sshwyR's blog

By sshwyR, history, 14 months ago, In English

Hello Codeforces! We (dqa2021, Xiejiadong, xryjr233, Z18 and me) are excited to invite you to take part in Codeforces Round #664 (Div. 1) and Codeforces Round #664 (Div. 2), which will happen on Aug/12/2020 17:35 (Moscow time).

Huge thanks to:

There will be 5 problems in Div.1 round and 6 problems in Div.2 round. You'll be given 2 hours to solve them.

The story of this round is about that man. Instead of displaying his name, I prefer telling one of his legends (or joke):

"I have a 'friend', who makes lots of money every day, earning a billion in the blink of eyes. With a wave of his hand, OIers all over the world will follow him."

You can post your guesses in the comments.

UPD: Score Distribution:

  • Div.1: 500 — 1000 — 1500 — 1750 — 2500
  • Div.2: 500 — 750 — 1000 — 1250 — 1750 — 2250

Good luck!

UPD: Congratulations to the winners!

Div.1:

  1. Benq
  2. ecnerwala
  3. nick452
  4. kefaa2
  5. neal

Div.2:

  1. C.S.T.T
  2. MyLoveKUN
  3. Rchen3
  4. Rainbow_sjy.qwq
  5. evilbuggy
 
 
 
 
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14 months ago, # |
  Vote: I like it +12 Vote: I do not like it

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

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    13 months ago, # ^ |
      Vote: I like it +60 Vote: I do not like it

    What's the point of declaring top 5 people in Div.2, after all they are just Div.1 guys who registered with their alternative account, isn't this unfair? for example just look at the person who is placed 3rd in Div.2, this guy was newbie, like seriously? This disrupts entire rating distribution.

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14 months ago, # |
Rev. 4   Vote: I like it +410 Vote: I do not like it

I am tester, please upvote for me! :)

( I'M CUTE, GIVE ME CONTRIBUTION )

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

Is the man is Bill Gates? :D

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

is that man Jack Ma? just a random guess

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14 months ago, # |
  Vote: I like it +149 Vote: I do not like it

We don't want stories please, as they come in the way of easily comprehending the problem statement. :(

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    14 months ago, # ^ |
      Vote: I like it +77 Vote: I do not like it

    Oh don't worry. The statement is not long and we just use the main character to describe it. The story itself is easy to comprehend :)

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

      Maybe you could highlight the formal statement or something to separate it from the story.

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

      For me who is afraid of number theory, this is good news to be happy about. If there is a math problem, I am also ready to change to the alternate.

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14 months ago, # |
  Vote: I like it +280 Vote: I do not like it

Tester — Planning to participate in the next div 1.
sshwyR — You already tested it.
Tester -

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

    MeToo

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    14 months ago, # ^ |
      Vote: I like it +78 Vote: I do not like it

    This contest was put on hold for a long time due to the difficult Di2A.

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      14 months ago, # ^ |
        Vote: I like it +70 Vote: I do not like it

      difficult Di2A

      Thank you , now excuse me while I go to sleep for 35 hours real quick

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

        Go for 40 , than that would be sacrilegious

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

          Two set violin = small pp

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

            Bass = pp large

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

              What? You bass gang, had some arguments before, but now is just butthurt, VIOLIN > bass. TwoSet responded to all Davie's videos, but Davie just ignored the "bass diss track" (which was asking for a livestream at the end) and the "davie fakes playing the violin". Brett played bass poorly, but it was REAL, but Davie, on the other hand, FAKED playing it. So what side is winning, huh?

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

      Oops, difficult Di2A. So scary....

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

      Is this difficult div2A? Then what about the contest, which was for Russian elementary school students, with a 1700 rating problem A. (it was like 10 contests before) :D

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

    Feeling sad for such testers but they will get contribution.

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

      Well I knew I wont participate in one div1 when I was testing. Anyways I did participated but in unrated mode. Just that I didnt knew I wont be participating on 12 Aug.

      Spoiler
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14 months ago, # |
  Vote: I like it +60 Vote: I do not like it

Another Chinese Round! This round has special and interesting stories.Good luck and have fun for every contestants. Some information in advance:Those stories may all be about one mysterious person.

[O)]-[(O]/

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

Is the man @MikeMirzayanov ? <3 i have not any guess about his income but i love him and want to see his name in the problem statement ..

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

Mukesh Ambani???

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

Boss Cai!!

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

I am almost sure this man is Jack Ma.

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

I want to know what is Polygon platform ? Strange word for me

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

    A platform for making problems

    https://polygon.codeforces.com/

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

    Polygon in basically the platform for creation of programming contest problems. It also supports problem statement writing, test data preparing, model solutions, judging and automatic validation.

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

    The mission of Polygon is to provide platform for creation of programming contest problems. Polygon supports the whole development cycle: - problem statement writing - test data preparing (generators supported) - model solutions (including correct and wittingly incorrect) - judging - automatic validation

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

I know! The man is Cai Rui[user:Boboniu]!

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

Intersting fact: tourist will set a new best rating ever on codeforces if he wins this round!

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

What are you talking about ???

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

Chinese Round?

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14 months ago, # |
Rev. 2   Vote: I like it +199 Vote: I do not like it

Fun fact: seeing MiFaFaOvO can't participate in this round, if he doesn't in the next global round, he will vanish from the live standings xD

And "poof": he is gone xD

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

    Excuse my ignorance, but what is the rule exactly you're talking about?

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      14 months ago, # ^ |
      Rev. 2   Vote: I like it +52 Vote: I do not like it

      Only the users which participated in a rated round within $$$6$$$ months will be included in the Rating Standing.

      The last rated round for him took place on $$$Feb^{17th}$$$.

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14 months ago, # |
  Vote: I like it +70 Vote: I do not like it

I guess it's CCF

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

I guess this man is BINOD- Richest man of NEPAL.
You can just google it out.

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

I do CP for fun.. What about you guys?

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14 months ago, # |
Rev. 3   Vote: I like it -13 Vote: I do not like it

Billie Jean is not my love

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14 months ago, # |
  Vote: I like it -118 Vote: I do not like it

BINOD !

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

    Please don't spam Codeforces at least. There are many who don't know the context and it gets annoying after a while.

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

It's Mr-Cai!!! Hahahaha

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

When I see Chinese round, I think of Mathforces.

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

so... Yet another maximizing profit problems ... or Yet another guessing name problems :v

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

Ronan Ryan?

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

interestingLSY Time for your picture about boboniu!

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14 months ago, # |
Rev. 3   Vote: I like it -14 Vote: I do not like it

Ah-oh, Retired_MiFaFaOvO as a tester, which means we can't see his competition with tourist this time. A huge pity!

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    14 months ago, # ^ |
      Vote: I like it -71 Vote: I do not like it

    Either way ,he will never compete against tourist ,after the time he gained 1st spot . He just prays for some bad luck for tourist ,so that he can remain on top.

    Before that ,i Never thought ,such legendary grandmaster would ever fear to directly compete against someone. But anyway that someone tourist is not ordinary guy.

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    14 months ago, # ^ |
      Vote: I like it +49 Vote: I do not like it

    But It's still an interesting and exciting contest

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14 months ago, # |
  Vote: I like it +69 Vote: I do not like it

He must be Zide Du(杜子德) from the CCF.

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    14 months ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    China Computer Federation or China Collecting-money Federation?

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

Is him, right ?

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

Please provide short statements for problems and understandable english,I am learning english .Thanks

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

    Copy to baidu translate (

    Website: fanyi.baidu.com

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14 months ago, # |
Rev. 11   Vote: I like it -64 Vote: I do not like it

[Deleted]

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    14 months ago, # ^ |
      Vote: I like it +39 Vote: I do not like it

    You take slander as freedom?

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

    You should know it's not polite to comment others like that

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    14 months ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    Freedom is not your excuse for slandering.

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

    Although it is your right to express your own ideas,but your last comment was unsuitable indeed in some way.And codeforces delete your comment for a great many people downvote you.I think it is reasonable.

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14 months ago, # |
  Vote: I like it +49 Vote: I do not like it

See this 89567737

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

What's happening to "timeanddate.com"? :|

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

    The website works well for me, probably it's because of your network problem.

    And I don't think these sorts of comments are meaningful here

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      14 months ago, # ^ |
      Rev. 5   Vote: I like it -29 Vote: I do not like it

      This comment has been edited. Don't dislike me anymore! D: leave me alone!

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

        Excuse me is there anything wrong?

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

How to Mukesh ambani here?

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

Boss Cai -- boboniu

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

Makes lots of money everyday. That must be dzd from CCF(China Collecting-money Foundation)!

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

I became psychopath because of the little pony, please make sure that the problems have short statements. with appreciation to your great story.

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

I really hope that the problems are not as hard as the quiz.....

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

He must be Zhang Aiqin ?

