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

题意：很简单，就是输出n^n的最后一个数字时什么。

思路：前面有过一个0-9的n次方的题目，HDOJ1097题，那一题中我用代码推出了循环节，这个题目，我用的循环节全为4了.

HDOJ1097题博客链接：http://blog.csdn.net/qq_26525215/article/details/50949847

import java.util.Scanner;

public class Main{

    static long db[][] = new long[10][4];

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        dabiao();

//      System.out.println(db[4][0]);

//      System.out.println(db[4][3]);

//

        long t = sc.nextLong();

        while(t-->0){

            int n = sc.nextInt();

            int f=n%10;

            int m=n%4;

            //System.out.println(m);

            m--;

            if(m<0){

                m=3;

            }

            System.out.println(db[f][m]);

        }

    }

    private static void dabiao() {

        for(int i=1;i<=9;i++){

            for(int j=0;j<4;j++){

                db[i][j]=dabiao(i,j);

                //System.out.print(db[i][j]+" ");

            }

            //System.out.println();

        }

    }

    private static long dabiao(long i, long j) {

        long m=i;//4,3

        for(int k=1;k<=j;k++){

            m=(m*i)%10;

        }

        m=m%10;

        return m;

    }

}