djm03178's blog

Codeforces Round #688 (Div. 2)
By djm03178, history, 2 months ago, In English

오랜만이에요, 코드포스! (Long time no see, Codeforces!)

I'd like to welcome all of you to Codeforces Round #688 (Div. 2)! The contest will start at 04.12.2020 16:05 (Московское время), and it is rated for all participants with ratings under 2100. Note the semi-unusual start time.

You will be given 6 problems and 2 hours and 15 minutes to solve them. The score distribution will be announced soon.

All problems are prepared by me, with a lot of help from the testers making me realize that my solutions are dumb.

Thanks to Green55, JooDdae, cs71107, YeongTree, DreamingLeaf, jh05013, Lemonade255, 39dll, alswhp, Ku_Top, sonjaewon, slah007, jooncco, and rkm62 for testing the round, and especially xiaowuc1 for helping polish English statements as well. I would also like to thank 300iq for round coordination, and MikeMirzayanov for the great Codeforces and Polygon system.

See you in the round!

UPD: The scoring distribution is 500 — 1000 — 1500 — 2000 — 2500 — 3500.

UPD 2: The round is finished. Thanks for your participation! I'm sorry about underestimating the difficulty of problem B, but I hope you still enjoyed the problems! The editorial will be posted in a minute.

UPD 3: The editorial is out!

UPD 4: Congratulations to the winners!

Div. 2

1: caoyizhong

2: Depth_First_Search

3: Misaka23334

4: TheWolfWhoCriedBoy

5: EzioAuditoreDaFirenze

Unofficial Div. 1

1: Geothermal

2: jiangly

3: neal

4: saketh

5: Pyqe

 
 
 
 
  • Vote: I like it
  • +757
  • Vote: I do not like it

Comments Comments (340)
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2 months ago, # |
  Vote: I like it +313 Vote: I do not like it

As the problemsetter, I want comments!

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    2 months ago, # ^ |
      Vote: I like it -96 Vote: I do not like it

    djm03178 orz....

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    2 months ago, # ^ |
      Vote: I like it -27 Vote: I do not like it

    G.O.D

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    2 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Ok I commented as per your wish.

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    2 months ago, # ^ |
      Vote: I like it -49 Vote: I do not like it

    So yummy~~

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    2 months ago, # ^ |
      Vote: I like it -7 Vote: I do not like it

    comments

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

    djm03178 (Good problemsetter)

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    2 months ago, # ^ |
      Vote: I like it +31 Vote: I do not like it

    As a participant, I will give you upvote.

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

    As a participant, upvoted and commented :)

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    2 months ago, # ^ |
      Vote: I like it -24 Vote: I do not like it

    // cout << "Hi" << endl;

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    2 months ago, # ^ |
      Vote: I like it -27 Vote: I do not like it

    sir make sure to add some or atleast a dp problem between A-D.

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

      the problems have bet set a long time ago. So I think this is not possible.

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

    I hope it is a good contest :D

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

      same i also hope it be like the educational 99

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    2 months ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    As a tester, I wish I could participate in this round as this round is very well prepared by djm03178. Good luck to all the participants!!

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

      rkm62, I see your name as a tester in many contests. As a CP enthusiast, can you please let me know what can I do to be a tester?

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        2 months ago, # ^ |
          Vote: I like it +12 Vote: I do not like it

        You need to know the problem setters (to get asked) or take part in many CF contests (to be asked by Mike if the problem setters cannot find enough testers).

        Hope it helps.

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

          Thanks a lot! Sadly, I don't really know any of the problem setters. I guess I'll have to wait patiently and hope Mike asks me sometime in the future.

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

        Theirs a pretty big discord server where a lot of people hang around. Imo it's the best place to get to know problem setters and probably ask to be a tester. Just remember to not be so annoying ;)

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    2 months ago, # ^ |
      Vote: I like it -14 Vote: I do not like it

    Why is there so many downvoted comments .

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    hii sir big fan sir!! :")

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    yes

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Dude C was easier than B... at least for me. I litterary wasted 2 hours trying to solve B. Props on C though pretest 1 is very helpful to skip a ton of mistakes.

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    yes... you got my comment as you are problem setter. thanks a lot. this round was awesome

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2 months ago, # |
  Vote: I like it +64 Vote: I do not like it

As a tester, Please enjoy problems!

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    2 months ago, # ^ |
      Vote: I like it +28 Vote: I do not like it

    why is this getting downvotes lmao

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      2 months ago, # ^ |
        Vote: I like it +32 Vote: I do not like it

      Even when I saw that comment first time, it already got about 70 downvotes, which made me surprised.

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      2 months ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      When I first saw this comment it got more than 70 downvotes and I didn't know why... To tell the truth, I haven't seen such kind of comments being downvoted before.

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    2 months ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    As a tester, I think he deserves upvotes. He did a great job as a tester.

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2 months ago, # |
  Vote: I like it +24 Vote: I do not like it

Wow.. I'm looking forward to participate in this round!

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

Korean Round!

