lperovskaya's blog

Yandex.Algorithm-2014: qualification round
By lperovskaya, 11 years ago, translation, In English

We’d like to invite you to take part in the qualification round of the Yandex.Algorithm competition that will start on May 25, 2014, at midnight. You need to start your qualification within 24 hours from the beginning of the qualification round. Qualification will last 100 minutes.

To advance to the next stage of Yandex.Algorithm you will need to solve at least one problem during your qualification.

Participants, already advanced to the elimination stage from warm-up round, are also welcome on the qualification.

Good luck!

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11 years ago, # |
  Vote: I like it +11 Vote: I do not like it

I have a question:

From the Rule, "If a problem is solved with a “blind” submission, the total penalty time is decreased by n seconds, where n = ((number_of_contestants_who_did_not_solve_this_problem) × (contest_length_in_seconds)) ÷ (total_number_of_contestants)."

Who will be counted as "number_of_contestants_who_did_not_solve_this_problem" and "total_number_of_contestants"? I wonder those who are eligible for the round but not participate in the round at all will be counted or not.

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

    It includes all people who have made a submission (even get 0 point in the end).

    You can confirm this by calculate on this standing.

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11 years ago, # |
  Vote: I like it -41 Vote: I do not like it

can someone explain how to solve the problems ?

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    11 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    I hope there are editorial after contest has ended

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11 years ago, # |
  Vote: I like it +17 Vote: I do not like it

This webpage has a redirect loop :(

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11 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Will the coming rounds be held and rated on Codeforces? I mean there was a previous round on CF in 2011...

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

sadly I missed it :( :(

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

    Actually, you have about 5 hours to start it

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

      BIG Thanks I thought it's saturday not sunday ,Thanks again

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11 years ago, # |
  Vote: I like it -15 Vote: I do not like it

Ah! it needs further registration, I thought codeforce handles are enough. Too bad!

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

Is there any way to see other participants solution after the contest?

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11 years ago, # |
  Vote: I like it -15 Vote: I do not like it

Why cant we resubmit blind submissions?

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

    Because it is no more blind. You have already been told that what you initially thought was incorrect, and you have been provided a chance to correct it.

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      11 years ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it
      You have already been told that what you initially thought was incorrect

      Who told me?:D

      and you have been provided a chance to correct it.

      What?

      It does not say after the submission if it has passed the tests or not (only sample tests are checked).

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

        What's the point of blind submission now?

        It does not say after the submission if it has passed the tests or not (only sample tests are checked).

        Do you want to check that your submission passes the tests from the problem statement? That is something you can do manually.

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          11 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it
          What's the point of blind submission now?
          It does not say after the submission if it has passed the tests or not

          while regular submissions are being tested immediately.

          Do you want to check that your submission passes the tests from the problem statement? That is something you can do manually.
          No and this is why the result of blind submission is useless.
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            11 years ago, # ^ |
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            I think I didn't got you correctly. Were you asking why yo can resubmit blindly during the contest? I was thinking about upsolving due to some reason. If your question is about the contest then I think the reason is to add more drama, you should be sure about first time you submit it.

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

In the final standings, the + indicates normal submission, while a tick indicates blind?

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11 years ago, # |
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Is there any editorial for the contest ??

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

    Of course not, its not even over yet.

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      11 years ago, # ^ |
        Vote: I like it -6 Vote: I do not like it

      I don't understand, the qualification round is not finished ??

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        11 years ago, # ^ |
        ← Rev. 4   Vote: I like it 0 Vote: I do not like it

        It's a "virtual" contest. You can start your participation in any moment between it's start and finish time. At the end, results of the participation of all participants will be merged and we will have final standings.

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        11 years ago, # ^ |
          Vote: I like it -8 Vote: I do not like it

        Contest already finished.

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

          Actually, no. What's over is the time window when one can start a virtual contest. A participant can start at 23:59 MSK and still have 100 full minutes to compete.

          The real end time is 01:40 MSK.

