AquaMoon's blog

By AquaMoon, history, 3 years ago, In English

Hello, Codeforces!

I'm glad to invite you to Codeforces Round 732 (Div. 1) and Codeforces Round 732 (Div. 2), which will be held on Jul/11/2021 17:05 (Moscow time). Note the unusual start time of the round.

The round will be rated for both divisions. Each division will have 6 problems and 2.5 hours to solve them. There may or may not be an interactive problem, so I suggest you should read the guide for interactive problems.

All problems were written and prepared by CoupDeGrace, kuangbin, mejiamejia, Sugar_fan, Melacau, Nanako, GOATWU, Cirno_9baka, Suiseiseki, ODT, box, Ynoi, syh0313, wh0816 and me.

And thanks to QAQAutoMaton, gamegame, starusc, interestingLSY, chenkuowen, JianfengZhu, Monogon, ijxjdjd, Schwarzenegger, sunsiyu, Nezzar, mtw, TheOneYouWant, gyz_gyz, kimoyami, njupt_lyy, ptd, coderz189, manik.jain, absi2011, gyh20, errorgorn, antontrygubO_o, oolimry, zhangzx123, wxhtzdy, tribute_to_Ukraine_2022, SonDinh, triple__a, Pecco, AzusaCat, dorijanlendvaj, Imakf, Forever_Pursuit, kartikeyasri23 for testing and good advice, isaf27 for his excellent round coordination and help with preparation and MikeMirzayanov for great systems Codeforces and Polygon.And you, for participating!

This is my first round ever. Great efforts have been put in during the past six months. We are sincerely looking forward to your participation. We hope everyone will enjoy it.

Good luck!

UPD1: Here's the score distribution:

Div 1: 500 — 1000 — 1500 — 2250 — (2000+2000) — 4000

Div 2: 500 — 1000 — 1250 — 1750 — 2250 — 3000

UPD2: Sorry to everyone! This is my first round and I spent lots of time to prepare it. I prepared at least 23 problems, and chose these eight ones for this round. I invited 35 testers to test, and chose problems according to their feedback. However, the difficulty is more than expect.And the pretest is not strong, leading to FST for many people. I blamed myself sadly. So I must say sorry to you, my dear friends. I will improve myself in the future. Wish you good luck and happy everyday. (๑•ᴗ•๑)

UPD3: Tutorial is available.

UPD4: Congratulations to the winners

Div1:

1.gisp_zjz

2.VivaciousAubergine

3.xtqqwq

4.Swistakk

5.Endagorion

Div2:

1.ki_msaga

2.jk-jung

3.RBurgundy

4.motomuman

5.flashsonic

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

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

First Upvote done.

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

orz! Looking forward for a great round!

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

I can confirm problems are awesome! Good luck in the round!

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3 years ago, # |
Rev. 2   Vote: I like it +426 Vote: I do not like it
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    3 years ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    though only the first 2 problems of div2 are different :)

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

Is Aquamoon a girl?AquaMoon

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

    Yes, I hope you like my problems!

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

      You are a very cute girl, I believe I must have good luck in this contest. Hope I can like your problems.

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

        Thank you! Wish you good luck and happy every day! (๑•ᴗ•๑)

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

        I believe I must have good luck in this contest.

        And you didnt even participate lol

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

      I just solved a single problem! :(

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

    Yes. She is a lovely girl.

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    3 years ago, # ^ |
      Vote: I like it +128 Vote: I do not like it

    Yes, but, unfortunately, this is not a dating site. :P

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3 years ago, # |
  Vote: I like it +68 Vote: I do not like it

Coordinators: How many testers do you want?
Authors: Yes.

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

    Most of the testers are my friends. It's me but not coordinators who invited them for testing.I want to make the problems better.

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

      Seems like you like making even blue as your friends. Can I be your friend in next round? :P

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        3 years ago, # ^ |
        Rev. 2   Vote: I like it +38 Vote: I do not like it

        Of course you can. Welcome!

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

          I am interested in testing too. Hope i can be your friend too

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

            Tip: If you want to be her friend, you can ask her discord using the cf station letter.

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

              what is cf station letter. Pardon my naiveness but i dont spend a lot of time on cf. Also google search showed irrelevant results

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

      It displays that you are quite passionate in making your first round likeable. I wish you good luck that everyone will like your contest!

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

      I think that this contest shows well why friends are not good testers

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

As a tester, problems are great!

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

2.5 hours... Any reason?

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

    Determination is based on the feedback of the testers.

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

      The photo below the goodluck!..is giving me good vibes ❤️

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

      How u practice your math skills like typical pnc problems and dp + pnc , so difficult to find intuition sometimes.

