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HDU1393_Weird Clock【水题】

2024-08-04 19:29:59   220人阅读

Weird Clock


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2636    Accepted Submission(s): 943

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
 
Author

DU, Peng


题目大意：有一块表，上边总共有0~59个刻度，每一刻度代表一分钟，

给你当前的分钟s，和一个数d，d表示一次能顺时针转动d次s分钟。即

一次能转d*s分钟，问：表能否转到0上边。若能转到0上，输出满足要

求最小的次数；否则输出"Impossible"。

思路：很奇葩的题目要求，按照要求求解即可，每次记得对6取余，因为

表盘是以60为环循环的。


#include<stdio.h>
#include<string.h>

int main()
{
    int s,d;
    while(~scanf("%d%d",&s,&d) && (s||d))
    {
        if(s >= 60)
            s %= 60;
        int count = 0,flag = 1;
        while(s)
        {
            s += s*d;
            s %= 60;
            if(count > 1000)
            {
                flag = 0;
                break;
            }
            count++;
        }
        if(flag)
            printf("%d\n",count);
        else
            printf("Impossible\n");
    }
    return 0;
}



HDU1393_Weird Clock【水题】


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