Codeforces LATOKEN Round 1 (Div. 1 + Div. 2)
By Dormi, 3 years ago, In English

Hello, Codeforces!

Welcome to the Codeforces LATOKEN Round 1 (Div. 1 + Div. 2) supported by LATOKEN, that will start on Jun/13/2021 18:35 (Moscow time). It will be a combined rated round for both divisions (and div 3). Please, note the unusual start time.

All problems were created by Aaeria, crackersamdjam, Dormi, Ninjaclasher, and Plasmatic.

We would like to thank:

You will be given 8 problems and 180 minutes to solve them. This contest features at least one interactive problem, so we recommend reading the guide of interactive problems before the contest.

We hope that statements are short and pretests are strong and that you find the problems interesting! Good luck, have fun, and we wish everyone high ratings!

The scoring distribution will be announced closer to the beginning of the round.

Thanks to LATOKEN, winners will get awesome swag:

  • 1st place = MacBook Air M1 + Job offering and Stock Options
  • 2nd place = Xiaomi Mi 11 + Job offering and Stock Options
  • 3rd place = Xiaomi Mi Notebook Pro 15" Enhanced Edition + Job offering and Stock Options
  • 4th-10th place = Job offering and Stock Options + Merchandise Pack
  • 11th-55th place = Merchandise Pack

Random 10 participants outside of top-55, who solved at least one problem and participated in rated Codeforces rounds before, will get a t-shirt!

A few words from the LATOKEN team:

We welcome you to the intellectual battle between the best coders from around the world! LATOKEN strives to make trading crypto as easy as having a social media account. We are always looking for the best of the best to join our team. As a result, LATOKEN is currently in Coingecko’s Top 15 with over 600 000 app installs within just one year from its launch and 1m+ users in total. The emerging blockchain based financial systems will open exciting opportunities, because crypto is fully global and mostly independent of conservative banks.

If you want to become a part of something bigger, work remotely with a team of international professionals and change the world, contact our recruiters for internships and employment opportunities in the LATOKEN team. Use our bot or fill out the form below and tell us a little about yourself.

Bot→
Contact Form →

UPD: The scoring distribution is 500 — 1000 — 1250 — 1500 — 2250 — (1750 — 1250) — 3250 — 3500.

UPD: Editorial

Congratulations to the winners:

  1. Benq
  2. tourist
  3. Radewoosh
  4. Endagorion
  5. ecnerwala
  6. heno239
  7. VivaciousAubergine
  8. gamegame
  9. Pyqe
  10. maroonrk

The prizes will be distributed in the next few days after we remove cheaters.

  • Vote: I like it
  • +638
  • Vote: I do not like it

Comments Comments (288)
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3 years ago, # |
← Rev. 2   Vote: I like it +116 Vote: I do not like it

Do the contest or TLE real life

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3 years ago, # |
  Vote: I like it +103 Vote: I do not like it

As a tester, the problems are great, as always. I hope you enjoy one of the first Canadian rounds!

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3 years ago, # |
  Vote: I like it +193 Vote: I do not like it

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3 years ago, # |
  Vote: I like it -24 Vote: I do not like it

Mandatory "tourist, What will you do with your MacBook Air M1?" comment.

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3 years ago, # |
  Vote: I like it +180 Vote: I do not like it

I was just wondering why 1st-3rd placed participants are not being offered the job while 4th-10th are XD.

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    3 years ago, # ^ |
      Vote: I like it +124 Vote: I do not like it

    FAANG already hired them.

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

    They are the real thugs, they can live with winning cash prizes in big coding contests.

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

He is Valentin Preobrazhenskiy not Valentine.

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3 years ago, # |
← Rev. 4   Vote: I like it +17 Vote: I do not like it 
"Random 10 participants outside of top-55, who solved at least one problem and participated in rated Codeforces rounds before, will get a t-shirt!"

Am I going to be the one of those 10 ? rounding up number of people solving atleast 1 problem to 20000 as we have 3 hours given probaility is ( 1/(20000)), . It comes out to be (0.00005)

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

    Well almost, you need to multiply 0.00005 by 10, because "Random 10 participants outside of top-55...". So that's 0.00005*10 = 0.0005. But I don't think it changes anything)

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3 years ago, # |
  Vote: I like it -21 Vote: I do not like it

be annoy of the suck statement at last several round.

statements are short look forward to this round.

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

    I was wondering why your rating graph is like this!!!

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

      yep, I was graduated several year, now i want CF agin!!!

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3 years ago, # |
  Vote: I like it -24 Vote: I do not like it

PurpleCrayon orz

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3 years ago, # |
  Vote: I like it -26 Vote: I do not like it

Unfortunately, the start time is 11.35 PM in my area :(

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    3 years ago, # ^ |
      Vote: I like it +47 Vote: I do not like it

    thank you for useful and interesting comment

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3 years ago, # |
  Vote: I like it +65 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +34 Vote: I do not like it

As a tester I must say the problems are nice and interesting. May you all get a positive delta.