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3 years ago, # |
Rev. 4   Vote: I like it +246 Vote: I do not like it

Good problems! The difficulty level is moderate, the knowledge covered is broad, the statements are clear and short and the solutions are more natural.

I would like to give upvote to the writers!

update near the end of the contest: Now you may know what's the true meaning I said.

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    3 years ago, # ^ |
    Rev. 2   Vote: I like it -133 Vote: I do not like it

    Great thanks to conscience problem writers!()

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    3 years ago, # ^ |
      Vote: I like it +141 Vote: I do not like it

    "The difficulty level is moderate"

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

      There's a famous problem setter in China.

      And he always set comments like this after the hardest contest in China.

      I just translate it into English and change something.

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

        you should've said this before the contest "/

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

          But someone didn't let me say.

          So I could only set a comment like that, expected someone would get the meaning.

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      3 years ago, # ^ |
      Rev. 2   Vote: I like it +85 Vote: I do not like it

      5g50e0

      Every asian rounds ever

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

    no nothing is true T-T, you played with us.

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

I'm a tester and I'm sure the problems are excellent!

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

Good lucky to everyone.Get a high rank!

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

As a tester, I hope you can get the ideal score :)

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

As a setter, I hope you enjoy the problems!

Subscribe to Technoblade to AK div1 :)

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

    And what about div2 guys? What should they do to AK round #732 div1 ?

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      3 years ago, # ^ |
      Rev. 2   Vote: I like it +59 Vote: I do not like it

      Work hard and improve yourself. Good luck! Wish you would be happy everyday.(๑•ᴗ•๑)

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3 years ago, # |
Rev. 6   Vote: I like it -66 Vote: I do not like it

After mathforces and messforces I am hoping for good codeforces Round .

This was just a sarcasm guys don't take too seriously?.

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

    It's like talking random shit which belittles the efforts of the setters and testers and then trying to push it aside as a joke. I can't see why you wouldn't get downvotes for this.

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

Good luck in the round! I do wish that you can enjoy our contest!

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

How is this not in the homepage yet?

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

    It's on the homepage now.⁄(⁄ ⁄•⁄ω⁄•⁄ ⁄)⁄

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

I literally had a heart attack seeing this much color at one place!!

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

    Bro thats serious, go to the doctor.

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

      i took this advice seriously and doc prescribed me 3 more cf rounds in this month
      >.<

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

As I tester, I can say that the problems are very interesting! Strongly suggest you to participate!

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

Wow, Chinese Round again!! So many familiar writers and testers, can't wait to participate in ;)

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

    Why is it called the China Round

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

As a participant , I can say that I dont know anything about this round.
Just look I see the post after 6 days !!!

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

I can't confirm as I'm mentioned in the blog for absolutely no reason, but I do believe problems will be great.

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

First time I've seen the word "behoove" in a sentence...interesting word!

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

    Sorry, I have fixed the issue. Sorry for my word. Sorry again. Could you forgive me? (╯﹏╰)

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

      There's nothing wrong with behoove? I just thought it was interesting..

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

Coming!

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

    Orz yyb !!! You're my idol !!!

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

      I have not worked out any problem for a long time, probably I cannot solve any problem in the round >_<

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

        I bet you will solve at least three problems (*≧▽≦*)

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

          I think that is impossible >_<. I have no ability to solve so many problems in a round now.

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

            Don't be too modest ! (✿ω✿)

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

              But that is the truth. QwQ

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

                I wonder how this idiotic talk is getting upvotes

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

                  This is to illustrate that contribution points reliably measure valuable contribution. :)

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

I am really excited for the problems! hope high rating for everyone .. including me :)

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

Aarghh!! Clashing with Wimbledon finals time (:

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

    Are you participating in Wimbledon finals?

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

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

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

I'm sure that AquaMoon's first attempt will be glorious

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

first time a round post where round coordinator is given a backseat and testers are racing to win the game~

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

This comment section is one of the most wholesome ones I've seen on codeforces thanks to AquaMoon!!

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

Aoligei, ganle!

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

Mathforces->My favorite genre

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

The mighty Chinese questioning team!

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

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

Looking at testers and setters I already feel like I'm going to learn a lot from this round. Thank you for efforts!!

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

An All-star Round

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

Best wishes to everyone):

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

It must be an interesting round.

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

I think Div.1 F is a data-structure problem. Because ODT is one of writers.

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

99% of the comments : "as a tester...."

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

As a tester...

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3 years ago, # |
  Vote: I like it -40 Vote: I do not like it

Is this rated?

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

A 'relevant to this blog' meme

Hnet-com-image-2

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

Should i start my journey with this contest ?is div1 easy than div 2 ? I have registered today itself ,i don't know anything about this site

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

    difficulty level is div1>div2>div3 If you are just getting started, div3 would be better. But there is no harm in giving div2 contest. Give div2 contests.

