VK Cup 2016 — Round 1
By Radewoosh, history, 9 years ago, In English

Hello everybody!

I'm glad to announce that Round 1 of VK Cup 2016 will take place this Monday, and me (Radewoosh) and Kamil Dębowski (Errichto) are the problemsetters!

There will be an official round for teams from VK cup, but if you are not eligible to participate in it, then you can compete (alone, not in team) in one of two additional editions (one for div.1 and one for div.2), so everybody is invited to take part in the competition! Just register in your category here. All three rounds will be rated. Div.1 and Div.2 editions will look like normal CF round, but will have common problems with official edition.

If you can't register before the round, then you will be able to do it during the contest (but not for the entire duration, you can cheсk it here). Let's thank Mike for this great feature!

We want to thank GlebsHP for help in preparing the problems and MikeMirzayanov because without him we wouldn't have such a great platform as Codeforces, where we all can train and develop our passion.

You will again help Limak, your favorite bear. This time it may be harder, because evil Radewoosh will try to disturb him.

We wish you good luck and great fun! Can't wait to see you during the contest! :D

UPD Scoring will be:

For VK: 500 — 750 — 1000 — 1500 — 2000 — 3000

For Div.2: 500 — 1000 — 1500 — 2000 — 2500

For Div.1: 500 — 1000 — 1500 — 2000 — 3000

UPD Editorial is ready.

UPD Congratulations to the winners!

In official VK:

1.Never Lucky: subscriber and tourist
2.SobolevTeam: Seyaua and sdya
3.LNU Penguins: witua and RomaWhite
4.Dandelion: Um_nik and sivukhin
5.uıɟɟnɯ ɐuɐuɐq ǝɥʇ ɟo uɹnʇǝɹ╰(º o º╰): enot110 and romanandreev

In Div.1

1.dotorya
2.kcm1700
3.JoeyWheeler
4.KrK
5.Swistakk

And in Div.2

1.osmanorhan
2.nhho
3.fudail225
5.agaga
4.alanM

Also let's thank qwerty787788 and AlexFetisov for testing problems, without them it would be much harder to prepare contest, so give them an applause!

Announcement of VK Cup 2016 - Round 1
  • Vote: I like it
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  • Vote: I do not like it

Comments Comments (165)
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9 years ago, # |
← Rev. 3   Vote: I like it +3 Vote: I do not like it

Standard important question — are people from one team allowed to use more than 1 computer / are they allowed to use more than 1 computer simultaneously (in case they are far from each other and they literally need to use two computers)?

EDIT: Not valid anymore.

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9 years ago, # |
← Rev. 3   Vote: I like it +8 Vote: I do not like it

To participate individually or To participate with your friend, that's the question!

UPD: To participate individually it is!

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

How would rating change be calculated with both teams and individuals in the same contest ?

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

    Yeah!I have increased 154, and I am very lucky as for having 7 successful hacking attempt.

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9 years ago, # |
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I'm sorry if this question seems silly or repetitive, but will 2 person teams be allowed for the unofficial version too?

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

    Blog was updated, now it seems like we have to compete individually.

    A shame, and I was looking forward to doing a rated contest as a team too.

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      9 years ago, # ^ |
        Vote: I like it +74 Vote: I do not like it

      We consider the possibility to make such an experiment in the nearest future! :-)

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9 years ago, # |
← Rev. 2   Vote: I like it +26 Vote: I do not like it

I am kinda afraid that all those contests with individuals + teams will make even greater mess with rating system than what it is right now. Hope I'm mistaken.

EDIT: Yay :)

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9 years ago, # |
  Vote: I like it +2 Vote: I do not like it

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

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

For how long this round will last?

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    9 years ago, # ^ |
      Vote: I like it +35 Vote: I do not like it

    Both official and unofficial version will be 2-hour long.

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9 years ago, # |
  Vote: I like it +23 Vote: I do not like it

I feel old when such young people are problemsetters.

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

Will all the problems be the same for the three contests?

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    9 years ago, # ^ |
      Vote: I like it -51 Vote: I do not like it

    when div1 and div2 contests are separated, it usually means that problems will be the same

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9 years ago, # |
  Vote: I like it +42 Vote: I do not like it

Will the problem statements be available in English?

