Biorhythms（生物周期）

Time Limit: 1000 ms Memory Limit: 10000 KiB

Problem 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**（周期 p， e，i ）** 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.***（结束条件）
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.（输出）

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.

一开始也没有思路，看看了别人的用的中国剩余定理 （好吧不会，面对困难不能放弃~ 奥里给~），

学渣的我，结合了好多例子终于看明白了。大致这样吧。 23 28 33 是三个周期。要我们求一个数 n 。
使得 n = p (mod 23) ; n = e (mod 28) ; n = i (mod 33) ; （也就是 n % 23 = p ）
p , e , i, d 是要求我们输入的 , p 对应 n % 23 的余数 ; e 对应 n % 28 的余数 ; i 对应 n % 33 的余数。 d 是开始的天数 (就是当 d = 10 ，若发生重叠的天数是15 那么 n = 15 - 10 = 5 天 )

做法
1) 例出下列式子。
n = p (mod 23) ; n = e (mod 28) ; n = i (mod 33) ;
2)
n = 1 (mod 23); n = 0 (mod 23); n = 0(mod 23);
n = 0 (mod 28); n = 1(mod 28); n = 0(mod 28);
n = 0 (mod 33); n = 0(mod 33); n = 1(mod 33);
（按照列看，会发现第一列的余数是 1 0 0 第二列的余数是 0 1 0 第三列是 0 0 1）
3）（按照列看）
28×33×a mod 23 = 1 ; (第一列) ;
23×33×b mod 28 = 1 ; (第二列)；
23×28×c mod 33 = 1 ; (第三列) ；
（仔细观察一下式子） a b c 可以用求出来，（for 循环一下~）
在这里 a = 6 ； b = 19 ； c = 2 ；
4）最后 n = (28×33×a×p + 23×33×b ×e + 23×28×c×i - n ) % lcm (23,28,33) ; (lcm 最小公倍数) ；

(表达有限…)

#include <iostream>// 1002 生物节律   acm
using namespace std;
int main()
{int p,e,i,d,n,j=1;cin>>p>>e>>i>>d;while(p!= -1 && e!= -1 && i!=-1 && d!= -1){n = ((5544*p)  + (14421*e) + (1288*i) - d) % 21252  ;if(n<=0){cout<<"Case "<<j++<<": the next triple peak occurs in "<<n+21252<<" days."<<endl;}else{cout<<"Case "<<j++<<": the next triple peak occurs in "<<n<<" days."<<endl;}cin>>p>>e>>i>>d;}return 0;
}

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