HDOJ 1061 Rightmost Digit（循环问题）

来源：互联网发布：cdn网络加速编辑：程序博客网时间：2024/05/21 03:54

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;    }}

0 0