By RussianCodeCup, history, 10 years ago, translation, In English

Hi, all!

The last possibility to get to the Elimination Round of Russian Code Cup this year is the Third Qualification Round, that will take place on Sunday, June 5, 16:00. Everyone who has not yet qualified is welcome to participate. Good luck, and see you at Russian Code Cup 2016!

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10 years ago, hide # |
 
Vote: I like it +40 Vote: I do not like it

Do you can check system faster? In the previous round we were waited about two minutes to know the results.

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10 years ago, hide # |
 
Vote: I like it +31 Vote: I do not like it

Curiously, did they fix the bug with problemset presentation? Or when the contest starts, we will see 1st-qualifiication's problems again?

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10 years ago, hide # |
 
Vote: I like it +5 Vote: I do not like it

Can someone who is not Russian participate ?? if yes Link to the contest website please .

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    10 years ago, hide # ^ |
     
    Vote: I like it +51 Vote: I do not like it

    http://www.russiancodecup.ru/en/

    The problems are quite interesting, but mail.ru reinvented the wheel developing their own system, which turned out to be the worst Online Judge ever, so expect huge bugs and slow testing during the contest. If you don't want a t-shirt from them, you'd better just wait for the problems to appear on the CF Gym.

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10 years ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

How can I submit solutions ? I am not seeing any submit button option.

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10 years ago, hide # |
 
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Is there some problem with the judge?

My code runs fine offline and on ideone, but gets runtime error on test case 1 in problem D. I asked the jury, and they confirmed that Case #1 is indeed sample case..

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10 years ago, hide # |
 
Vote: I like it -8 Vote: I do not like it

For D we needed this, right? http://e-maxx.ru/algo/tree_painting Found it 5 minutes before the end :/

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10 years ago, hide # |
Rev. 3  
Vote: I like it 0 Vote: I do not like it

What's wrong with this solution for D?

Keep a BIT on the DFS order for prefix sums from root to the node. For updates, subtract 1 from those which lie in the subtree of the current node, and to answer query use formula val[a] + val[b] - 2 * val[LCA(a, b)]. Only corner case is that both nodes are children of same node that has already been deleted where answer is 2. I kept getting WA on case 7.

Code
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    10 years ago, hide # ^ |
    Rev. 2  
    Vote: I like it +3 Vote: I do not like it

    There is no corner case. You just have to subtract 1 from everyone in the deleted node's subtree, including the node itself.

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

      Shit, I was doing exactly that initially, but had some bug so rewrote large part of it, and messed that up and came up with new logic :/

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10 years ago, hide # |
Rev. 2  
Vote: I like it +33 Vote: I do not like it

To D I copy-pasted both LCA and HLD summing on paths. But what is funny is that I didn't copy it from my library but from problem E from RCC week ago :P (however I could as well copy-pasted from library, but second round was closer in directory tree than library :P).

Week ago 120 minutes weren't sufficient for me to get a result sufficient for qualification. Today 8 minutes were enough :P.

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10 years ago, hide # |
Rev. 2  
Vote: I like it -15 Vote: I do not like it

What's wrong with my solution? I used the same idea with the editorial, but instead of using a segment tree with lazy propagation, I used a trick with BIT that allow me to do range update, single node query (the same trick with this editorial, problem C div 1).

My solution

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10 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

when will the editorial be out??