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

If you forgot to see the prizes section of the post.

Do visit again!

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3 years ago, # |
  Vote: I like it -21 Vote: I do not like it

very weird time :|

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

Just switched from Instagram to Codeforces, a few days back, and trust me Codeforces is much better.

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3 years ago, # |
  Vote: I like it -33 Vote: I do not like it

note the unusual start time

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3 years ago, # |
← Rev. 2   Vote: I like it +69 Vote: I do not like it

4th-10th place = *Job offering* and Stock Options + Merchandise Pack

I want this level of confidence in my life.

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

I will sleep during the round.

Spoiler
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3 years ago, # |
← Rev. 3   Vote: I like it -23 Vote: I do not like it

Codeforces is the best ❤️

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3 years ago, # |
  Vote: I like it -74 Vote: I do not like it

Is it rated?

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

Sorry I'm getting back to CF after many years and now I see there is also a Div3. Will this round also be rated for Div3 folks?

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

    Yes, there is pretty much no div3 as such. Anyone who's a div3 is also a valid div2

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

    As now there's also a Div4 you're Div4. But all contests of higher divisions than your division are rated for you. Unless round has separate contests for separate divisions. Because then you can only participate in your division contest.

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

      his ranks say otherwise, he'll soon reach expert

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

        Thanks! I hope so too. I aim to reach Div1 sometime in the next few months. Let's see how it goes.

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3 years ago, # |
  Vote: I like it -47 Vote: I do not like it

Just switched from Instagram to Codeforces, a few days back, and trust me Codeforces is much better.

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

I guess top gear isn't popular here :/

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3 years ago, # |
  Vote: I like it -12 Vote: I do not like it

wannahavealife 1000-7 ? ? ? ?

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

it will be unrated for div 3??

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3 years ago, # |
  Vote: I like it +156 Vote: I do not like it

Screenshot-20210611220724-857x462

Seems codeforces finally realized nobody reads this rules anyways, so they just removed it lol

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

This contest is rated for division 3 or not?

Any please replay

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

    Yes

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

      Thanks for your fast reply

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

        So are you gonna participate or not? Just curious about the effect that being rated causes on participation.

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

          I will definitely participate in this contest .

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

    Every div3 member is also considered a part of div2 too. So hopefully it would be rated for everybody

    It will be a combined rated round for both divisions.

    @authors can you please help with this ambiguous statment maybe write it "rated for everybody" ?

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

interesting interactive problem. I hopeeeeeee

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

RIP my hairs after 3 hours :(

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    3 years ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    Wait.. You guys still have hair?

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

      wait, you guys still have +ve delta

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

        cuz we guys don't use alts.

        Lol.

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

          wait, there are alts??

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

            It's clearly not working for you.

            Lol sorry XD.

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

              Out of my place to conjecture this, but is moongazer your alter ego's account? Seems like instead of discussing ethical issues inside you discussed them here!

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

                Unfortunately, I am not that insane.

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

Contest will clash with Roland Garros finals :(

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

Everyone here should know that Latoken is a scam exchange: https://www.cointelligence.com/exchanges_list/latoken/ (In Russian): https://telegra.ph/Kak-rossijskij-skam-LATOKEN-zarabatyvaet-dengi-i-stroit-totalitarnuyu-sektu-12-13

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

    I mean, it's dealing with crypto, and crypto is highly unregulated...

    But anyways, free money for CF!

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

Is this contest rated?

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

The scoring distribution is 500 — 1000 — 1250 — 1500 — 2250 — (1750 — 1250) — 3250 — 3500

(1750-1250) means?? can anyone explain please.

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

    There will be two parts to the problem. One easier version(F1-1750) and one hard(F2-1250).

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3 years ago, # |
  Vote: I like it -59 Vote: I do not like it

One More Macbook to tourist

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

For some reason there is a scoring distribution update in the English version of this blog post, but there is no one in Russian version.

UPD: It's ok now.

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3 years ago, # |
  Vote: I like it +83 Vote: I do not like it

For real, what's the point of making combined round when there are enough problems to split round for both divisions? Nobody from div1 wants to solve first few problems as well as nobody from div2 even reads last few problems

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

Hope! this round will go better for me than the Deltix round

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3 years ago, # |
← Rev. 2   Vote: I like it +27 Vote: I do not like it

The entire comment section in one not-somewhat-related meme:

picsart-06-13-05-14-58

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

Most People: This is definitely my chance to prove myself and get MacBook Air!! Tourist: This is definitely my chance to be first again!!

