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; } }
