HDU 1061 Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58270    Accepted Submission(s): 22084

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.
就是找一下规律，思路是这样的：
末位数      相乘后的末位数 
1               1 
2               4   8   6   2 
3               9   7   1   3 
4               6   4 
5               5 
6               6 
7               9   3   1   7 
8               4   2   6   8 
9               1   9
0               0 
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	int T,n,sd,digit;
	scanf("%d",&T);
	while(T--)
	{
		if(scanf("%d",&n) != EOF)
		{
			sd = n%10;
			if(sd==0 || sd==1 || sd==5 || sd==6 || sd==9)
				digit = sd;
			else if(sd == 2 || sd == 8)
			{
				if(n%4 == 0)
					sd = 6;
				else
					sd = 4;
			}
			else if(sd == 3)
			{
				if(n%4 == 1)
					sd = 3;
				else
					sd = 7;
			}
			else if(sd == 4)
			{
					sd = 6;
			}
			else{
				if(n%4 == 1)
					sd = 7;
				else
					sd = 3;
			}
			printf("%d\n",sd);
		}
	}
	return 0;
}