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

I am betting that tourist will not win this round. It will be some other LGM.

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3 years ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

it's div1 + div2 = div3 :)

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

today junior tesla is expecting to solve 4 problems and upsolve extra 1 problem. Lets see.

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

Thanks to Djokovic I can convince myself that my rating crashed because I was watching Roland Garros finale
while giving the contest.

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3 years ago, # |
← Rev. 5   Vote: I like it -7 Vote: I do not like it

Although the Problem C was very interesting here's a meme
IMG-20210613-220345

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3 years ago, # |
  Vote: I like it +34 Vote: I do not like it

Div 1+2 = rating loss

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

It takes eternity for queue to end

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

    Only to get "Idleness limit exceeded on pretest 1".

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3 years ago, # |
  Vote: I like it +12 Vote: I do not like it

queueforces

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

There are no running submissions.

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

Is this still rated because the queue has been stuck for a long time now ?

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

I'm waiting for the results of my submissions from last 10 min.

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

queue too long.

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

People who are in queue are gonna remain in the queue forever if the number of running submissions is zero.

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3 years ago, # |
  Vote: I like it +382 Vote: I do not like it

I always wanted to do it, and today I finally did

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

    you have plenty of time to play with these queue times :D

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

If someone wrongly pastes commented solution of A in problem B's submission, and locks problem B alone, won't this be a problem as others in room who have not solved A can see solution of A by solving only problem B ? And what if someone has commented previous problems solutions for all submissions ? How does codeforces handle this ? MikeMirzayanov

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

    Do you mean that somebody uploads their solution for A instead of a solution for B by mistake? But a solution for A is very unlikely to pass pretests for B. And it makes no sense to show the source code of failed submissions to hackers (I don't know if that's how it really works, because I have never tried hacking other solutions in non-educational rounds yet).

    Or do you mean that somebody submits a valid solution for B, together with a commented out solution for A in the same source file? Is anyone ever doing something like this? I do sometimes use my A solution's source code as an initial template for B. But by the time when I'm ready to submit it, no parts of A remain there.

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

      Is there any difference between submitting code with commented solution of other problem in it and pasting it in an open blog ( or social media, IDE, etc. ) ?

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3 years ago, # |
← Rev. 3   Vote: I like it +2 Vote: I do not like it

Anyone else having difficulty to understand the Problem C? Or am I the only one?

UPD: I got it. And I'm stupid. :(

UPD2: I think, I solved it.

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

    I perfectly understand it, the only problem is I am stupid and cannot figure out how to solve it ;)

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

      Same with me for Problem B. I guess it'll be a sad contest for many.

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

        Unfortunately it was that way for me with both B and C; I suppose I can kiss my pupil rank goodbye...

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

        I dunno why but when I first read problem B I was thinking of the 3 columns problem without any proof and then boom dunno how I got accepted L.O.L.

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

          Would you like to explain what is 3 columns problem?

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

            The subproblem I generate to consider problem B i guest?

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

            Consider the i, i — 1 and i + 1 position.

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

    I spent half an hour, read several times, and don't understand what it means by "only by swapping numbers in a column". and finally I ask a question to admin, and then I got it.

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

      Yes, I didn't notice the announcement first. I felt really dumb noticing it after few minutes.

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3 years ago, # |
  Vote: I like it +20 Vote: I do not like it

interactiveforces...

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3 years ago, # |
  Vote: I like it +76 Vote: I do not like it

ME: Get's TLE on D :( .No worries, let's stress test by generating some random trees. Google please give me a random tree generator?
GOOGLE: Sure

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3 years ago, # |
← Rev. 2   Vote: I like it +16 Vote: I do not like it

The person who explained B's sample test cases

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

How to solve E? I think the soln is something like, do individual blocks on k, for last (n % k), if (n % k) is:

  • 1 — then no answer exists
  • any even number — 2 more queries
  • 3 — 3 more queries
  • any odd number more than 3, remove 3 elements in 3 queries and remaining even in 2.

Is this roughly along the correct lines?

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    3 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    You can think that you have n columns and in one operation you add +1 to exactly k of them. You need to obtain odd sum in each column.

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

    How can you go for 11 4? I couldn't get this.

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

      if n % k is odd and k is even, the answer wouldn't exist.

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

        Thanks, then why the hell my solution is failing. Btw this is the only case where answer ceases to exist, right?

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

          Not sure about when n % k is odd and k is also odd. How are you solving this?

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

            Just make a move on first $$$n - n \% k$$$ elements and last n % k elements and this is same as k -> odd and n % k -> even.

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

              you might save some moves by saving up n-(n%k) indices by repeating them in previous queries. like

              n=13, k=5

              q1 : 1 2 3 4 13

              q2 : 5 6 7 8 13

              q3 : 9 10 11 12 13

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

                Yup I know and implemented this too, the above move needs to be done when n/k == 1.