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9 years ago, # |
  Vote: I like it +80 Vote: I do not like it

Will the problems be sorted by difficulty? In VK Cup 2015 the problems were not sorted.

If not, will the ordering be the same for all 3 versions of the contest?

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    9 years ago, # ^ |
    ← Rev. 2   Vote: I like it +46 Vote: I do not like it

    It says "Div.1 and Div.2 editions will look like normal CF round". So answer is yes i guess, at least for unofficial versions.

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      9 years ago, # ^ |
        Vote: I like it +72 Vote: I do not like it

      Why does this answer get downvotes, can anybody explain? I confirm what muratt wrote — yes, problems will be sorted. And you will see the number of points for each problem. Just a normal CF round (without any dynamic scoring).

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        9 years ago, # ^ |
          Vote: I like it +94 Vote: I do not like it

        (without any dynamic scoring).

        Yay!

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9 years ago, # |
  Vote: I like it +57 Vote: I do not like it

Looking forward to Bear and Forgotten Tree 3

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9 years ago, # |
  Vote: I like it +70 Vote: I do not like it

Hour 16. Still nobody asked whether this contest is rated or not. Something isn't right.

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    9 years ago, # ^ |
      Vote: I like it -17 Vote: I do not like it

    All 3 contests are rated

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      9 years ago, # ^ |
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      I know but by tradition someone asks anyway.

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        9 years ago, # ^ |
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        why dont you ask :)

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          9 years ago, # ^ |
            Vote: I like it +43 Vote: I do not like it

          I don't want negative contrbution. :D

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9 years ago, # |
  Vote: I like it +395 Vote: I do not like it

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9 years ago, # |
← Rev. 3   Vote: I like it -14 Vote: I do not like it

Ignore this comment

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9 years ago, # |
  Vote: I like it +252 Vote: I do not like it

Limak = reverse(Kamil).

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    9 years ago, # ^ |
      Vote: I like it +85 Vote: I do not like it

    Finally life makes sense!

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    9 years ago, # ^ |
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    I thought Errichto is his name lmao

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    9 years ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

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    9 years ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    That is not true origin of that bear's name. Errichto really likes trees (those growing from bottom to up :P) and "limak" is a name of type of tree, namely "water elm, Ulmus laevis (Eurasian elm closely resembling the American elm thrives in a moist environment)"

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    9 years ago, # ^ |
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    And why is Limak a polar bear (or any bear, for that matter)?

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

This will be my first VK cup....looking forward to it :-) Btw thanks to MikeMirzayanov for allowing registration during contest feature. It will be great.

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

There is no register option for the contests

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

If I had 500 points can I write it?

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9 years ago, # |
  Vote: I like it -115 Vote: I do not like it

For the love of tradition

Is it rated?

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9 years ago, # |
← Rev. 2   Vote: I like it -65 Vote: I do not like it

YOOOO Limak is back! HYPE THIS UP GUYS ♡♡♡

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9 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Can we participate in Div2/Div1 rounds if we haven't participated in the qualification rounds?

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9 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

Just 260 people registered for Div1 edition, from which not all will participate. Isn't it better to make the round (the div1 and div2 edition) combined?

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

    It won't be changed now.

    Btw. the situation wasn't that different one year ago. There were 438 div1 pariticpants with at least one submission. There is still a couple of hours left here so I guess there will be enough competitors to make it fun and interesting.

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      9 years ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      Either way it will be fun solving the problems :)

      I just taught if there are less participants it could mess up the rating a bit.

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      9 years ago, # ^ |
      ← Rev. 2   Vote: I like it +17 Vote: I do not like it

      couple of hours left

      I thought I was mistaken about the starting time when I read this, so I went to see the starting time and I turned out to be correct, so I used dictionary to translate "couple" to make sure I know the correct meaning of that word :D

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        9 years ago, # ^ |
          Vote: I like it +26 Vote: I do not like it

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          9 years ago, # ^ |
            Vote: I like it +4 Vote: I do not like it

          I used google translate but it only showed "two individuals of the same sort considered together." anyway when the comment is posted there was about 7 hours left, which is not that small and not near 2

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

        Couple: 2 (always)

        Pair: 2 (of the same type)

        A Few: 3-6

        A Handful: 3-6, but most often 5

        Several: 3-7 (and not identical)

        Some: 2-10+, can be used with continuous quantities

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          9 years ago, # ^ |
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          Gotta write this on a cheat sheet.

