hdu 1393 Weird Clock

解决奇特时钟问题的最小硬币数量

最新推荐文章于 2017-12-08 15:41:41 发布

原创最新推荐文章于 2017-12-08 15:41:41 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#模拟

ACM_模拟专栏收录该内容

26 篇文章

订阅专栏

本文探讨了一种特殊时钟的工作原理及如何使用特定硬币使其指针回到初始位置的问题。通过模拟和索引思想，解决了在给定初始时间和硬币类型的情况下，最小化所需硬币数量的问题。

题目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=1393

题目描述：

Weird Clock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2639 Accepted Submission(s): 945

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

题意:

判断最短转几个d次能刚好到60刻钟。

题解：

模拟，并用了索引思想，重复了不是60的刻钟永远到不了60.

代码：

#include<stdio.h>
#include<string.h>
bool visited[61] = {false};
int s=0,d=0;
int main()
{
    while(scanf("%d%d",&s,&d)!=EOF&&(s+d)>0)
    {
        memset(visited,false,sizeof(visited));
        visited[60]=true;
        visited[s]=true;
        int cnt=0;
        while(s != 60)
        {
            s=s+s*d;
            cnt++;
            while(s>60) s-=60;
            if(visited[s]) break;
            else visited[s] = true;
        }
        if(s==60) printf("%d\n",cnt);
        else printf("Impossible\n");
    }
    return(0);
}