SlavicG's blog

Codeforces Round #726 (Div. 2)
By SlavicG, history, 3 years ago, In English

Hello Codeforces!

mesanu and I are very glad to invite you to Codeforces Round 726 (Div. 2) which will take place on 18.06.2021 17:35 (Московское время). The round will be rated for participants with a rating lower than 2100. Division 1 participants are also welcomed to take part in the competition but it will be unrated for them.

The round was originally planned to be a division 3 round so the problems might be slightly easier than average Div 2.

You will be given 6 problems and 2 hours to solve them.

We would like to thank the following people:

The scoring distribution is: 500 — 750 — 1000 — 1500 — (1250+1750) — 2000.

UPD: Editorial

  • Vote: I like it
  • +690
  • Vote: I do not like it

Comments Comments (335)
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3 years ago, # |
  Vote: I like it +523 Vote: I do not like it

As a coordinator...

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    3 years ago, # ^ |
      Vote: I like it +93 Vote: I do not like it

    omg orz! monogon taking over cf!

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      3 years ago, # ^ |
        Vote: I like it +390 Vote: I do not like it

      Eventually I will fire Mike so I can be the real top contributor. (Don't tell him I said that)

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        3 years ago, # ^ |
          Vote: I like it +79 Vote: I do not like it

        monogonforces orz

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

        As a follower of Monogon, I am obligated to upvote this comment.

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        3 years ago, # ^ |
          Vote: I like it +71 Vote: I do not like it

        At this rate, that day isn't far when...

        WORLD WILL BE TAKEN OVER BY MONOGON
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        3 years ago, # ^ |
        ← Rev. 3   Vote: I like it +190 Vote: I do not like it

        In the future, if you're gonna need an icon just contact me. I'll do it for free just a few upvotes.

        Should I resume share or okay this?

        1623813407111

        Also,

        The guy who looks 34-35% similar to Mike - meme
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        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        ORZ 1-gonforces

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    3 years ago, # ^ |
      Vote: I like it +212 Vote: I do not like it

    Congrats for the successful steal of Anton's Job.

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

    whatever man just take my upvote :/

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

    in this contest i had solved problem A in first 8 mins and i was checking for the results and it was showing "in queue" so i thought to submit it on m1.codeforces so that it get tested fast but now i have seen that both of my solution got passed but the system have rewarded me the marks for the second one which i have submitted 2mins after and also it has given me -50 . can you plzz say something or do something on this plzzzz first time in 6 months i was able to solve 4 prblms in div 2

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

      According to the official rules, only the last submission with pretests passed is considered for system testing. When your submission was in queue you should've just waited and moved on to read the other problems.

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3 years ago, # |
  Vote: I like it +185 Vote: I do not like it

As a tester, I tested.

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3 years ago, # |
  Vote: I like it +77 Vote: I do not like it

newbie sens

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

As a participant, I hope to become purple in this round. Wish me luck !

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

    All the best, as a participant I hope to get my confidence back

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

    Good Luck! I hope you become purple after this round but I sincerely hope that it's not coming at my ratings' expense.

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

      Seems the opposite happened lol, you got rating at my ratings' expense :O. Anyways, good show.

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

As a tester, i'm proud of these guys.

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

    Just take my upvote

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

      Probably copying previous comments does not help. People who plan on doing this can probably take note. Hehe

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

5 months later, I'm still annoyed that you didn't take kostia244's suggestions for your new handle :(

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

As a tester, SlavicG ORZ

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

    inspirational

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

    It really motivated me SlavicG

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

      what about

      this guy?


      RunTimeTerror Motivation overflow

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

        tenkei retired

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

          He already gave us so much motivation and showed us everything is possible if we try honestly and embrace hard work. Hope he enjoy his retirement. We will miss him. Orz.

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

Score distribution suggests that the contest is All-Killable

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3 years ago, # |
  Vote: I like it -49 Vote: I do not like it

So blues are writing div2s now? whats next? greendians writing div1?

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

    Chup chapri. Dont write these useless messages.

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      3 years ago, # ^ |
        Vote: I like it -39 Vote: I do not like it

      keyboard_warrior Another bluedian defending your blue buddies? Hope even you get to write a contest next.

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        3 years ago, # ^ |
        ← Rev. 2   Vote: I like it +20 Vote: I do not like it
        • problemsetting is all about having original ideas, which does not strictly correlate with rating.
        • even if the problems were not worth it, the testers and coordinators and other people involved with this contest (a lot of whom have a ton of experience with contest-making i'd assume) would have caught that and we wouldn't even be seeing this announcement.
        • adding on top of the previous point, there are more than just the problemsetters who are there to make the whole contest experience better for us, like testers and such. this stuff takes time to make and improve, and i believe that these people spend a lot of their time doing awesome work to gift us these valuable contests.
        • like come on, don't be a jerk valuing people just by some random internet point. that's not cool at all...

        (edit: fixed some grammar mistakes)

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

          What about purples/blues(not me) who have really good problem ideas and not able to propose a contest because of the rating barrier and end up in submitting them in less kown websites or just moving on with dissatisfaction.

