HDOJ-ACM1097(JAVA) A hard puzzle

Posted 2020-07-26 xiezie

 

这道题就是HDOJ的1061的变形：

1061 ：求n的n次方的个位数 http://www.cnblogs.com/xiezie/p/5596779.html

1097 ：求n的m次方的个位数

 

因此，就不在这里赘述了

以下是JAVA实现：

import java.io.BufferedInputStream;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedInputStream(System.in));
        while(scan.hasNext()){
            int n = scan.nextInt();
            int m = scan.nextInt();
            int rightMostDigitOfN = n%10;
            ArrayList<Integer> rightMostDigits = arrays.get(rightMostDigitOfN);
            int size = rightMostDigits.size();
            if(size == 1){
                System.out.println(rightMostDigitOfN);
            }else{
                System.out.println(rightMostDigits.get((m-1)%size));
            }
        }
        scan.close();
    }

    static ArrayList<ArrayList<Integer>> arrays  = getRightMostDigitArray();

    static ArrayList<ArrayList<Integer>> getRightMostDigitArray(){
        ArrayList<ArrayList<Integer>> arrays = new ArrayList<>();
        for(int i = 0 ; i != 10 ; i ++){
            ArrayList<Integer> integers = new ArrayList<>();
            integers.add(i);
            int n = 2;
            while(true){
                int rightMostDigit = (int) (Math.pow(i, n)%10);
                if(rightMostDigit==i){
                    break;
                }else{
                    integers.add(rightMostDigit);
                }
                n++;
            }
            arrays.add(integers);
        }
        return arrays;
    }

}