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

          Hasn't tweety posted an image from dictionary that positively proves that "couple" is not always "two"? By the way I should not be an authority when it comes to English correctness, but I use "couple", "a few" and "several" (but not "pair") in more or less the same meaning.

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          9 years ago, # ^ |
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          I won't count on this holding in any formal problem statement. Specifically, "several" usually means "0 or more".

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

        Obligatory xkcd

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        9 years ago, # ^ |
        ← Rev. 2   Vote: I like it +56 Vote: I do not like it

        I can't believe noone posted this... :D

        XKCD 1070

        (https://xkcd.com/1070/)

        Edit: And after 9 HOURS, somebody posts it WHILE I'M WRITING? Seriously? :D:D

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        9 years ago, # ^ |
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        Thanks people, nice English lesson. You guys should do it more often. :P

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9 years ago, # |
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Is this contest writing in English?

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

What's the scoring distribution for the division 2 contest?

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

Ahhh... I want to get green today. :/ Huh !

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9 years ago, # |
← Rev. 3   Vote: I like it +80 Vote: I do not like it

When I'm on hacking at problem A, at first I'm on room 1, then I moved to room 10 unexpectedly! While reading other submission code, then I moved again to room 1! And again moved to room 10!

Not only that, sometimes I can't even open another submission (for problem A which I had locked before). And other than that, the notice of "The solution has been hacked or resubmitted" popped out most of the time, while they aren't!

Bug?

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9 years ago, # |
← Rev. 2   Vote: I like it +63 Vote: I do not like it

Am I the only one that travels through different rooms and cannot hack properly?

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    9 years ago, # ^ |
      Vote: I like it -27 Vote: I do not like it

    You can only hack in your room

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    9 years ago, # ^ |
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    I meant that after clicking my room, I am sometimes sent to room 6, sometimes to room 8 (where I am). Moreover, I can see some solutions from both rooms and some of them I cannot see at all.

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    9 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it

    Ok, now I moved to room #6 :]

    I am also getting "This solution has been resubmitted or hacked." a lot (just like athin), but it is also invalid...

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

Hackerfest again. Spent last 40 minutes F5ing the room... :-\

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

What was hack for C?

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    9 years ago, # ^ |
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    3 1 1

    The answer is -1.

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      9 years ago, # ^ |
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      Two Solutions in my room which looked "hacky" but I couldn't hack due to spending time on C too much fails in this case. (just tested now)

      With a rush of fear, I opened my code. And it gives -1 (Thanks Almighty)

      I only hope there isn't more like that on system tests.

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    9 years ago, # ^ |
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    I hacked 7 times with 3 1 1.

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    9 years ago, # ^ |
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    I hacked a solution with : 5 1 1

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    9 years ago, # ^ |
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    100 1 1

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    9 years ago, # ^ |
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    Also, any idea of pretest 5?

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      9 years ago, # ^ |
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      I had 3 WA before passing pretest 5. My code was failing on cases such as 6 3 3 (ans is not -1)

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        9 years ago, # ^ |
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        why ans is not -1 , can you give an example ?

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          9 years ago, # ^ |
          ← Rev. 2   Vote: I like it 0 Vote: I do not like it 
          1 2
2 3
3 4
2 5
2 6
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            9 years ago, # ^ |
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            isn't d=4 for this example.

            d = 4 -> 3 -> 2 -> 1 -> 2 -> 6

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              9 years ago, # ^ |
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              No, because the diameter doesn't have to go through the root(1). It's:

              d = 4 -> 3 -> 2 -> 6
              or
              d = 4 -> 3 -> 2 -> 1

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              9 years ago, # ^ |
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              you have gone through "2" twice.

