HDOJ 1393 Weird Clock(明确题意就简单了)

Posted 2020-10-03 wzjhoutai

Problem Description
A weird clock marked from 0 to 59 has only a minute hand. It won’t move until a special coin is thrown into its box. There are different kinds of coins as your options. However once you make your choice, you cannot use any other kind. There are infinite number of coins of each kind, each marked with a number d ( 1 <= d <= 1000 ), meaning that this coin will make the minute hand move d times clockwise the current time. For example, if the current time is 45, and d = 2. Then the minute hand will move clockwise 90 minutes and will be pointing to 15.

Now you are given the initial time s ( 1 <= s <= 59 ) and the coin’s type d. Write a program to find the minimum number of d-coins needed to turn the minute hand back to 0.

Input
There are several tests. Each test occupies a line containing two positive integers s and d.

The input is finished by a line containing 0 0.

Output
For each test print in a single line the minimum number of coins needed. If it is impossible to turn the hand back to 0, output “Impossible”.

Sample Input
30 1
0 0

Sample Output
1

明确题意之后就简单了，
思路：一个钟面仅仅有一根分针。对于一个数字d，把钟面上的分针指向的时间s往后拨s的d倍。问给定d，反复这种操作多少次能回拨到0。

若不能则输出Impossible。

注意：此处的s是不断更新的。！！。！

。！！！！！
由于钟面仅仅有60分钟。所以最多不会超过60次，直接暴力就可。

import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        while(sc.hasNext()){
            int s= sc.nextInt();
            int d = sc.nextInt();
            if(s==0&&d==0){
                return ;
            }
            int num=s;
            for(int i=1;i<65;i++){
                num = (num+(num*d))%60;
                if(num==0){
                    System.out.println(i);
                    break;
                }
                if(i>63){
                    System.out.println("Impossible");
                    break;
                }
            }
        }

    }

}