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2 months ago, # |
  Vote: I like it +47 Vote: I do not like it

Is it rated?

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2 months ago, # |
  Vote: I like it +11 Vote: I do not like it

Amazing writers Amazing testers and Amazing coordinator i wil remember the day 2020 December 4 forever.

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

    It's my ex-girlfriend's birthday on that day... Should I wish her excitedly, normally or not at all? The only thing I'm excited about is the contest and a little bit about the INDIA-AUSTRALIA cricket match.

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      2 months ago, # ^ |
      ← Rev. 2   Vote: I like it +55 Vote: I do not like it

      Then several cricket match is given and they have distinct costs and happiness

      Unfortunately, I only participate k match among them

      What is the best strategy I maximize the girlfriend's happiness and then minimize the costs I spend

      Wow amazing problem!

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        2 months ago, # ^ |
          Vote: I like it -17 Vote: I do not like it

        It's pretty simple, girlfriend's happiness index will always be zero no matter what algorithm you apply in your life :(

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      7 weeks ago, # ^ |
        Vote: I like it +12 Vote: I do not like it

      Everyhing can be "normal" again if you wish her "excitedly".Gd luck buddy go ahead.

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2 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Another Korean Round! This round will be amazing!

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

WOW i cant stand it

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2 months ago, # |
  Vote: I like it -18 Vote: I do not like it

This round is sexy.

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2 months ago, # |
  Vote: I like it -34 Vote: I do not like it

팬이에요 :fan:

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2 months ago, # |
  Vote: I like it +45 Vote: I do not like it

Hope they dont give out a contest full of "interesting observations" problems.

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

꼭 참가하겠습니다!(I will participate in this competition!) :god: :fan:

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    7 weeks ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    안녕하세요? 한국에 이사이트같은 것 없어요?

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

please don't make it like educational round 99 !

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2 months ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

Wait.... A negative rating tester??? Thats something new

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    2 months ago, # ^ |
      Vote: I like it +54 Vote: I do not like it

    A negative-rating tester whose atcoder rating is more than 2000 :D

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

Hope contest will bring some new learning

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

Oh friendly time for Chinese! But we Chinese will have a important contest (NOIP) right after this, hope for good luck after all!

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2 months ago, # |
  Vote: I like it +29 Vote: I do not like it

May the pretests be strong!

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

Where the difficulties of problems in Codeforces round 685 and other?

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

unusual start time

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

Hope to see some interesting problems with short statements and strong pretests :)

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

    hoping that queue should also be fast not as that of previous two contests !!

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

Looking forward to solving some amazing problems. :) //Hope I can solve more than 2 this time XD

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2 months ago, # |
  Vote: I like it -8 Vote: I do not like it

The contest will start at Friday, December 4, 2020 at 18:35UTC+5.5.

Most Important Note the unusual start time.

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    7 weeks ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    Learnt a new lesson "Don't remind anything if it is already posted on Contest blog". Thank You.

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

Oh， time is more friendly than others

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2 months ago, # |
  Vote: I like it -14 Vote: I do not like it

I will participate in this contest.

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

한국 시간 고려해서 semi-unusaul start time으로 맞춰주신 거 감사합니다! Thanks for considering Korean time!

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2 months ago, # |
  Vote: I like it -28 Vote: I do not like it

Is it rated?

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2 months ago, # |
  Vote: I like it -11 Vote: I do not like it

Why semi-unusual start time and not unusual? :thonk:

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2 months ago, # |
  Vote: I like it -10 Vote: I do not like it

Normal People be like, 'Hey' , 'Hi', 'Hello'.... and @djm03178 be like, "오랜만이에요, 코드포스! (Long time no see, Codeforces!)"...that's pretty cool and badass at the same time. ;)

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2 months ago, # |
  Vote: I like it -22 Vote: I do not like it

Can I get contribution without posting a meme

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

Hope a Great contest

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    2 months ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    Why I am getting downvote. what's wrong

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

      How do you know that contest is great before it even started ?

      Upd : This was a reply to his previous comment he edited the comment after that.

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

hope to become green

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

t's very sad that on the day of the round there is an RMI and I won't be able to raise the rating :(

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2 months ago, # |
  Vote: I like it +11 Vote: I do not like it

i hope this time queue should be fast not as previous two contests !!

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

(Long time no blue, Codeforces!)

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

Plot twist-Early announcement was made just to get more upvotes.

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2 months ago, # |
  Vote: I like it +14 Vote: I do not like it

hmmm strange 1-gon hasnt put any comment yet.

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

Hope this time we don't have to wait too much time for the rating.

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2 months ago, # |
  Vote: I like it -57 Vote: I do not like it

코드포스 is the wrong spelling

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    2 months ago, # ^ |
      Vote: I like it +42 Vote: I do not like it

    That's how Koreans call Codeforces in Korean communities.

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

I hope the problems will more enjoyable than last round .

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2 months ago, # |
  Vote: I like it -27 Vote: I do not like it

Notice the unusual timing.