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          9 years ago, # ^ |
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          1 2

          2 3

          3 4

          3 5

          3 6

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

          edges 1 2
          2 3
          3 4
          2 6
          2 5

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      9 years ago, # ^ |
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      5 2 2

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      9 years ago, # ^ |
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      8 6 6

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      9 years ago, # ^ |
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      try 10 5 5

      ans:

      1 2
      2 3
      3 4
      4 5
      5 6
      5 7
      5 8
      5 9
      5 10

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9 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

I am pretty sure some people disappeared and then appeared again in my room — http://i.imgur.com/05MyTRC.png, http://i.imgur.com/kTZ7xFB.png :)

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9 years ago, # |
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What is wrong with my code for DIV2 C : code

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    9 years ago, # ^ |
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    check out the case h==d. when n >= h+1 there is a solution and otherwise there is not

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    9 years ago, # ^ |
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    For this case : 4 2 2
    Your solution gives:

     1 2
     2 3
     1 4
     
    diameter != 2
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    9 years ago, # ^ |
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    As far as I can see
    You first create a path of length H and then a path of length D-H . then attach the remaining nodes to the root .
    If D==H you can't attach the remaining nodes to the root because that just increases the diameter . Instead you should attach them to node 2.


    Hope this helps!

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9 years ago, # |
  Vote: I like it -35 Vote: I do not like it

C...

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9 years ago, # |
← Rev. 4   Vote: I like it +10 Vote: I do not like it

EDIT: this post contained some doubts about div1-D, but everything is clear now.

Maybe someone can find this useful: a proof that c does not matter when choosing an optimal order.

We're maximizing the score, which is where xi is when we are done with problem i. To maximize this, we should clearly minimize , which does not depend on c. (To minimize this weighted completion time, we should sort the problems by the so-called Smith's ratio . We have freedom when these ratios are tied, but again this does not depend on c.)

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    9 years ago, # ^ |
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    Your proof is correct. However, in case of several optimal solving plans, some of them might produce paradox, while others do not.

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9 years ago, # |
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Will BigInteger work for Div2 D ? Got 5 seconds late in submitting solution :'(

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    9 years ago, # ^ |
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    got tl on test 8 with BigInteger. http://pastebin.com/zMQbDyL1

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    9 years ago, # ^ |
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    I don't think so but I think we can solve it visualising the bits in the coefficients everytime they get multiplied by 2 there is a left shift of 1. We can easily see whether the a_k will be an integer or not!

    I think we can similarly find when will the co efficient exceed k. I didn't even tried it just because I was so busy in hacking !

    +9 hacks :D

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

    I don't think you should evaluate the whole thing. I think that only the last 30 or so coefficients matter (something depending on log2(k)).

    My idea was to work only with the last coefficients, If I had more time I would have tried a binary search on each one of the last coefficients to know if the coefficient can make 2 a root. Can someone confirm this idea?

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      9 years ago, # ^ |
      ← Rev. 7   Vote: I like it +9 Vote: I do not like it

      It's not necessarily the last 30, consider the following polynomial:

      1 0 0 ... 0 0 0 ... 0 0 -2 1

      I solved it the following way. First, bring the polynomial into the following normal form: write where . This is a simple preprocessing step (push ai / 2 (integer division) to ai + 1).

      After this, let k be the smallest value such that bk ≠ 0. That is, |bk| = 1. This means we either add or subtract 2k. Clearly then, changing some coefficient k' > k will change the value of the polynomial by steps of size 2k' > |bk 2k|, so all these coefficients are irrelevant. Similarly, if k' < k - 30 then we cannot change ak sufficiently to shift the value of the polynomial by more than 2k. So we will consider changing the values k - 30, k - 29, k - 28, ..., k.

      Suppose we have fixed the coefficient we want to change, say k. We have already established that the current value of the polynomial is divisible by 2k (since k is smaller than or equal to the smallest index i such that bi is nonzero). Therefore, we certainly can change ak such that the polynomial becomes zero, the only question is whether we won't violate the constraint that |ak| ≤ K (K is the input value, k is the index we are considering).

      We can solve this problem in a really nice way: the current value of the polynomial is either positive or negative. This is only determined by the largest nonzero index in the bi. Suppose the current value is negative. Then we simply add K - ak to bk, propagate this, and check if the polynomial is now positive or zero. If it is, increase the answer by 1. You can implement this last check in such a way that you only have to consider the indices k, k + 1, k + 2, ..., k + 30.

      If we say that the 30 we used above comes from , then the final complexity is .

      Solution: 17002298

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        9 years ago, # ^ |
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        Thanks for the great explaination and for the counterexample !

