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[CHALLENGE] Power Of Two
The challenge is quite simple ... You'll have to do a function that tells whether or not a number (the input) is a power of two. But beware ! Two criteria will be judged as the challenge is easy in itself. Your function will have to be AS OPTIMIZED AS POSSIBLE and AS SHORT AS POSSIBLE. Good luck :)
29 Respostas
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x & (x - 1) == 0
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@sayan, a number of two is binary written as looots of 0s, and one and only one number 1, wherever in the number.
So if you reduce it by 1, it will have no common bits with its past self
Examples :
8 : 1000 & 0111 = 0000
7 : 0111 & 0110 = 0110
6 : 0110 & 0101 = 0100
@Roman, reaaaaally nice !
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here you go! hope you like it!
https://code.sololearn.com/cocQa8N0gEG7/#cpp
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@sayan, nice but log is expensive
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@sayan i like your idea in version 2 ššš
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baptiste
please explain that & thing up there .
idk that
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It's "and" bitwise operation. All numbers that are powers of two has only one 1 bit. & (x - 1) remove this bit, so result will be 0.
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version 2
no log
no imporing
https://code.sololearn.com/cAvL7fCAob5R/?ref=app
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x-1 and & are single instructions at machine level. floating log and int<>float conversion are 100x more expensive
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https://code.sololearn.com/cbxAg0Yv6Dav/?ref=app i wish i knew the bitwise thing! not very optimized but here my try on this
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@Baptiste Here is the answer for q^p for any q: http://cr.yp.to/papers/powers.pdf
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if p&(p-1)==0 is the fastest way to see if p is a pure power of two
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ooh...
ya but brother i did not know that thing
that binary operator...
baptiste yesterday explained me
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vcc
baptiste
its my 4 th version
in python
it can calculate up to "3 lakh " yes
check it out
https://code.sololearn.com/cT5WxLWt9jIk/?ref=app
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3 lakh ....
more more more
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Good ! You can use integer div for k (//)
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What is a lakh ?
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3 lakh=300k
Costuma ter perguntas como essa?
Aprenda de maneira mais eficiente, de graƧa:
