Find x and y satisfying ax + by = n in C++

In this problem, we are given three integer values a, b, and n. Our task is to find x and y satisfying ax + by = n.

Let's take an example to understand the problem

Input : a = 4, b = 1, n = 5
Output : x = 1, y = 1

Solution Approach

A simple solution to the problem is by finding the value between 0 to n that satisfies the equation. We will do this by using altered forms of the equation.

x = (n - by)/a
y = (n- ax)/b

If we get a value satisfying the equation, we will print the values otherwise print "no solution exists".

Example

Program to illustrate the working of our solution

#include <iostream>
using namespace std;
void findSolution(int a, int b, int n){
   for (int i = 0; i * a <= n; i++) {
      if ((n - (i * a)) % b == 0) {
         cout<<i<<" and "<<(n - (i * a)) / b;
         return;
      }
   }
   cout<<"No solution";
}
int main(){
   int a = 2, b = 3, n = 7;
   cout<<"The value of x and y for the equation 'ax + by = n' is ";
   findSolution(a, b, n);
   return 0;
}

Output

The value of x and y for the equation 'ax + by = n' is 2 and 1