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

    I have a dream, that one day authors may be judged by the quality of their problems, rather than the color of their handle.

    As a tester, I can say that the problems are nice, and I recommend participating :D

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

    Well, if you look at their number of solved problems you can easily say they solved enough problems to make their own problems. Also some red coders tested/coordinated it so there is no reason to not make a round. And those "greendians" are better than an unrated person with -53 contribution.

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

    mad?

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

    And what is your rating, mister alt user?

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

As a tester, I couldn't help much with review of problems except in the initial div3 phase. The round is well balanced and may positive delta come your way :blobheart:

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

All the best to all the participants,

may there be a positive delta... The scoring distribution is quite interesting.

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

Moldova power!

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

As a tester, I highly advise you to read all the problems!

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

    I often see this comment whenever we have a problem which is divided into two( easy & hard) versions . Let me guess E1 is a bit easy than D :) and one more thing we have to do a bit more reading :)

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

      I got it right , E1 was easier than D :)

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

    Don't they all suggest the same thing. I think the trick to acing the contest might be to not read all the problems and not get depressed to try the problems.

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3 years ago, # |
  Vote: I like it +42 Vote: I do not like it

These rules changed?

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

    I guess they can be count as a writer in the past because they made 2 unoffical div 4 contests and those rounds had a good amount of participants.

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

      No disrespect, but they are a bunch of some easy problems. So how do they count?

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

        they said this meant to be a div3 round,so those problems were enough to give them a chance to write a round.

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        3 years ago, # ^ |
          Vote: I like it +39 Vote: I do not like it

        To reds, a Div. 2 is a bunch of easy problems. To oranges, a Div. 3 is a bunch of easy problems. So how do they count?

        Honestly, I think the current rules were just a way to limit the amount of contest proposals without disenfranchising anybody. So if you were lucky enough to get something in before the new requirements, then you basically get grandfathered in.

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

          Div 2 is not rated for oranges and above. Div 3 is not rated for blues and above. But this round is rated til 2100.

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

            And since it was reviewed, coordinated, and tested by a bunch of people high rated, it should be reason enough to believe that the contest will have good enough problems for it to be rated for participants with rating upto 2100.

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

              Probably, still I'm also somewhat dubious about the ability of people to come up with the problem and solution ~500-1000 points above their own rating, and the hardest problem of the round should be at least that difficult

              No offence, it's just peculiar

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

                Rating doesn't measure your overall problem solving skills.

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                  3 years ago, # ^ |
                  ← Rev. 2   Vote: I like it +64 Vote: I do not like it

                  A more accurate explanation is probably that it is quite possible to make a problem that is 1000 points above your rating by making a problem from the solution, instead of actually solving a problem you came up with. And, even for problems that you came up with, you have a lot more time to try to solve them than when in a contest, and you also probably have more chances to solve a hard problem(the participants can't notice all of the hard problems that you didn't solve when problemsetting).

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              3 years ago, # ^ |
              ← Rev. 2   Vote: I like it +24 Vote: I do not like it

              Btw, all rounds have coordinators, but not all of them are equally good, so it doesn't really mean anything, just saying

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

                Well, I didn't want to drag this on, but let's do it.

                not all of them are equally good

                What is a good round / contest? one where you can solve 3 or 4 problems? you feel good after the contest, that maybe you solved every problem that you read during contest? Sorry to disappoint you that not every round can be good for you.

                Tbh, have you ever even thought about what it takes to prepare a round, or the effort put in by coordinators / testers. All you see is the "Codeforces Round #XYZ" blog on CF. Although, I have never authored a round myself, I want to make problems for a while, but I can't even think about them. Forget about preparation of statements, constraints, test cases, and many other things that I might not even know about right now. Even though some rounds are not well recieved by contestants, it doesn't trash all the effort that was put into it, nor does it mean that problems are shit. Hell,the problems were accepted by anton, and even as a joke, it is popularly known how high a bar he has set.

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

                  Nah

                  What is a good round / contest? one where you can solve 3 or 4 problems?

                  Rly, do not assume that the other person is stupid. Whether my performance is good or bad has nothing to do with the round being good or bad. Say, LATOKEN round was quite good even though I got -50

                  Model good round is a round with good problems, balanced problems, diversed topics, good problem statements and good editorials. Also without long queues and other technical issues but that is out of control of coordinators.

                  It's all nice and everything. Yet again, some rounds are better than the others despite all the hard work of the authors you mentioned

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

                  Have you realized that the first 5 requirements you stated are completely subjective? Sorry, this website doesn't cater to each and every one of your individual needs. The round is good, it was tested by a wide variety of colours, it was coordinated by reds and approved by Mike, I'm not sure what else you could ask for.

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

                  Well put, I couldn't have said it better.