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

    Answer doesn't exist only when k has more powers of 2 than n I think.

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

      n = 6, k = 4:
      1 2 3 4
      1 2 3 5
      1 2 3 6

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

    That's part of it, but note that it's not always optimal to do $$$\lfloor n / k\rfloor$$$ blocks, sometimes you might want to do more or less. When I say more, I mean you can wrap around and cover an index twice. As an example, the optimal strategy for $$$n = 10, k = 7$$$ might look like this:

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

    EDIT: After reading other comments it seems I did E in a very jank way, I did some strange bullshit with blocks of size $$$k$$$ that passed pretests so pray no FST.

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +32 Vote: I do not like it

    Make a graph with N + 1 nodes.

    Node i represents a state where i indices are used odd number of times.

    Here Node 0 is our starting point, and Node N is our desired end goal.

    Make an undirected edge from i to j, if you can perform an operation move from state i to state j. (can be bruted).

    Answer is shortest_path(0, N).

    Since we can move from one state to another state easily (again, brute), we get the answer after knowing the shortest path.

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

      Edit: Ignore

      Could you elaborate on how the number of edges can be bruted? Can't the degree of each node be upto O(k)?

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

        For a state i, brute the number of ones being converted to zeroes. (zeroes_to_ones + ones_to_zeroes = K).

        This way, you get all neighbours of i in O(k).

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

          Sorry I just realized my question was utter nonsense. I was asking how this didn't become $$$O(n ^ 2)$$$ even though degree could be upto $$$O(k)$$$ per node. I forgot $$$n \leq 500$$$ in the problem.

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

    If n%k equals 1 and k is odd, then answer exist. For example, for n = 6, k = 5:
    1 2 3 4 5
    1 2 3 4 6
    1 2 3 5 6
    1 2 4 5 6
    1 3 4 5 6
    2 3 4 5 6

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

    Using bfs from the final state and keeping count of number of positions used odd number of times will suffice

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

Today I got one of the best feeling that is solution D got accepted in last minute.

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

    I got WA on pretest 3. Can you give some hints?

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

      Were your queries more than N/2?

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

what was pretest 3 of E?

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

Thanks for including some easy problems for amateurs like me. :)

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

https://codeforces.me/contest/1534/submission/119394653 — This same case passes in my machine in 2 queries, i tried multiple other cases too. But CF says here it took more than 2 queries. Did i miss something here?

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

D problem... Killed more dreams than Fear every will ... :(

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

Solved F1 two minutes before the contests ends, what an exciting contest for me.

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

Can someone tell me, Why I'm getting ILE in Div2D ?
SUBMISSION.
I mean, I literally tried every combination of printf and endl. But couldn't get it worked. I usually use endl as a flusher but here it's not working.

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it 
    vi query(int r) {
    // cout << "! " << r << '\n';
    printf("! %d\n", r);

    The query format differs from what you print here.

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

+0 delta GG.

Should have not given the contest in the first place :(

Screenshot-290

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

    Hey buddy , do you know how can ishow the delta points for some reason i can't see it ?

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

I feel the problem statements could have been clearer/presented better, especially those with large statements.

At first glance, I saw "It can be proven that that if it is possible to recover the XOR sum under the given constraints, it can be done in at most 500 queries. That is, d <= 500", and went to write a code for atmost 500 queries..

Didn't help it was an interactive problem, where, if there's a WA I don't know if it's my IO format, or my solution is actually wrong.

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

how to solve D?

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +21 Vote: I do not like it

    Every tree is bipartite graph. You can ask only about nodes in smaller part. To know if node is in smaller part u can use this fact: distance between nodes from different parts is odd. So u just ask about first node, counting (cnt) nodes that have odd distance, and if cnt < (n+1)/2, first node is in bigger part, otherwise it is in smaller part. And then just greedy alg.

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

      can u explain these smaller/bigger parts in more detail. And how you are representing a tree as a bipartite graph. Would be appreciated

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

        1. So, in bipartite graph every node belongs to one of two parts. Part that contains more nodes than other is bigger part, other part is smaller part.

        2. So, as I said if distance between two nodes is odd, nodes belongs to different parts. So, ask about first node, every node that has even distance to the first one, is in the same part with first node, otherwise, if distance is odd, it is in different part

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    Here was my approach:

    Arbitrarily choose some node to be the root. Query that node. That gives you the rank of each node in the rooted tree. Our goal is to find the parent of each node.

    Iterate over the nodes in any order. Query each node whose parent is not yet known. We can use the result of the query along with the rank information to find that node's ancestors, siblings, and children.

    Notice that if we associate each query with the node and its parent, these pairs don't overlap. This proves that we won't exceed $$$\lceil\frac{n}{2}\rceil$$$ queries.