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

      Thanks bro .btw can you give me some tips to be a specialist over here

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

        A noob whose first problem solved on Codeforces involving fenwick trees :|

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

The score distribution seems very good! I hope I will be able to participate. If you need an extra tester in future contests, I'll gladly help :))

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

    Get to that level firstly man

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

      Hi, I hope to get there soon! It's going pretty good, no worries! :))

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

    Welcome! You can contact me in CF talks.

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

      At which rating we can be a tester and can i be a tester ?

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

        There is no requirement for rating, but you must be a good friend of the writer or coordinator (the core idea is that you are trustworthy).

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

I hope there are no math problems

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

CCP is proud of you! No more gang of four!

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

Sunday = contest day (Leetcode->Kickstart->CF) Back2back contests

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

    Also, Leetcode $$$\lt\lt$$$ Kickstart $$$\lt$$$ CF, gets better

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

Any tips for the newbies:)

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

What is the penalty for div 2?

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

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

Aquamoon responding to every comments leads to some fat contribution for her, ngl.

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

Standing on 1399 I can ask only one thing from you , I need atleast +1 rating change .

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

Q!!

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

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

AquaMoon is not good at Programming is a pun!

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

Why did i even start programming ?

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3 years ago, # |
  Vote: I like it +50 Vote: I do not like it

The difficulty gap between problem D and E (B and C in div 1) is so large. The contest has run 1 hours and half and no one accepted E in div 2.

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

    i quit after solving a, b, c.

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

      I quit after seeing problem E has no accepted submissions. And go to the standing board to see my current ranking dropping.

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

        I'm not even reading E, comment section is more exciting.

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

    Maybe not. D seamed crazy too to me at first. Maybe from some strange angle there's short solution too.

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

Wow. That's some really monster combinatorics :) I'd like to see where to read on this.

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

AquaMoon was so hard but the problem was interesting !!

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

    But still, problem statements where ones of the clearest I've seen on codeforces :)

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

We couldn't help this much. Tell AquaMoon to be good at programming.

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

Hackforces

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

AquaMoon aka 'not good at programming, yet better than you'

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3 years ago, # |
  Vote: I like it +87 Vote: I do not like it

Me seeing numerous solution of Div1A get hacked:

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

What is the hack test for Div1.A?

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

    1
    6
    2 1 2 1 2 1

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

      Oh, god. Two adjacent friends with same number but different directions can't switch their directions! Why did it seam so apparent they could... ah...

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

Did anyone actually manage to prove the solution for Div1B? I just did some guesswork based off the pascal's triangle. And also how weak is the pretest for Div1A because I see so many people getting hacked

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

    You can moves 1s in pairs. So consider 0s and one 1 from odd size blocks of 1s to be fixed. Now this is just normal stars and bars where you're trying to place the pairs of 1s among the 0s (odd block 1s are indistinguishable regardless of which side you place the block(s) on).

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

      Damn, I could not make that observation during contest, guess I have to work on some combinatorics.

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

    Note that the operation is equivalent to 'shift 2 1's to the right or left'.

    Now we can just move any even subsequence, so let's focus on the sequences of an odd number of 1's. Between these sequences, the number of 0's is constant.

    Great, so if we ignore the pairs, which we can move, we get a sequence like 0000100101 with no 2 consecutive 1's. Then we can basically ignore the 1's, we only get to choose which 0's to put each of the pairs of 1's between. Suppose there are Z 0's and P pairs. Using the stars and bars argument, there are exactly

    $$$\binom{Z + P}{P}$$$

    ways.

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

    Here's one intuitive way to justify the formula for Div1B: Greedily pair up adjacent ones. Then, you can (with the operation) permute the zeros and the pairs of ones arbitrarily, and the unpaired ones will have their locations determined by the zeros. If there are $$$p$$$ zeros and $$$q$$$ pairs of ones, there are clearly exactly $$$\binom{p+q}{p}$$$ different ways to do this.

    EDIT: I should have refreshed to check if this was already answered before typing this up. It's too late now, though!

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

    How did pascal triangle helped you in this "Aquamoon and Chess"?

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

      Just look at the 14th row of the pascal triangle and you will see the answers for the last three sample, 1287 and 715, so I assumed there is some kind of a combinatorial formula that immediately solves the problem and found the exact formula explained by others.

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        3 years ago, # ^ |
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        Seems I need to byheart the pascal triangle rows.

