Java Coding Test Course, Ax + By = C

Problem Description

Write a program to find all integer pairs (x, y) that satisfy Ax + By = C for the given integers A, B, and C.
The following conditions must be satisfied:

• 1 ≤ |A|, |B|, |C| ≤ 10^3
• x and y are integers.

For example, when A = 2, B = 3, and C = 6, possible solutions include (0, 2), (3, 0), (1, 1), etc.
However, there are cases where solutions exist infinitely, as well as cases where no solution exists.

Problem Solving Process

1. Problem Analysis

To solve the problem, first analyze it: the equation Ax + By = C seeks a combination of x and y.
The number and range of potential solutions vary depending on the signs of A and B.
For example, the case where both A and B are 0 is an exception; in all other cases, it’s essential to understand how one variable can change concerning the other.

2. Counting the Number of Solutions

To find the solutions to the problem, first check if C is a multiple of the greatest common divisor (GCD) of A and B.
If the GCD isn’t a divisor of C, then no solutions exist, which can be briefly explained as follows:

                gcd(a, b) = ax + by
            

Here, x and y are integers. This transformation shows that integer solutions can be found.

3. Java Implementation

Now, based on the above logic, let’s implement the program using Java.
I will explain each step and include the code as well.

3.1. GCD Calculation Function

public static int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
            

The above code is a function that calculates the greatest common divisor of two numbers a and b using the Euclid algorithm.

3.2. Finding and Printing Solutions Function

public static void findSolutions(int A, int B, int C) {
    // Calculate GCD
    int gcdAB = gcd(Math.abs(A), Math.abs(B));
    
    // Check if C is a multiple of GCD
    if (C % gcdAB != 0) {
        System.out.println("No solutions exist.");
        return;
    }

    // Logic for calculating x and y
    int x = C / A;
    int y = 0; // Initial y value
    // Print solution
    System.out.println("Possible solution: x = " + x + ", y = " + y);
    // Alternatively, it is possible to print other solutions using a loop
}
            

4. Complete Code

public class LinearEquationSolver {
    public static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    public static void findSolutions(int A, int B, int C) {
        int gcdAB = gcd(Math.abs(A), Math.abs(B));
        if (C % gcdAB != 0) {
            System.out.println("No solutions exist.");
            return;
        }

        // Fixed-point iteration for solution finding
        for (int x = -1000; x <= 1000; x++) {
            if ((C - A * x) % B == 0) {
                int y = (C - A * x) / B;
                System.out.println("Possible solution: x = " + x + ", y = " + y);
            }
        }
    }

    public static void main(String[] args) {
        findSolutions(2, 3, 6);
        // Additional test cases
        findSolutions(1, -1, 0);
    }
}
            

This code allows you to find solutions for various combinations of A, B, and C.
The part that outputs the solutions is found using a loop directly,
generally providing faster results when the numbers are small.

Conclusion and Summary

In this course, we examined how to find integer solutions to Ax + By = C.
We explained how to determine whether solutions exist using the GCD, and how to print possible solutions using a loop.
These techniques can be applied to a variety of problems and will be very helpful for advanced studies in algorithms.