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

                  They're not so much subjective as you're trying to make it sound

                  Say if the statement is long and confusing for a lot of people, it is a bad statement and there is little subjective about it

                  If one problem has difficulty 1000 and the next one is 2200, it is unbalanced round

                  Etc

                  Also I don't quite understand what you're arguing about, Do you disagree that some rounds are better than the others? Because that is pretty much the sole point of my messages here

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

                  As the round has neither started nor you are a tester, then how can you say anything about a round when you literally don't know anything about the round

                  I don't know if you read the testing part, but the round has testers from almost every color from pupil to LGM.

                  So, it would be better if u shut up now and give your opinions after the round has ended.

                  There is a comments section in editorial also

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                  3 years ago, # ^ |
                  ← Rev. 2   Vote: I like it +41 Vote: I do not like it

                  Lets just end this thread by a well known standard result : Quality of contest depends on your rank in that contest

                  QED

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

    What actually happened was some combination of Mike liking the problems but not thinking it was Div 3, and having Anton and Monogon coordinate.

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +39 Vote: I do not like it

    They wrote an unofficial div4 contest on gym before, so it isn't a violation of the rules really. They had experience in problemsetting and messaged me (who started coordinating after that blog) about it, as it says in the last paragraph.

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

As a tester, I can certify this is a great round with interesting and clear problems. Best of luck to all who are participating!

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

Both the problemsetters have a highest rating of expert. So how did they propose this round, as you have to be a master to write one?

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

    You can suggest problems if you are an expert. I don't know if it's any different than proposing contests.

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

As a participant I hope to get past C this round

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

I wish I could be expert. People how to get expert batch within a short time? Is it their outcome of lot of practice, How can I get expert batch? Please suggest me. (Sorry for unnecessary comment. But I need to know to develop my problem solving skill)

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

    Literally just solve questions that are challenging enough, so basically problems that are rated +200 ish from ur current rating and try to understand and apply new concepts u encounter from editorials n blogs then u should be set.

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

I didn't knew that Blues can also host contest at codeforces.

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

are the problem statement short mainly first 2 as i am unable to do after that in last rounds hope that i will be able to solve a,b,c this time.

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

    As a tester I think that reading all problems in 2 hours is very well doable.

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

      So we can assume there can not be much more than 1000 words in each statement. Very reassuring

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

        Yeah but tbh I've never found a contest where I wasn't able to read all problems. Reading them is not taking a large amount of time compared to solving them, regardless of the number of words

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3 years ago, # |
  Vote: I like it +79 Vote: I do not like it

If Monogon can write a good 3500, which is 900 higher than his rating, then can't SlavicG write a good 2600?

2600 is the upper end of div2 difficulty. I think this will be an excellent round!

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

I have no idea who Zoroland_ is, and how he/she copied my code for A,B,C problems. It maybe because i used idone in public settings during this contest. I will stop using idone at all after this.

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

Motivational speech: The round was originally planned to be a division 3 round so the problems might be slightly easier than average Div 2.

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

    In other words, if you perform badly then you're even worse than you thought.

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

Might be another speedforces coming up!!

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

Ah I see, the tables have turned now!

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

Thanks, and I hope the statement will be short and clear.

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

omg finally ssense ans slavic round, cant wait to participate!

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

Hope to solve problem C this time.

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

I wish I was never born

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3 years ago, # |
  Vote: I like it -69 Vote: I do not like it

Ban radewoosh,he is racist.

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

Anyone noticed that when you move your cursor to testers' handles, you can see text "Tester" or "orz Tester"?

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

    I see 1-gon >> yoyo master.

    This is how it works-
    <a class="rated-user user-red" href="/profile/1-gon" title="monogorz">1-gon</a>

    and without actually tagging the person-
    <a href="https://codeforces.me/profile/1-gon" title="master yoyo yoyo master master monogon monogon master 1-gon orz orz orz testing feature orz" style = "color:red; font-weight:700;">monogon</a>

    Result: monogon

    I hope 1-gon didn't recieve a notification for this.

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

      I didn't get a notification

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

        Thanks! You never disappoint me, my lord. And you are fast.

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

As a commenter , I wish +2 :) SlavicG

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

    I'm also in desperate need of +2

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

I think 1-gon is gonna takeover codeforces completely !

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

This is gonna be my first contest. Wish me luck guys :)

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

    All the best bro! Mistakenly I downvoted your comment so the best I could do is wish you luck!!

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

As a participant, "As a + something" jokes are not funny anymore...

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

    That's not true. "Anymore" implies there was a time when they were funny.

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

      So you're testing if they will keep upvoting those I guess

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

I want to be a tester. But I don't know any of the problem setters, and I don't just want to be friends for benefits just to become a tester. Is there any hope?

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3 years ago, # |
  Vote: I like it +36 Vote: I do not like it

Score distribution says the total score of E is 3000, whereas the score of F is 2000. Does this mean E2 is harder than F? Interesting.

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

The day is not far when MikeMirzayanov will sit and rest and 1-gon and other GMs and LGMs will take over Codeforces ...

