HDU-中国剩余问题-生理周期峰值

问题及代码：

/*  
*Copyright (c)2014,烟台大学计算机与控制工程学院  
*All rights reserved.  
*文件名称：HDU.cpp  
*作    者：单昕昕  
*完成日期：2015年1月27日  
*版 本 号：v1.0  
*问题描述：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. 
*程序输入：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 = 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.
*生理周期
Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 92557		Accepted: 28306
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”。
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.
*/
#include <iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int p,e,i,d;
    int x;
    int index = 1;
    while((cin>>p>>e>>i>>d)&&(p!=-1)&&(e!=-1)&&(i!=-1)&&(d!=-1))
    {
        p%=23;
        e%=28;
        i%=33;
        x=i;
        while(!((x-p)%23==0 && (x-e)%28==0))
        {
            x += 33;
        }
        x -= d;
        if(x<=0) x += 21252;
        cout<<"Case "<<index<<": the next triple peak occurs in "<<x<<" days."<<endl;
        index++;
    }
    return 0;
}

运行结果：

知识点总结：

中国剩余定理

p+23a=x；

e+28b=x；

i+33c=x；

其实就是求能被23除余p，能被28除余e，能被33除余i的最小的数。

这样我们可以用中国剩余定理（自行百度）：

假设：t1能被28、33整除，能被23除余1；

      t2能被23、33整除，能被28除余1；

      t3能被23、28整除，能被33除余1；

那么：t1*p能被28、33整除，能被23除余p；

          t2*e能被23、33整除，能被28除余e；

          t1*i能被23、28整除，能被33除余i；

so，t1*p+t2*e+t3*i就是一个能被23除余p，能被28除余e，能被33除余i的数，但是它不一定是最小的，需要减掉n个23、28、33的最小公倍数，这样就能得到x。但是我们要计算的是给定日期d到x的天数，就需要x - d > 0。

学习心得：

第一种是逐步满足法，方法麻烦一点，但适合所有这类题目。
第二种是最小共倍法，方法简单，但只适合特殊类型的题目。