Devil's blog

Codeforces Round #659
By Devil, 3 years ago, In English

Hello, Codeforces!

I'm very glad to invite you to the first Cuban round Codeforces Round #659 (Div. 1) and Codeforces Round #659 (Div. 2) which will take place on Jul/24/2020 17:35 (Moscow time).

All problems in this round were created and prepared by mnaeraxr, gilcu3, dcordb, jcg and me. We tried to make them interesting and diverse and hope that you will enjoy them!

You will be given 2 hours to solve 6 problems in both divisions, and we highly encourage you to read all of them :)

We would like to thank:

The score distribution will be announced shortly before the round (or earlier).

UPD: Here is the score distribution:

Div2: 500 — (500 + 750) — 1750 — 1750 — 2500 — 2500

Div1: 1000 — 1000 — 1750 — 1750 — 2000 — 2250

Good luck and have fun!

UPD: The editorial is ready Editorial

UPD: Congratulations to the winners!

Especially for tourist, Benq and Radewoosh who solved all the problems!!!

Div. 1:

  1. tourist

  2. Benq

  3. Radewoosh

  4. ksun48

  5. jiangly

  6. yosupo

  7. ecnerwala

  8. Egor

  9. Marcin_smu

  10. ainta

Div. 2:

  1. AhoCorasick

  2. okikust

  3. crystal302

  4. ServantSaber

  5. Kirill_Kudr22

  6. RunMIT

  7. tripPple_A

  8. CodeSlayer4425

  9. ZADaCHI

  10. ITO

 
 
 
 
  • Vote: I like it
  • +732
  • Vote: I do not like it

Comments Comments (536)
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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope for an interesting first cuban round :)

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

    This round was not at all interesting.

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +150 Vote: I do not like it

    It looks like someone forgot to mention that the contest was for Russian 5th to 8th graders. (No offence intended).

    Anyways, this just makes it more exciting. Thanks for this round.

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    3 years ago, # ^ |
      Vote: I like it +86 Vote: I do not like it

    New round to avoid : Cuban Round

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

    It became really interesting when I tried submitting the first problem and not to mention, when I started reading Problem B.

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

      I became depressed after reading problem A & B also C

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        3 years ago, # ^ |
          Vote: I like it +41 Vote: I do not like it

        C was the real saviour

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

        I was like A will never left me but today A left me.

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +98 Vote: I do not like it

    Who is this Koa the Koala and why does she want to cross the shore?

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

      Google's first result for "can koala swim?"

      "Koalas drown in swimming pools when they are looking for water to drink. Although koalas can swim, if there are no assisted ways for a koala to climb out they will eventually drown."

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    3 years ago, # ^ |
      Vote: I like it +41 Vote: I do not like it

    did someone forget this was a coding round and not a reading round?

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

    why is everyone downvoting because of the contest... :(

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

    I started reading the first problem , and before i could even see the test cases , the round was over...

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

      I suggest improving your reading comprehension before trying a competition. Good Luck!

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

    I would rather do Eva than doing these problems

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

    Score for B1 should be more than C. Currently B1 score is significantly low than that of C. And every one knows the level of these 2 questions.

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

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

    After seeing too many Red coders as testers, we should prepare ourself for such a lengthy, disordered(e.g. problem B and C order) round.

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3 years ago, # |
  Vote: I like it -82 Vote: I do not like it

Shitty memes incoming!!!!!!

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3 years ago, # |
  Vote: I like it -44 Vote: I do not like it

Lol you are from Havana University

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

what is cuban round ?

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3 years ago, # |
  Vote: I like it +302 Vote: I do not like it

As a tester, I finally get to make one of these posts.

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3 years ago, # |
← Rev. 4   Vote: I like it -55 Vote: I do not like it

This comment was downvoted by trolls

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    3 years ago, # ^ |
      Vote: I like it +78 Vote: I do not like it

    Come with me sir, we're gonna find out who tf asked.

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3 years ago, # |
  Vote: I like it +73 Vote: I do not like it

Last time with this much red coders, we saw something dangerous.

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3 years ago, # |
← Rev. 2   Vote: I like it -92 Vote: I do not like it

When last contest is rated and you want another contest to participate in and you are happy and confused at the same time!!!:

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

Will it be IQ test round? I like IQ tests.

