This year also solving one problem i.e prob B was sufficient to qualify.
I solved only prob B and got my 25 ponits to move to next round.10 for small
and 15 for large data.

On Sun, May 8, 2011 at 11:57 AM, Brian Watkins <[email protected]> wrote:

> This is the first time you've needed to solve more than one problem to
> qualify for a GCJ.  Back in 2003-2006 one problem always sufficed and
> in 2008-2010 one small and one large set was enough.  Recently it's
> been impossible to qualify without a large set; that made it
> interesting to wait for the end to see if your large passed.
>
> This time any three small sets guaranteed qualification and you could
> see long before the round finished that you had thirty guaranteed
> points, more than the required twenty-five.  For the first time, you
> could be sure you had qualified as early as a few minutes into the
> round.
>
> Problem A was a simple simulation but couldn't add up to enough points
> to qualify even with the large set.
>
> Problem B was the most complicated.  It required you to keep track of
> a lot of conditions and check for various combinations in a loop, but
> the underlying algorithm was trivial.  I skipped this one.
>
> Problem C was a matter of recognizing the even parity condition after
> the statement gave you a battery of hints.  The actual writing of the
> answer was very simple.
>
> Problem D was a probability computation mixed with an induction
> condition.  It reminds me of the form of the series expansion
> expression of (1+1/n)^n, though the answer is different.  Once you see
> that the average thumps to rearrange n numbers in order is exactly n,
> it's even more trivial than C to write.
>
> Problem D is the first ever contest problem that requires you to find
> a general O(n) sorting algorithm to solve it.
>
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-- 




Thanks & Regards,
Satyajit Bhadange

http://satyajit-algorithms.blogspot.com/

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