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    7 weeks ago, # ^ |
      Vote: I like it -21 Vote: I do not like it

    My first comment on Codeforces only getting downvotes. Noice :)

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2 months ago, # |
← Rev. 5   Vote: I like it -26 Vote: I do not like it

Now my contributions are in negative :D

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope I could solve A and B in time. Thnx for the contest.

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7 weeks ago, # |
  Vote: I like it +11 Vote: I do not like it

Good to see u again Lemonade255 !!!

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Only thing giving me hope is the unusual start time .May be something unusual happens for everybody in rating changes.
xoxo +delta*2

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

I hope there are more interesting greedy, math, constructive algorithms.It can exercise thinking :D

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Your effort is appreciated! May the CodeFORCES be with you...

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7 weeks ago, # |
  Vote: I like it -39 Vote: I do not like it

vaibhav garg

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7 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

my first contest after reaching expert! Super excited :D

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

    I feel great when I see people like you, as some people after reaching expert, stop giving contests in order to uphold their ratings.

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      7 weeks ago, # ^ |
        Vote: I like it +7 Vote: I do not like it

      It's likely that I might become a specialist after today's contest, but I love this quote from the Batman movie — "Why do we fall? So we can learn to pick ourselves up."

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

i wish to be a good contest

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7 weeks ago, # |
  Vote: I like it -33 Vote: I do not like it

Здравствуйте всем тот кто сдаёт эту соревнование !!! Удачи вам конечно я желаю !!! После соревнования если вы не смогли решать не которые задачи , тогда заходите по ссылку на Мой Телеграмм канал мы будем разбирать задачи ровно через час после завершения соревнования !!! Успехов вам !!!

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

This is my first time to participate in codeForces competition. What is the reward for this competition?Thanks!

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Explore the competitive programming, that would increase your problem solving skills. And after participating in this contest, you will definitely get a rating, rank among all the participants. So enjoyed it. Its a coding game.

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7 weeks ago, # |
  Vote: I like it -18 Vote: I do not like it

Удачи всем !!!

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I am really really excited for this contest. Participating in Div-2, for my kind of rating people is really beneficial.

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7 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

Hope the problem statements are as short as announcement

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

Wow... From the Round #620, we can see that djm03178 is a very talented problemsetter. I think this round will be great!

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Is there a way to cancel the registration? I can't participate suddenly！！！

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes, you can cancel it, but remember — no submissions -> contest unrated for you

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I should go to sleep before 11:00 tonight so actually this div.2 is the last round before I leave :)

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7 weeks ago, # |
  Vote: I like it -17 Vote: I do not like it

Yo Boiz i am sing song... Kaali kaali nain ka pakode kaale shoe leke gaddi kaali kaali teri ghadi wali da bache bache wich hit kudiyo da bradd pitt leke gaadi kaali kaali teri ghadi wali da teri mutt marida tenu pyaar krda....all black

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

1 hour left i hope it will be easy. I also hope i increase and u all increase in the rate good luck :D

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

As a participant, thanks djm03178 for the contest.

In conclusion, good luck to everyone ^_^

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7 weeks ago, # |
  Vote: I like it +2 Vote: I do not like it

Hope to perform good in my 100th contest :) Good luck everyone....

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

As a newbie, starting my 10 mins wim hof breathing session before the contest starts in 12 min to improve my chances :)

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7 weeks ago, # |
← Rev. 2   Vote: I like it -33 Vote: I do not like it

this contest is pretty hard

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7 weeks ago, # |
  Vote: I like it -12 Vote: I do not like it

https://www.scaler.com/event/coding-interviews-dynamic-programming?rcy=1&rce=e068e621bfd5

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7 weeks ago, # |
  Vote: I like it +8 Vote: I do not like it

I think djm03178 has done a typo while writing Div-*2* lol.

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

Is it Codeforces Round #688 (Div. 1.5)?

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7 weeks ago, # |
  Vote: I like it +10 Vote: I do not like it

Is this a DIV 1.5 round?

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7 weeks ago, # |
  Vote: I like it +26 Vote: I do not like it

I feel that in quest of making unique problems, problem setters are increasing the difficulty of the problem.

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

this is a very hard contest

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7 weeks ago, # |
← Rev. 2   Vote: I like it -26 Vote: I do not like it
The comment removed because of Codeforces rules violation
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7 weeks ago, # |
← Rev. 2   Vote: I like it -22 Vote: I do not like it

I think there was some other comment related pob B. what's going on? why codeforces deleted comment related pob B?

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    7 weeks ago, # ^ |
      Vote: I like it +61 Vote: I do not like it

    It is against rules to discuss problems before a contest ends. Please, wait for the end of the round.

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      7 weeks ago, # ^ |
        Vote: I like it -58 Vote: I do not like it

      Dude, I got banned on 2 days and can't participate on next cf round on FenWick because of writing comments "A < C < D < B for me", It's soo stupid. I think it's not prohibited to write ur opinion about level of problem not statement.

