若某数x分别被d1、、…、dn除得的余数为r1、r2、…、rn，则可表示为下式：
x=R1r1+R2r2+…+Rnrn+RD
其中R1是d2、d3、…、dn的公倍数，而且被d1除，余数为1；
R1 、R2…、Rn是d1、d2、…、dn-1的公倍数，而且被dn除，余数为1；
D是d1、d2、…、的最小公倍数；
R是任意整数，可根据实际需要决定；
且d1、、…、必须互质，以保证每个Ri(i=1,2，…，n)都能求得.
该问题可以用初等数论中的同余方程组的求解问题。利用同余的符号，可以将上述问题转化为下面的同余方程组：
x≡2(mod 3)；
x≡3(mod 5)；
x≡2(mod 7)；
不难看出上述同余方程组的解并不唯一，因为如果x是一个解，则x+3×5×7×k=x+105k也是该同余方程组的一个解，其中的k可以是任意整数。事实上，从3，5，7两两互质可知上述同余方程组的任意两个解相差105的倍数。所以，一旦求出“最小正整数解”x0，则每个解均可表示为x=x0+105k。
如何求出上述同余方程组的一个解呢？我们的祖先聪明的把问题转化为以下三个非常特殊的同余方程组的求解
a≡1(mod 3)； b≡0(mod 3)； c≡0(mod 3)；
a≡0(mod 5)； b≡1(mod 5)； c≡0(mod 5)；
a≡0(mod 7)； b≡0(mod 7)； c≡1(mod 7)；
显然，如果求出了a，b，c的一组值，则2a+3b+2c就是原同余方程组的一个解，再把这个解除以105，则相应的余数即为所求的最小整数解。经过简单计算可知a可以取70，b可以取21，c可以取15。所以2a+3b+2c=233，除以105后得余数为23，这就是所求的最小正整数解。
Biorhythms
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.
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
PKU
1 #include <iostream> 2 3 using namespace std; 4 5 int main() 6 { 7 int k=0; 8 int p,e,i,d; 9 cin>>p>>e>>i>>d; 10 while(p!=-1&&e!=-1&&i!=-1&&d!=-1) 11 { 12 k++; 13 int t=(p*5544+e*14421+i*1288-d+21252)%21252; 14 if(t==0) t=21252; 15 cout<<"Case "<<k<<": the next triple peak occurs in "<<t<<" days."<<endl; 16 cin>>p>>e>>i>>d; 17 } 18 return 0; 19 }