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 <aakashj....@gmail.com> wrote:
> what about precomputation and then binary search...?
>
>
>
> On Tue, May 24, 2011 at 6:37 AM, sravanreddy001
> <sravanreddy...@gmail.com>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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>
>
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> -Aakash Johari
> (IIIT Allahabad)
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