C. Line (cf)

This problem utilizes Diophantine Equations:

We can implement the above using recursion:

        
            void f(int a, int b, int &x, int &y)
            {
                // base case
                if(b==0)
                {
                    x = 1;
                    y = 0;
                    return;
                }
                int x1, y1;x1
                f(b, (a%b), x1, y1);
                x = y1;
                y = x1 - (a/b)*y1;
                return;
            }