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    3 years ago, # ^ |
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    Yes.First we will add 2 zeroes one at the beginning and one at the end.let x be the number of zeroes.let ai be the number of ones between the ith and (i+1)th zero for all 1<=i<=x-1 .Notice that the operation is same as subtracting 2 from one index(which has value atleast 2) and adding it to 2 to one of its neighboring index so the invariant is parity of ai.

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3 years ago, # |
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How to solve B?? Turns out this whole cute blog was a clickbait.

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

    The key observation here is that when swapping, the swapped characters' position do not change. So you go over all n strings and store each letter's position, then go over the remaining n — 1 strings to rule out positions, which leaves you all the letter positions of the missing string.

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

      Omg, this was easy, I got confused with swapping and shuffling in the statement and messed up pB.

      I thought strings could be shuffled after swapping. Smh.

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        3 years ago, # ^ |
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        Yeah, it happens. The statement is quite long :)

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        3 years ago, # ^ |
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        How did you solve D?

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          3 years ago, # ^ |
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          ((No. of consecutive ones-1) for all streaks of ones +no. of zeroes) choose (no. of zeroes)

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            3 years ago, # ^ |
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            Yeah, I just realized that too. Ran out of time to implement during contest :(

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              3 years ago, # ^ |
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              FSTed in C , I think I did very poorly, smh, so much weak pretest.

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                3 years ago, # ^ |
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                You should check my result, it'll make you feel a bit better ^^

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

    I took the sum of ASCII values of i'th(1<=i<=m) character of all the original strings and subtracted from it the sum of ASCII values of i'th(1<=i<=m) character of all the new strings.Then just accumulated the character for that value in a string.that's your output.Check my submission for clarity.

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

tilted off the face of this fucking planet, I can't solve C because I can't implement for fking shit, even with 50 minutes AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAVAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAASDFKLSANFLKAS;JFLSJKL

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    3 years ago, # ^ |
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    it is actually very difficult to upvote your comment in my computer because of the huge AAAAAA

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    3 years ago, # ^ |
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    There's a lone V in your AAAAA...

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3 years ago, # |
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Finally tourist will be back on rank 1 .

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

I tried to hack some of the solutions with test

1
7
3 2 3 2 3 2 3

I think the answer should be YES here, but the checkers says it's NO. Any ideas? :)

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

    After you convert the array to 2 2 2 3 3 3 3, the direction of each element is L R L L R L R. There is no way to convert that to all R

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      3 years ago, # ^ |
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      L R L L R L R -> R L L L R L R -> R R R L R L R -> R R R R L L R -> R R R R R R R ?

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        3 years ago, # ^ |
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        I dont think your final array is sorted

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        3 years ago, # ^ |
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        In the second-to-last step, you cannot make LR to RL.

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        3 years ago, # ^ |
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        When you swap an L with an R, nothing changes

        L R

        -> R L (swapped locations)

        -> L R (lefts become rights, rights become lefts)

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          3 years ago, # ^ |
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          Holy Moly! Thanks for explanation :)

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        3 years ago, # ^ |
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        I think you can't go from L R L L R L R to R L L L R L R by swapping only the first 2 elements. They still face the same direction.

        Note that when a person moves, they also carry direction with them. In this example, the first person is facing the L. After swapping positions with the second person, they will face R, so the direction for element 2 is still 'R'

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    3 years ago, # ^ |
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    Shouldn't it be NO? (Either that or my solution is going to FST)

    In essence a swap moves a person to a position of the opposite parity. So for an even number of swaps they must be on the same parity in the end. So count the number of people with each $$$a_i$$$ on each parity in the original and sorted arrays. They should be the same for YES.

    With 3 2 3 2 3 2 3, we have all 3s on odd parity and all 2s on even parity.

    After sorting we have 2 2 2 3 3 3 3 which definitely doesn't satisfy the property. Is my analysis wrong and is there some way of achieving this movement?

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    3 years ago, # ^ |
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    11 pretests passed. Kudos to the authors :))

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    3 years ago, # ^ |
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    I made the same mistake , think "LR" will be "RL" in a swap.

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3 years ago, # |
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When hackers get hacked :>

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    3 years ago, # ^ |
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    new 200iq contest strategy: write a bad solution so that you get hacked, realize that many other solutions can be hacked with the same test and gain more points than you lost

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    3 years ago, # ^ |
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    Give test please!

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

For D2E is this true — If we make a graph of array as nodes result is $$$2^{c}$$$ where $$$c =$$$ number of bipartite components?

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    3 years ago, # ^ |
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    I thought about counting in such components, wherein component nodes(arrays) are pairwise distinct in every position (component is a full graph) and the answer is a sum of C(size_of_component, n) for each component

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3 years ago, # |
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Not to complain but this round is WAY TOO HARD for me

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3 years ago, # |
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System testing is gonna slower than internet explorer.