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        7 weeks ago, # ^ |
          Vote: I like it +27 Vote: I do not like it

        Your opinion on the difficulty of problems is information that can affect other participants. Just never talk about problems until the round is over. It's very simple. By the way, I recommend solving problems during the round. Note that the read-only mode does not limit participation in the rounds.

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    7 weeks ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    The owner's comment makes it seem like the problem in its essence was being discussed. Perhaps that was the plan all along to delete the comments and make it seem like the approaches to the problem were being discussed. On the contrary, statements were made to connote the recent trend in absurd difficulties for Div 2B and there is nothing from the rules here that forbids that. If a specific rule bears that intent, I would humbly suggest that the English version is re-written in a way that the meaning is not lost during translation.

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      7 weeks ago, # ^ |
        Vote: I like it +17 Vote: I do not like it

      Can-do's and Can't-do's 5th

      "The contestants are forbidden to talk about subjects, related to the problems, with anybody, including other contestants. It is only allowed to ask questions to the jury via the system (see the 'Questions' section)."

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        7 weeks ago, # ^ |
          Vote: I like it -28 Vote: I do not like it

        Ignoring the terrible punctuation on the first sentence that confuses anyone that properly understands English, one is only left with the second sentence.

        "It is only allowed to ask questions to the jury via the system". This implies that there should be no discussions about the problems in themselves. This does not imply that someone cannot say that Div2B is hard publicly.

        Like I said if one can't say this, the rule needs to be extended correctly.

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope your rate increases :)

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7 weeks ago, # |
← Rev. 4   Vote: I like it -8 Vote: I do not like it

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7 weeks ago, # |
  Vote: I like it -68 Vote: I do not like it

If individuals cannot speak on how they feel on this platform, why is there even an option to comment? There was a harmless conversation about the ridiculousness of supposed DIV 2B and it was censored. At least reset my contribution to 0 and do it smoothly. Many individuals here are not used to force and imposition as common in many communist countries. In the modern world, there are guidelines for doing things and thus only overstepping comments should be reported and then censored. What was being discussed is that: As there is no benefit to doing CP, there is no point participating in these contests since some problem setters want to act as gatekeepers and keep creating overly difficult Div 2Bs.

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    7 weeks ago, # ^ |
      Vote: I like it -12 Vote: I do not like it

    We understand that sometimes problemsetters may make mistakes as we all are humans but why delete comments??

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I think codeforces is trying to avoid beginners. People are expecting a lot even from B.
All those confidence of solving A,B,C during contest is going down... day by day...

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

    yes and thats making me much sad :.(

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    7 weeks ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    This is a Div2 tho. I agree that the second problem may seem harder than usual but not that much. It was an easy to implement solution which needed a very nice observation. I think that it's better for the second problem to have something tricky in it, instead of being very easy and boring. Trust me, this kind of problems are good for beginners too because it teach them to think outside the box :). Also, there are div3 contests which kind of allows the beginners to solve the first 3 problems quick.

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7 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

Now I understand, contest timing > 2 hours then it is going to be Div 1.5

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Today's Contest was really tough.

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7 weeks ago, # |
  Vote: I like it +16 Vote: I do not like it

In my opinion, B felt much harder than C.

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I understood B after thinking for some minutes, but couldn't apprehend C at all. What did I miss?

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7 weeks ago, # |
  Vote: I like it -51 Vote: I do not like it

Dumb problemsetters, they never learn...

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

I'm curious how many people just solved D quickly by just simulating to get the expected value of (k 0s) followed by a 1 and observing the answer from that (though it is pretty easy to obtain even without that).

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

    Transforming the tries to coin flips, it totally sounds like a common probability problem, which of course it is heh: https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads

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

      Yeah, one of my friends just told me about that, but simulating is really trivial as well (and faster if your math is weak like mine) with a simple code like this, and its pretty intuitive why it should work after you see the result.

      Simulation Code
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    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    I derived expected value for covering different size gaps between successive checkpoints. For $$$g = 1$$$, ie consecutive checkpoints, expected value is $$$2$$$. For $$$g = 2$$$, using expected values, we get $$$6$$$ and then $$$g = 3$$$ we get $$$E(g) = 14$$$, using three equation. and then I thought maybe it is $$$E(g) = 2^{g+1}-2$$$ but dont know the general proof

    Edit:

    Ok link by robinz62 was useful, thank you

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    7 weeks ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Lol You don't even need to simulate take a look at 1265E - Beautiful Mirrors

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

Got terribly stuck at B. Can anyone please give me a hint now since the contest is over?

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    first try to find how can you find min value of given array without any changes

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    7 weeks ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    The key observation is that changing a number at begin is like removing it from the array.

    So we need to find which number, if removed, contributes most.

    The contribution per number is the sum of differences of the adjacent numbers, and the number itself.

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

    If you want to see the approach or idea for solution of B:

    Approach
    Solution
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7 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

how to solve D??i feel dumb for 1 hr after solving c.

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

    Note that You need expected 2^(k+1)-2 moves to do 10000...0 where there are k digits.

