Radewoosh's blog

By Radewoosh, history, 6 months ago, In English,

Hello coders! I hope that you are enjoying the New Year as much as me. To make its beginning even greater, Codeforces is going to host a contest and I will be an author of all tasks. Hello 2019 will take place on Friday.

Using the opportunity, I want to thank to:

  • lewin and mnbvmar for testing the round.
  • mnbvmar for indescribably helpful discussions about problems.
  • _kun_ and KAN for round coordination and help with preparation.
  • MikeMirzayanov for such great platforms (you know which ones :P).

The round will consist of 8 problems and you will be given two and a half three hours to solve them. Yes, the round will be rated.

There will be no interactive problems, but if you want you can read this document anyway, it's always good to learn new things.

Good luck and see you during the contest!

UPD1: Editorial

UPD2: I'll be on the Discord channel after the contest, so you will be able to ask me about the problems.

UPD3: You're probably wondering what the statements will be about. I hope that it will be another great year for Codeforces. As it's the community that creates it, I decided to write statements about the people who already have or had their part in Codeforces' history. As I wanted to be objective, the statements will be about 8 people who triumphed the most times in CF rounds. Using the opportunity, I want to invite these 8 people to take part in the contest. Let's say that the first person who will guess the set in the comments wins some free contribution. Good luck!

UPD4: The round will be 3 hours long.

UPD5: The drain will be adjusted and the scoring will be 500-1000-1500-2000-2750-3000-3500-4000.

UPD6: The round is over, congratulations to the winners!

  1. V--o_o--V
  2. ecnerwala
  3. tourist
  4. lych_cys
  5. Petr

And to the first-to-solvers!

UPD7: Editorial

 
 
 
 
  • Vote: I like it
  • +1909
  • Vote: I do not like it

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6 months ago, # |
  Vote: I like it +550 Vote: I do not like it

You accidentally uploaded the ediotiral, but i am honest and don't look there

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    6 months ago, # ^ |
      Vote: I like it +453 Vote: I do not like it

    Well, thanks for that, but you shouldn't worry so much :P

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    6 months ago, # ^ |
      Vote: I like it -12 Vote: I do not like it

    you changed your handle from numb. So many quora links are gonna be dead now

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

      I have seen a new name in my friend list and wondering about who is this guy... Finally I understood...

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      »
      6 months ago, # ^ |
        Vote: I like it +17 Vote: I do not like it

      If you change your handle the links refered to the lastwill redirect to the new profile page

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

    I wish everyone as honest as you.World Would be Better Place to Live

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6 months ago, # |
  Vote: I like it +22 Vote: I do not like it

Nice editorial tho.

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

I can neither confirm nor deny that editorial without Radewoosh's approval

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6 months ago, # |
Rev. 2   Vote: I like it -225 Vote: I do not like it

lol

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6 months ago, # |
  Vote: I like it +44 Vote: I do not like it

That editorial, haha, so funny!

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6 months ago, # |
Rev. 3   Vote: I like it +70 Vote: I do not like it

this.handle+" 2019!";

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6 months ago, # |
  Vote: I like it +239 Vote: I do not like it

Thanks for fast editorial.

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6 months ago, # |
  Vote: I like it +90 Vote: I do not like it

I really hope noone is enjoying the New Year as much as you.

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6 months ago, # |
  Vote: I like it +228 Vote: I do not like it

Can you please reupload the editorial? It is not available in my country and now I feel at disadvantage.

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6 months ago, # |
Rev. 2   Vote: I like it -24 Vote: I do not like it

One of the best editorials ever seen. Nice one.. Haha:P

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

Why aren't you blue anymore? Also, what is the blog gonna look like when you upload the real editorial?

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6 months ago, # |
  Vote: I like it +10 Vote: I do not like it

too excited about the first contest in this year and as well as a contest by Radewoosh.

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

Awesome editorial! :P

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

-

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

By the way, why isn't the Rating of the problems in Goodbye 2018 shown yet?

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

could you explain in more detail div2a? including official solutions also helps.

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

Nice editorial :P

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6 months ago, # |
Rev. 2   Vote: I like it -30 Vote: I do not like it

LOL HAHABA SHIT JOKE=))))

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6 months ago, # |
  Vote: I like it +184 Vote: I do not like it

Me: Oh, an editorial... let's to see!

(wild video appears...)

Me:

