POJ1006-Biorhythms

Biorhythms

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 126206

 

Accepted: 39900

Description

Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical, emotional, and intellectual cycles, and they have periods of lengths 23, 28, and 33 days, respectively. There is one peak in each period of a cycle. At the peak of a cycle, a person performs at his or her best in the corresponding field (physical, emotional or mental). For example, if it is the mental curve, thought processes will be sharper and concentration will be easier.
Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak.

Input

You will be given a number of cases. The input for each case consists of one line of four integers p, e, i, and d. The values p, e, and i are the number of days from the beginning of the current year at which the physical, emotional, and intellectual cycles peak, respectively. The value d is the given date and may be smaller than any of p, e, or i. All values are non-negative and at most 365, and you may assume that a triple peak will occur within 21252 days of the given date. The end of input is indicated by a line in which p = e = i = d = -1.

Output

For each test case, print the case number followed by a message indicating the number of days to the next triple peak, in the form:

Case 1: the next triple peak occurs in 1234 days.

Use the plural form ``days'' even if the answer is 1.

Sample Input

0 0 0 0

0 0 0 100

5 20 34 325

4 5 6 7

283 102 23 320

203 301 203 40

-1 -1 -1 -1

Sample Output

Case 1: the next triple peak occurs in 21252 days.

Case 2: the next triple peak occurs in 21152 days.

Case 3: the next triple peak occurs in 19575 days.

Case 4: the next triple peak occurs in 16994 days.

Case 5: the next triple peak occurs in 8910 days.

Case 6: the next triple peak occurs in 10789 days.

 

翻译：

Description

人生来就有三个生理周期，分别为体力、感情和智力周期，它们的周期长度为23天、28天和33天。每一个周期中有一天是高峰。在高峰这天，人会在相应的方面表现出色。例如，智力周期的高峰，人会思维敏捷，精力容易高度集中。因为三个周期的周长不同，所以通常三个周期的高峰不会落在同一天。对于每个人，我们想知道何时三个高峰落在同一天。对于每个周期，我们会给出从当前年份的第一天开始，到出现高峰的天数（不一定是第一次高峰出现的时间）。你的任务是给定一个从当年第一天开始数的天数，输出从给定时间开始（不包括给定时间）下一次三个高峰落在同一天的时间（距给定时间的天数）。例如：给定时间为10，下次出现三个高峰同天的时间是12，则输出2（注意这里不是3）。

Input

输入四个整数：p, e, i和d。 p, e, i分别表示体力、情感和智力高峰出现的时间（时间从当年的第一天开始计算）。d 是给定的时间，可能小于p, e, 或 i。 所有给定时间是非负的并且小于365, 所求的时间小于21252。 

当p = e = i = d = -1时，输入数据结束。

Output

从给定时间起，下一次三个高峰同天的时间（距离给定时间的天数）。 

采用以下格式： 
Case 1: the next triple peak occurs in 1234 days. 

注意：即使结果是1天，也使用复数形式“days”。

 

分析：

题目中给出的p表示从第0天开始后的第P天出现了第一次体力高峰期；

            e表示从第0天开始后的第e天出现了第一次情商高峰期；

             i表示从第0天开始后的第i天出现了第一次智商高峰期；

又因为体力、感情和智力周期，它们的周期长度为23天、28天和33天。因此要求哪天能达到体力高峰期，也就是从p开始，每次+23天；感情的高峰期是e开始每次+28天；智商的高峰期是从i开始每次+33天。

这道题目最开始的想法是直接暴力，从day=1开始，每次加一天，用三次判断，分别判断day%23,day%28,day%33是不是同时为0，但是考虑到这样做会有相当大的时间损耗，所以后来优化了算法，先找到第一次达到体力高峰期是哪天，随后就可以每次只用搜索day+23的天数去判断是否符合下面两个条件，接下来就是再找到同时符合体力与情商高峰期的一天，又因为23和28的最小公倍数就是（23*28），此后只需要判断day+466是否符合第三个条件，如果符合就输出答案（day-起始的天数d）并跳出循环，这样可以优化时耗。

 

下面给出AC代码：

#include<iostream>#include<stdio.h>using namespace std;int main(){int p,e,i,d,t,a;t=0;while(cin>>p>>e>>i>>d){if(p==-1&&e==-1&&i==-1&&d==-1) break;t++;a=d+1;for(a;a<21252;a++)   if((a-p)%23==0) break;for(;a<21252;a+=23)if((a-e)%28==0) break;for(;a<21252;a+=644)if((a-i)%33==0) break;printf("Case %d: the next triple peak occurs in %d days.\n",t,(a-d));//一开始把输出放在if条件下，后面有被注释掉的那个语句，虽然样例能过，但是会wa。。//if(a>=21252) printf("Case %d: the next triple peak occurs in %d days.\n",t,(21252-d));}return 0;}

这里解释一下为什么会wa：

题意里是说，最后的结果要<21252，但是，不保证相遇天数小于21252。

之所以上述方法直接printf可以过，是因为(a-i) (a-p) (a-e)在这里处理了，会让a-i的值小于21252.

如果直接暴力枚举这里的条件也要+d。

所以应该改为if(a>=21252+d) printf()；

这样就可以吧余数造成的影响给削掉了。

附上一组poj上的边界测试数据：

 

特殊测试数据

24 29 34 0           1

24 29 34 1           21252

24 29 34 2           21251

0  0  0  0           21252