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

I think I missed D by printing a edge multiple times :(

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

The only time I am in top 500 is during system testing (T_T)

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

How to solve B?I had some proof that decreasing a number to the maximum of two sides will work but got a wrong answer on pretest 2

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

    you only need to decrease a number if it is greater than both its adjacent element.

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

      I did so but still a wrong answer.

      Spoiler
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        3 years ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        I think this is becoz , u have not dealt with n=1 separately for input 1 1 2

        the answer should be 2

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

    Erase a block off of a histogram if it is strictly higher than the previous histogram and the one after it. If you do so, you will pay $1 for each erased block and decrease ugliness by 2.

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

    Make sure you only do this on bars that are strictly higher than both of its adjacent bars. Also be careful with the left and right end bars as they only have one adjacent bar.

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

      I take care of both this situations but it still doesn't work.Can you take a look at my submission if you have time?

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

    Handle n=1 case separately. For n=1,ans will be arr[0]

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

      Yep I see my codes doesn't work there . Thanks!

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

    Suppose we have our original array orig and and our new array arr. We want to minimize:

    $$$\sum |orig[i] - arr[i]| + \sum_{} |arr[i] - arr[i - 1]|$$$

    So.. if we update $$$orig[i]$$$ to some $$$arr[i]$$$, we need to make sure that it is going to decrease

    $$$|arr[i] - arr[i - 1]| + |arr[i + 1] - arr[i]| + |orig[i] - arr[i]|$$$

    as compared to

    $$$|orig[i] - orig[i - 1]| + |orig[i + 1] - orig[i]|.$$$

    And from here, you can formally prove that its optimal to update a value iff it forms a mountain.

    And this makes some intuitive sense too: obviously, updating a valley is a bad idea. And if we update something that is neither a valley nor a mountain, then its pretty pointless and suboptimal.

    Also case when $$$n = 1$$$ needs some special case (or at least, my implementation needed special care).

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

      Padding the array with one extra 0 element in the beginning and one extra 0 element in the end removes the need to handle special cases.

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

how to solve problem C ?

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

    count the number of cycles formed from $$$a,b$$$ and then $$$answer = 2^{#cycles}$$$

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

    2^(number of connected component)

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

    Try to inverse some column, u will notice that now there is same number in first row, then try to inverse it, and repeat it until it has solved configuration. Every column than u have visited belongs to one cycle. Every cycle is independent from other cycles. Every cycle have 2 states. So answer is just 2^(amount of cycles)

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

    In C, the columns build components. That is, if you swap one column, you also have to swap another one, then another one and so on, until a circle is complete.

    Find the number of such circles, and ans=2^number since foreach circle we can swap it, or not swap it.

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

    C gave me deja vu for this problem: 1465C - Peaceful Rooks

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

    https://www.youtube.com/watch?v=xwi_JeZbEu4&t=1s

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

What was pretest 3 in problem D? I failed it 7 times!

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

https://codeforces.me/blog/entry/91617?#comment-803516 YAY I was right. All Tourist fans owe me a buck xD. Congrats Benq

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

Anyone tried randomization in D and got it accepted?

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

When Ninjaclasher forces you to do a problem with your own name in it

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

Thank you for the round! The problems were interesting.

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

why G was so big constraint? My O(NlogN) solution get TLE (TT)

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

Hey Guys Do you know how can i see the delta points columns in the contest ?

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

    install an extension called CF-predictor

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3 years ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

Does anyone know why I can't view other people's submissions? Apologies if the reason was mentioned somewhere already and I missed it.

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

finally the rating loosing streak is back. So excited to have more negative deltas. Wish me luck.

Thanks for this round and making me feel how dumb I am.

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3 years ago, # |
← Rev. 3   Vote: I like it +21 Vote: I do not like it

Apparently many people solved $$$E$$$ with some casework. You can do also like: (which I think is intended solution).

Solution
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3 years ago, # |
  Vote: I like it +83 Vote: I do not like it

Big ups to the authors for the nice problems, and bigger ups for including the answers to the possible sample queries in the interactive problems where it's reasonable. That's such a nice thing to do <3

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

will the number of cycles and number of connected components be equal in C?

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

    Yes, basically all connected components will be pure cycles

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

      2^(Cycles) is the answer

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

      Please can you check my submission 119395523? I calculated 2^(connected comp) but it's giving WA on tc4.

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

        In your solve function, for clearing the g[i], i should be from 0 to n, i.e., for(int i=0;i<=n;i++) (you did i<n instead of i<=n)

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

Is there a way to see how many people failed system test for a particular problem ?

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it -13 Vote: I do not like it

    They will never show that as people will backfire for weak pretests. Although this contest had strong pretests. No one on my friend list got FST.

