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
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
ZIs a constant multiplier,IIs a constant increment, and
MIs a constant modulus. For example, supposeZIs 7,IIs 5, and
MIs 12. If the first random number (usually calledSeed) Is 4, then we can determine the next few pseudo do-random numbers are follows:
As you can see, the sequence of pseudo-random numbers generated by this technique repeats after six numbers. it shoshould 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 valuesZ,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
Seed!
Input
Each input line will contain four integer values, in order,Z,
I,M, AndL. The last line will contain in four zeroes, and marks the end of the input data.
LWill be lessM.
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
When the third group outputs 501, it is too careless to submit the statement without seeing it... the question also implies that the loop does not necessarily start from the first one--I am still too disconnected.
# Include <stdio. h>
Void main ()
{Int Z, I, m, l, I, a [10000], cycle [10000], K, sum, num = 0;
While (scanf ("% d", & Z, & I, & M, & L ))
{If (Z + I + M + L = 0) break;
Sum = 0; k = 0; ++ num;
For (I = 0; I <m; I ++) A [I] = 0;
While (A [l] = 0)
{A [l] = 1; ++ sum;
Cycle [Sum] = L;
L = (z * L + I) % m;
}
If (L! = Cycle [1]) {While (cycle [k + 1]! = L) ++ K ;}
Printf ("case % d: % d \ n", num, Sum-k );
}
}