湘大OJ第1490题 Generating Random Numbers

　　湘大OJ第1490题，Generating Random Numbers（题目链接）。

Generating Random Numbers

Description

Anyone who considers arithmetical methods of producing random digits is,
of course, in a state of sin.
- John von Neumann

As a talented mathematician, von Neumann invented a method to generate random numbers sequentially. The method is quite simple: Given A, B, M, and the first element of the sequence x[0], then we will get \(\displaystyle x[i] = (x[i-1] * A + B) % M\), for all \(i > 0\).

As an even more talented mathematician, you found his method are problematic. The sequence is not really random, that is, it always repeated previous numbers after several steps. For example, assume A = 3, B = 5, M = 7, and x[0] = 2, then we will get x[1] = 4, x[2] = 3, x[3] = 0, x[4] = 5, x[5] = 6, x[6] = 2, x[7] = 4, x[8] =3, … We can see that x[6] is the same as x[0], then the same sequence begins to repeat again and again. Now you want to write a program to find that, for a given A,B,M and x[0], what is the smallest i so that x[i] equal to a previous number.

Input

The first line of the input contains an integer T (T ≤ 100), indicating the number of test cases. Then T cases follow. Each case contains one line with 4 numbers A, B, M and x[0]. You can assume 1 ≤ M ≤ 30000, 0 ≤ A, B, x[0] ≤ M.

Output

Output one line for each test case, indicating the smallest i that x[i] is repeated.

Sample Input

2
3 5 7 2
3 5 7 1

Sample Output

6
1

Source

“开启时代杯”湘潭市第二届大学生程序设计大赛 (Internet)

　　这一题没什么难度，暴力枚举，直到相等。

　　C语言代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

typedef int COUNT;

int main (void)
{
    int n;
    int a, b, m, x0, x;
    COUNT i;
    scanf( "%d", &n );
    while ( n -- )
    {
        scanf( "%d%d%d%d", &a, &b, &m, &x0 );
        x = x0;
        for ( i = 1 ;  ; i ++ )
        {
            x = (x * a + b) % m;
            if ( x == x0 )
                break;
        }
        printf( "%d\n", i );
    }
    return EXIT_SUCCESS;
}