杭电OJ题 1393 Weird Clock解题报告

最新推荐文章于 2019-06-25 23:49:15 发布

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本文探讨了一个奇特的钟表问题，钟表仅有一分钟指针，并且需要特殊硬币才能移动。每种硬币都有一个数值d，表示在当前时间的基础上顺时针移动d分钟。目标是找到最少的d硬币数量，使得从初始时间回到0点。通过实例输入和输出，解释了解决该问题的方法。

Weird Clock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1771 Accepted Submission(s): 643

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

————————————————————————————————————————————————————————
考虑无法达到题目要求，即输出“Impossible” 的条件是当出现重复出先以前出现过的时间的时候即为无法达到要求，我设置了一个长度为60的数组记录是否当前的时间是否已经出现过。

/****************************
 *Name:Weird Clock.c
 *Tags:ACM water
 *Note:当第二次出现相同结果时表示不可能实现题目要求，输出Impossible
 ****************************/

#include <stdio.h>

int main()
{
      int s, d, t, save[60], i;
      while(scanf("%d%d", &s, &d) != EOF && (s || d)) {
	    for(i = 0; i < 60; i++) {
		  save[i] = 0;
	    }
	    if(d == 0 && s) {
		  printf("Impossible\n");
		  continue;
	    }
	    t = 0;
	    while(s != 0 && !save[s]) {
		  save[s] = 1;
		  s = (s * (d + 1)) % 60;
		  t++;
	    }
	    if(s == 0) {
		  printf("%d\n", t);
	    } else {
		  printf("Impossible\n");
	    }
      }
      return 0;
}