eh2arch's blog

Codecracker 2015
By eh2arch, 9 years ago, In English

Update2: The contest duration for round 2 has been extended to 3 hours!

Update: The details of Round 2 have been added! Please check :) To upsolve questions of Round 1, head over to our site and try the questions on the practice contest.

Presenting, the flagship event of MUKTI -- CODECRACKER! hosted by National Institute of Technology, Durgapur.

Codecracker is a competitive coding event, drawing participation from over 50 countries. The contest will be governed by ACM-ICPC rules.

Cash prizes worth 25000 INR on offer. The competition is cut-throat. Do you think you have what it takes?

Come. Code. Challenge yourself against the best and prove your mettle. Log on to: http://codecracker.in/index.php/login

Round1: Timings: 21:30 IST Duration: 2.5 hours

Round2: Timings: 21:30 IST Duration: 3 hours

The winners will be decided upon the basis of cumulative performances in the both the rounds

Log on to https://www.facebook.com/groups/codecracker for constant updates!

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9 years ago, # |
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how to register? i only see login

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9 years ago, # |
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The registrations will open soon :)

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9 years ago, # |
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The registrations are open now!

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9 years ago, # |
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nice contest...`` But ~~~~~ std::ios::sync_with_stdio(false); cin / cout (for i/o) ~~~~~ are not accepted by ur online compiler...

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9 years ago, # |
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What was the solution for Colors problem?

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    9 years ago, # ^ |
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    At first, let's solve problem for even n. We can see that answer for n = 2 is k·(k - 1). Let's add one sketch on the left side and one sketch on the right side. For each possible coloring of 2 central sketches there are k2 - 3k + 3 ways of coloring added sketches ((k - 2)·(k - 2) ways, if color of the leftmost sketch is differ from the color of third sketch, and k - 1 more ways, if they are the same). By induction we can prove formula: ans[2x] = k(k - 1)·(k2 - 3k + 3)x - 1.

    If n is odd, we can remove central sketch, solve problem for even n and multiply answer by k - 2 (it is number of possible colors for central sketch).

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      9 years ago, # ^ |
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      I am the author of this problem . This problem has a very simple solution . Suppose there are N dots in a line and we are told to obey the required conditions while colouring . Now we can detach the second half and reverse it and put it just below the upper half . Now the problem boils down to colour each dot such that no adjacent colours are same .This can be done by simple combinatorics . We can make separate cases for even and odd .

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      9 years ago, # ^ |
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      Shoot, I had this but still failed...

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9 years ago, # |
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Can participate in Round 2 without participating in Round 1? I missed round 1 "_"

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    9 years ago, # ^ |
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    Yes , you can participate in Round 2 . The cumulative scores of both rounds will decide the winner so you still stand a chance .

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9 years ago, # |
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The details for Round 2 of the competition have been updated! The contest will be of 3 hours duration to cater to the more challenging and interesting problem set this time around!

Don't miss out on the fun :)

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