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

Is this Gennady Korotkevich? Lol

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

How many number of common problems ?

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

Richie Rich?

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

It could be suneo's dad from doraemon xD

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

is it jeff bezos ?

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14 months ago, # |
  Vote: I like it -11 Vote: I do not like it
My guess
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    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Asia's richest man, Worlds 4th. He is definitely very rich but I don't think it the right answer. Suneo's dad and Richie Rich are much more likely candidates.

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

please what's the difficulty fo this contest because i'm a beginer and i want to take a part

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

    It's better for you to test your skills in Division 4 or 3. However, this is Division 2 so you can still be able to maybe solve the first problem or who knows, maybe even the second one! Good luck to you!

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

I'm the man y'all are talking about. admin please don't put me on read only again plix.)

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

Olers?

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

Is that man Mark Zuckerberg??

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14 months ago, # |
Rev. 2   Vote: I like it +13 Vote: I do not like it
The Man is:
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14 months ago, # |
  Vote: I like it +9 Vote: I do not like it

Is the man Mike Mirzayanov ?

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

JPow?

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14 months ago, # |
  Vote: I like it +41 Vote: I do not like it

If there are unnecessary stories, please, do mark them in Italic or any other way. I've faced a lot of issues about bad storytelling in Codeforces. :)

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

boboniu LGM.
rating 3117 in Codeforces
rating 10000 in Hearthstone War Chess.
%%%%%%%

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

ALIBABA

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

is it Mukesh Ambani ?

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

What are "Olers"?

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

    The people like us. Take part in OI. OI, Olympics Information. OIer, the people learn OI. OIers is plural form of OIer. Understand? :)

    I'm not good at English. I hope you can understand xD

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

    OIers are coders who participate in programming competitions (mostly used in China I guess). OI = Olympiad Informatics

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

    It's what Chinese called people who participate in Olympic in Informatics contests.

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

    The goalkeeper of RCD Espanyol, click the link for more details.

    https://www.transfermarkt.de/oier-olazabal/profil/spieler/8176

    Since Wu Lei (the best football player in China) is also in this team, we use the word "Oier" to show our respect to RCD Espanyol so that they pay Wu Lei on time.

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

I think he is clb(boboniu)/cy/kel

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

I love Money :B

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14 months ago, # |
Rev. 11   Vote: I like it -12 Vote: I do not like it

Its Jack Ma I guess- cofounded ALIBABA- China's richest person!

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

i guess Jack Ma .

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13 months ago, # |
  Vote: I like it -15 Vote: I do not like it

tourist? Pretty sure it's him :)

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

I guess he's kkksc03, head of Luogu(an oj in China). For every monthly contest in Luogu, if you want to watch the video tutorial, you must pay 10¥ first. And what's more there's lots of OI online lesson made by Luogu in every summer and they're not free! So he can make lots of money in this way.

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

    Actually, according to himself, the money is not paid to him directly and a lot of money is spent in maintaining Luogu's website. And the lessons' money will be given to the teachers. So I'm very sure that it's not him.

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

Any hint about no of common problems or score distribution ?

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13 months ago, # |
  Vote: I like it +19 Vote: I do not like it

Score Distribution?

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13 months ago, # |
Rev. 2   Vote: I like it -35 Vote: I do not like it

Binod

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

Obviously,the man is DUZIDE:P

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

Codeforces Round #664 ~ Made in China

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

I guess the person is a CODER. :>

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

Score Distribution?

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

    Sorry for the late publishment of score distribution :)

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

I have a request for MikeMirzayanov.. can we have TEAM contest at codeforces.

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

how are you talking about me??

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

It's €€£

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13 months ago, # |
Rev. 3   Vote: I like it -34 Vote: I do not like it

.

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

sshwyR how many problems will be common in both divisions?

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    13 months ago, # ^ |
    Rev. 2   Vote: I like it -12 Vote: I do not like it

    I guess there are 4.

    D2A->D2B->D1A/D2C->D1B/D2D->D1C/D2E->D1D/D2F->D1E

    upd: When I saw the scoreboard I knew there are only 3 but I forgot to update QAQ

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

    3

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

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

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

Today's tester busy with increasing their contribution .

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

Div2 Problem E with 1750 (^_^)...it should be solvable

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

iam not a tester, but please upvote for my pp :).......

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

The man must be, President of IOI: Richard Forster

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

Zhengru IOI will be internationalized.

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

Karuna! :D

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

An off-topic question: how can I view the past contests right before contests?