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

    Number of steps to cross 1 is 2 , 1 0 is 6 , 1 0 0 is 14 , 1 0 0 0 0 is 30 . relation is s = 2*s+2 . Odd number of steps is never possible since to crossing any stage must be even . Now find biggest s which is close to n and keep doing that until n reaches 0 . For example for 20 it will be 14 + 6 i.e 1 0 0 1 0

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Damn.. submitted D in the last 10 seconds. I hope it holds.

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Congrats dude. It passed.

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

      Thanks, but it needs to pass system tests as well. Fingers crossed.

      Edit: Cool, it passed.(submission)

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7 weeks ago, # |
← Rev. 2   Vote: I like it -44 Vote: I do not like it
Spoiler

Why didn't this work for C?I think there is a carazy mistake there.

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I think problem D needs more explanation.

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

D, how can we get an integer number of expected tries at all?

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

    I guess the expected value converges to an integer when total tries is infinity.

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

      If there is one stage, then

      1/2 + 2 * 1/4 + 3 * 1/8 + ...

      I do not see for any number of stages how I could make such sequence ever result in an integer sum, by adding some checkpoints :/

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

        See this

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          7 weeks ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          This is helpful link, thanks.

          So my above formular is kind of right, and simply sums up to 2.

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

        probability you will cross that level is 1/2 . Now i ask you a question , how many times you will win if you play at that level 2 times ? Answer is 1 . Else probability won't be 1/2.

        You can similarly generalize that beating that level n times will take 2*n matches .

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Beating n consecutive levels without a checkpoint is equivalent to getting heads in a coin flip n times in a row. Having that in mind you can read https://www.codechef.com/wiki/tutorial-expectation.

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Probability to cross a level is 1/2 . Thus total number of moves require to cross is 2 . Suppose i crossed a level and fell again to it . Then again expected number of moves required to cross it again will be 2 and thus total 4. Hence not only it will be integer but also even.

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      7 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Ah... ok, got it. Thanks. I was simply completly wrong ;)

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      7 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Probability to cross a level is 1/2 . Thus total number of moves require to cross is 2 .

      How? How can you relate probability with number of moves?

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

        I should have used "expected" in place of "total" . But i think most people would have got it.

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7 weeks ago, # |
  Vote: I like it +21 Vote: I do not like it

Why guys why?? :(

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

How to solve D ?

It's either too much math or too much pattern finding

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

    If you have a range with 1 at the beginning and k-1 0s after, then this range increases the expected value by 2^(k+1)-2. So while remaining length is > 0 you can simply take largest k for remaining length and put 1 and k 0s at the answer.

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      7 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      What's the proof for that? I tried to prove it during contest but couldn't complete the proof.

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        7 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        https://codeforces.com/contest/1265/problem/E

        See this problem and its editorial

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          7 weeks ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          :( Now I really regret not upsolving that contest.

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

        I found that off google by searching "expected number of coin tosses before getting n in a row", where there's a convenient codechef link with the closed form

        the proof does seem kind of messy, so I suspect most people might've done the same

        EDIT: I realize it's quite obvious now if you write the recurrence relation $$$E_k = E_{k - 1} + \frac{1}{2} + \frac{1}{2}(1 + E_k)$$$

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          7 weeks ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          I firstly had a formula like 2^0 * k + 2^1 * (k-1) + 2^2 * (k-2)+...+2^k * 1 + 2^k. And I wrote bruteforce for k to 100 and saw the pattern.

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      7 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      You mean\begin{equation} {2^{k+1}-2} \end{equation} How do you prove it ?

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

        Let $$$E_x$$$ be the expected number of the state where you have beat x stages.

        It's easy to get $$$E_k=0$$$ and $$$ E_x = \frac{1}{2}E_0 + \frac{1}{2}E_{x+1}+1 , x < k$$$

        After calculation, you will get $$$ E_0 = \sum_{i=1}^k \frac{1}{2^i} + \sum_{i=1}^{k} \frac{1}{2^{i-1}}$$$

        And it means $$$E_0 = 2^{k+1}-2$$$

        That's the answer of k stages like 1 0 0 ... 0 0

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        7 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        There is one more beautiful way to proove it.

        1. note that when you are at a 1-cell and you have $$$s - 1$$$ 0-cells till the next checkpoint you either succeed instantly with the $$$1/2^s$$$ probability, or you loose somewhere and return to the 1-cell. This situation is equal to flipping a coin that gives heads with the probability of $$$p = 1/2^s$$$. It's a known fact that the expectation of the number of throws is $$$\frac{1}{p} = 2^s$$$
        2. now let's realize that the expectation of number of times you visit a cell after a checkpoint halves as you go further. It becomes quite obvious if you think about revisiting the 0-cell as about going back into this cell instantly after loosing a game in a cell after it. Indeed, if you loose a game you return to the last checkpoing and so steps after it you return to the 0-cell we are talking about.
        3. hence, the expectation of revisiting the cells of the 10...0 block with k zeros is $$$2^{k + 1} + 2^{k} + \ldots + 2 = 2^{k + 1} + 2^{k} + \ldots + 2 + 1 + 1 - 2= 2^{k + 2} - 2$$$

        Voila.

