Question

How to round an integer up to the nearest multiple of 'multiple' in C#

Answer 1

using System;

class RoundUpProgram
{
    /*
        roundUp(n, multiple)
        --------------------
        Rounds the integer n *up* to the nearest multiple of `multiple`.

        Mathematically:
            result = ceil(n / multiple) * multiple

        We use integer arithmetic for efficiency:
            (n + multiple - 1) / multiple  → smallest integer ≥ n/m
    */
    static int roundUp(int n, int multiple)
    {
        if (multiple <= 0)
        {
            // Defensive programming: avoid undefined behavior.
            // In real-world code, you'd throw or handle this differently.
            return n;
        }

        // Efficient integer rounding up:
        return ((n + multiple - 1) / multiple) * multiple;
    }

    static void Main()
    {
        Console.WriteLine("roundUp(53, 20)  = " + roundUp(53, 20));
        Console.WriteLine("roundUp(68, 30)  = " + roundUp(68, 30));
        Console.WriteLine("roundUp(7, 100)  = " + roundUp(7, 100));
        Console.WriteLine("roundUp(119, 100) = " + roundUp(119, 100));
        Console.WriteLine("roundUp(781, 100) = " + roundUp(781, 100));
        Console.WriteLine("roundUp(1026, 100) = " + roundUp(1026, 100));
        Console.WriteLine("roundUp(11689, 1000) = " + roundUp(11689, 1000));
    }
}


/*
run:

roundUp(53, 20)  = 60
roundUp(68, 30)  = 90
roundUp(7, 100)  = 100
roundUp(119, 100) = 200
roundUp(781, 100) = 800
roundUp(1026, 100) = 1100
roundUp(11689, 1000) = 12000

*/

 

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