--- Kazufumi-MIT-Mitani <[EMAIL PROTECTED]> skrev:
> "Daniel C. Sobral" <[EMAIL PROTECTED]> wrote
>
> > That technique is O(ln(n)), where n is the number in question.
> >
> > Frankly, for numbers up to 32, a table will wield the best results,
> > and might actually be smaller than some of the suggestions given so
> > far.
>
> Counting n as bit, it is O(n) :p
No it isn't. You're only looping a maximum of 32 times, not 2^32 times.
> Unrolling the loop,
>
> if(n<=2^0) 2^0
> else if(n<=2^1) 2^1
> else if(n<=2^2) 2^2
> else if(n<=2^3) 2^3
> :
> :
> else if(n<=2^31) 2^31
>
> is suitable for this problem?
IMHO, the best solution so far, if a lookup table isn't suitable, is the binary
search version that someone posted.
Regards, Tommy
===
Tommy Hallgren
Briljantg. 31, SE-421 49, G�teborg
Tel.: 0709 - 312 404 (GSM)
Tel.: 031 - 47 65 28 (Home)
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