350 - Pseudo-Random Numbers

 Pseudo-Random Numbers 

Computers normally cannot generate really random numbers, but frequently are used to generate sequences of pseudo-random numbers. These are generated by some algorithm, but appear for all practical purposes to be really random. Random numbers are used in many applications, including simulation.

A common pseudo-random number generation technique is called the linear congruential method. If the last pseudo-random number generated was L, then the next number is generated by evaluating (  , where Z is a constant multiplier, I is a constant increment, and M is a constant modulus. For example, suppose Z is 7, I is 5, and M is 12. If the first random number (usually called the seed) is 4, then we can determine the next few pseudo-random numbers are follows:

As you can see, the sequence of pseudo-random numbers generated by this technique repeats after six numbers. It should be clear that the longest sequence that can be generated using this technique is limited by the modulus, M.

In this problem you will be given sets of values for Z, I, M, and the seed, L. Each of these will have no more than four digits. For each such set of values you are to determine the length of the cycle of pseudo-random numbers that will be generated. But be careful: the cycle might not begin with the seed!

Input

Each input line will contain four integer values, in order, for Z, I, M, and L. The last line will contain four zeroes, and marks the end of the input data. L will be less than M.

Output

For each input line, display the case number (they are sequentially numbered, starting with 1) and the length of the sequence of pseudo-random numbers before the sequence is repeated.

Sample Input

7 5 12 45173 3849 3279 15119111 5309 6000 12341079 2136 9999 12370 0 0 0

Sample Output

Case 1: 6Case 2: 546Case 3: 500Case 4: 220

随机数表的产生，水题。给你Z，I，M，L四个数，（Z*L+I）mod M = 新的L。问的是什么时候产生循环，而这个循环并不一定是从第一个开始的。所以开个数组记录一下。

直接贴代码吧

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;int main (){    int Z,I,M,L,i,t=1;    while(cin>>Z>>I>>M>>L)    {        if (Z==0) break;        printf("Case %d: ",t);        t++;        int a[10010]={0},p=1,b[10010]={0};        a[L]=1;        b[p++]=L;        while(1)        {            L=(Z*L+I)%M;            if (!a[L])            {                a[L]=1;                b[p++]=L;            }            else break;        }        for (i=0; i<p; i++)            if (b[i]==L) break;        cout<<p-i<<endl;    }    return 0;}