if u have many test cases, this approach is helpful.

On Tue, May 24, 2011 at 6:42 AM, Piyush Sinha <[email protected]>wrote:

> its equal to calculating the Fibonacci numbers till we get a number
> which is closest to and lesser than N...anything better??
>
> On 5/24/11, Aakash Johari <[email protected]> wrote:
> > what about precomputation and then binary search...?
> >
> >
> >
> > On Tue, May 24, 2011 at 6:37 AM, sravanreddy001
> > <[email protected]>wrote:
> >
> >> @Anyone worked on this before?
> >> I'm thinking of a time complexity of *(log N)^2 --> Square of (log N)*
> >> I've to prove on this..
> >>
> >> If someone have time.. can you *prove that, the T'th fibinocci number is
> >> always greater than 'N'*
> >> *where T = (log N)^2 *
> >>
> >>
> >>
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-- 
-Aakash Johari
(IIIT Allahabad)

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