Uva-350 Pseudo-Random Numbers

 Pseudo-Random Numbers 

Computers normally cannot generate really random numbers, but frequentlyare used to generate sequences of pseudo-random numbers. These are generatedby some algorithm, butappear for all practical purposes to be really random. Random numbersare used in many applications, including simulation.

A common pseudo-random number generation technique is called the linearcongruential method. If the last pseudo-random number generated wasL,then the next number is generatedby evaluating ( , whereZ is a constantmultiplier, I is a constant increment, and M is a constant modulus.For example, supposeZ is 7, I is 5, and M is 12. If the firstrandom number (usually called theseed) is 4, then we can determine thenext few pseudo-random numbers are follows:

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

In this problem you will be given sets of values for Z, I, M, and theseed, L. Each of these will have no more than four digits. For each suchset of values you are to determine the lengthof the cycle of pseudo-random numbers that will be generated. But becareful: 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 theinput 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 numbersbefore 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

题意：

给　L[n] = (Z * L[n-1] + I) mod M，求L[n]的循环周期。L不一定从初值处开始循环。Z，I，M，L都是不大于四位的整数。

解析：

直接模拟当相同的数字出现两次，跳出循环。

#include <stdio.h>#include <string.h>const int N = 10010;int flag[N];int main() {int Z,L,I,M;int cas = 1;while(scanf("%d%d%d%d",&Z,&I,&M,&L) != EOF) {if(Z==0 && L==0 && I==0 && M==0)break;memset(flag,0,sizeof(flag));int cut = 0;int tmp = L;while( ++cut ) {tmp = (Z * tmp + I) % M;flag[tmp]++;if(flag[tmp]==2)break;}printf("Case %d: %d\n",cas++,cut-1);}return 0;}