Rightmost Digit

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.

题目的大意是：
比如输入应该4，就4*4*4*4=256，如果是3，就3*3*3=27，
这里只取最后一位，256就取6,27就是取7.

说一下解题思路：
这道题目刚刚拿到手，我二话不说直接用了粗暴的方式for循环，麻蛋的，这里的N取值(1<=N<=1,000,000,000).很明显就是不行的，必须找规律，看懂下面的那个数组是怎么出来的就行了。

#include<stdio.h>#include<string.h>//4,8,6,2  9,7,1,3  6,4,6,4  9,3,1,7  4,2,6,8  1,9,1,9int a[10][4]={{0,0,0,0},{1,1,1,1},{6,2,4,8},{1,3,9,7},{6,4,6,4},{5,5,5,5},{6,6,6,6},{1,7,9,3},{6,8,4,2},{1,9,1,9}};int main(){    int T,N;    scanf("%d",&T);    while(T--){        scanf("%d",&N);        printf("%d\n",a[N%10][N%4]);    }    return 0;}