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

      I found a way, just go to the status bar and check all rejected submissions keeping the tests greater than the number of pretests.

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

What if i tell you, that in problem A you just need to check whether given matrix satisfies:

RWRWRW                   
WRWRWR        
RWRWRW 
WRWRWR

or

WRWRWR
RWRWRW
WRWRWR
RWRWRW
WRWRWR

Realized this in post-contest discussion with my friend.

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

Finally Tutorials are here. I was thinking of finally offering some goat to the devil. Thanks to problem setters for fast update.

PS: BEFORE YOU COMPLAIN NO TUTORIAL TO BE SEEN IN THE BLOG. TRY TO FIND IT FIRST

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

Fst on D coming :(

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

Now it's time to wonder if I got a great place thanks to competing drunk or despite competing drunk.

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

https://codeforces.me/contest/1534/submission/119394653

Can somebody try this out on their machine. I submitted this in contest and it gave WA on pretest-1 and i tried it on my machine multiple times and it worked well. I'm really frustrated as I don't understand how the same code in 2 machines can give different output (my strategy isn't random, the program will always query the same nodes for a particular tree everytime it runs).

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

    When your local submissions behave differently from codeforces submissions, it often indicates that your code may have undefined behavior. In your case, I think you are trying to access an element out of bounds of the done array: see 119406377 where the assertions I added fail. I think it is due to the line fire2.pb(mp(x,b+1));, where b can be up to n-1

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

      Thankyou so much for taking out the time to look into this and getting back to regarding the issue. I definitely learnt something new today. I understand why this happened now. Have a nice day :)

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3 years ago, # |
  Vote: I like it +28 Vote: I do not like it

Thanks for the great problems and strong pretests!

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3 years ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it

Looks like highest rating record on codeforces is going to be broken.

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

Two questions:

F: Does anybody remember a similar situation when solving a small version and then trying to upgrade it was a mistake? Normally it's rather a good way, but there the small version required only SCC while the big one doesn't (which may lead you to really long code).

H: Isn't an interaction with $$$10^5$$$ flushes rather slow/dangerous/generally not CF-friendly?

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

    I finally found a problem with a good upgrade in this F. Not like F1 has an interesting idea, just a straightforward "build graph, do SCC", where the main point of failure is building the graph in a wrong way, but there's an element of double checking — if F1 passes, you can expand on it to solve F2, while if your SCC is wrong and pretests for F1 don't catch it, then pretests for F2 might because it's way more than SCC!

    There's a good reason to go for SCC in F2 even if you don't have F1. As soon as you have it abstracted as a directed graph, compressing to a DAG gives you plenty of guarantees for the next step — actually figuring out the solution. It's not hard to write either, just reproduce those 20 lines of Tarjan from whichever online resource you want.

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

      Well, I took my code to F2, erased the whole thing with SCC, did some obvious stuff and it worked, so I don’t find SCC useful here at all.

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

        Well, I used SCC and didn't have to erase and rewrite, so I think it was a good path to take.

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

          What’s your solution then?

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

            Mark vertices that you need to cover, make scc, mark respective vertices in scc, unmark those that are reachable from another marked.

            Now you have set of marked vertices none of which is reachable from the other.

            I claim that you can sort them in some order that any source in SCC will cover segment of marked vertices, just sort them by column id of any marked representative in scc vertices.

            Now you have to figure out minimum and maximum reachable marked vertex from each source and cover array with minimum number of segments, which can be done greedily without actually finding segments: take first unmarked vertex, go through back edges and mark every node as visited, from each marked node go down and mark all nodes as visited as well. You will reach all sources that cover first marked vertex, which is precisely what greedy on array does. Then take next marked unvisited vertex and so on.

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

            First, rephrase the whole thing as: you've got a directed graph where some vertices are marked (the $$$a_i$$$-th sand in $$$i$$$-th column), find the smallest set of vertices such that each marked vertex is reachable from at least one chosen vertex.

            I compress to DAG, note that only topmost components should be chosen, then note that from the way the graph is constructed, no two topmost components can share a column, so they're easy to sort. Also, paths in the graph are actual paths in the grid, so the set of topmost components from which a given vertex/component can be reached is a contiguous interval.

            With a simple graph traversal, for each component, I find the leftmost and rightmost of the top components which are above it (minimum and maximum of their numbers in sorted order) and the problem reduces to picking the smallest number of points that cover all given intervals, which is well-known.

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

    I thought there should be a natural extension in F. So I continued with the SCC formulation and reduced it to finding a minimum size set of vertices in a DAG such that every leaf is reachable from some vertex in the chosen set. Then I tried to solve it with some (wrong) flow models. Does anyone know if this problem admits a good solution in general?