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    3 years ago, # ^ |
      Vote: I like it +424 Vote: I do not like it

    As a tester, I can confirm you need a brain to solve these problems.

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

      Writing 'As a tester' was really important for that confirmation :p

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

      MagentaCobra How much problems did he reject this time xD :)

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        3 years ago, # ^ |
          Vote: I like it +112 Vote: I do not like it

        I'm pretty sure he rejected ~15 problems

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

          After the contest can we have the rejected problems for practice?

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            3 years ago, # ^ |
              Vote: I like it +436 Vote: I do not like it

            No, they are recycled to be rejected in next contest again.

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

      After the contest, I was convinced that I have no brain :(

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

      Man...I should have read this before the contest:(

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

      As a human being, I can confirm that everyone has a brain smh.

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

As a tester, I’m glad I tested.

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

    Seems like we are again going to have round 657 with less than few hundreds people solving C

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

    [Enjoy your contribution] .

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

    As a participant, I should've known what you were saying.

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    3 years ago, # ^ |
      Vote: I like it +62 Vote: I do not like it

    Now we know what happens when DEVIL himself wishes Good luck

    Capture)

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

There is an amazing team behind this contest. I'm looking forward to participate. Good rating everyone

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

As a non-tester, I'm glad that I'll be able to participate in it <(")

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

    So, how did the contest go for you ? XD

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

      My ratings didn't go down, so it was good I guess :\

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

So happy

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

    Still holding to this opinion after the contest? ¯_(ツ)_/¯

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

      It's irony, the contest is so happy, it's like yesterday's contest.

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

Imagine getting back from retirement for a Cuban round, writers are all amazing. Looking forward to it, see you in the standings, altho I prolly don't pass a,b :). Удачи

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

The Div1 part of the contest is shown in Calendar on saturday, too. I think this is not intentionally.

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

We still remember the last red round

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

After a long time Um_nik became as a tester. I hope this round will be great.

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

It's a good opportunity for me to become PUPIL again :)

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

Congratulations! :)

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

Yes, i have heard of Cuban. Krasnodar is the best city.

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    3 years ago, # ^ |
      Vote: I like it +55 Vote: I do not like it

    No no no, not that Cuban. This Cuba

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

      Why do you downvoting me? That was just a joke. Of course i know what cuban means. Why do you upvoting some captain obviousness who didn't get the joke and tried to "explain" me what Cuban is?

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

        Some people just don't get sarcasm, man. I upvoted you and am glad that your comment is now receiving upvotes too.

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

    .

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3 years ago, # |
  Vote: I like it +53 Vote: I do not like it
Div2 A
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3 years ago, # |
  Vote: I like it -23 Vote: I do not like it

As a participant, I am all scared looking at so many Reds

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

You can do it! who want to become green, blue, ...

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

Ok

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

    shut up!!

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

      Your DP says that you stopped caring a long time ago then why is it bothering you. Everything gonna be alright!

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

Very excited for my first contest! I have covered all the basic data structures in past 1 month . Now I am looking to take part more frequently . Any suggestions on how to be consistent and any further resources ??

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

In codeforces round When I submit the problem and it gets pretests passed in very first submission. Ooh that feelings can't explain in words. When I am writing this I am crying with smiling face

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

    I think you haven't experience that feeling when you submitted it WA 10 times and the 11th time is got Pretest passed. Literally crying with smiling face.

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

As a tester, I wish every participant good luck and high rating on this (very nice) round!

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

I Hope Cuban round will be as good as Cuban Cigar. :)

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

A very interesting score distribution!

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

    score and difficulty level no match at all !

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

Perfectly balanced, as all things should be

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3 years ago, # |
← Rev. 2   Vote: I like it +60 Vote: I do not like it

( ° ͜ʖ °)

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    3 years ago, # ^ |
      Vote: I like it +168 Vote: I do not like it

    He should have waited for CF round #666.

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

      How about we reserve CF Round #666 to be held on 13th November, 2020 (Friday the 13th) and ask Devil to hold it?

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

        Only 7 contest in 4 month? I will lose my sanity

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

          Maybe we can try "some" educational rounds!

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

      Round 666 is coming up :)

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

    lets make it 666 again

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

    The cursed contest.

