Implement a function that divides a by b and writes the quotient and the remainder through q and r, choosing the pair whose remainder is never negative.

• a is the dividend and b the divisor. q and r point to two int the caller already owns: write the quotient through q, the remainder through r, and return nothing.
• The pair you write must satisfy a == b * q + r with 0 <= r < |b|. The second half is the whole exercise. The remainder is never negative, whatever the signs of a and b.
• C division does not hand you that pair. / truncates toward zero, so -7 / 2 is -3 and -7 % 2 is -1. Here the answer is q = -4 and r = 1, because 2 * -4 + 1 is -7 and 1 sits in range. Start from the truncated pair and repair it when the remainder came out negative.
• A negative b does not flip the remainder rule. div_mod_floor(7, -2, &q, &r) writes q = -3 and r = 1, since -2 * -3 + 1 is 7. For a positive b your quotient ends up rounded down, for a negative b rounded up, and in both cases it is the remainder that decides.
• An exact division needs no repair and leaves r at 0. -8 by 4 gives -2 and 0; -8 by -4 gives 2 and 0.
• An a of 0 gives 0 and 0 for every b.
• b equal to 0 is outside the domain, and so is a == INT_MIN together with b == -1, whose quotient has no int to live in. Nothing calls you with either, so do not spend code detecting them. Every other pair is fair game, INT_MIN and INT_MAX included.
• The tests print what you wrote as the quotient then the remainder on one line, separated by a single space.

Implement a function that divides a by b and writes the quotient and the remainder through q and r, choosing the pair whose remainder is never negative.

• a is the dividend and b the divisor. q and r point to two int the caller already owns: write the quotient through q, the remainder through r, and return nothing.
• The pair you write must satisfy a == b * q + r with 0 <= r < |b|. The second half is the whole exercise. The remainder is never negative, whatever the signs of a and b.
• C division does not hand you that pair. / truncates toward zero, so -7 / 2 is -3 and -7 % 2 is -1. Here the answer is q = -4 and r = 1, because 2 * -4 + 1 is -7 and 1 sits in range. Start from the truncated pair and repair it when the remainder came out negative.
• A negative b does not flip the remainder rule. div_mod_floor(7, -2, &q, &r) writes q = -3 and r = 1, since -2 * -3 + 1 is 7. For a positive b your quotient ends up rounded down, for a negative b rounded up, and in both cases it is the remainder that decides.
• An exact division needs no repair and leaves r at 0. -8 by 4 gives -2 and 0; -8 by -4 gives 2 and 0.
• An a of 0 gives 0 and 0 for every b.
• b equal to 0 is outside the domain, and so is a == INT_MIN together with b == -1, whose quotient has no int to live in. Nothing calls you with either, so do not spend code detecting them. Every other pair is fair game, INT_MIN and INT_MAX included.
• The tests print what you wrote as the quotient then the remainder on one line, separated by a single space.