;;;;;;;;;;;;::;,.,xOOOOOOOkdoc;,,;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;,,'''',,,,,,,,,'',,:cloo:'.
;;;;;;;;;;:ccccc,'lOOOOOOOOOOOxoc;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;,,''''''''''',;cloxkOOOOl..
;;;;;;;;;:cccccc:,;xOOOOOOOOOOOOOxoc;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;,''''''';:loxkOOOOOOOOk:..
;;;;;;;;;:cccccc::,ck0OOOOO0OO00OOOOxl:;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;,'',;codkO0OOOOOOOOOOOo'..
;;;;;;;;:cccccc:::;,ck0OOOOOOOOOOOOO0Okoc;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;:cloxOOOOOOOOOOOOOOOO0x;..'
;;;;;;;:ccccccc::;;,,:xOOOOO0OOOOOOOOOOOkdc;,;;;;;;;::::ccccccccc:::::cldkOOOOOOOOOOOOOOOOOOO0k:..''
;;;;;;;:cccccc::;;,,,';dO0OOOOO0OOOOOOOOOOkdodddxxxxkkkOOOOOOOOOOkkkxkkO0OOOOOOOOOOOOOOOOOOO0kc'''''
;;;;;;;::::::::;;,,,''',cxOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOxc,,,,''
;;;;;,;;;;;;;;;,,,,'''''',lkOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOko:;;;;,,,
;;;;,,,,,,,,,,,,''''''''''.,lkOOkxkOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOkoc:c:::;;,,
;;;;'''''''''''''''''''''''..;lodkO0OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOkxk0Okxocccccc:::;,,
;;;,'............'''''''''''..;xOOOOOOOOOOOO0OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOkxdol::::cccc:::;;,,
;;,'..............''''''''''';dOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOko;;::::::::;;;;,,
;;'................'''''','',dOOOO0OOOxdoccxOO0OOOOOOOOOOOOOOOOOOOOxdoclxOOOOOOOOOOOo;;;;;;;;;;;;,,'
;,'................'''''',,,lOOOOOOOOoo00c.;xOOOOOOOOOOOOOOOOOOOOOll0O:.:k0OOOOOOOOOOo,,;;;;;;,,,,''
,'..................''',,,,ckOOO0OO0Ol,::,.,d0OOOOOOOOOOOOOOOOOOOOc,::,.;x0OOOOOOOOOOkc',,,,,,,,,,''
,'..................''',,';dOOOOOOOO0kc,,,;oOOO0OO0OO0OOOOOOOOOOOOkc,,,:dOOOOOOOOOOOOOd,',,'''''''''
'...................''',,,lOOOOOOOOOOOOkxxkOOOOOOOOxdddkOOOOOOOOOOOOkxxOOOOOOOOOOOOOO0kc''''''''''''
'...................''',,;xOOOkkkOOOOOOOOOOOOOOOOOxo:;:dOOOOOOOOOOOOOOOOOOOOkkkkOOOOOOOd;'''''''''''
'.................'''',,,ckxollllldkOOOOOOOOOOOOOO0OkkOOOOOOOOOOOOOOOOOOOkdlllllodkOOOOOc'',''''''''
''''...........'''',,,,,;odccccccccokOOOOOOOOOOOOOOO0OOOOOOOOO0OOOOOOOOOxlcccccccclk0OOOd;,,,,,,,,''
'''''''',''',,,,,,,;;;;;;ddcccccccclxOOOOOOOOOOOOkxdddddooddkOOOOOOOOOOOdccccccccclx0OOOOl,;;;;,,,,'
',,,,,,;;;;;;;;;;;:::::;:xOdolclllokOOOOOOOOOOOOxllodddxdddllkOOOOOOOOOOkocccccccldOOOOOOd:;:;;;;;,,
,,,;;;;:::::::::::::::cc;lOOOkkkkOOOOOOOOOOOOOOOdlodxxdxddxdcdOOOOOOOOOOOOkxdoddxkOOOOOOOOl:::::;;,,
,,;;::::cccccccccccccccc::dOOOOOOOOOOOOOOOOOOOO0kolldxxxxxxoldOOOOOOOOOOOOOOOOOOOOOOOOOOOOd::c::;;,,
,,;;:::cccccccccccccccccc:cxOOOOOOOOOOOOOOOOOOOOOOkxoooooooodOOOOOOOOOOOOOOOOOOOOOOOOOOOOOkc:c:::;,,
,,;;::::cccccccccccccccc:c:cxOOOOOOOOOOOOOOOOOOOOOOOOOkkkkkOO0OOOOOOOOOOOOOOOOOOOOOOOOOOOOOo;:::;;,,
,;;;;::::ccccccccccccccc:::;ck0OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOx::::;;,,
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6 months ago, # |
  Vote: I like it +19 Vote: I do not like it

Your editorial sucks! The video isn't available in my country!

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6 months ago, # |
  Vote: I like it +173 Vote: I do not like it

memerickcf

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6 months ago, # |
  Vote: I like it +63 Vote: I do not like it

clicked on the Editorial

Really Radewoosh? Thanks for being the first one doing that to me in 2019 :D

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

The most unique editorial I have ever seen. Hoping for a nice contest now (paradoxical I guess, waiting for the contest after editorial)

Also, Thanks a lot for suggesting us to learn about new things :)

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

Looking forward to it, hopefully I will finally become green again :P

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6 months ago, # |
  Vote: I like it -64 Vote: I do not like it

Radewoosh you can't beat tourist

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

Not gonna lie, I thought I badly missed the contest..

Until I swiped left my mobile screen..

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

I've been looking forward to a round prepared by a single expert, but, however..

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

wulalalala is it rated?

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

what is the rating of the contest ?

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

    If the question is, Is it rated? then answer is YES. or, if the question is Rated for which division? then answer is Everyone who will attend the contest.

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6 months ago, # |
  Vote: I like it +46 Vote: I do not like it

Fastest editorial in my codeforces history. Thanks Radewoosh

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

Well-written editorial ^^ but I think you forgot to link the statements that go with it :\

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6 months ago, # |
  Vote: I like it +5 Vote: I do not like it
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6 months ago, # |
  Vote: I like it +12 Vote: I do not like it

I've already read so many editorials that I know this gXcQ by heart.

