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 theseed) 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 4
5173 3849 3279 1511
9111 5309 6000 1234
1079 2136 9999 1237
0 0 0 0

Sample Output

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

    题意：每次输入四个数据Z,I,M,L,根据伪随机数的原理，求种子轮回周期的大小。式子已经在题目中给出：L = (Z*L+I) mod M,不过题目有一个陷阱就是题目给出的L并不一定就是初始的种子，需要自己判断。（可以设立一个数组把所有出现过得种子全部储存下来，如果再次出现就是初始种子）

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int a[100000000];

int main()
{
    long long int Z,I,M,L;
    int k = 0;
    while(scanf("%lld%lld%lld%lld",&Z,&I,&M,&L)!=EOF)
    {
        if(Z == 0 && I == 0 && M == 0 && L == 0)
        {
            break;
        }
         for(int i=0;i<M;i++)
         {
             a[i] = 0;
         }
        k++;
        int pl = L%M;
        a[pl] = 1;
        int count = 0;
        long long int sum = 0;
        while(1)
        {
            count++;
            Z = Z % M;
            sum = (Z * L);
            sum = sum + (I%M);
            L = sum % M;
            if(L == pl)
            {
                break;
            }
            if(a[L] == 1)
            {
                count--;
                break;
            }
            a[L] = 1;
        }
        printf("Case %d: %d\n",k,count);
    }
    return 0;
}