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

The person wrote this notice got lots of contribution :)

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

    when you give monogon contribution, he repays you with contribution as well

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

I hope that all those who work hard will get good rating

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

SlavicG Guys just have a look at his profile .... Respect ++ man :)

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

.

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

Im so proud of you guys for making this round !!! Way to hyped for tomorrow :yayy:

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

It's coinciding with WTC Finals ToT

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

Did cf change their rule? I thought yellow was necessary to set problems!

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

I think either D or E is going to be an interesting game problem.

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

    Well done myself!!

    Anyone here who wants to get a prediction of their future cf rating?? ASk me!! (or better, I should write a blog abt it :D)

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

      What about mine

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

        Around 1450 after this contest and then you're gonna reach expert in a month or so.

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

      Can you please predict my future?

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

    :D

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

    thanks for div3)

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

I am curious to know 'who invented orz ?' tag that person :o

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

    who invented orz?

    Answer: orz's parents.

    And if you want to know about them just ask orz. Maybe he'll ask them to join codeforces (just like SecondThread asked his mom) and then you can tag them. ✌️

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

Hello! I would like to propose the next Div. 1 please. I know I am unrated and have not given any contests but please do not judge me by my rating (or the lack of it lol) and give me a chance to write problems for an official contest. Thank you.

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

    Tagging Monogon for better reach.

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

      You should either get the required rating to propose it, or demonstrate that you have sufficient problemsetting experience.

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

expecting my first codeforces contest!

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

i hope i can solve atlest 1 problem todaY all right i solved a and almost b.thank you for this contest author.boosted my confidence

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

Hi All, My rating is 1309, Am I not allowed to appear in this contest.

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

Long Queue?

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

Queueforces :(

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

I have submitted problem A 4 minutes ago.It's still in queue..

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

waiting for my submissions to deque :)

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

In problem A is each appended number also limited by $$$10^4$$$ or any non-negative numbers are okay?

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

In Problem B, can Anton entre the same room twice?

(According to the Visualization, he can't, but the problem didn't say it.)

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

wa2forces...

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

I think this round should be unrated.

The reasons:

  • queue forces.

  • F is similar problem ,it was solved 200+! (include Low level players (like me).).

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

    lol what

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

    The same problem should be banned from such kind of contest!Strongly support this proposal.

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

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

    actually it should be the other way around

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

      How? Its pretty intuitive that the optimal string you use in E is going to be a repetition of prefixes, so just try all of them, felt as easy as A-C at least in my opinion.

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

        D is to realize the "good" and "bad" states — if you just throw out some prime factorizations on paper you'll realize that as long as you have an even power of 2, or at least 1 "2" and another number, it's a good state and Alice wins. (if i get fst, rip)

        Else bob.

        Anyways for E, im bad at cpp so I used java and had a bunch of stupid bugs

        Finally, I think C is the hardest problems :P

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

          Yes, and I brute first 1000 values of D and solved it the same way. But it actually took time to do that. At least for me E1 felt far more obvious.

          C would actually be tricky if they didn't give away both the cases in the samples lol. In terms of difficulty I'd say:

          B < A < C < E1 < D < E2 < F

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

            geniosity orz, thinks B is easier than A

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

              My solution to B was just printing "1 1 n m", sure it took me a tad bit longer than A, but even I found it easier :P

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

            How did you solve E2 , can you just give a basic idea

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

              I gave the rough idea in this comment

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

                Can u please explain how u even brute forced D,I tried but i couldn't.

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

            how did you brute force 1000 values? logic?

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

    I still don't know how you're supposed to reach the solution of D, I just brute forced for small values and observed the general and edge case (and then proceeded to submit only the edge case, not the general one and get confused when it got WA)

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

      I think the general idea is: plot out some initial values, make a conjecture, then try to prove your conjecture.

      Conjecture: For any $$$n \gt 0$$$, the first player has a winning move iff $$$n$$$ is even and not an odd power of 2.

      Proof by strong induction. The base case $$$n = 1$$$ is obvious. For the inductive step, let $$$n \gt 1$$$ and suppose the claim holds for all $$$m \lt n$$$. We can show the claim is true for $$$n$$$ by case analysis:

      Case 1: $$$n$$$ is an odd prime. Then there are no moves and the claim holds.

      Case 2: $$$n$$$ is an odd composite. In this case all divisors of $$$n$$$ are odd. Suppose the player plays some divisor $$$a$$$, and let $$$b = n / a$$$. Then the new value is $$$m = n - a = a \cdot (b-1)$$$. Since $$$a$$$ and $$$b$$$ are odd, $$$m$$$ is even but not a power of 2, so by our inductive hypothesis $$$m$$$ is a winning value. So $$$n$$$ has no winning move and the claim holds.

      Case 3: $$$n$$$ is an even number, but not a power of 2. Then $$$n$$$ has an odd proper divisor $$$a$$$. Let $$$b = n / a$$$ Then $$$b$$$ must be even so $$$m = n - a = a \cdot (b-1)$$$ is odd, and by our inductive hypothesis a losing value. So $$$a$$$ is a winning move for $$$n$$$ and the claim holds.

