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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