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3 years ago, # |
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What the hell is wrong with you? Wasn't half a year enough to prepare well? Why does this bold note in D even exist? I don't see any reason to make this task "interactive" and put that garbage in the statement. Did you want to make your long awful statements even more unreadable? Well, then you succeeded. Also, please, learn such mathematical constructions as "intervals" and "segments" so you don't have to cripple your statements with these countless 10^-18. Pls stop creating problems.

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    3 years ago, # ^ |
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    I think this is because it's quite hard to check if an input is valid. And, maybe codeforces doesn't have a way to provide a different input format as a hack unless the problem is interactive.

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    3 years ago, # ^ |
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    They are making the problems interactive to support the hacking format.

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    3 years ago, # ^ |
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    Prob hacks. Well, if you use intervals instead the answer has to be defined as limit and that's another can of worms.

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    3 years ago, # ^ |
      Vote: I like it +262 Vote: I do not like it

    Hello. It is very unpleasant to read your comment, so I can't ignore it.

    About the interactive problems Div2B and Div1D: it is impossible to ensure, that the input is valid because it is NP-hard problem. So, validator can't be written and we have some ways how to deal with it:

    • Generate tests without validator and disable hacks. Such problems already appeared on Codeforces.
    • Make this problem interactive. In this case, everything can be validated. For the participants this variant doesn't change anything.

    It was my proposal to use the second option, maybe it was not the best, we will consider this experience in the future. Anyway, if the paragraph about the interaction was unclear to you you had an option to ask the question to us.

    About the intervals. Everything was defined formally. There were no mistakes in that definitions. So what can be told here? We know the difference between intervals and segments.

    If you couldn't solve the problems it is very stupid to take out your anger on authors. I have personally seen how much work was done to make this round possible and know how much sadness you can make by your comment.

    Of course, in this round, there are some things, that we could do better. And thanks to everyone, who will write his or her good or bad opinion, but not in such form.

    And according to the style of your comment, I propose you stop writing such comments after the round.

    I hope I will see many rounds from AquaMoon and her friends because in my opinion the problems were very interesting and had very beautiful solutions.

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

      I also liked the round, thanks for your efforts in preparing it.

      One thing I want to point out here. Maybe it would be better to let the participants know why this problem is "interactive". It was a little annoying reading a restriction that comes out of nowhere.

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      3 years ago, # ^ |
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      what the hell are you talking about? why if problems are disgusting and statements are ugly people can't talk about it in comments?

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        3 years ago, # ^ |
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        He never asked not to talk about it. You know you can discuss about them without insults to authors right?

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      3 years ago, # ^ |
      Rev. 2   Vote: I like it +101 Vote: I do not like it

      Hello. It is very unpleasant to read your answer to this comment, so I can't ignore it.

      First of all I want to clarify that I missed the round and had a chance only to read the problems in the middle of it so I can not honestly say anything about the quality of the problems even so I may or may not like them. I want to thank authors and testers for their efforts in preparing this round, but certainly there were some issues which could be fixed for the sake of future contests.

      About Div1D: I believe that most of the displeased contestants were confused by the fact that problem was interactive for some reason and that it was included in the middle of the statement without any proper explanation. I think it should be considered a bad tone to say that if you had any questions(about obviously somewhat unclear statement) you should have asked them lol. From my and my friends experience of problem setting, it is considered to be authors' and coordinators' responsibility to make problem statements clear and easy to understand for the majority of participants, as it is your desire to attract as many people as possible to take part in your round.

      About Div1E: I guess that the main objection here is that using intervals of time with +- 1e-18 thingies sounds a bit strange and obfuscates the problem statement in a certain way. For sure, your definition was formally correct, but again it is best to write it more clearly. These intervals just look ugly and it's unpleasant to read a problem statement like that.

      It also looks very petty to me to make remarks like "If you couldn't solve the problems it is very stupid to take out your anger on authors". It's very bold to make such assumptions, and from what I understood despite the offensive form of his comment the main problem in this round in his opinion was not his lack of ACed problems during the contest or even the quality of the problems themselves, but the problem statements. Even the most offensive phrase is just a reference to an old comment from the famous nuttela coder if I recall correctly.

      There were lots of rounds and lots of displeased people in the comment section with messages way more rude than this one(for example under my global round). I agree that it was wrong to put all the blame on the authors, however, decent problems statements are mainly coordinators responsibility and your a bit childish response made it look even worse than the comment itself.

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

        Maybe late, but here is an explanation on Div1E.

        Cirno_9baka told me the problem like this: there are some squares in a plane, and there is a line (y=x), the line can't cross the squares( but not include the bound), and change the line need some cost, find the min cost.