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

Wait, do we have subtasks in div2(asking because of the score distribution/and that it isn't mentioned)

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

    I suppose because in the format given like (750 + 750 ) suggests so

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

    It is written there will be 6 problems, so I think (500+750) => B1 and B2, B1 being subtask:)

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

I think that you should also author the seventh round following this one.

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

hello darkness, my old friend

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

Will Tourist top the tables today??

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

Hope,this first cuban round will be very interesting :D

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

Interesting score distribution! GL & HF!

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3 years ago, # |
← Rev. 5   Vote: I like it -12 Vote: I do not like it

ups

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

A sub-task for B? This is going to be wild.

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3 years ago, # |
  Vote: I like it +66 Vote: I do not like it

My schedule this summer,

EAT -> PARTICIPATE IN CF CONTEST -> READ EDITORIAL -> SLEEP

REPEAT

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

good luck to you ！！！

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

MikeMirzayanov Just curious to ask when running 2 contests (Div1 and Div2) you use the same server fot both the contests or 2 different servers (with Div1 having less specification server because of load)

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

Another Anton Round!

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

tourist registered for the contest, I think he'll take his position today!

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3 years ago, # |
← Rev. 2   Vote: I like it -7 Vote: I do not like it

i am having 1940 rating. and I don't want it to decrease for next 2 month so just wondering if someone can tell me how much maximum rank can ensure me 0 decrease? like any idea? is around 1000 fine

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

    There is a way:copy code and get "skipped".But your account might be blocked

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

photo-2020-07-24-17-23-42 .

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    3 years ago, # ^ |
      Vote: I like it +64 Vote: I do not like it

    The meme is as hard to understand as solving Div.1F

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

    incase anyone didn't get the meme carefully read the second line of the announcement!

    See Here
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3 years ago, # |
  Vote: I like it +99 Vote: I do not like it

The new strategy to decrease server load is working!

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3 years ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

Seems Red Coders doesn't think of beginner coders !!

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

Again a Div1 type contest :(

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

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3 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Glad I didn't submit even a single code!

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

div2 feels like div1 :(

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3 years ago, # |
  Vote: I like it +144 Vote: I do not like it

Last contest was such a class. Back to unnecessary complicated questions. All problem setters should learn something from Monogon and Ashishgup.

I know I am going to get downvoted to hell, but needed to get the truth out.

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

    +1,I love Monogon's problem and this contest is quite bad.

    Some of Ehab's math problem are also good.

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

    Since ^ comment is getting some attention, I want to add if we can come up with a way to voice our opinion about the round? Maybe upvote/downvote a contest or individual problems?

    I am not against harder problems but there's a difference between good kind of hard and bad kind of hard.

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3 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

UPD: Figured what was wrong on pretest 3. Sorry for spreading hate.

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

I'm a newbie. I didn't even understand the questions.

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

my confidence is indirectly proportional to length of questions

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

    My confidence is directly proportional to length of questions. I cannot solve and more people cannot solve.

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

Reading div2 B1-B2 problems is more difficult than doing a plank.

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3 years ago, # |
  Vote: I like it +48 Vote: I do not like it

Worst score distribution i have ever seen in Codeforces!!!

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

Div-2 contest doesnt seem to a div-2 type.No balance in problems at all.

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

Looks like Devil's plan at work here.

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

If the frequencies of these types of contest increases, it will inculcate the feeling of fear in beginners' mind towards competitive coding. No offense to anyone.

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

you asked for harder questions, so there you go :P

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3 years ago, # |
  Vote: I like it +60 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +27 Vote: I do not like it

Me after reading Div2 A, B1, B2:

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

Not worth complaining about the difficulty of the problems. Just improve next time and learn from your mistakes.

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

Do not trust someone whose account name is Devil

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

Kudos to all problem setters and team. What a contest. I realized that need to learn more practice more.

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

    There are better ways to learn that fact.

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

i was able to solve the first one phewwww!

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

    That's an achievement. Not kidding at all. You should be proud of yourself. I am unable to figure out where my code is failing for past one hour.

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3 years ago, # |
← Rev. 2   Vote: I like it -31 Vote: I do not like it

[Deleted]

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3 years ago, # |
  Vote: I like it +53 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +123 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +135 Vote: I do not like it

Another difficulty staircases

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

    Yeah, and I bet most of users solved problem B by guessing solution, not even having proof...

