I need the easiest and shortes way, to round an integer to the next "30-series".
For example:
431 --> 450
459 --> 480
298 --> 300
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@Reimeus what code should I write, if I have no idea how to do it? – xMarcoGP May 31 '17 at 19:29
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For 450 as input, do you need, 450 or 500 as output? – Arnaud Denoyelle May 31 '17 at 19:31
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@ArnaudDenoyelle 450 – xMarcoGP May 31 '17 at 19:32
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Even if you don't know how to do it, you should include what you have so far, since this feels like a homework question: https://meta.stackoverflow.com/questions/334822/how-do-i-ask-and-answer-homework-questions – nevets1219 May 31 '17 at 19:33
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@nevets1219 it's not an homework question. I don't have anything concrete so far, else I'd have included it in my question. – xMarcoGP May 31 '17 at 19:35
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Well a simple search yields multiple results: https://stackoverflow.com/questions/9303604/rounding-up-a-number-to-nearest-multiple-of-5 – nevets1219 May 31 '17 at 19:40
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Thanks @nevets1219 – xMarcoGP May 31 '17 at 19:43
1 Answers
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It depends on the language you are using. A simple example could be
def round2multipleOfThirty(x):
if x % 30 < 15:
return x - x % 30
else:
return x + 30 - x % 30
which works in python. But do you actually want a round function? Your example numbers are always rounded up to the next greater multiple of 30. My function provided is a round which would provide
431 --> 420
459 --> 450
298 --> 300
domna
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Python REPL was just at hand ;) But it should be easily translatable into Java the same way and there is another answer dealing with the exact question in Java. – domna May 31 '17 at 19:44
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1Yes, but be aware that for instance python 3 automatically converts divisions to float. But this is not default in other languages and therefore in general an easier and better answer. – domna May 31 '17 at 19:46
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@ArnaudDenoyelle: Even easier: `x + (-x) % 30` (which works for floats like `x=0.5`, too). – Mark Dickinson Jun 01 '17 at 07:19
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@domna: Your solution won't translate well to a language like Java or C++, because for negative `x`, `x % 30 < 15` will always be true in those languages. – Mark Dickinson Jun 01 '17 at 07:21