I liked my solution well enough to post it. It really makes sense this
way.
(in PYTHON):
if L * C >= P: return 0
count = 0
while(True):
count += 1
P = P / C
if L >= P: break
return math.ceil(math.log(count,2))
On May 23, 5:10 am, 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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