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

      You'd bet that but you'd be wrong. If you actually could guess the solution, then it wasn't too hard to prove three or four cases.

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    3 years ago, # ^ |
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    In my opinion

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

I thought WA on pretest 2 sucks the most, but today I learnt, WA on pretest 3 sucks the most. Getting WA on pretest 3 in C :(

Nvm I removed an if case and it worked. :)

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3 years ago, # |
  Vote: I like it +27 Vote: I do not like it

tourist returns to #1 !!!

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

Wait... Has Codeforces increased the difficulty of Div.2???

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3 years ago, # |
  Vote: I like it +63 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +120 Vote: I do not like it

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

The language of Div.2 B1 was very bad. :(

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

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

I happy for tourist, he will return to it's normal position :) tourist

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

    You should not add someone's handle unnecessarily

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3 years ago, # |
  Vote: I like it -44 Vote: I do not like it

Monogon's Div.1 contest was MUCH better than today's contest. problems were unbalanced and Div.1 B was terrible problem...

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

TFW you not only fail to solve div1A, but can't understand why your solution to div1A is wrong.

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

    Same! But I suppose I'm somewhat relieved/surprised to see a grandmaster say this! No offense ofc.

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +34 Vote: I do not like it 
    aabb
xyxy

    I still have no idea how to solve that problem, although something did pass pretests eventually. Had to move to more trivial problems like E instead :)

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

      Go through all letters in increasing order. For each letter, get the minimum of letters that this letter needs to be converted to (lets aabb — xyxy: a converts to x and y, take the smaller one), then convert all the letters (a) to the lexicographically smaller one (x or y) (note you need to skip letters that are already ok).

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

      What do you mean?

      For that one it's aabb bbbb xbxb xyxy

      (according to my code + my manual attempt)

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

      So the transitive reduction of a DAG isn't a subgraph?

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

        Transitive reduction requires that there is a path from a to b if and only if there was a path from a to b in the original graph. This problem doesn't have the "only if" requirement.

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

          24 (Dadgum.)

          I thought that adding more edges can't hurt the number of operations...

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      3 years ago, # ^ |
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      Acc. to my code, aabb->xxbb->xxxx->xyxy

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    3 years ago, # ^ |
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    Yes! Me too. WA on pretest 3...

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

    I also had a wrong pretest 3 at first and then I implemented a (in my mind) equal solution in a different and it worked lol (solution is back down). What I did it first was: go through letters in increasing order, if a letter is not good then convert it to a letter that also exists and needs to be converted to the same letter (ignore if it's already good, convert directly only if there are no such cases). Idk why this approach is any different than my other, could be just a program error since it's a bit more complex.

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

    I spent two whole hours battling pretest 3, only to no avail.

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3 years ago, # |
  Vote: I like it +46 Vote: I do not like it

It feels like the contest is based on some olympiad for students of 5-8 grades

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

I have a strange question. Who is Koa the Koala?

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3 years ago, # |
  Vote: I like it +27 Vote: I do not like it

Kind request to stop such bullshit Div2

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

lol, there was a duplicate test case in problem B. ~~~~~ 5 2 3 1 2 3 2 2 ~~~~~

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

    I was unable to reach sample test case left the problem after reading the problem statement only.

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

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

bad problems , bad statements . didn't enjoyed this contest .

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

This contest is not worth 2 hours of 20k Div-2 participants.

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

thanks antontrygubO_o for amazing coordination of this round.

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

Anybody else having had problems opening the page of their room?

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

Devil $$$2B$$$, Devil $$$1C$$$.

Devil Statement, Devil difficulty.

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

Can someone explain why this idea gives WA3 in Div2-C / Div1-A?

If any letter needs to be changed to an alphabetically smaller one, output -1.

Otherwise, build a directed graph (e.g. a matrix) where graph[a][b] = 1 iff we need to change a to b at least one index, a<b.

Then, drop the unnecessary edges. An edge a->b is unnecessary iff there is another path with more nodes that connects a and b. Note that the path will contain strictly shorter (in the alphabetic sense) edges, so we can start with an empty graph and add edges of alphabetic difference 1, 2, 3, ... 19, at each point checking that the letters are not already connected.

The answer is the number of edges in the final graph. I have no idea what I'm missing.