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

119382434 Can someone tell me why I'm getting runtime error?

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

    In practice mode (i.e. after the contest), runtime errors on codeforces can be caught by the C++ diagnostics compiler automatically. I resubmitted and it shows that your char arrays a and c are too small: 119406134

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

      Thanks for the information. Is there any way to overcome this error?

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

        Make the arrays bigger, e.g. char a[n+1][m+1], c[n+1][m+1]. Alternatively, use 0-based indexing in the rest of your code.

        An array int a[50] only has valid positions a[0], a[1], ..., a[49].

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3 years ago, # |
  Vote: I like it +36 Vote: I do not like it

Benq has raised the bar of highest cf rating to 3796 now. Beating Tourist . Congratulations Benq.

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

https://codeforces.me/contest/1534/submission/119344612 Why did this fst? Can't seem to figure out.

Edit -> Don't use global variables when you don't have to.

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

Hi. Can someone explain me why I got WA on test 4 in B? The idea is correct but I can't find the mistake in my solution. Thanks

119386786

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

    hist[i+1] causes an undefined behavior because the max n is 4e5 and the size of hist is 4e5, enlarge hist by 1 element and soln will pass

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

      I gave hist a size of 4 * 100005 which is 4 * (1e5 + 5) so that shouldn't be the problem. Also, I just tried with hist[(int) 4e6] and again got WA on test 4.

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

        ah, I'm blind, sorry, you don't fill it with zeros for every testcase, that's the reason, it is global

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

good to be back

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

can some one tell why I am getting WA on this submission my friends solved literally the same idea and they got AC I literally can't find any wrong I am getting wrong on pretest 2 my program is printing 01234501234512345623456734567845678956789100 how can the program prints smth like this plz help https://codeforces.me/contest/1534/submission/119401086

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

    I made a few little changes in the cases where n = 1 and n = 2, and it passed: 119404096

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

Since the highest ever rating title was most likely taken (at least for now) from tourist by Benq, do you think he will take his revenge and first reach the 3800 mark?

Any bets who will be first over 4000?

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

    Neither of them.

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    3 years ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    I suggested this in the past:

    Spoiler

    But now that Benq got the highest max rate on Codeforces history, The competition has become strong!

    but you forgot Radewoosh, He also get a rating of 3700+

    now (Benq VS tourist VS Radewoosh), The trio with a rating of 3700+

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3 years ago, # |
  Vote: I like it +83 Vote: I do not like it

To not keep you waiting, the ratings updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!

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3 years ago, # |
  Vote: I like it -21 Vote: I do not like it

become newbie!

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

Really liked C. I am not sure if this was already commented by someone on problem C, but here is what I did. Let the first row be the permutation $$$a$$$ and let the second row be $$$b$$$. We have the property that $$$a_i \ne b_i$$$ for all $$$1 \leqslant i \leqslant n$$$. Now, since this does not change the answer, apply the permutation $$$a^{-1}$$$ on both arrays.(by this I mean changing indice $$$i$$$ to $$$a^{-1}_{i}$$$) Now we have the identity permututaion and some derangement of it at the bottom. Let this derangement be $$$c$$$. Note that $$$c$$$ is equal to $$$b \cdot a^{-1}$$$. The important observation is that if you swap $$$i$$$ with $$$c_i$$$, then continuing you must swap $$$c_i$$$ with $$$c_{c_i}$$$ and so on until you "cycle" back to $$$i$$$. This inspires to think of the cycles in the permutation $$$c$$$. So, if there are $$$X$$$ cycles, we have $$$2$$$ possible states for each of them. Either you go through that cycle once or you do nothing.(note that going through a cycle brings you back to where you started). Thus the answer is $$$2^{X} \pmod {10^9+7}$$$. So, we can say more about the graph in the tutorial, each vertex has degree $$$2$$$.

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

    In my C solution 119368225, the array formed of upper/lower pairs simply gets sorted by the value of the upper element in each pair (since this doesn't change the answer). After this is done, the upper elements became the identity permutation and the bottom elements became a derangement of it. The rest is the same.

    The only difference is that my solution has $$$O(n \log n)$$$ time complexity because of applying sort. And yours is a smarter $$$O(n)$$$ solution, achieving the same end result. My reasoning was "let's sort the input data and see if the problem becomes more simple" and indeed that's what happened! I overlooked the fact that this degraded time complexity, but humongous sizes of the input data typically have i/o as the performance bottleneck (to the extent that C++ people are forced to use the sync_with_stdio voodoo magic). While sorting the input data rarely or even never becomes a real performance problem.

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

      Nice idea to escape using too much space. Really liked that sorting idea! I never thought of using that to make the code simpler! Thanks for the help!

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

Still don't understand why my solution on D catches: "Idleness limit exceeded on test 1". I even do cout.flush() after endl. Can anyone help me?

