HDOJ1097 A hard puzzle

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46236    Accepted Submission(s): 16971

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(0<a,b<=2^30)

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

计算A的B次方的最后一个数字。

这种题目通常都是有规律的，这里就是以4为一个周期。

但是实际操作中B的计算次数应该是B%=4+3。

import java.util.Scanner;

public class Main{
  private static Scanner scanner;

  public static void main(String[] args) {
    scanner = new Scanner(System.in);
    while (scanner.hasNext()) {
      // n^m的最后一位
      int n = scanner.nextInt();
      int m = scanner.nextInt();
      n %= 10;
      m = m%4 +3;
      int res = n;
      for (int i = 0; i < m; i++) {
        res *= n;
        res %= 10;
      }
      System.out.println(res);
    }
  }
}