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

    i did the same thing, WA on tc 3.

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

    So on the test AB, DC you will print 3, because A->B->C->D, right?

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

      My answer is 2. The original graph does not contain a->b or c->d, and I only ever remove edges.

      Edit: Okay, I see it. The entire problem is that I only remove edges. For example, in aabb -> cdcd the optimal solution is to turn aabb -> bbbb and then two more operations. My solution prints 4 because it cannot add the edge a->b.

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

        Ah ok, AABB CDCD, the answer is 3, because AABB->CCBB->CCCC->CDCD

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

      shouldn't it be 2? coz A -> C and B -> D ?

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

    Got it

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

    aabb cdcd

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

    I did something weird, but it passed pretests somehow... basically ignore the fact that the graph is directed, answer = 20 — number of connected components.

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

    this problem can be solved by a simple dfs finding number of vertices in every component.

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

      infxx Can you please elucidate? I tried something similar but it didn't work. 87922305

      The major difference I see is that you are making the graph undirected(right?). Why so?

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

    Norrius What is the fix for this approach?

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

      It's really easy: just find the sizes of (undirected) connected components with DFS/BFS, let's say they are $$$c_1, ..., c_k$$$. Then the answer is $$$\sum_i (c_i - 1)$$$, because there is some spanning tree in each of the components, and you know how many edges a tree has.

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

        I don't quite get the intuition behind this fix? Like, how did you get to this? And why does that work? I get this idea that we need to find the minimum spanning tree(taking all edges' weight as 1) for our original directed graph(but there's an issue in it). But I do not get it that how making the graph undirected solves our issue?

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3 years ago, # |
  Vote: I like it +374 Vote: I do not like it

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

well

That was different

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

    each participant's state during contest was exactly like your profile pic ...

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

For Problem D in div 2., can we solve the problem for each bit independently?

Edit: NVM I just proved it to my self

Proof: if you have an odd number of bits, set at a particular position for all numbers in the array, the winner will be decided in this round Other wise you have an even number of bits, and your opponent will get the same result as you for that bit position.

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3 years ago, # |
  Vote: I like it +48 Vote: I do not like it

1384B1 - Koa and the Beach (Easy Version) is my new favorit of shittiest problem statement ever.

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

    Pictures would have helped.

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

      And some other story, mayby one which makes some sense or none at all... I needed to ask twice before understanding that the girl cannot swim if the water is to deep. How can somebody come up with such wired picture? This is simply the opposite of logic. I mean, why?

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

    Yeah, adding "More formally, " just made it harder to understand.

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3 years ago, # |
  Vote: I like it +67 Vote: I do not like it

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

Too time-consuming when reading Div2B problem :))

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3 years ago, # |
  Vote: I like it +91 Vote: I do not like it

So laughy to hear, when problems are nice in easy contest, and bad in hard contets. LOL. bad != hard.

Problems were cute && hard, div2 guys just cant solve them and find it is bad && want unrated — it is so stupid && two-faces

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

    The problems were pretty good, it's just the fact that C was a lot easier than B in div2 (You can look by yourself). I find B as a very nice problem, but it took me more than 15 minutes to understand the statement.

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

      Yeah, you are right. I am just angry about people who wrote that problems were bad and shitty

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

        I understood you. I'm also finding this annoying. If you understand the statements of the problems, they are enjoyable (in my opinion). I just hope that in the future, the authors will write the statements more concisely and the problems will go in increasing difficulty.

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

    Finally a really accurated opinion. I left this community two years ago, when i retired from regionals. Now two years without basically coding at all, i return for the first cuban round, and i managed to solve A, B1, C and get a positive score given by the fact that when i retired there were approx 4000-6000 participants per round in Div2. I found out that now there are nearly 20k, but 16k of them at least, only know to cry about problems that a retired person can solve with 0 warn up, thanks so much codeforces, for leaving me this bad memory as about the first round written by my country.

    To the person who i'm replying, really my most sincere congratulation, for not being one of those crying babies. Thanks!

