Fanurie's blog

By Fanurie, 3 years ago, In English

Hello!

We are glad to announce the start of the sixth edition of FIICode!

The format will remain the same as last year (3 qualifying rounds and the final one). All rounds will be held online on CSAcademy.

Because the final is online, we decided that this year the contest will be national, meaning that only the first 25 Romanian / Moldovan participants will be able to officially participate in the final round and win the prizes. But those who cannot officially compete in the final are of course welcome to participate. Also, the difficulty of the qualifying rounds will be similar to a Div.2 Codeforces contest (slightly easier).

The problems are created and tested by bicsi, Juve45, denis2111, lungualex00, cristian1997, sebi110, vladd, Fanurie, antohir, RazvanPanaite, IulianOleniuc, ApetriiRadu, pauldiac and smoc_georgemarian.

To officially participate you can register here.

The first round takes place Saturday, April 17, 18:05 Romanian time.

LE: First round starts in 1h

The second round takes place Tuesday, May 4, 15:05 Romanian time.

The third round takes place Friday, May 21, 18:05 Romanian time.

LE: Third round starts in 2h

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3 years ago, # |
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Excited to participate after a long time at CSAcademy.

But the round clashes with TCO21 Algorithm Round 1A. Is it possible to move the contest time an hour early?

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    3 years ago, # ^ |
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    We apologize for the overlap, but we think that some of the participants would not know in time if the contest would start an hour earlier so we will not change the time of the contest.

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3 years ago, # |
  Vote: I like it +95 Vote: I do not like it

Once every year CSA rises from the dead for FIICode.

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3 years ago, # |
  Vote: I like it +58 Vote: I do not like it

Hey, since this is held on CSA I want to bring a few issues into light in hopes that they can be fixed and will not ruin FIICode.

I have done a few virtual contests on CSA lately and noticed the following:

  • When I submit something, I don't see the test cases being judged like I used to (except sometimes I do), instead I see "submission received, estimated queue time: [some number]", with "some number" sometimes being absurdly high. There is no way to see the result besides going to the submissions tab.
  • If I click on the link to submissions from some other page, I have to also manually refresh the page to see the data actually refreshed — IMO, clicking the link should do that.
  • Most importantly, the verdict shown in the submissions tab is incorrect sometimes. It sometimes shows a wrong solution as "accepted"; you have to open the list of test cases to actually see that one of them failed.
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    3 years ago, # ^ |
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    Thanks for the notifications, we are looking to solve them today.

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3 years ago, # |
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Auto comment: topic has been updated by Fanurie (previous revision, new revision, compare).

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3 years ago, # |
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Where would the contest link be given? Fanurie

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3 years ago, # |
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Auto comment: topic has been updated by Fanurie (previous revision, new revision, compare).

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

How to solve endgame?

Not sure if this soln was intended.
But for Scrambled Eggs, this assignment problem can be converted into a bipartite graph matching problem with given nos on one side and primes factors on another side. Then it asks for a match of size K. Same nos on the left side can be compressed.

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3 years ago, # |
  Vote: I like it +19 Vote: I do not like it

The editorials for all the Round 2 problems have been posted on CSAcademy. We can further discuss solutions here.

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    3 years ago, # ^ |
      Vote: I like it +19 Vote: I do not like it

    I read the editorial of Endgame. Suppose there are segments $$$[0,1], [1,2], [3,4], [4,5]$$$, It seems editorial says $$$\{[0,1],[1,2]\}, \{[3,4],[4,5]\}$$$ are chains. I think this is not true because they are not maximal. $$$\{[0,1],[1,2],[3,4],[4,5]\}$$$ satisfies the condition then this is a chain.

    Sorry if I misread the statement.

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      3 years ago, # ^ |
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      A chain is a maximal set of intervals with the property that for any interval I in the set, there is an interval J in the set (I \ J) such that their intersection is non-empty or I is the only interval in the set (the intersection is considered non-empty even if the intervals intersect in a single point). This is the definition from the statement.

      Thus {[0, 1]} is not really a chain because it is not maximal, we can expand the set by adding [1, 2] to it.

      EDIT: your example is not a chain because it is not connected.

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        3 years ago, # ^ |
          Vote: I like it +16 Vote: I do not like it

        It seems there is no restriction for connectivity.

        In $$$\{[0,1],[1,2],[3,4],[4,5]\}$$$, when $$$I=[0,1]$$$, there exists $$$J=[1,2]$$$, etc.

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          3 years ago, # ^ |
            Vote: I like it +22 Vote: I do not like it

          It seems you are right. I read the statement again and it looks like the definition does not imply connectivity as I previously thought. When I first read the statement I thought "hmm maximal set of intervals...hmm every interval I has another interval J such that their intersection is non-empty, this must mean that the chains are actually connected components". This is one of the rare problems which in order to solve it you have to read the statement wrong:)

          All in all, this looks like an honest mistake on the part of the authors.

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      3 years ago, # ^ |
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      We are very sorry for the mistake in problem D of round 2. We should've mentioned that the intervals in a chain must be connected. We hope it affected you as little as possible.

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    3 years ago, # ^ |
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    I've solved Endgame with another approach. Instead of segment tree I've use stacks, and made two passes (one from left to right, and one from right to left). While iterating from left to right standing at position $$$x$$$ we need to know what is the maximum number of chains can be made from the segments that have both ends strictly to the left from position $$$x$$$. We can keep all those chains in stack (actually, we need to save only rightmost point of each chain). Similar for right-to-left pass.

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

Auto comment: topic has been updated by Fanurie (previous revision, new revision, compare).

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3 years ago, # |
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Auto comment: topic has been updated by Fanurie (previous revision, new revision, compare).