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

Fastest editorial ever!!! Also that's one of my favorite songs to cheer me up too.

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

Is it just me, or does Radewoosh look really similar to Rick? o.O
Radewoosh vs Rick

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

Never gonna give you uuuup! I like this editorial :)

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6 months ago, # |
  Vote: I like it +20 Vote: I do not like it

How to solve D ?

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

Exceptional Editorial :P

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6 months ago, # |
  Vote: I like it -80 Vote: I do not like it

Me: Wow! An Editorial, before the contest! (Let's see... I don't want to miss the chance today!)

Clicked the Editorial

Hope I can dance like him after the first contest of the year!

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6 months ago, # |
  Vote: I like it -35 Vote: I do not like it

The past Hello contests:

Happy New Year!!

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

Wow, early editorial. When are you gonna update the ratings?!.

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

Goodbye 2017;Hello 2018.:D

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

I'm gonna listen to this editorial everyday

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

Thanks for the video editorial , really educational !

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

Just hope, everyone has high ratings with this new year. Tough competition though :P

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

OMG, Radewoosh contribution +200. That`s so much!

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

Fuck U Radewoosh.My New Year Resolution was not to Open Youtube(Personal Reason).And Bcoz of u it is broken now.

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6 months ago, # |
Rev. 2   Vote: I like it +10 Vote: I do not like it

this will get me out of master :^D

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

Everyone says that he hopes to be green, gray and other color. I hope to become Double Legendary Grandmaster SuperJava

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6 months ago, # |
  Vote: I like it -13 Vote: I do not like it

lol

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6 months ago, # |
  Vote: I like it -21 Vote: I do not like it

Is it rated??

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

ArepitaDePernil will never tell a lie and hurt you!!

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

Hope to reach green this contest :)

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

it would be cool if pressing the downvote button would also redirect u to smth funny

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

Lets see if this post gets more upvotes than Hello 2018. Radewoosh you are my favorite.

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6 months ago, # |
Rev. 5   Vote: I like it +6 Vote: I do not like it

I think that the set of people will be: tourist Petr Radewoosh Um_nik rng_58 OO0OOO00O0OOO0O0…O PikMike Vovuh

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

Challenge in UPD3 is pretty simple using Codeforces API but I'm too lazy so let me guess. Let me spare them these mentions: Petr, tourist, Radewoosh, OO0OOO00O0OOO0O00OOO0OO, Um_nik, rng_58, V--o_o--V, YuukaKazami

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

I wonder what is the number of participants who won at least one rated(for div1) contest. Too lazy to check using API and/or by opening results, of course.

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

UPD3 challenge: tourist, Radewoosh, OO0OOO00O0OOO0O00OOO0OO, Um_nik, V--o_o--V, LHiC, Petr, encerwala

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

UPD2: I'll be on the Discord channel after the contest, so you will be able to ask me about the problems
I can't enter the server ((

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6 months ago, # |
Rev. 2   Vote: I like it +302 Vote: I do not like it

Egor it was very tricky from you to hide on 158th place in overall standings with these 11 wins of yours :P

Petr tourist Radewoosh OO0OOO00O0OOO0O00OOO0OO Um_nik rng_58 V--o_o--V Egor — be aware that Radewoosh invites you to take part in this contest :)

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    6 months ago, # ^ |
      Vote: I like it +65 Vote: I do not like it

    And we have a winner!

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

    Seems cool, he's just after tourist and Petr, not sure about OO0OOO00O0OOO0O00OOO0OO, but all in all, 11 is too much!!!

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    6 months ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    I'm not saying that V--o_o--V is not cool, but among his 8 wins 2 are schoolboys competitions and 3 are VKCup rounds, so 5 wins with huge restrictions on participants.

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

    Radewoosh invites himself to take part in his own contest? :thinking:

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    6 months ago, # ^ |
      Vote: I like it +36 Vote: I do not like it

    includes unofficial participation

    gimme some of the left over contribution

  • »
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    6 months ago, # ^ |
      Vote: I like it +81 Vote: I do not like it

    Anyone wants to create Codeforces version of this?

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      6 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      Something like this?

      Link

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

        Thank you, looks nice :)

        Maybe we shouldn't mix d1/combined matches and d2/d3/unrated matches?

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

          You're welcome. I will try to modify the UI a bit later separating divisions into different tables. :)

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    6 months ago, # ^ |
      Vote: I like it +86 Vote: I do not like it

    Only reason I failed last several rounds is to deceive everyone in anticipation of this

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

I just want to tag dreamoon sorry_dreamoon :P.

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6 months ago, # |
  Vote: I like it +15 Vote: I do not like it

In UPD 3 , I think they are (in sorted order from number of contests winning)

1 — tourist

2 — Petr

3 — OO0OOO00O0OOO0O00OOO0OO

4 — rng_58

5 — Egor

6 — Um_nik

7 — V--o_o--V

8 — Radewoosh

9 — YuukaKazami

and since you said that you invite first 8 to take part in contest...so I think that maybe you mean 9th which is YuukaKazami

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    6 months ago, # ^ |
      Vote: I like it +52 Vote: I do not like it

    Unfortunately there are two persons on 9th place, so I'm also inviting myself to take part (and I will, but from the other side :P).

    And sorry, but our contributionboy was faster.