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

Anyone else with wrong answer on pretest 3 in C? What was it??? :/

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

    me :(

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

    I guess that was the first real test with many test cases. Test 2 contained some corner cases I think

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

    Same here. Not sure what it was, but I know I did not handle a case like: aabb cdcd (Answer should be 3: converting a->b, then b->c and b->d)

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

C was DSU or DFS/BFS/ Spanning Tree loved solving it <3;

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

    It was easy greedy. You overkilled it. Can you briefly say about dsu solution?

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

      I converted all conversions into the form of A to B in a directed graph if at the end of all conversions character A was required to be converted into B . Now if A>B then always answer is "-1" , this is self explanatory. Now I added all the edges to the graph , if A-->B , Now any successful completion or conversion can can be done efficiently if for every pair A to B to which conversion is required we follow exactly one path and no more than that , this conversion can be seen as path from node A to node B in the graph. Now for a connected components if we want minimum paths to follow and connect all characters which are need to be converted into one path, we need to find just a spanning tree of the obtained graph. Thus we can obtain it easily using any technique like BFS/DFS/DSU . Thus it was a good problem of Spanning tree, I think. May be I over-killed it.

      Consider the small case graph where aab-->bcc A->B , B->C, A->C. since the spanning tree of above graph will be A->B->C , which is minimum possible answer for this conversion process.

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

        Wow, that's an elegant approach. Thank you :)

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          3 years ago, # ^ |
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          Yours Welcome , What is the greedy approach by the way.

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

            Created a directed graph.

            Then I traversed from the character 'a'. Assuming our current character is 'C'th.

            Using one operation, I changed all character 'C' to the minimal character over all his sons, and now add the new childs to the minimal character.

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

            AM_I_Learning What is wrong with my solution 87922305?

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

        Very interesting approach, thanks for sharing.

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    3 years ago, # ^ |
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    Can you please share your approach

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3 years ago, # |
  Vote: I like it -101 Vote: I do not like it

Time for antontrygubO_o to resign. What a disaster.

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

In my opinion Div 2 C (Div 1 A) was waaaaaaaay easier than Div 2 B.

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

New type of contest to avoid : Cuban Round

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3 years ago, # |
← Rev. 2   Vote: I like it +62 Vote: I do not like it

Div2D/Div1B solution is really nice:

Notice that for a given bit, if it occurs in an even number of elements, then the players will either both have this bit in their sums or both not have the bit (either both have an odd number or both have an even number of the bit) regardless of the actions taken in the game. Thus, we can look for the highest bit that occurs with odd frequency. If there is no such bit, then it must be a draw. Someone will have this bit in their sum, and someone else will not, and the other bits do not matter since this is greater than the sum of all the lower bits.

So, if we call this bit b, then we can replace all of our numbers with either 1 or 0 depending on whether or not they have bit b. Now we simply need to solve the case where there are only 0's and 1's in our array, which is left as an exercise to the reader.

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

    It wouldn't help me, unfortunately. I got almost all the conditions. I've checked if the number of 1's equals 1 instead of count % 4 == 1. I just considered 1 and 3 and did not notice that for 5, the first person also gets the odd number of elements...

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

    But I just stuck in how to solve 0-1 array!!!Please tell me!

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

      First, think about the case with no 0's. Who wins if there is just one 1? Three 1's? Five 1's?

      Now, add a single 0. How does that change things? How about a second? A third?

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

bad contest with long statments

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

Nice way of reducing the load of the server and taking care of the long queues :P

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

I feel lucky for div.1 participants for not having to solve such B.

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3 years ago, # |
  Vote: I like it +164 Vote: I do not like it

Awesome problems! From which AtCoder round did you steal E?

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

    Jajaj thanks. It is nice to hear some good feedback. We (setters) are really glad you like the problems.

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3 years ago, # |
  Vote: I like it +100 Vote: I do not like it

I think I finally kinda understand people who whine about terse (few details) editorials.

Yesterday I tried to solve AGC 027 E. I found an easy classification about what intervals we can compress into a single element and what the result of that compression is. But I failed to come up with the DP that would count the subsequences.

Today I see 1383E - Strange Operation which is obviously very similar (basically, rules about collapsing intervals are simpler). I decide to read the editorial of 027E in the hopes that the DP might be similar. And what do I find from the editorial?

"We can count such $$$t$$$ by a simple DP. This solution works in $$$O(\left|s\right|)$$$ time." (end of editorial)

In other words "lol that's trivial" :/

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

    That's one sure way to make us solve the exercises "left to the reader", just put them in another contest, right?

