PKU 1006 Biorhythms （中国剩余定理 * *）

Biorhythms

http://poj.org/problem?id=1006

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 97029		Accepted: 29832

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.

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

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.

Source

East Central North America 1999

 

 

中国剩余定理（转）

 

就是要求下面那个联立同余方程组的解：

(n+d)%23=p;           //这里的p（以及下面的e，i）应该使小于23（28, 33）的，可数据并不是这么回事，疑问呐？？？？？？？

(n+d)%28=e;

(n+d)%33=i ,求n 。

这道题用到的是近世代数里面讲过的中国剩余定理。这里就说点题外话，关于中国剩余定理的来历：

民间传说着一则故事——“韩信点兵”。
秦朝末年，楚汉相争。一次，韩信将1500名将士与楚王大将李锋交战。苦战一场，楚军不敌，败退回营，汉军也死伤四五百人，于是韩信整顿兵马也返回大本营。当行至一山坡，忽有后军来报，说有楚军骑兵追来。只见远方尘土飞扬，杀声震天。汉军本来已十分疲惫，这时队伍大哗。韩信兵马到坡顶，见来敌不足五百骑，便急速点兵迎敌。他命令士兵3人一排，结果多出2名；接着命令士兵5人一排，结果多出3名；他又命令士兵7人一排，结果又多出2名。韩信马上向将士们宣布：我军有1073名勇士，敌人不足五百，我们居高临下，以众击寡，一定能打败敌人。汉军本来就信服自己的统帅，这一来更相信韩信是“神仙下凡”、“神机妙算”。于是士气大振。一时间旌旗摇动，鼓声喧天，汉军步步进逼，楚军乱作一团。交战不久，楚军大败而逃。
在一千多年前的《孙子算经》中，有这样一道算术题：
“今有物不知其数，三三数之剩二，五五数之剩三，七七数之剩二，问物几何？”按照今天的话来说：一个数除以3余2，除以5余3，除以7余2，求这个数.
这样的问题，也有人称为“韩信点兵”.它形成了一类问题，也就是初等数论中解同余式.这类问题的有解条件和解的方法被称为“中国剩余定理”，这是由中国人首先提出的.
好，我们言归正传，是不是发现我们要求的方程组和孙子算经历的方程组非常大相似？

一般地，中国剩余定理是指若有一些两两互质的整数，则以下联立同余方程组对模 有唯一解：

孙子算经中的计算过程总结出来应该如下表示：

因为n%3=2,n%5=3,n%7=2且3，5，7互质 使5×7被3除余1，用35×2=70； （这个称为35相对于3的数论倒数） 使3×7被5除余1，用21×1=21； 使3×5被7除余1，用15×1=15。
（70×2+21×3+15×2）%（3×5×7）=23

同样我们可以得到本体的计算过程：

使33×28被23除余1，用33×28×8=5544； 使23×33被28除余1，用23×33×19=14421；
使23×28被33除余1，用23×28×2=1288。 （5544×p+14421×e+1288×i）%（23×28×33）=n+d
n=（5544×p+14421×e+1288×i-d）%（23×28×33）

当然为了节省提交程序所运行的时间，我们可以预先写个小程序把这三个数论倒数预先求出来，然后再程序里直接应用结果就可以了。

#include<stdio.h>

#define N (23*28*33)

int p,e,i,d;

int main(){
    int a=0,b=0,c=0;
    while(a%23!=1)
        a+=28*33;
    while(b%28!=1)
        b+=23*33;
    while(c%33!=1)
        c+=23*28;
    int cases=0;
    while(scanf("%d %d %d %d",&p,&e,&i,&d)){
        if(p==-1 && e==-1 && i==-1 && d==-1)
            break;
        int ans=(p*a+e*b+i*c-d)%N;
        if(ans<=0)
            ans+=N;
        printf("Case %d: the next triple peak occurs in %d days.\n",++cases,ans);
    }
    return 0;
}