      Case 4: $$$n = 2$$$. $$$n$$$ is an odd power of 2$ and there are no moves, so the claim holds.

      Case 5: $$$n = 2^k$$$ for some $$$k \gt 1$$$. Then any valid move is of the form $$$a = 2^i$$$ for some $$$1 \le i \lt k$$$. Let $$$j = k - i$$$ and let $$$b = n / a = 2^j$$$.

      • If $$$j = 1$$$ then $$$m = n - a = 2^{k-1}$$$, which by our inductive hypothesis is a losing value when $$$k-1$$$ is odd.

      • If $$$j \gt 1$$$ then $$$m = n - a = a \cdot (b-1)$$$ is even, but not a power of 2, so by our inductive hypothesis this is a winning value and $$$a$$$ is a losing move.

      So when $$$n$$$ is an even power of 2 then $$$a = 2^{k-1}$$$ is a winning move, otherwise there is no winning move, and the claim holds. This completes the induction.

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

    That's obvious coz D:1500 and E1:1250

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

    i was solving e1 and e2 for an hour)

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

lot's of FST's coming :(

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

Weak pretest in E1 — my AC code fails on n=8,k=8 [dcdadcdb]

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

What's the logic for D? I just observed the pattern from a brute-force solution and got an AC without any logic.

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

    Due to induction. For numbers > 9 and not power of 2, you can see that it if it's even then Alice will win else bob. If even number has odd factor, then we can subtract odd factor from it and resultant number will be odd and use answer of previous numbers.

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

      rananjay23 Can you explain a little bit more the logic of D?

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

        I started writing down numbers from 1 to 21 and found following : Bob wins for 1,2,3,5,7,8,9,11,13,15,17,19,21 Alice wins for 4,6,10,12,14,16,18,20

        For even n, if it's not power of 2, then there is some odd divisor say o>1. Hence n-o is odd and we know result of it . Example : n=22, then 22-11=11 and we know who wins if n is odd and less than 21.

        Similar reason you can find for odd numbers > 21.

        For even numbers with power of 2, It's always alternate i.e for 8 bob will win, for 16 Alice, for 32 Bob. Here also explain (just subtract divisors and see) in similar way.

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

    I wrote out a proof here: https://codeforces.me/blog/entry/91835?#comment-805791

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

The pretests for E1 are trash!

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3 years ago, # |
← Rev. 2   Vote: I like it +32 Vote: I do not like it

F is a easy version of https://www.luogu.com.cn/problem/P6185

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

Nice Round. Thanks for this :)

Not good for me. Solved E1 after 4 WA's just because of min/max string silly mistake. Rank dropped from 500 in first half to 1400 till end. And left with no time to try E2

Moral:- Don't submit again and again by changing small part just erase whole solution do that again. Hopefully that silly mistake will be resolved.

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

The problems are cute, but don't you think this is closer to Div2.5 than Div2? My first full solve of a Div2 and its around what I would have got for a fairly fast n — 1 problems in a Div2 normally.

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

    Yess, the authors have mentioned in the above blog:

    The round was originally planned to be a division 3 round so the problems might be slightly easier than average Div 2.

    which is probably why it felt easier than the rest :)

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

Found the OEIS sequence for D in the last 3 minutes. Completed my code 10 seconds after the contest got over due to a stupid mistake.

Now I feel sad.

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

    Wait same, but I got when $$$\omega(n)^{\phi(n)} \equiv 1 \pmod{n}$$$, which is apparantally wrong.

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

      I got this one
      OEIS

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

        That one doesn't seem to be the answer though. I'm not sure but I get that the answer is

        Odd -> Bob win

        Even we have two cases

        1. 2^k with odd k -> Bob win

        2. Otherwise -> Alice win

        This passes the pretest but I need to wait for the system test for confirmation.

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

I didn't get A for most of the contest because I'm an idiot, but this was a fun contest!

For E, what you do is take some prefix and concatenate it until it's at least length $$$k$$$. In E1, you can brute force every prefix. In E2, is the approach some kind of z-function thing?

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

Weak pretests for E1. :( . whereas good pretests for E2 :( .
Test case on which your E1 might fail
12 30
dbdbdadbdbdb
Correct Output
dbdbdadbdbdadbdbdadbdbdadbdbda

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

Was E2 = E1 + suffix array for the reversed string?

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +9 Vote: I do not like it

    I thought to do that, but then I passed pretests by greedy...

    And now it failed. lmao.

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

    I used rolling hash values in combination with binary search to find out which part of the subsegment of the string is different from the prefix string.

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

    Not really, its a simple greedy.

    Clearly as we proceed from the starting of the string (after the first character) its optimal to take any $$$s_i \lt s_0$$$ since it will lead to a lexicographically smaller result than appending the string which will add $$$s_0$$$ again at that point.

    But what about if $$$s_i = s_0$$$? Then we have to check further characters. We can see that it will produce a better answer if in the first position where $$$s_{i + k} \ne s_{k}$$$, we have $$$s_{i + k} \lt s_{k}$$$.

