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2014年09月05日 ⁄ 综合 ⁄ 共 1052字 ⁄ 字号小中大 ⁄ 评论关闭

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16539 Accepted Submission(s): 6355

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

2
3
4

Sample Output

7
6

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

Author

Ignatius.L

这是我用递归写的此题，代码如下：

#include <stdio.h>
int main()
{
	int f(int n,int s);
    int n,m,s,x,j;
    scanf("%d",&m);
    while(m--)
    {
		x=0;
       scanf("%d",&n);
	   s=n%10;
	   j=f(n,s);
	   printf("%d\n",j);
    }
    return 0;
}
   
    int f(int n,int s)
	{
		int j;
		if(n==1)
		{
			return (s);
		}else if(n==2)
		{
			return (s*s%10);
		}else
		{
			if(n%2==0)
			{
				j=f(n/2,s);
				j=j*j;
				j=j%10;
			}else
			{
				j=f(n/2,s);
				j=j*j*s;
				j=j%10;
			}
		}
		return (j);
	}