0_equation.c

#include <stdio.h>

// soliton 1: 
void equation(int n) {
	for (int a = 0; a < 10; a++)
		for (int b = 0; b < 10; b++)
			for (int c = 0; c < 10; c++)
				if ((a * 10 + b) + (c * 10 + a) == n)
					printf("A = %d, B = %d, C = %d\n", a, b, c);
}


// soliton 2: 
#define TRUE(a, b, c, n) ((a * 10 + b) + (c * 10 + a) == n) 

void    equation(int n)
{
	int a = 0;
	int b, c;
	while (a <= 9)
	{
		b = 0;
		while (b <= 9)
		{
			c = 0;
			while (c <= 9)
			{
				if (TRUE(a, b, c, n))
					printf("A = %d, B = %d, C = %d\n", a, b, c);
				c++;
			}
			b++;
		}
		a++;
	}
}

/*
	Assignment name  : equation
	Expected files   : equation.c
	Allowed functions: printf
	-------------------------------------------------------------------------------
	The goal of this exercise is to find all possible answers to the following
	equation :

		AB + CA = n

	where A, B, and C are individual digits [0-9] and n is an integer.
	Note that here AB is not the product of A and B. It is merely a two digit number
	with A being the first digit (tens place) and B the second digit (ones place).

	Implement a function that, given an integer n, prints on the standard
	output all the possible values of A, B, C for which the equation is true.

	Your function must be declared as follows:

	void    equation(int n);

	If a solution could not be found, nothing is printed.
	Examples:
	For the value n = 42, the output would be :
	$> ./equation 42
	A = 0, B = 2, C = 4
	A = 1, B = 1, C = 3
	A = 2, B = 0, C = 2
	A = 3, B = 9, C = 0
	$>
	For the value n = 111, the output would be :
	$> ./equation 111
	A = 2, B = 9, C = 8
	A = 3, B = 8, C = 7
	A = 4, B = 7, C = 6
	A = 5, B = 6, C = 5
	A = 6, B = 5, C = 4
	A = 7, B = 4, C = 3
	A = 8, B = 3, C = 2
	A = 9, B = 2, C = 1
	$>
	For the value n = 0, the output would be :
	$> ./equation 0 | cat -e
	A = 0, B = 0, C = 0$
	$>

	Note:
	- The displayed output will always be sorted in ascending order beginning with A, then B and then C.
	(as shown in the examples above)
*/