    We can maintain this by just keeping a counter for this k when we reach an equal element and resetting it when we get a smaller character (or breaking if its a larger character). Additionally we have to be careful to store this temporary string during the equal part somewhere else and only add it once we reach this k-th position.

    Code — 119871715

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

      "We can see that it will produce a better answer if in the first position where si+k≠sk, we have si+k<sk"

      What about the prefixes which are ending between i and i+k? Should we not check for them as well and figure out the best prefix in this window? What's the reason behind claiming that repeating (i+k)th prefix is the best option?

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

https://oj.uz/problem/view/BOI13_pipes

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

I often take Virtual Practice on CF and am used to facing a lot of hack tests. So since I saw the small number of the pretests, I have been very worried to be hacked!

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

How to prove for D?

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

I solved E (E1+E2) using 2 Pointer approach. 119902053 Edit :- Got WA on Test 93, Weak Pretests :(

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

    why didn't you just `n=min(n,k);` instead of `if(k<=n)do(k);` else do(n);

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3 years ago, # |
  Vote: I like it -36 Vote: I do not like it

A lot of plagiarism happened in Round; Many posted solutions on youtube. Better to make round Unrated.

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

Missed E2 by 1 sec :(

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

How to solve D ?

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

    i actually wrote a recursive solution and then i could see the pattern of even odd.

    Next i just validated for even odd. Found that for power of two that are odd , answer is false.

    Hope that helps

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

meet a new cheater This is how kedos123 bypasses Plagiarism testing.

I am watching him from so many contest , He has done this today and in previous contest, and I am sure he must have done it multiple times before as well. People like Master_Jiraya are spoiling the sport. I don't understand where would cheating take them in life. They will never get anywhere in life but always remain what they are i.e cheater. He should be banned from the platform as soon as possible . MikeMirzayanov sir pls ban him and skip his solutions .

todays submission[submission:119865102] 119855748 saw his submission time vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++; vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++; vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;vikings++;bb--;vikings--;bb++;

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

What's the idea behind E2? Can it be solved using Z function?

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

    Rolling hash with binary search got me pass the pretests

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

    I used same idea for E1 and E2. I just checked for the character which exceeds any character in the prefix.

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

Was the TL for E2 really tight? My O(N) soln kept giving TLE on case 12.

https://codeforces.me/contest/1537/submission/119909827

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

    Try this input: 50 100 zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfeda

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

Can E2 be solved without suffix array??

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

    I solved it without suffix array. Waiting for the system test.

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

      Can you please explain your solution?

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

        I took a random string. Let's say-"pabccepabcdf" I ran a loop over the string length. At the instant, where the character is greater than the 1st character ("p" in this case), I printed the string till the last character. At the moment, the character is equal to the 1st character, I checked further characters, if any of them is greater than the corresponding character in the prefix of string, I printed the string till the point I found a matching character for the first character.

        For example:- In the example, I have taken, the moment I found the 2nd p, I started checking the further characters with the corresponding character in the prefix. "a" with "a", "b" with "b" and so on. I found that in the d in pabcd is greater than the corresponding c in "pabcc". So, I stored the prefix "pabcc" and printed it required number of times.

        UPD: It passed. Submission

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

          So overall you figured out that "pabcce" is a better prefix than "pabccepabcd" and then you broke the loop.

          But what about the prefixes in between this matching window? "pabccep", "pabccepa", "pabccepab", "pabccepabc" Why are you not comparing with these as well? to figure out the best prefix in this window

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

            "pabcce" is lexicographically smaller than "pabccep".

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

              In this case, yes. But you are not comparing these prefixes in your solution in general(you are just ignoring them). I guess there is a way to prove that all these prefixes will not lead to a better result as compared to the current prefix.

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

                Yes, it can be proved that those will not lead to a better result. I can't give you formal proof but I will try as much as I can.

                It is sure that there isn't a character greater than "p", else we would have used break and printed the loop with prefix just before that point. So, those prefixes must have a second occurrence of "p". The character just before "p" will be surely smaller than "p". So, that prefix will be lexicographically smaller

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

Problem D have some ideas in common with a problem which was part of the recently first round of the Tournament of Towns Mathematical Olympiad

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

D is very similar to MIT PRIMES 2019 G5

Even the names are the same...

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3 years ago, # |
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AE1BE2DC would have been a nice ordering (for me, at least).

I think that's weird, isn't it?

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

Sometimes Java is worse.

problem C . I used LinkedList. Costs me TLE . like seriously dumb Changed to ArrayList and AC . lol

Problem D . Create a set of long for checking powers. Got WA — because i was comparing int with the long values in set Took me hell lot of time to bebug what went wrong

Good problems. Sadly i coudnt perform well :(

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

E is a standard application for Z-array.

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    3 years ago, # ^ |
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    i tried using lps as in KMP, by i guess was missing some cases.

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      3 years ago, # ^ |
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      Try to append z character n times in the given string and then apply the same thing. This worked for me but I used Z algo.

