radhakrishnancegit's blog

Announcing Kurukshetra 2012 Online Programming Contest

By radhakrishnancegit, 13 years ago, In English

Hi,

We would like to invite you all to participate in Kurukshetra OPC 2012 (An Individual Contest ), a 4-hour acm-icpc style online programming competition, which is organized as a part of Kurukshetra 2012 , the Technical fest of College Of Engineering Guindy, Anna University,India. Start time : Tuesday ,31st Jan 2012,9 pm IST . [ Find when it starts in your time zone Time in other Zones]

Contest Page: http://www.spoj.pl/KOPC12/

Problem Setters: innocentboy2,sundargates

Contact: radhakrishnancegit@gmail.com or opc@kurukshetra.org.in

Problems : 6-7 problems with mix of easy (2), medium (3) and hard (1-2) levels.

Prize Money in INR
First Place: 15000 INR
II place : 5000 INR
III place : 3000 INR

We are expecting that a few of you will finish all problems in around 1.5 hours.Considering the coders of our college we extended the duration of the contest to 4 hours

Please ignore this if you came to know about this contest by other means .Sorry for that .

Thanking you.

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Comments Comments (6)
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13 years ago, # |
  Vote: I like it +1 Vote: I do not like it

A gentle reminder. Contest Started :) Happy Coding :)

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13 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Nice contest! I liked it :) Thank you.

I had some hard time with "Combinations" and I still can not solve it. Maybe someone could tell me the idea behind this problem? :)

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    13 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    If x=(nC1)2+2*(nC2)2+3*(nC3)2+4*(nC4)2+............+n*(nCn)2, than x=n(2nCn)-x. Using precalc we find all n! and (n!)^-1 (n<=2000000) modulo p, then only find in O(1) n*(2nCn)*(2^-1).

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13 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Has anybody solved F using DP in O(n^3)? Is the problem in input, should we read using char[]? Thanks.

UPD There is a DP solution in O(n^2). The large input is not guilty.

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13 years ago, # |
← Rev. 4   Vote: I like it +3 Vote: I do not like it

I am very amazed how this wrong greedy solution for problem C get accepted:

For digits 9 --> 1, put them in smallest places that no numbers were put before.

That solution failed the test case: 0 0 0 2 0 2 0 2 0, and failed every random test case I generated (against the modified D's bipartie matching algorithm I used to calculate the best length of the number).

So, either the test case is super super weak, or the author's solution is wrong.

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    13 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I would like to apologize for wrong test data in problem C.As we discovered this after the end of the contest we are going to remove the problem from the contest along with the scores and penalties.Greedy fails .

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