hdu_1370Biorhythms(互素的中国剩余定理)

Biorhythms

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2408    Accepted Submission(s): 1053

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 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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

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

1

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.

题意：

有3个循环周期，周期天数分别为23、28、33。对于某一年，已知某年这3个周期的某一峰值分别是当年的第p、e、i天，

问从第d天开始到最近一个满足3个周期都达到峰值的日期还有多少天。

可以简化方程为

23x + p = a;

28y + e = a;

33z + i  = a;

然后可以写成

a===p(mod 23)

a===e(mod 28)

a===i(mod33)

这样就是转化成中国剩余定理的标准形式，注意中国剩余定理要满足模值要互素，正好在这个题中23 和28和33是互素的

介绍一下中国剩余定理

假设有式子

x===a[i](mod b[i])

0<i<k 的n个式子，式子中要求任何两个b[i]之间是互素的

然后它的解的形式是

m= b[0]*b[1]*b[2]*……*b[n];

去M[t] = m/b[t];

M[t]^-1 为用扩展欧几里得求出的M[t]对于模m的逆元

有ans = a[0]*M[0]*M[0]^-1+a[1]*M[1]*Mt[1]^-1+……+a[n]*M[n]*M[n]^-1;

有一个很好的解释的网页：http://baike.baidu.com/link?url=EIpCzOqfxyMH041vk-ek0PSq1rNN7fPjIBx7sc-loDuinvxd--T7F4ieqK4d2M1stOgry6H11JDZFbGQXmAoZg1uMLhaCOvtYfqr9NP2FGNtK6tNmr2Qa8QsS3cWkktY35hASkSLVwYJivB0_tpfN73mNEq2iaGckg2uuYXtPpYYAhLfItL-aVSZ2PWw12aIAwzQ-HxYH5lxprhIUx2BF_

下面是代码：

 //互素的中国剩余定理
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int mp[] = {,,};
 int s[];
 int exgcd(int a, int b, int &x, int &y)
 {
     if(b==) {
         x = ;
         y = ;
         return a;
     }
     int ans = exgcd(b,a%b,x,y);
     int tm = x;
     x = y;
     y = tm-a/b*y;
     return ans;
 }

 int chinese_reminder()
 {
     int n = ;
     int ans = ;
     int x,y;
     for(int i = ; i < ; i++) n*=mp[i];
     for(int i = ; i < ; i++){
         exgcd(n/mp[i],mp[i],x,y);
         ans+=s[i]*x*(n/mp[i]);
     }
     ans = (ans+n)%n;
     return ans;
 }
 int main()
 {
     int ans;
     int T;
     scanf("%d",&T);
     int cnt = ;
     int d;
     while(~scanf("%d%d%d%d",&s[],&s[],&s[],&d))
     {
         if(s[]==-&&s[]==-&&s[]==-&&d==-) return ;
         printf("Case %d: the next triple peak occurs in ",cnt++);
         ans = chinese_reminder();
         if(d>ans) printf("%d days.\n",-d+ans);
         else printf("%d days.\n",ans-d);
     }
     return ;
 }