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    6 months ago, # ^ |
      Vote: I like it -12 Vote: I do not like it

    you forgotten: 0 _ Wael.Al.Jamal

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

I cant understand the editorial, anybody can explain with more details? :D

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

It sure will be the best 2019 contest so far.

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6 months ago, # |
  Vote: I like it +40 Vote: I do not like it

Change the start time to 20:19 ! :D

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    6 months ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Taking part in a contest begins at 22:35(UTC+8) is usual for me now, but if it lasts for 3 hours...

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

I had thought the editorial is not standard but I really can't stop myself from clicking it!

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

I hope there won't be solutions in the problems.

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

Even Problems In an Odd Year will Make My Day !!!

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

One of the Best Editorial ever.

Thanks for the Motivation.

Another Editorial if you wanna take a look.

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

finally.. someone generous. Three hours !

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

Really enjoyed this editorial.

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6 months ago, # |
  Vote: I like it +13 Vote: I do not like it

If you can't see the editorial, you can try this. It's the same video.

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

The 8 people will not includes OO0OOO00O0OOO0O00OOO0OO because it's too hard to write.

QAQ

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

Nice editorial. Hahaha.

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

Outrageous editorial!

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6 months ago, # |
  Vote: I like it -68 Vote: I do not like it

Why those stupid early contest times for this and Goodbye 2018? I'm not a shut-in with nothing to do except programming contests at any time!

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

    Actually it's a little too late for Asian. It's hard to meet everyone's need. So it's better to enjoy those you can participate in and stop complaining about everything.

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      6 months ago, # ^ |
        Vote: I like it +17 Vote: I do not like it

      It's hard to meet everyone's need.

      Yeah, which is what two consecutive big rounds with the same starting time don't do. If you look at the contest list, the last 8 rounds including this one (some div1) had an early starting time.

      There are countless things I'm not complaining about, so how about you don't try to dismiss a valid point? Which rounds can I enjoy if it's been almost a month since one didn't start in the early afternoon?

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6 months ago, # |
  Vote: I like it +22 Vote: I do not like it

PM 11:30 to AM 02:30 in my country 8ㅅ8

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

First time I have watched any Editorial before contest..... it' like Boost up..hahahaha

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

welcomeBack();

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6 months ago, # |
Rev. 2   Vote: I like it +78 Vote: I do not like it

Adjusted drain, please. Nobody wants their C-D to be worth more than G-H.

@ MikeMirzayanov, _kun_, KAN

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    6 months ago, # ^ |
      Vote: I like it +44 Vote: I do not like it

    Pls pls pls!!! It lasts 3 hours (not 2.5), so it is even more important!

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

    it is adjusted

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

I believe it will be very cool!

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6 months ago, # |
  Vote: I like it +15 Vote: I do not like it

I Have a Farsi exam tomorrow, I know Im going to fail but F*** that, I waited a year for this.

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

Very exclusive tutorial.Following the tutorial most of the contestant including me become legendary grandmaster without participating any contest :D

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6 months ago, # |
  Vote: I like it +4 Vote: I do not like it

Never gonna give you up =))

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6 months ago, # |
  Vote: I like it +88 Vote: I do not like it

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6 months ago, # |
Rev. 2   Vote: I like it +43 Vote: I do not like it

Score distribution? Updated Now

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

Good luck to all! Wish you to see lot of "Happy New Year!" verdict!

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6 months ago, # |
Rev. 4   Vote: I like it +1 Vote: I do not like it

Before start, delayed by 10 mins :|

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6 months ago, # |
  Vote: I like it +17 Vote: I do not like it

why delay 10 min ?

  • »
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    6 months ago, # ^ |
      Vote: I like it +29 Vote: I do not like it

    Finally my time travel machine worked, I moved everyone 10 minutes in the past

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

10 minutes delay means now I will skip the last 30 minutes of the round, not a good thing :(

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6 months ago, # |
  Vote: I like it +22 Vote: I do not like it

PING : 600000ms

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

I'm confused because the contest was suddenly delayed a few seconds before the start of the competition.

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

JEBAITED XD

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

Is it another DDOS attack???

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    6 months ago, # ^ |
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    Maybe expecting more participation, I guess.

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What an Anti-Climax (Delayed by 10 mins)

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

There should be a notification when a contest is delayed. We press OK and suddenly get shocked to see delay :/

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6 months ago, # |
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Radewoosh, You want to crack the 12K record?

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6 months ago, # |
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UDP5 : 10 min delay.

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6 months ago, # |
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Trying to get 12k registrants and beat Goodbye 2018 with that delay? haha

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I think the delay was for beating the record for the most registered people in a contest.

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About +600 participants. Good luck :)

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It's great idea from you to put a link to your song in the blog, nice advertisement man

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Starting new year with 10 minutes delayed contest, it gonna be legen wait for it dary year.

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6 months ago, # |
  Vote: I like it +46 Vote: I do not like it
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  Vote: I like it -30 Vote: I do not like it

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6 months ago, # |
Rev. 2   Vote: I like it +118 Vote: I do not like it
Setters be like:
A: lets make an A-level problem
B: lets make a B-level problem
C: lets make a C-level problem
D: let them lose their mind :O
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6 months ago, # |
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What is TC4 of Prob C?

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6 months ago, # |
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Radewoosh unbalanced contest

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Why does E have only 1 correct submission? Overkill or wrong official solution?

