| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 107569 | Accepted: 33365 |
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.
翻译：
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 107569 | Accepted: 33365 |
Description
Input
当p = e = i = d = -1时，输入数据结束。
Output
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.
解决思路
中国剩余定理，本题难点不在编程，而是分析题目并转化为数学公式
要引入本题解法，先来看一个故事 “韩信点兵”：
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。
再看我们这道题，读入p,e,i,d 4个整数
已知(n+d)%23=p; (n+d)%28=e; (n+d)%33=i ,求n 。
两道题是一样的。但是韩信当时计算出结果的？
韩信用的就是“中国剩余定理”，《孙子算经》中早有计算方法，大家可以查阅相关资料。
“韩信点兵”问题计算如下：
因为n%3=2, n%5=3, n%7=2
我们先找一个最小的数能被5和7整除，并且除以3的余数为1，这里，我们可以找到最小的数是70。有什么好处呢，大家知道70 * 1 % 3 = 1，那么70 *
2 % 3 = 2，70 * a % 3 = a,这样可以方便找到140这个除以3余数又为2的数，同时又能被5和7整除。
同理，我们可以找到21是能被3和7整除但是除以5余数是1的数，那么63便是除以5余数是3. 15能被3和5整除，但是除以7余数是1，那么30便是除以7余数是2的数。
那么 140 + 63 + 30 =
233，必定是一个除以3余数是2，除以5余数是3，除以7余数是2的一个数，而我们又很容易知道233加上或者减去3,5,7的最小公倍数，得到的数同
样满足这个性质。韩信已知士兵人数在2300~2400之间，所以只需要233 + i × lcm(3,5,7)= 233 + 105 * 20 =
2333.
同样，这道题的解法就是：
假设过n天后，三个生理周期同时到高峰。
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;
当问题被转化为一条数学式子时，你会发现它无比简单。。。。直接输出结果了。
源码
1 /* 2 poj 1000 3 version:1.0 4 author:Knight 5 Email:S.Knight.Work@gmail.com 6 */ 7 8 9 #include<iostream> 10 using namespace std; 11 12 int main(void) 13 { 14 int p,e,i,d; 15 int time=1; 16 while(cin>>p>>e>>i>>d) 17 { 18 if(p==-1 && e==-1 && i==-1 && d==-1) 19 break; 20 21 int lcm=21252; // lcm(23,28,33) 22 int n=(5544*p+14421*e+1288*i-d+21252)%21252; 23 if(n==0) 24 n=21252; 25 cout<<"Case "<<time++<<": the next triple peak occurs in "<<n<<" days."<<endl; 26 } 27 return 0; 28 }