Thanks! But I am still not able to understand that how you got f(1)=2 and f(2)=3?
On Sat, Sep 12, 2009 at 12:39 PM, bas <[email protected]> wrote: > > Let f(k) be the expected number of packs we need to buy if we still > don't have k cards (or we have C-k cards). > To solve the problem we need to find f(C). Note: f(0)=0. > > In the first example C=2; > When we buy the first pack, we will get one of the cards we need with > the probability 1. > f(2)=1*f(1)+1 > means that we expect to buy that many packs to get all the cards we > need. But we still don't know f(1). > For k=1 we have one card we need, and we need to find the other one. > When we buy this next pack, we will get the card we already have with > a probability 1/2 and the card we need with prob 1/2. > f(1)=1/2*f(1)+1/2*f(0)+1 > Then: f(1)=2; f(2)=3 > > > > > > On Sep 11, 9:54 pm, qasim zeeshan <[email protected]> wrote: > > Hi All, > > First of all congrats to all who made it to next round. > > > > I have seen some codes for Problem C. but didn't understand. Can anyone > > please explain both example test cases for Problem C. > > > > Thanks > > > > -- > > Qasim Zeeshan > > > -- Qasim Zeeshan PUCIT, PU Alumni '04 Software Engineer CambridgeDocs Pakistan A Division of Document Science | EMC USA http://qzeeshan.blogspot.com/ --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