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9 years ago, # |
← Rev. 5   Vote: I like it +43 Vote: I do not like it
shitpost warning

Explanation: I code a solution with complexity because I see TL is 4s, but it TLs by a small amount. It seems that intended solution should run in well under 1s.

Edit: I coded an optimized version after, and get WA. The bug? 

K=min(K, C*5);
instead of 
B=min(B, C*5);
I don't even...
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9 years ago, # |
  Vote: I like it +5 Vote: I do not like it

writing for(int i = n — 1 ; i > 0 ; i -- ) instead of for(int i = n — 1 ; i >= 0 ; i -- ) is the worst reason to get your A failed :(((

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

Are there any kind of optimizations being done on this code: http://codeforces.me/contest/658/submission/16992448 ? I was gonna try to hack it as it seems like bruteforce (couldn't though, short by few seconds :| ). But it passes main tests too

Edit: nvm i didn't read k <= min(6, n) carefully...

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    9 years ago, # ^ |
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    1 ≤ k ≤ min(6, n)

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      9 years ago, # ^ |
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      Damn, i feel stupid now. I was somehow interpreting it like max and coded even considering that only :\

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        9 years ago, # ^ |
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        I didn't see that too, so i used priority_queue and got accepted, i was expecting a TLE :3

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    9 years ago, # ^ |
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    it's not brute force !


    k <= min(6,n) so in worst case the set's size will be 6. so answering queries is O(1) .

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    9 years ago, # ^ |
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    he is using set and k <= 6 so it seems ok

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9 years ago, # |
  Vote: I like it +21 Vote: I do not like it

I'm afraid there were some problems with the hacking system today. When I tried to open a solution of my roommate, it said "This solution has been hacked or resubmitted..." However, after I reloaded the room page for several times, the hacked verdict didn't appear on the standings. Also, according to the standings, at this time, there were only 2 successful hacks, but the above problem happened with 5-6 solutions.

Did anyone have the same problem? It wasted me a lot of time to hack :((

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    9 years ago, # ^ |
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    I did. I explained below what happened to me =/

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

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

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

what is D 7th test?

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    9 years ago, # ^ |
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    100 1000
    -62 57 -27 -67 49 -10 66 -64 -36 -78 62 -75 -39 75 -47 -36 41 -88 62 -43 22 29 -20 58 40 16 71 -2 -87 12 86 -90 -92 67 -12 -48 -10 -26 78 68 22 -3 66 -95 -81 34 14 -76 -27 76 -60 87 -84 3 35 -60 46 -65 29 -29 2 -44 -55 18 -75 91 36 34 -86 53 59 -54 -29 33 -95 66 9 72 67 -44 37 44 32 -52 -34 -4 -99 58 7 -22 -53 11 10 10 -25 -100 -95 -27 43 -46 25
    According to the checker

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9 years ago, # |
  Vote: I like it +248 Vote: I do not like it

I am Batman!

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9 years ago, # |
← Rev. 2   Vote: I like it +20 Vote: I do not like it

Just one comment about C. It seems to be the type of problem where there are two very distinct solutions:

  1. The intended one which iterates over the remainder of the value in O(n) or O(nlogn)

  2. Mine (and some others) that iterates over the value itself in O(nlognlogmaxT)

The issue is that the first step of the solutions are not similar. I thought of the slower solution first, and after getting TLE 8, I did the usual of trying to optimize out one of the logarithms, which failed. I never really stepped back and redid all of my thinking.

How should I deal with this situation in the future? Anyone have any tips?

(also I had the same hacks problem everyone else was referring to — getting put in random rooms, and especially bad, sometimes the other room still had my score in it and I just thought CF was slow when the solutions didn't load)

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    9 years ago, # ^ |
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    One trick I use is to look at submission times of other contestants. If they submitted very fast, solution is probably pretty simple (easy implementation). I thought about solution 2 as well, but it seemed too messy and I doubted other contestants could implement it in such a short time.

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

Did anyone else got moved to another room during the challenge phase? While hacking I kept getting a message "the solution has been hacked or re-submitted" only to find out that after the refresh I'm in a room I'm not assigned to... This kept happening until the end of the contest, I had to refresh several times each time to get to my room and actually be able to open a solution. I suppose the new feature with after-registrations is somewhat buggy.

