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 > > > -- > > You received this message because you are subscribed to the Google Groups > > "google-codejam" group. > > To post to this group, send email to [email protected]. > > To unsubscribe from this group, send email to > > [email protected]<google-code%[email protected]> > > . > > For more options, visit this group at > >http://groups.google.com/group/google-code?hl=en. > > -- > You received this message because you are subscribed to the Google Groups > "google-codejam" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group > athttp://groups.google.com/group/google-code?hl=en. -- You received this message because you are subscribed to the Google Groups "google-codejam" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-code?hl=en.