        Then change it to the statement with the backgroud is a hard work ( The problem had been changed because no writters could solve the previous version. ) How to define we can walking on the bound?

        We have discuss about both closed interval and open interval, if it is closed interval, than the boundary is not able to be walked.

        It seemed that open interval can solve the problem, but there is still a problem, how to control the time?

        We can stop the time, but if we let the time go, and then, the person will die immediately.

        Finally, we thought to define it with +-1e-18 is the most clearly definition for the problem.

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

      The statement of Div2B is hard to understand, which is bad given that it is a more or less trivial problem.

      The main points are:

      • missleading explanations about interactivity
      • the information "what is asked for" is in the last paragraph instead of the first
      • dispite of hard to read and to much information, still you tried to tell a "funny" story, making it even harder to understand (using unique words like a "stolen" string, "complete disarray")

      For me it was like 20 minutes text understanding but only 2 minutes problem understanding. Since A was not very clear either I did work on B before submitting A. And after having spent that 20 minutes I was out, not submitting anything.

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

      it is impossible to ensure, that the input is valid because it is NP-hard problem. So, validator can't be written and we have some ways how to deal with it.

      Hi ! I have a doubt regarding this. Can you please help me understand why getting a validator in a non-interactive version is np-hard and how making it interactive solves it ?

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    3 years ago, # ^ |
      Vote: I like it +334 Vote: I do not like it

    talant's comment is very rude. It is also arrogant and aggressive. I'm sad that high-rated participants allow themselves to lead the discussion in this way. This tone divides the community and multiplies the negative. The same facts could be stated without aggression.

    Let's read the text again.

    What the hell is wrong with you?

    Intolerable form and transition to personality. Great start to the discussion.

    Wasn't half a year enough to prepare well?

    Well, why is it here? It is solely a matter for the writers and the coordinator with what intensity they work.

    Why does this bold note in D even exist? I don't see any reason to make this task "interactive" and put that garbage in the statement.

    The word "garbage" has been added to make the authors more unpleasant.

    Did you want to make your long awful statements even more unreadable? Well, then you succeeded.

    Again all sorts of unjustified rudeness and amplification: "awful", "even more unreadable". Well, you're just manipulating! The problem is quite normal, except for this unusual phrase. There is nothing at all bad about it.

    Also, please, learn such mathematical constructions as "intervals" and "segments" so you don't have to cripple your statements with these countless 10^-18.

    What an arrogant tone. And very disgusting. In addition, the international mathematical community (in contrast to the Russian mathematical school) did not seem to agree on the terms interval and segment. See https://en.wikipedia.org/wiki/Interval_(mathematics)#Note_on_conflicting_terminology

    Pls stop creating problems.

    No, I prefer you just stop solving problems here, please. I really hope that the authors will not listen to you, but will simply draw some conclusions and please us with another round. They made significant efforts to make the community better. And you have no moral right to write to them in that tone.

    Let me summarize. I believe that it is imperative to give feedback on the rounds and problems. Including pointing out defects and mistakes. But this must be done in the form of civilized communication of educated people. You offend people. It's always easier to attack and criticize. It's just mean to behave like that. Preparing problems is very difficult, it is not an easy job. We must support the writers, be grateful to them. This is an adult and mature position. And mistakes and defects must be pointed out correctly.

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

C was beautiful =)

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Hack test(problem div2C/div1A)

1

10

5 7 5 7 5 7 3 3 4 7

Output: NO

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    3 years ago, # ^ |
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    Is "NO" what a correct program would output in this case?

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    3 years ago, # ^ |
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    I think the answer is 'YES'. My logic:

    Feel free to correct me if I'm mistaken somewhere.

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      3 years ago, # ^ |
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      every 5 is on odd position and 1 need to be on even position at the end so answer is no

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        3 years ago, # ^ |
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        Eh? The original three 5s are on odd positions, and 2 need to be on even positions (a.k.a the 'L's), which can be swapped to 'R's. Same goes with the four 7s.

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      3 years ago, # ^ |
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      You can't do L R L -> L L R.

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        3 years ago, # ^ |
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        oh yeah, i see. thanks a lot

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        3 years ago, # ^ |
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        why not..? Can you please explain...?

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          3 years ago, # ^ |
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          If you swap RL the R on the left becomes an L when it get's to the right, and the opposite occurs with the L, making RL -> RL

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Thanks for this beautiful contest.

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Can't believe the solved count of problem D in Div2

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

    Solution was leaked...lol

    A,B,C,D...Some uploaded all of them at 9:40 India Time

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

    They sky rocketed in last 20 mins

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    3 years ago, # ^ |
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    Thats why u shd never underestimate the power of telegram.
    If someone told u ans=ncr(zeros+ $$$\Sigma$$$ consecutive ones/2,zeros), you would have solved it too.
    No cases of plagiarism, no penalty, <500 rank, absolute win.