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

When will be the problems added to practice?

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

Come on with the practice...I want to see if my almost in time finished source for C is correct.

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9 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

An irrelevant question: What does VK stand for?

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

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9 years ago, # |
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Why do you keep upsolving closed for a while after contest? :(

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9 years ago, # |
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Spent about an hour trying to get a Java BigInteger solution to Div 1 B to work, with no avail. I always TLE'd. I've been thinking about switching to Java, because of BigInteger among other things, but this has really dissuaded me. Is it possible to optimize this? Are there advantages to Java?

After I gave up and wrote a C++ solution, it ran in 200 ms :/

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

    I think the complexity of a BigInteger solution is O(n2) since adding two BigIntegers takes O(n) and there are O(n) additions. BigInteger's can't exploit the fact that much of binary expansion of the numbers is 0.

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      9 years ago, # ^ |
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      That makes sense. Is there a workaround or is this method just bound to fail?

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        9 years ago, # ^ |
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        I don't think there is a workaround. BigInteger isn't designed for bit manipulations, it's just used for large numbers.

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9 years ago, # |
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http://codeforces.me/contest/658/submission/17008837 Why is this answer accepted?? 4 4 4 gives wrong ans.

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9 years ago, # |
  Vote: I like it +39 Vote: I do not like it

When are you going to update the rating?

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    9 years ago, # ^ |
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    its almost 2 am in my country. I can't sleep because of that. Have to catch a bus 5 hours later. I think Im gonna miss it -_-

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9 years ago, # |
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I wish that Wild-Card 1 will not be like the year before.

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

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

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9 years ago, # |
  Vote: I like it +52 Vote: I do not like it

tourist — 4126

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    9 years ago, # ^ |
      Vote: I like it +70 Vote: I do not like it

    The rich get richer...

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      9 years ago, # ^ |
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      But you ain't poor either! See me! I feel sad for my performance everyday ! :/

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      9 years ago, # ^ |
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      Until the final he probably would be 5000+ :P

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    9 years ago, # ^ |
      Vote: I like it +40 Vote: I do not like it

    WTF The rating change is ridiculous...

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    9 years ago, # ^ |
      Vote: I like it +49 Vote: I do not like it

    I know the new rating system is supposed to be better but what the hell is going on with tourist's rating? He used to get +40 for a first place and now he has +259, even though his rating is far beyond anyone else's. I don't really mind it, I'm just wondering what changed so drastically :D?

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      9 years ago, # ^ |
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      And his teammate is not a random weak guy but subscriber... +259 is not reasonable.

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      9 years ago, # ^ |
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      It's partly because of the team contest, but this is happening even for regular rounds, the rating system is being quite wacky around the top of the standings...

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        9 years ago, # ^ |
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        Before change about 50-60 was what he would get upon being first. After change it's +112, +124, +172 — while competing alone. Additionally, why does it matter if it's a team competition or solo when he is first?

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          9 years ago, # ^ |
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          Competing against teams is like competing against participants with higher rating. Beating them should reward you more.

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            9 years ago, # ^ |
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            But you are also competing in a team so aren't you also a participant with higher rating?

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              9 years ago, # ^ |
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              No. His team has exactly the same rating as tourist alone, lol.

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9 years ago, # |
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tourist getting 4k is like Leo getting Oscar.

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9 years ago, # |
  Vote: I like it +114 Vote: I do not like it

I'll just leave this here.

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9 years ago, # |
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why kicked in Vk cup round 1 (div 2) ?

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9 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Sorry if my questions is a little bit ridiculous,but I have to ask: are you really think that,same rating change for both team mate is correct? Was it the best way for distribution ratings?

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    9 years ago, # ^ |
    ← Rev. 3   Vote: I like it +14 Vote: I do not like it

    I think the team rating should be independent from individual rating, just like tennis lol

    A team contest is not that common though...

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9 years ago, # |
  Vote: I like it -32 Vote: I do not like it

I think it's time to create a new color above "legendary grandmaster" for tourist

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    9 years ago, # ^ |
      Vote: I like it +205 Vote: I do not like it

    Nothing personal, but I'd rather say it's time to fix formulas

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