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For Div-2 Coders, it was a 1/2 hour contest.

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very funny statements

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6 months ago, # |
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The test case 2 of D is killing me for past 1 hour. For some reason, I can't seem to figure it out (in fact n=6, k=3 provides me with the given output). However, as there has been nobody raising that issue, the only explanation is that I have forgotten all my maths.

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

    It was strange for me too The problem with me in this problem that I didn’t even understand what is the statement Also no explanation for the output of the second test . When i asked the author he repeated the statement for me I thought it is only my problem

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

    Well I had the same problem at the beginning The problem is that you take into account the number of divisors of a divisor of N to calculate the probability for each k

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6 months ago, # |
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oh god I understood C wrongly wasted 1 hour until I read it again it's way easier than what I was trying to solve :/

I will never read problems quickly again

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6 months ago, # |
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[DELETED]

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6 months ago, # |
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Maybe, one day, Science will be able to answer why div2 coders like me keep on refereshing the standing page for 2+1/2 hours, even when they are done with the problems in 1/2 hour (-_-)

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6 months ago, # |
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The strange thing that my rank keeps increasing if that i didn’t solve a problem after the first 40 min

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Yeahhh!!!

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    6 months ago, # ^ |
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    without taking part in the contest how did you solve A, B, C?

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    6 months ago, # ^ |
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    i dont know how many times i saw this in 2018...

    hope to not see this in 2019 anymore

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    6 months ago, # ^ |
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    After 20 minutes: 3 accepts, 3 submits

    After 3 hours: 3 accepts, 4 submits

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

Please make me expert :|
UPD: I became! <3

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6 months ago, # |
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How to solve D?

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6 months ago, # |
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Does D uses the fact that number of divisors are number of ways to select powers of its prime factorization?

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    6 months ago, # ^ |
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    Well given that N is up to 10^15, you can check the divisors of N in O( sqrt(N) )

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Hello 2̶0̶1̶9̶ depression.

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People being unable to solve D (at any points during contest time) be like:

  • Got at least one failed submission -> damned.
  • Solved ABC too slow -> damned.
  • Cannot get proper hacks (mostly on B) -> damned.

And I got the first and third case. Won't really blame (for a master, this performance of mine was way disappointing), but still, my point still stands. This contest is too unbalanced.
I know, most Div.1+Div.2 combined recently look quite similar that way, but still, this one took the gap to the highest magnitude. Like, a single mistake is a suicidal sign.

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    6 months ago, # ^ |
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    I got them all. C statement was not clear and wasted a lot of submissions on it.

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    6 months ago, # ^ |
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    It may include misinterpreting the problem.

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      6 months ago, # ^ |
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      Well, that would be the worst of any start of a year.

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In E is just greedy O(n logn sqrt(n)) passing?

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    6 months ago, # ^ |
    Rev. 3   Vote: I like it +33 Vote: I do not like it

    If by greedy you mean remove the longest monotonic subsequence, then I got WA with this approach. It is known to achieve , but I can only construct cases in which I need , so there is something missing.

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      6 months ago, # ^ |
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      I found that N = k2 + k + 1 needs f(N) ≥ k + 1 and that f(6) = 3, so this is obviously not the exact bound, but it's very close to the other bound from Erdos-Szekeres.

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    6 months ago, # ^ |
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    Solution is either greedily pick largest subsequence or split the whole sequence into the minimum amount of increasing or decreasing sequences.

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2o50s2

me.

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problem D , what maximum number of divisors n can have ? and if less than 1e3 can we use dp ?

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How to solve F?

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ok fellas, how do we solve D?

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    6 months ago, # ^ |
      Vote: I like it +50 Vote: I do not like it

    There's a straightforeward dp solution. Compute all dp[i][j] = expected value if starting value is i and you have j turns. (i are all potential divisors of n).

    However this is too slow. You can speed it up, by realizing that this function is multiplicative. Therefore you compute dp[p^e][k] for each prime factor p of n and multiply the results.

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

      How to find prime factors of n? I mean n is big and I couldn't find out whether n is prime or n is pq for some p and q being prime numbers.

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

        n isn't that big. In my case I simply computed all primes <= 4*10^7 using a sieve (which runs in <0.1 sec), and tried dividing n with all primes. But I also expect that a simple for loop until sqrt(n) will also be fast enough.

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          6 months ago, # ^ |
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          :(. I thought maybe 4 * 10^7 does not fit. Thanks.

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      6 months ago, # ^ |
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      Can you elaborate a bit on proofs of this function being multiplicative? I don't really getting it mathematically.

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        6 months ago, # ^ |
        Rev. 3   Vote: I like it +23 Vote: I do not like it

        Expected value of product of two independent random variables A and B is the product of their expected values.

        Proof for discrete A, B:

        (Here a and b are values that A and B can get.) P(A = a, B = b) = P(A = a)P(B = b) from definition of independence.

        At every step the power of every prime changes independently, since if our current number is y = x * pa

        , then the number of divisors of y where the power of p is a given value <= a is always just the number of divisors of x. Since all these probabilities are equal, they do not depend from x.

        Let A_{p,k} be a random variable indicating the value of the power of p after k steps. Our answer is since as we just saw Ap, k's are all independent

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

          sigh I'm just bad to not being able to draft this out.
          Thanks for the help, it's crystal clear to me! ;)

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        6 months ago, # ^ |
        Rev. 7   Vote: I like it +46 Vote: I do not like it

        You can do a induction over k.