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

    Could anyone explain the solution to Div2D ? Editorial is not helpful at all.

    What's the intuition behind the nCr formula?

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      3 years ago, # ^ |
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      Let n be the no. of zeroes and m be the no. of pairs of 1 . Observation : we can replace atmost 1 one from a pair of one's with any zero in the string . So answer is (n+m)C(n) , since there are total n+m things and we have to choose n of them which will be zero

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Already figured out D2D is calculating combinatorial number but didn't have the time to implement it... sad :(

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3 years ago, # |
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thanks for the contest

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3 years ago, # |
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I regret not trying to hack during the contest

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

Wow what a contest (--_--)....**** the test cases :|

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

Anyone faced an issue with B problem, thinking about the answer when n = 1 Submission 1 Submission 2 Ac I got two Idleness limit exceeded on pretest 3 because of this.

Great problems anyway!

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3 years ago, # |
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Couldn't solve problem A

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how to solve D?

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    3 years ago, # ^ |
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    Notice how the pawns travel in pairs. If a pair of traveling pawns meets an isolated pawn, then the odd pawn gets shifted one to the left and the pair continues on its journey:

    11010
    01110
    01011
    

    This is equivalent to if the odd pawn didn't exist at all.

    So we can iterate over the array and count the number of pawn pairs $$$p$$$ and zeros $$$z$$$. Then the total arrangements are $$$C(p + z, z)$$$.

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

    The answer is nCr, where r is the count of number of zeros in the string and n is r + (No of consecutive 1s)/2.

    For ex, 100111100

    Here, r is 4, and n is 4 + (1/2) + (4/2) = 6. So answer is 6C4

    You'll have to find nCr % p using some efficient technique (like here : https://www.geeksforgeeks.org/compute-ncr-p-set-3-using-fermat-little-theorem/)

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      3 years ago, # ^ |
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      Can you explain why it works?

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        3 years ago, # ^ |
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        Sure.

        Observe that if you have 0110, you can move this "11" as a single entity. So, you can have 1100 or 0011. Similarly, if you have 011110, you can move the two "11" as separate entities, like 110110 or 011011 and so on..

        Basically "11"s move together and 0 can move anyways as it pleases.

        Now, 011110 can basically be treated as collection of 4 elements where two 0s and two 11s are identical. This means you have 4!/(2! * 2!) possible ways to permute, which is same as 4C2.

        You can try multiple examples with variations like 01110 or 10110 (i.e. what if you have odd number of consecutive 1s) and you'll see that the observation still holds

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

          Thanks for this explanation! To add on the last point about odd number of consecutive 1s, if we fix the positions of all 1 pairs, then all the remaining single 1's position will be fixed as well, hence they do not really matter, right?

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            3 years ago, # ^ |
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            Yes. Once we choose certain "11"s as pairs that we're gonna permute, the remaining single 1s won't matter as they'll be fixed.

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          3 years ago, # ^ |
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          Thanks for this. Editorial was too bad for this problem.

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    3 years ago, # ^ |
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    The binomial solution is also pretty heavily motivated by analyzing the simples. Two of them are 13C5 and one of them is 13C4 which leads one to think whether we can break the structure down in order to get that form. Once you look for something along those lines, the solution becomes way easier to find.

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Man, Div 2 C really gave me hell. I can't think clearly about permutations this early in the morning.

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    3 years ago, # ^ |
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    And I didn't even solve it correctly in the end, FST my life. :(

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Why was B such a half baked interactive problem tho?

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

    I have the same question

    It actually caused some timelimit issues in python for me, so I would like to know what was the reason to set up the problem as interactive

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

      Can you tell me whether this TLEd due to time limit issues for py or anything can be optimized?:/

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        3 years ago, # ^ |
        Rev. 4   Vote: I like it +1 Vote: I do not like it

        The biggest issue is ans += chr(curr) actually. In pypy strings are trully immutable and this operation is O(n)

        But even without it you have input of 10^5 strings which is somewhat expensive and pypy is not good with io. I usually bypass it by having the entire input read in the beginning which is not possible in interactive problems, That was my issue.

        The solution is the same though. I think your code may pass if you resubmit in python3, not pypy

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

    Because it's impossible to write a validator if we don't use the hack format.

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

I think the word Pre-Tests is self-explanatory. Why did you even care to run pre-tests on Div1 A if they were so weak.

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

HackForces

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

5g4ypa

I accidentally submitted for A at 00:54. ╥﹏╥ I swear my hand autopilotted me to submit A.

