HDOJ 1061 Rightmost Digit（循环问题）

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;
    }

}