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    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it
    My approach passed 8 preteset . don't know correct or not.
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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

was there an easy way to solve c or it was brute force???????

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    yes brute force but n*n*40 time compexity so it is same as reading input

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7 weeks ago, # |
← Rev. 2   Vote: I like it -8 Vote: I do not like it

How to do D...

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7 weeks ago, # |
  Vote: I like it +8 Vote: I do not like it

Apart of the long statements and the difficulty of problem B, I think the problems were amazing! Thanks for the authors.

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7 weeks ago, # |
  Vote: I like it +18 Vote: I do not like it

I hate you Gildong !!

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Where to see the solution？？？？？？？？？？？

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

can someone help me with problem C?

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

    triangle space is base*height/2 since they want it multiplied by 2 so it's just base*height so you just need to find the longest line for each number and you can calculate the max space

    I got the idea but sadly could not code it in time

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    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it -24 Vote: I do not like it

    First insert positions of all d points in a vector of pair. Then fix any 2 points from that and rest is just math. Let 2 points be (a,b) and (c,d). By just math I mean the height of the triangle can be abs(a-c) or abs(b-d). There will be 8 possible combinations and the max out of them will be the answer. You dont need to worry about the 3rd point as we are fixing it in the 8 possible combinations taken.

    Have a look at htis photo: https://ibb.co/6WppWZL

    Code : 

    //Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
const ll maxn=100005;
const ll mod=1e9+7;

int main(){
	fast;
	ll t,n,i,j,k,d;
	cin >> t;
	while(t--){
		cin >> n;
		int gr[n][n];
		for(i=0;i<n;i++){
			string s;
			cin >> s;
			for(j=0;j<n;j++){
				gr[i][j]=(s[j]-'0');
			}
		}
		vector<ll> av;
		for(d=0;d<10;++d){
			vector<pi> v;
			ll ans=0;
			for(i=0;i<n;i++){
				for(j=0;j<n;j++){
					if(gr[i][j]==d){
						v.pb({i,j});
					}
				}
			}
			for(i=0;i<v.size();++i){
				for(j=0;j<v.size();++j){
					ll a = v[i].ff;
					ll b = v[i].ss;
					ll c = v[j].ff;
					ll d = v[j].ss;
					
					ans=max({ans,a*abs(b-d),c*abs(b-d),abs(n-1-d)*abs(a-c),abs(n-1-b)*abs(a-c),abs(n-1-c)*abs(b-d),abs(n-1-a)*abs(b-d),b*abs(a-c),d*abs(a-c)});
				}
			}
			av.pb(ans);
		}
		for(ll x : av)
			cout<<x<<" ";
		cout<<"\n";
	}
}
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7 weeks ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

B was way difficult !! i tried around all approaches but was getting none of them correct

Thought of sorting and making all elements equal to median of the sorted array that didnt even work , making all elements equal to minimum , maximum That also didnt work , taking absolute difference between the adjacent elements that didnt even work !!

I think B was really a difficult question !!

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

Very strict time complexity for C

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    7 weeks ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    its not strict. You are saving index of each character. NOW, if all 4*10^6 are same character then you complexity will be (4*10^6)^2 (as you are traversing each possible pair of each character) which is definitely not in time limit.

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

      are u seriously? I think, (4*10^6)^2 = 1.6e+13 is way too much, and also t <= 2000. O(n^2) is not good for this.

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

      Continuous maximum value of n could be 2000. Hence the value 4*10^6

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        7 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        But then matrix is n*n and u r saving indices of character so your set contains 4*10^6 elements. Moreover u r traversing ur set in two loops making it near 10^13...

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    My O(n*10*4*2)ish 100378697 got AC with 2997ms. I can't even.

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

What should I do, if someone asked me to send him the solution of the problem until the end?

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    7 weeks ago, # ^ |
      Vote: I like it -6 Vote: I do not like it

    humiliate him, tag that cheater in your comment

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

      Sulaimon is a cheater then

      (It means: "Give me the solution to the second problem in Div. 2 contset, please")

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

        After you have uploaded your image on any site(like imgur), open that image on that site,
        right click -> open image in new tab, then copy the URL(should end with JPG/PNG), use that URL to upload image

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Thank you for the round, I really need to work hard.

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7 weeks ago, # |
  Vote: I like it +2 Vote: I do not like it

Problem C gave me PTSD

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Long and difficulty problem statements !!

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

I hate problem B

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

what the fucking B

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7 weeks ago, # |
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Damn that B problem will nerf my rating by far.. Like you could skip one of the numbers each time so I decided 3 cases skip the 1st one, skip the smallest one and skip the biggest one but I would not get the AC. I would get WA on test 5. Can someone give me a test case where these cases don't work?

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain the solution of problem B...?

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    just erase every element of the array one by one and calculate the result. its just the explanation you have to do it in optimize way

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Look at this comment.