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3 years ago, # |
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My Sol. for B is (1 1 n m) But i see large codes of ppl?

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    3 years ago, # ^ |
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    It's correct,

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      3 years ago, # ^ |
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      how? can you please explain?

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

        The maximum number of steps would be 2*[(n-1) + (m-1)] (first testcase was a hint). No matter what the starting position is, If the yo-yo's are at (1,1) and (n,m), You will always get maximum number of steps.

        Proof-
        There are only 2 ways
        1. (i,j)->(n,m)->(1,1)->(i,j)
        2. (i,j)->(n,m)->(i,j)->(1,1)->(i,j)

        Calculate the number of steps for these, you will always get maximum number of steps.

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

          How did you think of or get 2*[(n-1) + (m-1)]. Sorry to ask but i am a newbie.Also thanks so much for the solution. I get it.

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            3 years ago, # ^ |
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            During the contest, I thought of the following problem-
            In a n*m grid we have to go from point A to B and then come back to A. Maximum number of steps would be 2*[(n-1) + (m-1)] for A=(1,1) and B=(n,m). This is very clear I guess.
            Now if I throw in a point C in between, you can see (1,1)-C-(n,m)-(1,1) will always return in 2*[(n-1) + (m-1)] steps. I then verified the claim against sample test cases.

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    3 years ago, # ^ |
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    same as my solution! it also can be n,1 and 1 , m it will also produce same answer.

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

After bricking C and getting it in the last 4 seconds, read D. These alice and bob problems always get on my nerves. Any suggestions on how to improve on game theory?

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

    Lol same, im not too good in game theory but here i just brute-forced and coded the pattern

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      3 years ago, # ^ |
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      can you please share your core on brute force -><- I tried writing it but i'm not getting the expected output. Edit: got it :)

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3 years ago, # |
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In problem C, what should the answer be for the test case below

Given = 1 1 3 3 6 6

Shouldn't the answer be 3 6 1 6 1 3 , coz the absolute diff b/w first and last is min =0 and also the difficulty level is 3

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    3 years ago, # ^ |
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    1 3 3 6 6 1

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      3 years ago, # ^ |
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      My bad, i didn't count the equality.

      It becomes tricky if the author changes the condition from h[i] <= h[i+1] to h[i] < h[i+1].

      Then for the test case i mentioned, your sequence will give difficulty = 2 whereas mine difficulty = 3. Am i sounding naive ??

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        3 years ago, # ^ |
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        Yeah, I made the same mistake. I was trying to solve the h[i]<=h[i+1] problem.

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    3 years ago, # ^ |
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    3 6 6 1 1 3 or 1 3 3 6 6 1, etc give more difficulty (note that difficult sequence was supposed be non-decreasing and not strictly increasing)

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

My solution for D.If n is odd, Alice will have to subtract some odd divisor of it, so landing on an even number, bob can subtract the same number again and keeps doing this, so bob wins, as he always starts at an even number other than 2. Special case 2^k, Alice can subtract n/2, not allowing bob to copy. If Alice doesn't subtract n/2, a new number will have an odd divisor allowing bob to subtract a random odd divisor and get the odd number scenario, so Alice is forced to subtract n/2 every time, so odd powers of 2 bob wins, even powers of 2 Alice wins.

PS: posted before editorial

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

Why is my solution is still in the queue while other's solution with high submission time got evaluated?

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    3 years ago, # ^ |
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    Yeah same here bro

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    3 years ago, # ^ |
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    not just you,I think will be okay after full system test

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

Did not solve D, but this is my post contest explanation.

  • If the current number is prime, you lose.
  • If the current number is odd (and not prime), you can only go an even number, and the even number is never a power of two.
  • If the current number is even, unless it is a power of two, you can go to an odd number.
  • When your number is even, you want to force the opponent to take an odd number, so that the opponent will give you an even number.

For powers of two there is only one way the interaction can progress.

Otherwise

  • If the first number is even, Alice can stay in even numbers and Bob will only receive odd numbers, and eventually Bob will lose.
  • If the first number is odd, Alice has to concede and let Bob get the first even number (that is never a power of two) and Bob will win.
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    3 years ago, # ^ |
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    "If the first number is even, Alice can stay in even numbers".

    Why would Alice want to stay on even numbers? How do you know what numbers would be beneficial?

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    3 years ago, # ^ |
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    Why does game theory exist :(

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3 years ago, # |
  Vote: I like it -72 Vote: I do not like it

Make this round unrated! Really long queues and horrible protests for E1, E2

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

If you are FSTing E (or worried of FST), try the following case:

3 4
bab

The correct answer is baba.

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    3 years ago, # ^ |
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    It is okay with me if I FST one problem. But when I FST 2 problem variants from the same mistake, it is actually sad :/

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

    while(s.size()<k) s = s+s;

    n = s.size();

    adding this will pass all testcases

    if solution failed in pretests than may be i came to this edition

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    3 years ago, # ^ |
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    What does fst means?