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

I will be orange without the pA hack. I am going to quit cp. (Nah just kidding but I hope the pretests are stronger :(

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

Solution for div1D

For finding the culprit, find the consecutive sums of coordinates of that aren't in AP.

With this we get the culprit, and the number $$$D$$$ to be added to one of the coordinates in that culprit time $$$T'$$$.

Now observe sum of squares of coordinates for a time moment.

They are of the form $$$f(t) = a + tb + ct^2$$$

Find a, b, c by finding the sum of squares for any three unchanged moments and solving the linear equation.

We now know what the sum of squares of coordinates will be at $$$T'$$$. It's just $$$f(T')$$$.

Now find a coordinate in $$$T'$$$ which if incremented by $$$D$$$, makes the sum of squares $$$f(T')$$$. This number is unique under the constraints. (brute force)

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

Who can estimate the difficulty of C div1?

I guess 2900?

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3 years ago, # |
Rev. 4   Vote: I like it +8 Vote: I do not like it

What the hell is the interactive solution? Why does not my solution work? I' ve added fflush(stdout) as tutorial says.

https://codeforces.com/contest/1546/submission/122130911

UPD: see my comment below.

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

    Dude, you have to flush in a loop, not outside the loop

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

    You only flushed for the last output, do it inside the loop

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

    You had to flush after every use of printf, your fflush(stdout) is outside the loop.

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

    UPD: problem is, as mentioned several times: flush must be inside loop, like my another solution: https://codeforces.com/contest/1546/submission/122130422

    The reason why does the second solution fails is that test case input does not send new line at the end of input. I'm reading input as scanf("%s\n",s). My opinion is that it is a bug of a test case.

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

      You don't actually ever have to flush the output, I have no idea why people in this thread and the problem statement said that you have to. The only problem was there not being a newline at the end of the input, which is really weird.

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

      It is very interesting, why the solution doesn't work.

      I checked now, that we send new line at the end of the input, so there is no bugs with a test case.

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

        May be a bug with the test system?

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

          It is very unlikely, because I think test system just creates a pipe between the solution and interactor and they communicate through it.

          Maybe scanf works differently when it reads the data from pipe.

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

Div.1 become speedforces, sad to see my negative delta...

1E and 1F just stand there, and no one can defeat even either of them! lol

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

Find a hack case in the last minute and lose 400 pts :(

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

Lots of WA on problem A because of my carelessness.
(._.) (._.) (._.)

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3 years ago, # |
  Vote: I like it +206 Vote: I do not like it

How do you manage to have 998244353 testers yet have weak test cases and imbalanced af contest.

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

    Maybe because 998244353 = 0 (mod 998244353)

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

    "How do you manage to have 998244353 testers yet have weak test cases"

    It's very easy: just don't stress test the testers' solutions. My solution for d1A would've gotten hacked.

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

Forgot that B was "interactive". Waiting for my Idleness limit exceeded on test 5 :D

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3 years ago, # |
  Vote: I like it +90 Vote: I do not like it

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

I was lucky to be late for the contest.

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3 years ago, # |
  Vote: I like it +279 Vote: I do not like it

As a tester, I dont know why they thought that most testers only solving AB was good difficulty balance.

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

when can I enter virtual participation in this contest?

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

    I believe it's just after the system finish testing all incontest submission so it should be in half an hour approximately

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

There is a decades-long technique to store secret data in the answer file.
A pity Polygon does not support it, and the authors have to resort to technical interactiveness.

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

    What's happened?

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

      Umm, nothing?..

      What didn't happen so far is supporting the technique in Polygon. Generally, pre-generated answer files are against best practices. So, such feature would be abused by novice or lazy problem authors, if it existed. At least that's the rationale I know.

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

        Can you explain how this relates to today's contest? Is this related to why Div 2B was interactive?

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

          Yeah. Ask yourself the question: why is the problem set as interactive?

          Anyway, it's not without upsides: the current technique (with interactiveness) allows for hacking, without changing anything in the hacking interface. I'd like it to matter more though, the common and very loud opinion is that hacks are bad somehow...

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

It's sad that just "knowing" a formula can get you solving Div. 2 D problems.

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

    Can you elaborate pls!

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

      The link in the spoiler describes it very well.

      Here in our problem, we have:

      $$$N = n - 1 - odds$$$
      $$$K = \sum\limits_{G ∈ Ones} ⌊\frac {|G|} 2⌋$$$

      and the answer becomes:

      $$$\binom{N - K + 1}{N - 2K + 1}$$$

      Where $Ones$ contains the sets of consecutive ones in the input string and $$$odds$$$ is the number of sets in $$$G$$$ which have an odd size.

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

    Wont it be Scenario 2? Edit: It can be solved with Scenario 3.

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

    Come on! It's