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7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Any Explanation on how to solve B ?

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7 weeks ago, # |
  Vote: I like it +15 Vote: I do not like it

Struggled on B for almost 2 hours but solved D in less than 10 minutes :( Feeling very terrible

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Story of my Codeforces career. And the funny thing is you need guts to skip B and go do D. Because come you did not solve D guys that were persistent on B will get B and you will get less score than them even if you get B later.

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7 weeks ago, # |
  Vote: I like it +2 Vote: I do not like it

Div2 B took me 20 minutes to understand the problem, 40 minutes to crack, 20 minutes to get the idea, 10 minute to implement, 20 minute fixing bug. And after that what happened? The contest has been finished!! BTW, problem b was amazing!

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yeah it was good but the level of question was like C.

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    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    could you please explain the solution?

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

      Correct me if I am wrong, If u don't change any elements in the array, the minimum operation is the sum of all abs(a[i],a [i+1]) where i is 0 <=i <size of the array. Now you have to greedily find changing which element would give you the minimum operation. It is long to write. Though keep in mind you have to subtract the max dif and also the changes to its adjacent element after changing, I am too lazy to write it after this struggle

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

    Deleted

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7 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

This round was a nightmare and a reminder of how much I need to practice

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7 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

How to not brick on this contest's B?

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7 weeks ago, # |
  Vote: I like it +28 Vote: I do not like it

Unpopular opinion: D is easier than B and C.

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7 weeks ago, # |
  Vote: I like it -8 Vote: I do not like it

This was a good set of problem set.

I Enjoyed solving them

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7 weeks ago, # |
  Vote: I like it +17 Vote: I do not like it

This kind of contest is bad for mental health.

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7 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

Not a good round, the statements wasn't clear enough.

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

:D I used 8 ifs for each for in problem C and get TLE :D

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7 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

Unpopular opinion I liked the contest B was C level but it's a nice problem the solution is not stupidly obvious .. solving C isn't too hard but understanding it may take some time

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7 weeks ago, # |
← Rev. 5   Vote: I like it +10 Vote: I do not like it

solution to problem B.

here
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    7 weeks ago, # ^ |
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    Please where did you get the idea that changing one element of the array implies removing it. If all the elements before the element you changed are not equal, I don't think you have removed it.

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      7 weeks ago, # ^ |
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      Look at this comment.

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      7 weeks ago, # ^ |
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      Removing that number means, you can convert that number to its next number which will skip the step to convert the suffix to that number. You can directly covert the suffix to its previous number.

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    7 weeks ago, # ^ |
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    Thanks a lot. It was a good editorial. Easy to understand.

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7 weeks ago, # |
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While solving problem B , I observed that for given array total operations required would be sum of absolute difference of consecutive elements. How to prove it concretely ?

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    7 weeks ago, # ^ |
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    Start from last element add absolute diff of last and second element... Then last 2 elements becomes equal and equal to 2nd last. Go to third last element then add the abs diff bw third last and second last and go on.... resulting in sum of ablolute diff bw adjacent elements..

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

    In an array where all elements are equal, the sum of absolute differences of consecutive pairs of elements is $$$0$$$. The reverse also applies — if that sum is $$$0$$$, all the elements must be equal.

    When incrementing / decrementing the suffix starting from index $$$i$$$, note that all the differences between elements in the suffix and the differences between elements outside the suffix remain the same. The only place where that difference changes is in the pair of elements where one is inside the suffix and the other is outside it — indices $$$i - 1$$$ and $$$i$$$.

    Since each operation changes the difference (and, performing optimal operations, decreases the absolute difference) by exactly $$$1$$$, number of operations required is greater than or equal to the sum of the absolute differences of consecutive elements. The shown algorithm can do it in that number of operations, thus the sum of the absolute differences of consecutive elements is the minimum number of operations required.

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

After CF predictor predicts me -50,

What my mouth says: Rating is not the matter. The idea u learnt is more.

What my mind says: Rating is the matter,dude.

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7 weeks ago, # |
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Please explain problem B in more detail.

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

    Let we have numbers $$$a_1, a_2, a_3... a_n$$$. Let make them all equal to m.

    Then to make $$$a_1$$$ equal to m we need $$$|a_1 - m|$$$ steps, and all other leemtns would increase by $$$m - a_1$$$.

    Then, to make $$$a_2$$$ equal to m, we need $$$|m - (a_2 + (m - a_1))|$$$ steps since $$$a_2$$$ is $$$a_2 + (m - a_1)$$$ now. $$$|m - (a_2 + (m - a_1))| = |a_1 - a_2|$$$.

    And total addition to elements is equal to: $$$m - a_1 + (m - (a_2 + (m - a_1))) = m - a_2$$$. From here we can see, that total addition to elements after $$$i$$$-th step are going to be $$$m - a_i$$$. So, the ansewer for number m is equal to $$$|m - a_1| + |a_1 - a_2| + |a_2 - a_3| + .. + |a_{n-1} - a_n| $$$.