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13 months ago, # |
  Vote: I like it +8 Vote: I do not like it
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13 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Too much if else... :(

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

Normal Div1:

Try to solve more problems

Horrible Round #664 Div1:

Try to solve the first problem faster

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    13 months ago, # ^ |
      Vote: I like it +27 Vote: I do not like it

    No,you should try to avoid FST instead of solving problems faster. It's my final standing: xjpbmbrm.png

    (I was rk128 before the system test...)

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

If I accept no question am i still rated?

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    13 months ago, # ^ |
      Vote: I like it +28 Vote: I do not like it
    if(submit any problem solution) Yes
    else No
    accept or not doesn't matter
    
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13 months ago, # |
  Vote: I like it +10 Vote: I do not like it

why they set so hard

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

Was this the "Single Test Case" Contest?

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

Really nice problems and clear statements. Enjoyed taking this contest :D

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

Codeforces rounds are great, they teach us how to look at a problem in a simple manner, without overcomplicating it. orz Great contest.

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

Well well well, once again someone posted solutions on youtube during the contest, you can even see his username in the video, i wonder if these guys are shameless or what Video link Username: Khayrulmithu

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

    I don't know why but that video sounds exactly like an airline pilot's announcement.

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

    And he got it wrong on test 10 :D

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

I really liked d1 B, but...

Starting from any vertex u

This would be much more understandable if it said from all vertices. Wasted lots of time solving the wrong problem.

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

    That could confuse people into thinking that it's possible to start from multiple vertices simultaneously, I think the clearest is "for all vertices u, starting from u, ..."

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

    I'm pretty sure that "any" was actually used correctly here

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

What was the point of swapping X and Y in Div2B?

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

Does this works for Div-2 E — Make a graph $$$g[i][j][k]$$$ = how many edges exist such that its head is on node with $$$i$$$ outgoing edges and tail is in node with $$$j$$$ outgoing edges and it is $$$k'th$$$ smallest one. Then we brute force every $$$c_i$$$. That will be $$$O(k!k^2)$$$

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13 months ago, # |
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The whole round is like an ingenuous play of MiFaFaOvO to get back his first place in the rating (though unsuccessful). Chinese people are really good at the art of war. My respect.

(please don't take this very seriously)

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    13 months ago, # ^ |
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    "There are two ways to go higher, self improvement, and sabotaging literally the entire community"

    He has made his choice.

    (Jk no offense pls :)

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    13 months ago, # ^ |
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    Seems his plan worked

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13 months ago, # |
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Why the "non-empty" in C? I guess it has to do with printing the output but it reduced problem quality by like 30%.

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    13 months ago, # ^ |
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    could u plz xplain how to solve todays div2 C problem

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      13 months ago, # ^ |
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      You do realize that I haven't read Div 2C?

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        13 months ago, # ^ |
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        ok..if someone other can help..bcz..i am very curious about it..that why my logic was wrong..even on sample test cases ..i know i was completely wrong.can not wait till editorial..xD

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      13 months ago, # ^ |
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      This is very easy implementation. DIV2 C LINK

      Just coz I am a newbie don't think that the solution is worthless. :)

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    13 months ago, # ^ |
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    Might as well allow output of negative length to boost problem quality?

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      13 months ago, # ^ |
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      Empty strings make sense, strings of negative length don't.

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    13 months ago, # ^ |
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    OH SHIT I missed it. Even my starting answer is an empty string. Luckily my code goes for large strings first when improving the cost in binsearch, so I got AC.

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13 months ago, # |
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.

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13 months ago, # |
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Is C checking if a cube like shape fits all of the given points by binary search / math?

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    13 months ago, # ^ |
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    Yes!But I am too weak at geometry,I come up with this idea quickly but don't know how to compute the area covered by several polygon...

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      13 months ago, # ^ |
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      Maintain the minimum and maximum values of $$$N$$$, $$$B$$$, and $$$N - B$$$.

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        13 months ago, # ^ |
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        Yes,but does it turn into a geometry problem??

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          13 months ago, # ^ |
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          Depends on your definition of "geometry" I guess. I didn't really use any "geometric" techniques. But yeah, you can visualize the things as hexagons.

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          13 months ago, # ^ |
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          No need to compute the area. Just find a point inside it.

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            13 months ago, # ^ |
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            Yes,But how to compute the point inside it?

