Description
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
Output
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.
分析：
首先我们要知道，任意两个峰值之间一定相距整数倍的周期。假设一年的第N天达到峰值，则下次达到峰值的时间为
N+Tk=N+Tk(T是周期，k是任意正整数)。所以，三个峰值同时出现的那一天S，S应满足S=N1+T1∗k1=N2+T2∗k2=N3+T3∗k3。N1,N2,N3分别为为体力，情感，智力出现峰值的日期，T1，T2,T3分别为体力，情感，智力周期。我们需要求出k1,k2,k3
三个非负整数使上面的等式成立。想直接求出k1,k2,k3貌似很难，但是我们的目的是求出S，可以考虑从结果逆推。根据上面的等式，S满足三个要求：除以T1余数为N1，除以T2余数为N2，除以T3余数为N3。这样我们就把问题转化为求一个最小数，该数除以T1余N1，除以T2余N2，除以T3余N3。这就是著名的中国剩余定理，我们的老祖宗在几千年前已经对这个问题想出了一个精妙的解法。依据此解法的算法，时间复杂度可达到O(1)。
传说西汉大将韩信，由于比较年轻，开始他的部下对他不很佩服。有一次阅兵时，韩信要求士兵分三路纵队，结果末尾多2人，改成五路纵队，结果末尾多3人，再改成七路纵队，结果又余下2人，后来下级军官向他报告共有士兵2395人，韩信立即笑笑说不对（因2395除以3余数是1，不是2），由于已经知道士兵总人数在2300~2400之间，所以韩信根据23，128，233，------，每相邻两数的间隔是105（3、5、7的最小公倍数），便立即说出实际人数应是2333人（因2333=128+20χ105+105，它除以3余2，除以5余3，除以7余2）。这样使下级军官十分敬佩，这就是韩信点兵的故事。
韩信点兵问题简化：已知 n%3=2, n%5=3, n%7=2, 求n。
“韩信点兵”问题计算如下：
因为n%3=2, n%5=3, n%7=2 且 3，5，7互质 （互质可以直接得到这三个数的最小公倍数）
令x= n%3=2 ， y= n%5=3 ，z= n%7=2
使5×7×a被3除余1，有35×2=70，即a=2；
使3×7×b被5除余1，用21×1=21，即b=1；
使3×5×c被7除余1，用15×1=15，即c=1。
那么n =（70×x+21×y+15×z）%lcm(3,5,7) = 23 这是n的最小解
而韩信已知士兵人数在2300~2400之间，所以只需要n+i×lcm(3,5,7)就得到了2333，此时i=22
已知(n+d)%23=p; (n+d)%28=e; (n+d)%33=i
使33×28×a被23除余1，用33×28×8=5544；
使23×33×b被28除余1，用23×33×19=14421；
使23×28×c被33除余1，用23×28×2=1288。
因此有（5544×p+14421×e+1288×i）% lcm(23,28,33) =n+d
又23、28、33互质，即lcm(23,28,33)= 21252;
所以有n=（5544×p+14421×e+1288×i-d）%21252
本题所求的是最小整数解，避免n为负，因此最后结果为n= [n+21252]% 21252
那么最终求解n的表达式就是：
n=(5544*p+14421*e+1288*i-d+21252)%21252;
#include<iostream>
using namespace std;
int main()
{
int p,e,i,d;
int time=1;
while(cin>>p>>e>>i>>d)
{
if(p==-1 && e==-1 && i==-1 && d==-1)
{
break;
}
else
{
int n=(5544*p+14421*e+1288*i-d+21252)%21252;
if(n==0)
{
n=21252;
}
cout<<"Case "<<time++<<": the next triple peak occurs in "<<n<<" days."<<endl;
}
}
return 0;
}