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 was
L,then the next number is generatedby evaluating ( , where
Z is a constantmultiplier, 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 firstrandom number (usually called the
seed) 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 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 给的4个数分别是 Z I M 还有初始L;,然后算(Z * L + I) % M;得出的结果作为下一个L,直到算出的L和第一个L一样时,就是算循环周期。 第一次就是从头到尾模拟一次。直到出现和第一个L相同时。但是超时了。。 后来看出了,不需要出现第一个L,只要期间出现重复的数就循环了。。 AC代码:#include<stdio.h> #include<string.h> int main () { int z,i,m,l; int temp ,count; int res[10005]; int t = 1; while (scanf("%d%d%d%d",&z,&i,&m,&l)) { count = 0; memset (res , 0 ,sizeof(res)); if(z == 0 && i == 0 && m == 0 &&l == 0) break; while (1) { l = ( z * l + i) % m; if (res[l]) break; count++; res[l] = 1; } printf("Case %d: %d\n",t++,count); } return 0; }