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I'm trying to design a function int smallestDivisibleAfter(int number, int divisor) such that it returns the smallest number greater than or equal to number that is divisible by divisor (which is non-zero) . Here all inputs and outputs are assumed to be non-negative.

Examples:

smallestDivisibleAfter(9,4); // Returns 12
smallestDivisibleAfter(16,2); // Returns 16

I came up with the code number + divisor - number % divisor. However this ceases to work when number % divisor == 0, since then smallestDivisibleAfter(16,2); // Returns 18 instead of 16.

In addition, number - 1 + divisor - (number - 1)% divisor does not work since int will be replaced by unsigned long long when I put this code into action.

What is the best solution here?

Paul Hankin
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Henricus V.
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2 Answers2

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If you want to avoid jumps, try: 

number - number % divisor + divisor * !!(number % divisor)

The !!x just converts the number to a boolean with 0 if x==0 and 1 otherwise.

Juan Lopes
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((number / divisor)  + 1) * divisor)

This will return the next greatest multiple of divisor after number.

If you ever plan on using floats: floor(number / divisor).

user6190774
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  • Integer division doesn't result in a float; there's nothing to floor(). – m69's been on strike for years Apr 12 '16 at 00:25
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    A small suggestion: it's better to gain confidence in your answer before publishing it. The goal of answering isn't to guess the right answer as quickly as possible, it's to provide a correct and convincing answer that other people can read and understand. – Paul Hankin Apr 12 '16 at 00:27
  • Yeah I just edited that, originally I didn't bother with floor then added it in casehe decides to use floats – user6190774 Apr 12 '16 at 00:27
  • This is not correct because if `number` is already divisible by `divisor` it will return `number + divisor`. – Lars M. Jun 22 '19 at 07:42