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3 years ago, # |
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This blog had > +700 before the system testing LOL

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3 years ago, # |
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E1 and E2 could've had better pretests. My solution failed system testing after passing 40+ pretests

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

fst in two contests in a row. fml.

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3 years ago, # |
  Vote: I like it +19 Vote: I do not like it
:(
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3 years ago, # |
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greedy forces lol

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3 years ago, # |
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What a nightmare system testing that was xD !!

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

I missed z character in z algo :XD
WA on testcase 93 : https://codeforces.me/contest/1537/submission/119898534
AC : https://codeforces.me/contest/1537/submission/119915549

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

Thanks for the great problems. Really nice round.

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

I think round should be unrated because i have negative delta

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3 years ago, # |
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Nice job guys xD .. But problem was good, no doubt !!
link

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3 years ago, # |
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Was tester's solution wrong? https://www.youtube.com/watch?v=seaskdKf2sI

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3 years ago, # |
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can you guys check why I have wrong answer on test 2 problem D https://www.ideone.com/6MA7rt

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3 years ago, # |
  Vote: I like it +165 Vote: I do not like it

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

To not keep you waiting, the ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

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3 years ago, # |
← Rev. 2   Vote: I like it -14 Vote: I do not like it

Can someone please explain why I got a -20 delta in today's round despite submission for problem-A getting AC ?

Here is my submission: 119905644

MikeMirzayanov sir, please look into this.

UPD: Issue resolved.

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

E1 & E2 can be passed using a greedy approach as well. U can have a look at mine sol which has a time complexity of O(N).

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

Has anyone managed to pass E2 using binary search and hashing. I know in that 2s TL it won't pass as it will be nlog(n) and is giving TLE. But now I saw binary search in the tags. So just curious to know has anyone passed using binary search.

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

Not so sure but it seems like the idea for F is somewhat kinda old.

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

I spent most of the 2 hours solving the wrong problem in C. I thought you had to climb to gain difficulty. And the rest of the contest didn't look so hard...

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

Problem F is nice.

Tried to cover all the observations in my solution video :) Explanation Link

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

I have to say that pretest and finaltest in problem E is too weak and I have seen a stronger problem F somewhere else.

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3 years ago, # |
← Rev. 2   Vote: I like it -20 Vote: I do not like it

Hi everyone!! I am really new to codeforces... But I wanted to know that does cheating occurs in every codeforces round?.... Yesterday, I saw after the contest, that answers to problems A, B, C were posted on YouTube before the contest started.... This is so sad on the cheaters part as well as codeforces that the questions of the contest are leaked out before the beginning of the contest.

Please answer SlavicG mesanu

cheating

leaked questions

MikeMirzayanov , please do something .. Don't let the questions get leaked before contest.

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

    I find it hard to believe the tasks were leaked before the contest. Can you provide some links?

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      3 years ago, # ^ |
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      mesanu https://www.youtube.com/watch?v=WdlsTRQ5n4Y

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

        His ideon code was created 16 hours ago. Even with error that puts it in the timeframe of the contest, not before.

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          3 years ago, # ^ |
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          wow... so you mean to say he solved all 5 quesions just within 15 minutes of the beginning of the contest and uploaded 5 you tube solutions , with links to the code within 15 minutes..?? does it make any mesanu

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            3 years ago, # ^ |
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            I also want to point out that there was a guy guessing the day before the contest that problem D or E would be on game theory.
            It bugged me because i don't recall the same person in any other contest guessing the problems. Maybe he does and I just didn't see it before.

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              3 years ago, # ^ |
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              Game theory is a popular topic and considering we had at least one game theory problem in our last 2 rounds, it's not that out bizarre that someone guessed it.

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3 years ago, # |
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Some one please help in Problem F2, or what are the prerequisite for it?

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3 years ago, # |
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Does anyone know the reason for a 45+min long queue in non-contest time now?

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3 years ago, # |
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You can tell the answer to this test 1 5 4 4 5 5 5 in task C?

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3 years ago, # |
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Can somebody tell me in problem D.Deleting Divisors? if I take n=12 then Alice will do 12-6=6 (as 6 is a divisor of 12), now n=6 and this is bob's chance now bob will do n=6-3=3 now the chance is of Alice and n=3 so Alice will not be able to win then bob's wins the match but the test cases are showing that Alice wins the match for n=12

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    3 years ago, # ^ |
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    Why would Alice choose 6? Players play optimally so Alice can choose to substract 3 from 12 making n = 9 which is losing for Bob.

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3 years ago, # |
← Rev. 4   Vote: I like it +15 Vote: I do not like it

Codeforces Round #726 (Div. 2) — (×)

Former Problem Round — (√)

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3 years ago, # |
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wow very strong pretests on E2

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3 years ago, # |
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Attention!

Your solution 119862059 for the problem 1537F significantly coincides with solutions All_Forward/119856937, royo_sea/119862059. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.ml/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

A code from 2020-03-17 ： https://www.luogu.com.cn/blog/AC-Evil/solution-p6185

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