nyojab problem

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。

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int main()
{
	int i,n,m,k,ans,t,vis[15],a[15];
	while(scanf("%d%d",&n,&m)==2){
		k=n%10;ans=1;t=0;
		memset(vis,0,sizeof(vis));
		if(m==0){
			printf("1\n");continue;
		}
		else if(n==0){
			printf("0\n");continue;
		}
		for(i=0;i<m;++i){
			ans=ans*k%10;
			if(vis[ans]==0){
				vis[ans]=1;
				a[t]=ans;
				t++;
			}
			else break;
		}
		if(i==m)printf("%d\n",ans);
		else
		 printf("%d\n",a[(m-1)%t]);
	}
	return 0;
}        

NYOJA^B Problem

标签：nyojab problem

原文地址：http://blog.csdn.net/r1986799047/article/details/43485971