Because the range is defined by multiplication instead of sum:
A-d <=x<=A+d                 : sum
A/C <=x<=A*C                : multiplication


2010/5/23 Saurabh <[email protected]>

> Hi Bharath,
>
> Thanks for your reply. But I couldn't understand one thing. Why is
> taking the geometric mean the most optimal method ?
> Could you explain a bit more on this ?
>
> Thanks.
>
> On May 23, 5:19 pm, Bharath Raghavendran <[email protected]> wrote:
> > You are trying to find a value of a between L and P such that u r sure
> that
> > a contestants can access at a time but C*a contestants cannot.
> >
> > Its basically a binary search for a satisfying element between L and P
> with
> > a small modification that instead of taking (L+P)/2 each time, go for
> their
> > geometric mean. This will maintain the ratio of first and last between
> the
> > two halves and hence will be most optimal.
> >
> > On 23 May 2010 17:40, Saurabh Aggarwal <[email protected]> wrote:
> >
> >
> >
> > > Hi Guys,
> >
> > > Could anyone please explain the logic behind the "Load Testing" problem
> ?
> > > From the solutions I gathered that they have found the number of bits
> in X,
> > > where
> >
> > > L * (C^X) > P
> >
> > > But couldn't understand why this was done ?
> >
> > > Thanks.
> >
> > > --
> > > Regards,
> > > Saurabh Aggarwal
> >
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-- 
grato,

Gustavo Pacianotto Gouveia

LTI - Laboratório de Técnicas Inteligentes
Escola Politécnica da Universidade de São Paulo
  [email protected]
  [email protected]
  [email protected]

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