        For n = a·b and , and D(x) being the number of divisors of x.

        We have:

        Because of the induction hypothesis we know that

        dp(d1·d2, k - 1) = dp(d1, k - 1)·dp(d2, k - 1).

        Therefore:

         = dp(a, kdp(b, k)
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          6 months ago, # ^ |
            Vote: I like it +3 Vote: I do not like it

          This is exactly what I was searching for. Thank you so much!

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

          You mean: "dp(d1·d2, k - 1) = dp(d1, k-1)·dp(d2, k-1)." in the line after: "Because of the induction hypothesis we know that", correct?

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

        It's also possible to prove it by noticing that performing one step on n is equivalent to calculate s / c where s is the sum of divisors and c is the number of divisors. These functions are known to be multiplicative. From there, you can use induction.

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  Vote: I like it +14 Vote: I do not like it

What's the intended complexity of E/G?

I have O(Nsqrt(N)log(N)) in E (I hope it passes with some pruning) and a heavy O(NKlogK) in G (most probably it's too slow).

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

    I'm really curious about E. What's the threshold for f()?

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      6 months ago, # ^ |
        Vote: I like it +85 Vote: I do not like it

      f(n) = 1 for [1, 2], 2 for [3, 5], [6, 9], [10, 14], [15, 20],  and so on.

      How to decompose a seqeunce of, for example, length [55, 65], into 10 parts? If the length of LIS is at least 11, remove it and repeat the process. Otherwise let dp[i] be the length of the LIS that ends at position i. Since there are at most 10 distinct values of dp[i] in this case we can decompose it into 10 parts (If we only look at elements with dp[i] = c for some constant c, this is decreasing.)

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        6 months ago, # ^ |
          Vote: I like it +44 Vote: I do not like it

        Yup. And we can consider the permutations of the following form:

        1, 3, 2, 6, 5, 4, 10, 9, 8, 7

        (concatenated decreasing sequences of lengths 1, 2, ..., k). It's possible to prove that this permutation needs k subsequences in any partition.

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    6 months ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    We have in E. We came up with plenty of LIS implementations and even the slowest ones fit in a second or so.

    G can be solved in O(NK).

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      6 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      So what was the intended solution? Because greedily picking the longest sequence seems to be O(N1.5logN) but gives WA.

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      6 months ago, # ^ |
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      Apparently my LIS implementation is exceptionally slow then.

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How to solve D? I could see that the expected value was a multiplicative function, but values for prime powers were not having a nice form either.

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

    The expected value of f(n, k) is the product of f(aipi, k), where ai is a prime factor of n and pi is the maximum value such that aipi is the factor of n.

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      6 months ago, # ^ |
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      Yes, that is same as saying f is multiplicative. But how to compute f(a_i^p_i,k)?

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

        Consider the naive DP approach to the original question. Denote dp[i][j] as the expected value when i replacements are done and the current value be j.

        Then for all t that is a factor of j, and n is the number of factors of j.

        The base case is dp[k][t] = t.

        Just use the same approach for each set of prime products, and of course you are not to use the values as indices.

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

    You can just DP the values for prime powers

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      6 months ago, # ^ |
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      You mean using the recurrence, (a+2)f(p^(a+1),k) — (a+1)*f(p^a,k) = f(p^(a+1),k-1)? Wouldn't that be too slow?

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

        EDIT: fix one-off in indexing

        Let DP[i][j] = f(pi, j). We can init DP[i][0] = pi. Then, For all j from 1 to k,

        Calculating the values is trivial in O(nk) by precalculating the mod. inverses, and looping i up, keeping track of the sum.

        Our answer for this prime is just DP[n][k].

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          6 months ago, # ^ |
            Vote: I like it +10 Vote: I do not like it

          I think you mean DP[s][j-1] on the right. It is the same as I have written. (i+2)*dp[i+1][j] — (i+1)*dp[i][j] = dp[i+1][j-1], if we keep track of the sum in your formula.

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      6 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      Ok I see it may not be slow, since n has atmost log(n) prime factors, this will be done in less than log(n)*k*O(1) iterations

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

Problem D, if n = 6 and k = 2 then why does 1 has 9 ways? I could just figure out 4 ways:

  • 6 6 1
  • 6 1 1
  • 6 3 1
  • 6 2 1
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    6 months ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Probability for all of them is different!

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

    You are not taking in account the probabilities.

    6 6 1 — 1/4 * 1/4 = 1/16
    6 1 1 — 1/4 * 1 = 1/4
    6 3 1 — 1/4 * 1/2 = 1/8
    6 2 1 — 1/4 * 1/2 = 1/8
    Sum = 9/16

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      6 months ago, # ^ |
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      Oh, i forgot about the probality. Thank you

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    6 months ago, # ^ |
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    For n=6 and k=2, 1 has only 4 ways.

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The system testing that's still pending but running at the same time and gives failed systest on A for everyone seems broken.

UPD: Now it gives AC, but the rest is still wrong.

UPD2: Now it's ok, but it said "system testing, standings frozen" for a while.

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6 months ago, # |
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Could someone please explain/provide a hint for solving D.