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      13 months ago, # ^ |
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      Several polygons? I think it reduces to finding the minimum side length of the square in a shape like this that can fit all the points, which should be fairly easy. But I spent most of the contest thinking I had to minimize the sum of distances instead of the maximum, so I couldn't implement it in time.

      cube

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    13 months ago, # ^ |
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    Rather diamond-like (with top-left and bottom-right edges cut). Math seems awful there.

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13 months ago, # |
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How to solve Div2C? Also, where does the N^2 check all pairs approach go incorrect?

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    13 months ago, # ^ |
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    Final Answer can be at most $$$2 ^ {9}$$$, so just brute force for every possible value, and pick the smallest.

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      13 months ago, # ^ |
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      Thanks.

      Do you think there a polynomial-time solution in case final answer is large?

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        13 months ago, # ^ |
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        I have O(n^2 log m), where m is the size of the largest input integer.

        Figure out how many leading zeroes we can get by picking greedily the pairs where the bitwise-and has the most leading zeroes, using O(n^2) time. The next bit has to be a 1 in any solution, so we can add that bit the the answer. Since the bit will be a 1 in any case, we can ignore is from now on: we set it to zero and find out again how many leading zeroes we can now get. Repeat until we can get all zeroes.

        https://codeforces.com/contest/1395/submission/89696109

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      13 months ago, # ^ |
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      Well, I did that but, the o/p for test 3 came out to be 179. How did you do it?

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      13 months ago, # ^ |
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      Your implementation is too elegant :)

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    13 months ago, # ^ |
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    I solve this way
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      13 months ago, # ^ |
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      Now I feel like an idiot,spend half an hour in implementing an dp solution..:/

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        4 months ago, # ^ |
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        Hey can we solve problem C it using Dp??
        I tried some approach but its giving wrong ans.

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      13 months ago, # ^ |
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      can you explain why you did if((a[i]&[b[j])|k)==k) cnt++,break;

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      13 months ago, # ^ |
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      Can you explain this approach?

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        13 months ago, # ^ |
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        My english is so bad.sorry.if you understand its my pleasure <3
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      13 months ago, # ^ |
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      Very clean aproach, fix the answer and check there are N pairs a[i]&b[j] that has the same bit positions. I spend 1 hour until I reach my dp aproach...

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    13 months ago, # ^ |
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    This is very easy implementation. DIV2 C LINK

    Just coz I am a newbie don't think that the solution is worthless. :)

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13 months ago, # |
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Is 1C a geometry problem? I thought it just need to binary_search and compute the area covered by several polygon..

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13 months ago, # |
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I passed the pretests in Div2B with a backtracking solution, how is that a thing?

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13 months ago, # |
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Hints for $$$Div 2 E$$$ Please?

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    13 months ago, # ^ |
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    Assume you have selected a tuple. Of course, you can now reduce the initial graph to one such that each node has out-degree of 1. Realise that for each node to be in a cycle in this resulting graph (and thus be possible to reach back to it in finite time), each node must also have an in-degree of 1.

    For each node (let it be Q), traverse over all nodes that point to it, let's call these set of nodes P. You can easily determine that for which value in the tuple will these nodes in P be pointing to Q. Also, this means that these values can not co exist in the tuple, since that would mean that multiple nodes point to Q in the resulting graph. Thus, you make a list of all these invalid combinations.

    Iterate over all K! tuples, and for each check if they consist of an invalid combination. If not, then it is valid.

    Sadly, was a bit late while implementing in contest :(

    Here's my AC code: https://codeforces.com/contest/1395/submission/89731507

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      13 months ago, # ^ |
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      What's the upper bound of the number of those invalid combinations?

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        13 months ago, # ^ |
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        By an invalid combination, I mean pairs $$$((i, j), (k, l))$$$ such that value $$$j$$$ at index $$$i$$$ in the tuple, and value $$$k$$$ at index $$$l$$$ in the tuple, are incompatible. So the maximum number of such invalid combinations are $$$K!(K!+1) / 2$$$.

        You just have to store all these pairs in a $$$K*K*K*K$$$ matrix.

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          13 months ago, # ^ |
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          Thanks a lot.

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13 months ago, # |
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1250 not enough points for Div.2D

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13 months ago, # |
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Long Long in problem A just ruined my whole contest.I almost wasted 50min behind it.

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13 months ago, # |
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In Div2C, Sample test 3, what combinations get the final result as 147? The minimum I could get was 177.

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    13 months ago, # ^ |
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    so true... There should be a greedy approach to solve this problem...

