本文探讨了一道有趣的编程题目，要求不超时地计算N的N次方的个位数。作者分享了初始的简单思路，即直接计算并取模，但发现这种方法会导致超时。接着，作者引用了百度论坛上的建议，将0到9的数字分成三类，通过分析乘法规律来减少计算次数。总结出个位数为4的数自乘会停在6，个位数为9的数会停在9，而2、3、7、8会循环变化。文章提供了优化算法的思考过程。
Rightmost Digit
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 13 Accepted Submission(s) : 5
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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).
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
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