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    6 months ago, # ^ |
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    Great Hint: In this problem E(XY) = E(X)E(Y),  gcd(X, Y) = 1.

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    6 months ago, # ^ |
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    Not sure if this will get AC because I didn't manage to get it working in time but here goes:

    1. get divisors of n by looping from 1 to sqrt(n) (O(n^1/3) can be used for number of divisors) O(10^8)

    2. build directed graph with divisors as vertices and edges going from a number to its divisors O(10^10)

    3. use DP to calculate probabilities O(10^4*10^10)??

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6 months ago, # |
  Vote: I like it +21 Vote: I do not like it

never expected that bitset.count() is so fast. is it supposed to be O(N/128) or its constant is low like everything else in stl?

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    6 months ago, # ^ |
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    I thought it's supposed to be O(1)

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      6 months ago, # ^ |
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      In GCC it's O(N) (proof).

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

        Depends on flags. With -Ofast -march=native, it's O(1): godbolt

        Bitset .count()'s "complexity" then of course is n / 64 * O(__builtin_popcount()) = O(n / 64) (probably even faster due to vectorization)

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        6 months ago, # ^ |
          Vote: I like it +18 Vote: I do not like it

        I am not care about contribution, but it's anyway annoying to get downvotes for the correct answer for non-trivial question. Dear green's and lower, do you understand that with such behavior you demotivate people to give answers for questions and to share their experience? Imagine that once your question can be ignored because of that and you will be left alone with your problems.

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6 months ago, # |
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What was wrong with the solutions to G that failed on test #4?

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Was it possible to do D without using the prime decomposition of N ? With a DP over its divisors

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    6 months ago, # ^ |
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    But when you know divisors, you know the prime decomposition too.

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      6 months ago, # ^ |
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      I meant without the optimisation involving the prime decomposition but I just saw its impossible I'm still wondering why the fonction is multiplicative

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Cool C and F. Awesome intro-statement for Um_nik problem.

B bad for beginners, E with bad limits — it's terrible that many people come up with and thought it would be too slow (or: limits wouldn't be so tight then). G is quite standard, and H is tedious to implement.

7/10, the contest could be better, but it was fine.

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    6 months ago, # ^ |
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    why is B bad for beginners?

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      6 months ago, # ^ |
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      Implementation, zero thinking. And then there is a jump to much harder to solve problem C.

      It's very hard to choose good easy problems. I try to avoid bitmasks (or similar) for example, because some people might not know them. More or less, I remember times when I started programming and I try to invent such problems that are solvable by thinking for such a kid. In my easy problems, one of the best ones was: Given a graph with N, M <= 2000, count triangles (triples of pairwise connected vertices).

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        6 months ago, # ^ |
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        Are you saying that it's too straightforward? Or that you need some knowledge to solve it?

        In my programming contest experience, iterating over subsets of a set (for example, using recursion or building the subsets iteratively) was one of the first things necessary for me to learn.

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        6 months ago, # ^ |
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        Explicit bitmask is not required though: one can do the knapsack DP.

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    6 months ago, # ^ |
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    How do you solve F? I figured bitsets would easily do queries 1,2,4 but spent over an hour unsuccessfully trying to work out a bitset solution that works with query 3.

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      6 months ago, # ^ |
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      I stored bitset "is the number of elements divisible by x odd". The third operation is then just bitwise and.

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      6 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      Yeah, bitsets work.

      Set bit i of set j to 1, if there is an odd amount of elements in the set divisible by i. Then join is xor, multiplication is and.

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    6 months ago, # ^ |
      Vote: I like it +36 Vote: I do not like it

    I think E is very cool (Though I agree that constraints should be lowered. I see no bad polynomial solutions).

    Is G really standard? For me O(NK2) or KlogK with FFT was quite standard, but then got stuck for like an hour to improve it (and still I have no idea).

    I don't like H neither, sorry.

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      6 months ago, # ^ |
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      For G, I think you can improve the NK2 to NK + N / K * K2 by simply computing a subtree dp only up to the subtree size.

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

      Well, in E there are some heuristics which use randomization. After guessing the function you can erase random LIS or LDS and if you've ended with too many sequences — repeat. I wanted to cut it.

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      6 months ago, # ^ |
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      For G, you can get AC with O(NK2) solution. Example: 47935104.

      Polynomials multiplication is good for SIMD optimizations. And compiler can do all the work for you, if your code would be simple enough :)

      Check https://godbolt.org/z/2r_bdG for nice comparison of source and assembly codes. You may notice, that lines 95 and 90 (most nested part of source code) are assembled into operations with 256-bit registers (ymmD).

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        6 months ago, # ^ |
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        Yeah, but I don't like it when knowing how to write a fast solution to a contest problem, or knowing that some solution is fast, depends on knowledge of hardware features like SIMD or cache. It can even break problems where the author doesn't realise that some bruteforce can be very fast.

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

    E in ? Radewoosh, go fuck yourself with such limitations.

    I would spend less time on this problem if I would code it right after coming up with solution.

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      6 months ago, # ^ |
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      Hahaha, I really agree with you on expected complexity, however I think it would be great if he followed the footsteps of your iconic comment https://codeforces.com/blog/entry/49932?#comment-339270 and replied "Also, I've consciously chosen TL to fail people like Um_nik" or ""I would spend less time on this problem if I would code it right after coming up with solution." — that's exactly why you didn't get it accepted" :D

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        6 months ago, # ^ |
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        Goddamn, did you remember that comment for like 2 years? holy sheet.

