hdu 1097

A hard puzzle

Problem Description
lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b’s the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

Input
There are mutiple test cases. Each test cases consists of two numbers a and b

Output
For each test case, you should output the a^b’s last digit number.

Sample Input
7 66
8 800

Sample Output
9
6

题意：给出a,b两个数，输出a^b结果的最后一位数？（水题，虽简单，但要细心）

0-9阶乘规律：
0: 0
1: 1
2: 2 4 8 6
3: 3 9 7 1
4: 4 6
5: 5
6: 6
7: 7 9 3 1
8: 8 4 2 6
9: 9 1

import java.util.Scanner;public class Main {    static int[][] num = new int[10][10];    static {        //数字规律存储num        for(int i=0;i<=9;i++){            for(int j=0;j<=9;j++){                 if(j==0){                     num[i][j] = 1;continue;                 }                 int k = i*num[i][j-1]%10;                 if(k == num[i][1]) break;                 num[i][j] = k;            }        }    }    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        int a,b;        while(in.hasNext()){            a = in.nextInt();            b = in.nextInt();            a %= 10;            if(a == 0){                System.out.println(0);            }else{                int length = 0;                for(int i=1;num[a][i]!=0;i++)                        length++;                int res = b%length;                if(res==0){                    System.out.println(num[a][length]);                }else{                    System.out.println(num[a][res]);                }            }        }    }}