    Now we want to minimize that sum. We can always choose $$$m = a_1$$$ to make first number 0. After that we just want to minimize the sum we have.

    To do that let assume we change $$$a_{j}$$$. After changing $$$a_j$$$ to $$$a_{j-1}$$$ (actually we want $$$a_j$$$ to be in interval from $$$a_{j-1} $$$ to $$$a_{j+1}$$$). After that $$$|a_{j-1} - a_j| + |a_j - a_{j + 1}|$$$ is going to become $$$|a_{j + 1} - a_{j - 1}|$$$. So we want to find such $$$a_j$$$ that $$$(|a_{j-1} - a_j| + |a_j - a_{j + 1}|) - (|a_{j + 1} - a_{j - 1}|)$$$ is maximum and change it to $$$a_{j-1}$$$.

    That's all.

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      7 weeks ago, # ^ |
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      Why should we bring this number to zero: "We can always choose m=a1 to make first number "0"?

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        7 weeks ago, # ^ |
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        See, nothing, except $$$|m - a_1|$$$ depends on m. As soon as we want to minimize the whole sum, we should make $$$m = a_1$$$. Otherwise $$$|m - a_1|$$$ is going to be bigger than 0, but making $$$m = a_1$$$ will make sum smaller.

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          7 weeks ago, # ^ |
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          And what m should we equate them with, and did I understand correctly that each a[i] = |a[i]-a[i-1]|. That is how to find it m?

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            7 weeks ago, # ^ |
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            No, you get it wrong. We just say that m is first number. We've proved that ansewer is $$$|m-a_1|$$$ + sum of absolute value of difference between $$$a_i$$$ and $$$a_{i+1}$$$ elements. Then, we want to find one number $$$a_j$$$ and make it equal to $$$a_{j-1}$$$ to minimize the whole sum

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              7 weeks ago, # ^ |
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              Thank you so much!

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7 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

Benefits of hard contests:

1- No long queues. 2- System testing starts and ends in 30 mins. xD

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7 weeks ago, # |
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C was easy but implementation was bit long i solve in 5 minutes implement in 80 minutes

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    7 weeks ago, # ^ |
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    Same! I also felt it was bit of implementation heavy.

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7 weeks ago, # |
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why is B being rejudged?

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    7 weeks ago, # ^ |
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    I think Mike added some more multi-tests and rejudged. Seems that my tests weren't strong enough :(

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7 weeks ago, # |
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Nice problemset! Hope there will be more problems at the same difficulty as this in the future for the middle-ratings like me. Thanks for your contributions =) very much

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7 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

The scoring distribution is 500 — 1000 — 1500 — 2000 — 2500 — 3500.

REAL scoring distribution is 500 — 2000 — 1500 — 2000 — 2500 — 3500.

LOL

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

It seems that AceKing and I were the only people in the whole contest who passed the pretests for C, but TLEd on the main tests :( that's quite a sweet spot we managed to hit with our execution time!

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    7 weeks ago, # ^ |
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    How to find this who all failed systests without exploring the whole ranklist obviously?

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      7 weeks ago, # ^ |
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      filter the submissions

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        7 weeks ago, # ^ |
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        I didn't see any filter for that. Are you sure?

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      7 weeks ago, # ^ |
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      As Forward. says, I clicked status, filtered on problem C, verdict Time Limit Exceeded. The results are displayed in reverse chronological order, so I looked specifically for entries submitted in the time range of the contest, but which got TLE verdict after the end of the contest. There were only 2 of us.

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        7 weeks ago, # ^ |
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        Okay seems fine but I was looking for something which can show me all types of failed sys-tests (WA/TLE/RE etc) on one page. Thanks anyways.

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          7 weeks ago, # ^ |
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          I don't believe that exists.

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        7 weeks ago, # ^ |
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        Not 2, there is also me who MLE-ed — you should filter by "rejected" and tests >= 10 (I think this is the number of pretests for C).

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          7 weeks ago, # ^ |
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          In my comment, I specifically said "TLEd" and referred to execution time. Unlucky that you MLEd, though.

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7 weeks ago, # |
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Today was the first time I needed to fix an MLE from pretests and fail system tests due to MLE :/

https://codeforces.com/contest/1453/submission/100368040

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7 weeks ago, # |
  Vote: I like it -55 Vote: I do not like it

Today contest very bad contest. I downvote today contest. Do not make tomorrow contest. I not give contest. djm03178 do not make contest again.

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7 weeks ago, # |
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B was probably more difficult than usual...

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

I think the 1st standing in div2 did not finish the competition independently for the submission time of his code. If he did, I apologize for this comment.

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7 weeks ago, # |
  Vote: I like it +5 Vote: I do not like it

I did a very stupid mistake in B.
For n = 2 answer will be 0 but I don't know what came to my mind, I printed their difference.
Sadly that case wasn't in the pretests.

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

The contest was very good and problem A was easy and standard.

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7 weeks ago, # |
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when will the rating changes will be published ???

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7 weeks ago, # |
  Vote: I like it 0