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          6 months ago, # ^ |
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          You would be surprised how my mind is capacious when it comes to anything related to competitive programming. We just really liked this comment so it stuck in our minds — probably many of you remember gags like sorry_dreamoon or notorious coincidence even though they were long ago as well.

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6 months ago, # |
Rev. 2   Vote: I like it -72 Vote: I do not like it

 Dear CF, is it very cold there?

Why are you freezing up?

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6 months ago, # |
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How do you solve C? I tried my solution many times but it kept failing on test case 4.

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6 months ago, # |
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what is hack for problem B?

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    6 months ago, # ^ |
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    For example 6 179 179 179 179 2 6

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    6 months ago, # ^ |
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    I hacked with

    4
    170 160 150 120

    and with

    5
    170 160 150 140 100

    The answer to both should be YES.

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      6 months ago, # ^ |
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      how can we reach 0 in first test case?

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        6 months ago, # ^ |
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        170 + 160 + 150 — 120 = 360 = 0

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6 months ago, # |
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MikeMirzayanov During this contest I was unable to hack in Qutebrowser (version 1.1.1 based on Chromium 56.0.2924.122) and in Firefox (version 64.0). I was able to view the code, but when clicking Hack the input field for the hack didn't load and I only saw a spinning wheel. Only with my third browser, Chromium (version 71.0.3578.80), it worked.

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6 months ago, # |
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I'm sorry if this is not the right place to do it, but during the contest there was a participant in my room who obfuscated his code (as you can see here: 47912509). I couldn't find any report button so here I am. Is there anything that can be done? Thanks in advance!

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

In problem C, what's the answer for the case:

4

(()

)

((

))

The pairs 1 — 2 and 3 — 4 form the same bracket sequence (()). So the answer is 1 or 2?

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    6 months ago, # ^ |
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    1 — 2 does not form a bracket sequence.

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    6 months ago, # ^ |
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    If second string is ")" answer is 1. If second string is "))" answer is 2. You don't care if some pairs form the same bracket sequence.

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      6 months ago, # ^ |
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      "The task is to connect some of them into ordered pairs so that each bracket sequence occurs in at most one pair"

      Reading this i think that each bracket (form by a pair) has to be unique. In this way the problem became significantly harde.

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        6 months ago, # ^ |
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        When it says "each bracket sequence occurs in at most one pair" does not mean that the result should appear once. Means that you can pick every element and add it to one unique pair and not to multiples.

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    6 months ago, # ^ |
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    Ans:2 we need the highest number of pairs of correct sequence. No need for finding unique sequence of the pairs.

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6 months ago, # |
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Problem D :
dp[n][k] = (1/x)*dp[divisor_1][k-1]
            + (1/x)*dp[divisor_2][k-1]
            + ..... + (1/x)*dp[divisor_x][k-1].

Here x = number of divisor of n.
dp[n][k] will be like this for example : dp[4][1] = {1*(1/4) , 2*(1/4) , 4*(1/4)}. and so on.

PS : Here one optimization is instead of taking divisor as dp state we can take divisor count as first dp state as only number of divisor matters for expectation.
For example 4 will have 3 divisor and so does 9.
dp[4][1] = {1*(1/3), 2*(1/3), 4*(1/3)}
dp[9][1] = {1*(1/3), 3*(1/3), 9*(1/3)}
We only need to count dp[3][k] -> here 3 is count of the divisor and not number it self.

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6 months ago, # |
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for F its my solution of F 47939459 get MLE, but any other AC solution's memory (bitset <7001> bs[1e5+1]) just like my solution, whats wrong with my solution?

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    6 months ago, # ^ |
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    I assume you mean problem F (not G).

    Your int gc[7001][7001]; takes ~186.97 MiB, and bitset <7001> bs[100001] takes ~83.54 MiB (these figures are not exact). Summing up, they take ~270.52 MiB. Sorry to hear that.

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20 minutes delay on a problem --> 1000 positions in the standings. Gotta love this difficulty distribution.

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When you solve A,B,C and your C get runtime because your vector size is 1e5, but need 5e5

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6 months ago, # |
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Wow, Editorial has been set before the contest. Thanks!!

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6 months ago, # |
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Can someone help me with the Problem C? my submission1 failed, I changed unordered_map to map, and instead of for(auto x: m), I used an iterator to loop, The solution is accepted. I couldn't figure out what was the issue I had?

Submission1

Submission2

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    6 months ago, # ^ |
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    int b = m[-1*x.first]; This one potentially changes the map (if key -1*x.first wasn't in the map). Which breaks iteration over it (if it's unordered_map).

    Spend 2 hours debugging same issue :(

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      6 months ago, # ^ |
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      Thanks, I created a blog for this before checking your reply, btw it (unordered map) was passing with clang++17 diagnostics. Do you have any idea why is it so?

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        6 months ago, # ^ |
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        Well, I don't think clang++17 diagnostics (or any automatic tool) can catch all possible bugs in the code.

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6 months ago, # |
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My screencast will be available here: https://youtu.be/UxtEMlc0sHM , including getting problem H correct 30 seconds after the end of the contest (just checked in upsolving), which would be just enough for the first place :) Oh well.