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    13 months ago, # ^ |
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    I think you applied the dp approach. I did the same thing at first and waste a lot of time. just brute force the answer

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    13 months ago, # ^ |
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    I think this was the combination

    Spoiler

    these are indices of the element from array a and b

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    13 months ago, # ^ |
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    I also got the same error during the contest but now I understand. Well, let's say our solution consists of two parts, 1 and 2. I am not going to explain the second part assuming you already know it.

    But in the first part, u must be choosing a minimum c1 among all possible c1 and then doing part 2 of the solution but we are doing a mistake here. Actually, part 1 of the solution has n possibilities. Choosing minimum c1 is just one of the possibilities. A second possibility, for example, can be choosing a minimum c2 and then doing part 2 of the solution. Similarily choosing a minimum c3 among all possible c3 can be the third possibility and so on. So there are n possibilities.
    So the solution is O(n^3) and not O(n^2). wicardobeth

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    13 months ago, # ^ |
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    This is very easy implementation. DIV2 C LINK

    Just coz I am a newbie don't think that the solution is worthless. :)

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13 months ago, # |
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i will fst in d,One detail was not considered,shit.

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13 months ago, # |
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That was implementationforces, I like that. Should be enough to become blue again :)

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    13 months ago, # ^ |
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    What was the solution to D?

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      13 months ago, # ^ |
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      Greedy.

      Edit, more understandable: We devide $$$a[]$$$ into set $$$a[]$$$ and $$$b[]$$$, where all $$$b[i]<=m$$$ and all $$$a[i]>m$$$

      Sort both sets biggest element first.

      Then choose consecutevly to use i elements from a[]. Foreach i we can calculate how much elements from b[] can be used then, and using prefix-sum calc in O(1) the funfactor of that i.

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        13 months ago, # ^ |
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        If only I had two more minutes to fix my implementation

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        13 months ago, # ^ |
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        Figured that, but couldn't implement. Thanks, will wait till I can see your submission.

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    13 months ago, # ^ |
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    I am used to see you in blue ;)

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13 months ago, # |
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How to solve div2 D?

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    13 months ago, # ^ |
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    1) Make 2 lists: one of all greater than m (let us call it gr), second all less than m(let us call it sm). 2) Sort gr reverse and sm normally. 3) Now compare highest unused of gr with sum of d consecutive of sm. If gr highest is greater than simply add that to ans(in other words muzzle d smallest elements of sm). Take care of these things: 1) Do the above only till you have atleast 1 element left in gr, cause you can add it in the end. 2) If you encounter condition like less than d elements remaining in sm, then dont compare with gr greatest element as you can add all remaining along with greatest, if greatest goes in the end. 3) In the end try to put all remaining elements of gr in the end of the permutation seperated by d elements each. For details see the code: 89725326 If I fail system test ignore this comment :D

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13 months ago, # |
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Can anyone please tell me what's wrong in my code? Problem A https://ide.codingblocks.com/s/308173

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13 months ago, # |
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Can someone tell me why this solution to 1C is wrong?89722745

I can't find the mistake :(

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    13 months ago, # ^ |
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    Oh now I find it

    I forgot one line if(zmin>zmax) return 0; :(

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13 months ago, # |
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Boboniu stands nowhere in front of Binod

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13 months ago, # |
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Did anybody else got more than one "Wrong Answer in Pretext 2" in Div2 A

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13 months ago, # |
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Any Idea of Test 16, Div1 B ?

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How to solve Div-1 B??

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13 months ago, # |
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Need help, why did my solution for Div2B did not pass ? I have used dfs- https://codeforces.com/contest/1395/submission/89714645

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    13 months ago, # ^ |
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    .

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      13 months ago, # ^ |
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      yes, but the first think that came to my mind was this solution so I implemented it. But my question is that, what's wrong with this DFS approach, should work right ?

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        13 months ago, # ^ |
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        Because the DFS might not find a suitable move e.g it moves in a corner and both adjacent cells are already visited then it backtracks and moves from some previous suitable position.

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    13 months ago, # ^ |
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    I solved it first printing the (short)path sx, sy to 0, 0 which is moving up until the border and moving left until the border, and also save this path to an array.

    Then move to the end of row and move 1 row down and back to the beginning of the second row.

    Print there values while moving if they are not in the array you saved before.

    Then do this steps until reach (n-1, m-1).

    Here is an image of clarification, start from (sx, sy):

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    13 months ago, # ^ |
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    Dfs is wrong because it uses backtracking(after using a path you go back to the last intersection in that path and complete the search.) although in this problem you can't go back to the same square.