    Can you give a hint to the idea for problem E? I've observed that no zeros from different contiguous components can be in the final string, but I couldn't find a way to count all possible final strings beyond the multiplication when we don't delete any 1's.

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

      In the end I'm not sure how much a more detailed editorial would have actually helped.

      Yeah it's basically "count # of distinct subsequences" with that restriction (and some additional fiddling with initial and final zeroes, which is too tedious to think about). For each subsequence let's count only the "leftmost".

      Suppose you have just ended a block of 1-s and would like to add a block of $$$k$$$ 0-s to your subsequence. Then you move find the first block of 0-s that's at least $$$k$$$ long and jump to the end of that.

      Suppose you have just ended a block of 0-s and would like to add a block of $$$k$$$ 1-s to your subsequence. Then you simply take the next $$$k$$$ 1-s in the original string and jump to that position.

      If you build a DP based on that you get $$$O(n^2)$$$ complexity, but it's relatively easy to optimize from there.

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

Were pretests strong ?

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

    Noooo...A had weak pretests...got WA on test 15 in system testing...this has to be one of the worst contests in recent times...

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

Now I am eagerly waiting for someone to upload a video editorial of B and C and comment here !!!

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

Don't know why but I thought 

5
1 1 1 1 1

is "LOSE" in div1B for the whole contest :)

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

what the hell was test case 3 for Div2 C ?

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

    Not certain (I didn't get C, but know one reason mine probably failed on it), but I bet it was something like: aabb cdcd Optimal is to convert a to b, then b to c and b to d, for 3 moves. I know I was still converting a-c,a-d,b-c,b-d.

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

Don't you think the Div.1 difficulty gap is terrible?

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

in Problem D I added the following if statement 1.5 minutes before the end and it worked and i was ecstatic XD

Spoiler

I really hope my Python won't TLE because there was a point I thought the pretests didn't pass because I didn't have enough bits so I went all the way up to 50 (even though 30 is enough). So I spend 20*n work for no reason what so ever and it just might cause TLE (on python it's super sensitive).

EDIT: It didn't TLE! But it was sure close. one pretest was 936ms XDDD

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

DIV 2 A before: You have to come up with something

DIV 2 A now: You have to write a long implementation

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

    This submission is small

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

    What are you talking about? it was like 11 lines of code and I was wasteful. I only used the letters 'a' and 'z'. You just have to use the last string, copy it until the prefix length, and add 'a'/'z' depending on situation.

    Python solution:

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

    Usually there was nothing to come up for in past rounds.

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

Another Russian olympiad for students of 5-8 grades :D.

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

Highly Undistributed Difficulty Level (Div2)

The Ranks of Participants who solved only A lie from 2745->infinite.

It was FastSolveProblemA Forces. And was someone like me who made a wrong submission on A and faced -50 penalty,it was like like Facing the Real Devil.

Ranking dropped down Dead.

We need better adjusted ProblemSets.

Ciao Rating !

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

I wanted to drown myself trying to solve div2B

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3 years ago, # |
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Damn! this round was soo tough than normal div 2 rounds

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

Bring back the good old days when DIV-2 was actually DIV-2.

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3 years ago, # |
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How is B1 easier than C, also why is points for B1 not even one third of C ?

Such a long statement for B1. B1 should atleast be of 1000 points.

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

tourist comback Top rated

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3 years ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

My idea for div1C

Let's make a directed graph G with 20 vertices (= letters) and edges from v1 to v2 if v1 needs to be changed to v2. Then we need to get a topological sorting S for G. Since it contains loops, pick the smallest possible set of vertices which are parts of any loop (so that if we remove these vertices, all the loops will be broken), put them in front of S, delete their inbound edges and sort the rest of the graph.

The answer will be size(S) — 1.

I don't know how to find loops in a directed graph and how to find that set of vertices :/

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

    Hm, I'm not sure I totally understand your solution, but maybe something similar would work with condensation SCC graph? I tried to think of something like that but couldn't make it work.

    EDIT: Actually, I read your idea more closely. Wouldn't this imply that on the complete graph the answer is 19 when it should be 39 (I think)?

    I had a feeling there might have been some solution involving bitmasks since $$$|\alpha| = 20$$$ rather than 26.