Getting Remainder While Dividing

Unity Engine
1

I need the values of the 3rd grade division ie. 8/3 = 2r2

I ALSO potentially need the value of one higher ie. 8/3 = 3r-1

The result I would use would be based on the smaller absolute value of the remainder, so in this case I would use 3r-1.

Potential Results would be:
7/3 = 2r1
6/2 = 3r0
13/8 = 2r-3
etc.

Here is some code I quickly wrote. I feel like this is a bad way of going about this. Any advice?

    public int lowerNumber;
    public int higherNumber;
    public int positiveRemainder;
    public int negativeRemainder;

    public void DetermineNumberAndRemainder(int u, int c)
    {
        positiveRemainder = u % c;
        lowerNumber = Mathf.FloorToInt(u / c);

        if (positiveRemainder == 0)
        {
            Debug.Log("ME(" + u + "," + c + ") = "
                + lowerNumber + "r" + positiveRemainder + ".");
            return;
        }
        else
        {
            higherNumber = Mathf.CeilToInt(u / c);
            negativeRemainder = higherNumber * c - u;

            Debug.Log("ME(" + u + "," + c + ") = "
                + lowerNumber + "r" + positiveRemainder
                + " OR " + higherNumber + "r" + negativeRemainder + ".");
        }
    }
2

modulo, represented by the % symbol in C#:

I’m not sure what more you need. integer division gives you the value, modulo the remainder.

You don’t even need this Mathf.FloorToInt or CeilToInt. If you divide ints, you do integer division. The result is the int without remainder.

3

Wait… 13/8 = 2r-3?

OK… you want to go backwards… weird.

Try something like:

        private static void DivWithRemainderBackwards(int u, int c)
        {
            int value = u / c;
            int remain = (u % c);
            if(remain != 0)
            {
                value += 1;
                remain -= c;
            }

            Debug.Log(string.Format("{0}r{1}", value, remain));
        }

(not sure if you wanted to support negative u,c values)

4

I think he also wants to make the decision based on what’s closer to 0, the positive remainder of the usual remainder operation and the result that’d overshoot and yield a higher quotient and negative remainder.

13/8 = 18 + 5
VS
13/8 = 28 - 3

choose the latter, as the absolute value of -3 is closer to 0 than 5.

So with a small tweak, it’s something like

public struct Result
{
    public int X;
    public int Y;
}

public Result PutAppropriateNameHere(int u, int c)
{
    // TODO: handle c == 0

    var quotient = u / c;
    var remainder = u % c;

    if ((c - remainder) < remainder)
    {
        quotient = quotient + 1;
        remainder = remainder - c;
    }

    return new Result { X = quotient, Y = remainder };
}

Only written for positive input values…

1 Like
5

Yeah, I wasn’t sure what conditions they were looking for it to be like that…

Wasn’t sure what “potentially” meant.

1 Like
6

Thank you Sud! Your code looks nice and clean.