By lperovskaya, 11 years ago, translation, In English

Hi!

The next stage will consist of three elimination rounds that will be held on:

Online round 114 Jul 2013, 21:00 MSK; Materials; Analysis

Online round 218 Jul 2013, 13:00 MSK; Materials

Online round 322 Jul 2013, 05:00 MSK; Materials

Each contest round lasts 100 minutes. Links to the rounds are avaliable on the round names above.

The scoring model for this stage will be Grand Prix 30.

To advance to the final round in St. Petersburg, Russia, you don't have to participate in all online rounds. You will need to place within one of these groups:

*top 4 in one of three rounds among contestants who have not yet advanced to the final round;

*top 9 in the cumulative results of two out of three rounds among contestants who have not yet advanced to the final round; or

*top 4 in the cumulative results of three rounds among contestants who have not yet advanced to the final round,

All finalists, another 75 best participants of elimination stage and another 75 random participants of the elimination stage, who will solve at least one problem will get a T-shirt!

You can always read the full schedule and regulations on the official website.

Good luck! Next stop: Saint Petersburg!

UPD: Results of the online round 1 are now official and we congratulate Kenny_HORROR for impressive victory using "all open" strategy.

First finalists are: Kenny_HORROR, Petr, tourist, eatmore.

UPD2: Congratulations to tourist for the online round 2 victory. We are glad to invite vepifanov, yeputons, ivan.popelyshev, peter50216 to the final round!

UPD3: Based on their scores in round 3, winger, KADR, ainu7, s-quark are invited to the final round. Congratulations!

Rightful place among finalists goes to

based on two out of three elimination stage rounds : burunduk3, misof, niyaznigmatul, cerealguy, Mimino, liympanda, romanandreev, watashi, RAVEman;

for the overall performance in the stage: Nerevar, ilyakor, dreamoon_love_AA, pparys.

Dear finalists! Congratulations and see you soon in St. Petersburg. I will contact you shortly on the matter of your final round participation. Please, do not ignore my emails and answer promptly =)

UPD4 Materials of the second test round are published: Test round 2; Materials

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

I apologize for the problem with the submission link.

If you receive the decline response from server use the link below

http://contest.yandex.com/contest/Contest.html?contestId=308&tab=submit

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    11 years ago, # ^ |
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    There was also a problem with sending questions to jury. I wasn't able to send it: 'There was an error. Pleaes try to send your question once again.'

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11 years ago, # |
  Vote: I like it -14 Vote: I do not like it

Will it be unrated?

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

I'm trying to submit a problem after the contest (for practice). I get the error message "You can send only UPSOLVING submission after contest end". Anyone know what I'm doing wrong?

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    11 years ago, # ^ |
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    You should submit using the new interface to get upsolving, so if it doesn't work as it doesn't for me — you're out of luck :)

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11 years ago, # |
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Yesterday I wait for the start at http://contest.yandex.com/contest/ContestList.html site, I was logged in and this site was in english. After the start the problems appear in russian, I looking for a button to switch english, but I didn't find.. So I went to the codeforces site and find a link http://algorithm.contest.yandex.com/contest/308/problems/ , it works fine, and here is a button what I looking for to switch english..

Is there any option to switch in english in the first site? Why do you have two different site? I'm just courious :)

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

I wonder how to decide another 75 best participants. Grand Prix 30 only affect top 30 , so only 30*3=90 top participants can get positive score at most. Therefore this system is not enough to decide top 100 who get T-shirt.

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    11 years ago, # ^ |
    Rev. 2   Vote: I like it +37 Vote: I do not like it

    There are also another two scoring criteria: number of solved problems and penalty time.

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      11 years ago, # ^ |
      Rev. 2   Vote: I like it +8 Vote: I do not like it

      Well, there is a flaw in this scheme :). I mean that these criteria are incomparable when applied to different contests.

      UPD. Yes, I've made a mistake. We don't compare them directly, but their sum.

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

I submitted only 1 problem of blind type in Round 1, it failed by int overflow.

Today, I also submitted only 1 problem of blind type in Round 2. Guess what? It failed by long long overflow. (problem C: if we sum the areas up, it will exceed long long) :(

Maybe only "All open" strategy is good for me.

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11 years ago, # |
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I want to know where I can find the solution to a problem.Some question I have not thought.

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11 years ago, # |
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Will you make editorial for these rounds, or possible to see other contestants solution?

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11 years ago, # |
Rev. 2   Vote: I like it +19 Vote: I do not like it

It was 5 seconds too late to submit E (and get accepted). See you next time, Russia :(

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11 years ago, # |
  Vote: I like it +56 Vote: I do not like it

Looking through years:

  • ACM ICPC ~2 problems per hour
  • TopCoder SRM 2.4 problems per hour
  • Google Code Jam 2.4 problems per hour
  • Codeforces Round 2.5 problems per hour
  • Russian Code Cup 3 problems per hour
  • Yandex Algorithm 3.6 problems per hour
  • ...
  • A competition in 2030 10 problems per hour expected

Yes, my hobby is extrapolation :)

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

Re-comment in English Version can anybody explain the solution of Problem A on Online round 3? getting a lots of wa .

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

    I will try to explain my solution.

    By the linearity of expectation, the expectation of the sum is equal to the sum of expectations. (i.e, the sum of the value of each candy multiplied by probability that you will eat it).

    There are two types of candies.

    For bad candies, you know that you will eat exactly one or zero of them(the latter is true if there are no bad candies at all). So, if there is at least one bad candy, you will eat each of them with equal probability of

    .

    The corresponding expectation coming from bad candies is the sum of the values of bad candies multiplied by p_bad.

    Another part of the expected sum is the sum for good candies. Let's iterate over all the good candies we have and calculate the appropriate expectation (the value of the candy multiplied by the probability that you will eat it).

    For i-th good candy, consider all possible numbers p of candies you can eat before it. It is clear that all candies you can eat before the i-th good candy must be good (otherwise you had to stop before). Therefore, the probability that you will eat it after eating other p candies is equal to the number of arrangemens of all candies of the form

    [good_candy1] ... [good_candyp][good_candyp + 1][all others]

    divided by the number of all possible arrangements with ith good candy sitting on (p + 1)st place.

    The number is equal to , where good_num is the total number of good candies.

    By summing over all possible p, and multiplying the sum by 1/n (the probability that the i-th candy will be at place p+1) you get the expectation that you will eat candy i.

    You need to be careful with special cases when p is greater than good_num, when there are no bad candies, etc.

    Here is the code: http://pastie.org/8165899

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

Any news about T-shirts? :)