My last submission: https://codeforces.me/contest/1534/submission/119510328

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

Can anyone please tell me what is wrong with my code 119381067

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3 years ago, # |
← Rev. 13   Vote: I like it -30 Vote: I do not like it

[DELETED]

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

Dear contest team,

I received a message that two of my solutions (B, D) significantly coincide with what user "gamegame" submitted. If you take a look at them, they are significantly different, especially D. I wrote and compiled my code locally without using any online resources. This round was my best contest in years, and I would be very sad if the rating changes were rolled back even though I didn't share my code with anyone or copy gamegame's code in any way.

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

When will you announce random T-shirt winners?

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

    Why do you care about t-shirt winners when you didn't participate?

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

Please announce random tshirts winner

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3 years ago, # |
  Vote: I like it +2 Vote: I do not like it 
`Attention!

Your solution 119352589 for the problem 1534A significantly coincides with solutions sanduio/119352187, Shunya_/119352589. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.me/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.`

Dear contest team(Dormi), (I'm posting it pretty late that's the reason I'm tagging you, sorry), I've received this message two days back however I saw it today, I just want to explain the whole thing rest it completely depends on you, apparently the guy (sanduio) whose code is similar to mine has been cheating in all the contests (he should've been blocked by now), please just have a look at his submissions, Your text to link here..., on that particular day I was as usual using cp helper extension in my VSCode however while running the test cases the antivirus was displaying that the trojan virus has been found and repaired in xyz.cpp file so to avoid that during the contest I shifted to ideone, Honestly I didn't know about that the code is public on Ideone before today. I understand it's my mistake that I violated one of the rules(which I honestly didn't know) however I'd like you to see his profile and his submissions so that you get the clear picture of the whole case (I'm not the one who copied), and please return back my ratings if possible, they are important for me. (This is the first time I'm posting a comment and honestly I don't understand much about blog and comment system, sorry if there's any error and sorry for being so careless). Thank you.

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

MikeMirzayanov

problem ratings of this round are not updated.

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3 years ago, # |
  Vote: I like it +93 Vote: I do not like it

Congratulations to prize winners! You will be contacted in a few weeks regarding the prizes.

  • Benq — MacBook Air M1 + Job offering and Stock Options
  • tourist — Xiaomi Mi 11 + Job offering and Stock Options
  • Radewoosh — Xiaomi Mi Notebook Pro 15" Enhanced Edition + Job offering and Stock Options

Job offering and Stock Options + Merchandise Pack:

List place Contest Rank Name
4 1534 4 Endagorion
5 1534 5 ecnerwala
6 1534 6 heno239
7 1534 7 VivaciousAubergine
8 1534 8 gamegame
9 1534 9 Pyqe
10 1534 10 maroonrk

Merchandise Pack:

List place Contest Rank Name
11 1534 11 ksun48
12 1534 12 jiangly
13 1534 13 neal
14 1534 14 Marcin_smu
15 1534 15 ko_osaga
16 1534 16 djq_fpc
17 1534 17 PinkRabbitAFO
18 1534 18 Um_nik
19 1534 19 Maksim1744
20 1534 20 SSRS_
21 1534 21 rama_pang
22 1534 22 ainta
23 1534 23 antontrygubO_o
24 1534 24 LayCurse
25 1534 25 ugly2333
26 1534 26 gop2024
27 1534 27 uwi
28 1534 28 LJC00118
29 1534 29 fastmath
30 1534 30 hitonanode
31 1534 31 Xellos
32 1534 32 Siberian
33 1534 33 Martin53
34 1534 34 never_giveup
35 1534 35 noimi
36 1534 36 dreamoon_love_AA
37 1534 37 fuppy
38 1534 38 natsugiri
39 1534 39 Arpa
40 1534 40 kostia244
41 1534 41 EzioAuditoreDaFirenze
42 1534 42 mango_lassi
43 1534 43 Rewinding
44 1534 44 kotamanegi
45 1534 45 KostasKostil
46 1534 46 dreaming_away
47 1534 47 SecondThread
48 1534 48 tatyam
49 1534 49 Golovanov399
50 1534 50 fsouza
51 1534 51 --d
52 1534 52 Suiseiseki
53 1534 53 conqueror_of_tourist
54 1534 54 Misericorde
55 1534 55 _overrated_

T-shirt:

List place Contest Rank Name
126 1534 126 sd0061
277 1534 278 BrayanD
810 1534 814 Sad_reacts_only
1068 1534 1080 MaytaCapac
2162 1534 2181 erfanul007
2484 1534 2515 I_love_Avicii
2976 1534 3016 SU_N_NY
4137 1534 4204 awesomemaths
4514 1534 4592 sejdemese
6367 1534 6510 vineeth143
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