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    11 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    I put together a short editorial in case someone needs it: http://www.rivsoft.net/blog/yandex-algorithm-2014-qualification-round/

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11 years ago, # |
  Vote: I like it -28 Vote: I do not like it

tourist solved all problems in 41 minutes! That is insane!!

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

    They were mostly implementation problems, though.

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11 years ago, # |
  Vote: I like it +44 Vote: I do not like it

Damn, why no any reminding email :(!? I missed such a good contest, because I missed qualification round :(.

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    11 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Did you participate in Warm Up round?

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

      No, I didn't have time to (lot of work to studies) and since it was "Warm Up" I assumed that problems presumably won't be amusing. Were reminding e-mails sent to Warm Up participants only? If so, that isn't a very clever idea.

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        11 years ago, # ^ |
          Vote: I like it +24 Vote: I do not like it

        It is pity, because Warm Up round was also an opportunity to qualify for main rounds (in the same way as qualification round — by solving at least one problem).

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

Seriously, why no email reminder? I missed it too, it was a really good contest and I wanted to participate :( :(

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11 years ago, # |
  Vote: I like it +29 Vote: I do not like it

I believe the statement of problem F was incorrect.

Currently the statement says: "The sequence starts with an opening round bracket “(”, ends with a closing round bracket “)” and holds only English letters (one or more) in between the brackets. Examples: (blush), (facepalm), (SmileSmileSmile)."

However, there was no "(one or more)" in the beginning of the contest. Without "(one or more)", I think "()" satisfies this condition (and I guess this is the reason of so many failures).

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    11 years ago, # ^ |
    ← Rev. 2   Vote: I like it +18 Vote: I do not like it

    We considered to retest all failed on test 4 submissions before fix with "one or more" was added with understanding when () is a smiley. Retesting will be done soon.

    Update: Retesting complete; all solutions failed on test 4 before statement fix were rejudged.

    But anyway, i am not sure if there is no difference between "holds only English letters" and "does not contain any characters, other than English letters" (something like that must be used for set of English words + empty set). For the first version of statement more likely string must be non-empty

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

C# solution for F:

// (づ*ω*)づﾐ★゜・。。・゜゜・。。・゜☆゜・。。・゜゜・。。・゜゜・。。・゜゜・。。・゜
Console.Write(Regex.Replace(Console.ReadLine(), @"([^A-Za-z0-9\s()]\)|\([A-Za-z]+\))", m => "<S>" + m + "</S>"));
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    11 years ago, # ^ |
    ← Rev. 2   Vote: I like it +10 Vote: I do not like it

    In java: out.println(in.readLine().replaceAll("(\\([a-zA-Z]+\\)|[^a-zA-Z0-9 ()]\\))", "<S>$1</S>"));

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      11 years ago, # ^ |
        Vote: I like it +13 Vote: I do not like it 
      print(re.sub(r'(\([a-zA-Z]+\)|[^a-zA-Z0-9\s()]\))', r'<S>\1</S>', input()))
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        11 years ago, # ^ |
        ← Rev. 2   Vote: I like it +29 Vote: I do not like it

        in perl:

        s/(([a-zA-Z]+)|[^ a-zA-Z0-9()]))/<S>$1<\/S>/g, print while (<>)

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

What was the approach to solve problem E?

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

    Change the sign of all elements in odd positions. Now you need to find the segment with the largest sum (in absolute value). If you calculate partial sums (from the start of the array to the current position), then the sum of any segment is the difference of two partial sums (corresponding to the segment's start and end). So all you need to do is go through the array once, keeping track of the current sum, and the minimum/maximum value of this sum so far.

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

The rules say that the participants with the highest cumulative scores will advance to the finals. Does this mean that we have to participate in all 3 rounds? They are scheduled in such a way that in any time zone at least one contest will be in the middle of a night.

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

    It doesn't — you can participate however you want. Still, the more you participate, the better your chances, since once excellent result gives you much more than 3 good ones.

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      11 years ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it

      That's good to know. The rules say it uses total points, number of solved problems and total penalty, but doesn't say which formula is used, so I can't tell how important it is to participate in all 3 rounds. Also, I thought the first round was yesterday, guess I'll wait for it today then as its the only one that happens at night for me.

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