Rhodoks's blog

By Rhodoks, 2 years ago, In English

Hello Codeforces!

We are glad to invite you to our Codeforces Round Codeforces Round 810 (Div. 1) and Codeforces Round 810 (Div. 2) which will be held on Jul/24/2022 17:35 (Moscow time). This round will be rated for participants of both divisions. Participants in each division will be offered 5 problems and 2 hours to solve them. The two divisions will share 3 problems.

The problems are prepared by me and zxyoi. We hope that everyone will enjoy this round!

We are sincerely thankful for the help provided by:

We tried our best to have detailed, clear, and short statements. I think that anyone can find some interesting problems in this contest. We suggest to read all the statements.

The score distribution will be announced later.

Wish you good luck and high rating!

UPD

For some reason, we have removed one of our problems. So now participants in each division will be offered 5 problems and 2 hours to solve them.

The score distribution is:

Div2: $$$500-1000-1500-2000-3000$$$

Div1: $$$500-1000-1750-2000-2750$$$

We have some more testers now, let's thank them!

UPD2

We adjusted our score distribution slightly.

UPD3: the Div. 1 part of the round is declared unrated.

UPD4

Sorry for the late editorial.

UPD5

Congratulate to winners:

Div1

  1. nantf

  2. ko_osaga

  3. Rebelz

  4. The_Noble_Brahman_Bison

  5. VivaciousAubergine

Div2

  1. csyakuoi

  2. Linkus

  3. yaoxi

  4. RGB_ICPC1

  5. bajablast

  • Vote: I like it
  • -3431
  • Vote: I do not like it

| Write comment?
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2 years ago, # |
  Vote: I like it +40 Vote: I do not like it

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

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    2 years ago, # ^ |
    Rev. 3   Vote: I like it +37 Vote: I do not like it

    Have a look here before downvoting this blog. I don't think Rhodoks has anything to do with whatever happened with div1 E.

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

      I don't suppose you can differentiate it. Maybe Rhodoks deserves it, maybe he doesn't. That's not the point. The point is the contest is flawed, so it's announce deserves to be down voted.

      I think whoever copied the problem already got enough condemnation from codeforces coordinators.

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

      It is sad when someone work with you makes a mistake and drags you down to the water .

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

A question: Why do contest organizers encourage contestants to read all the statements when a huge part of users can't even have an idea how to solve at least one problem in the contest or even understand it?

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    2 years ago, # ^ |
    Rev. 2   Vote: I like it +338 Vote: I do not like it

    The problem is to problem setter what child is to parent. Of course, you are not willing to let some of your children be ignored.

    And I am sure you can understand our statements easily :)

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      2 years ago, # ^ |
        Vote: I like it +60 Vote: I do not like it

      Still the vitosevskich's point is valid. For majority of the users reading the hardest problems before solving the easier ones is a bad strategy, and thus majority of the users won't come to the hardest problems.

      I guess that «We suggest to read all the statements» may well be replaced with «You don't have to solve the problems in the order they are present in the problemset» (which is only useful for those who is writing their first contest, or for everyone else as a reminder).

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      2 years ago, # ^ |
        Vote: I like it +397 Vote: I do not like it

      So are you a kidnapper?

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

        They were just babysitting the problems! stop blaming them for no reason!

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

    there are also instances where for example D < C generally, but only reading C and getting stuck on it would make your delta way lower than what would've happened if you read D and found it easy, getting AC, and so on

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

      instances where for example D < C generally

      which we have a very clear example recently

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      2 years ago, # ^ |
      Rev. 3   Vote: I like it +36 Vote: I do not like it

      It's not only when D<C for everyone. Sometimes, D can be easier than C only for you. For example, I came from an IMO background, so I usually find number theory problems much easier than other problems of the same rating. In particular, in Codeforces Global Round 17, I solved D (number theory) much faster than C, which contributes to my increase in rating.

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

    After solving problems upto my usual comfort level, often I find myself making progress on a problem number that I don't usually solve, but the topic is something I'm strong at compared to others. And whenever I solve it, I feel good that I read that problem.

    For eg. if you're very good at geometry than people at your rating (most/all people will likely have such a topic), if you find a geometry problem at E when you can only solve problems till C, then it's a good bet to read all problems.

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    2 years ago, # ^ |
      Vote: I like it +35 Vote: I do not like it

    You should have your answer now XD

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

Just curious, the number of testers is a bit low? How did you accurately estimate the difficulty value of each problem?

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

    It will increase soon. The testing is in progress and the final score distribution will be decided according to their feedback.

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    2 years ago, # ^ |
      Vote: I like it -25 Vote: I do not like it

    sir why are you impersonating human god mejiamejia???

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2 years ago, # |
  Vote: I like it +150 Vote: I do not like it
As a tester...
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2 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Good luck for everyone ❤

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

Changed my mind. Worst round ever

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2 years ago, # |
  Vote: I like it +52 Vote: I do not like it

Chinese rounds are always frightening, because of the maths involved.

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

    Can't improve if you shy away from Chinese rounds.

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    2 years ago, # ^ |
      Vote: I like it -114 Vote: I do not like it

    Well, it may not be that many math problems in our round. Just give it a try! :)

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

      the "may not be" is killer xDDD

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

        I also think taking the Chinese round will be a big boost and I feel that good thinking skills in maths are pretty good

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2 years ago, # |
Rev. 2   Vote: I like it -44 Vote: I do not like it

w

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

    I think you commented on the wrong blog

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

    I see you're using a macro for "MAX", and you have encountered a well known issue with it. If you expand the macro, you get

    return query(ql, mid, left, mid, 2 * index) > query(mid + 1, qr, mid + 1, right, 2 * index + 1) ? query(ql, mid, left, mid, 2 * index) : query(mid + 1, qr, mid + 1, right, 2 * index + 1);
    

    meaning you recompute one of the queries, resulting in a TLE, an easy fix would be to use the builtin std::max function. Also next time comment on the appropriate blog :p

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2 years ago, # |
  Vote: I like it +131 Vote: I do not like it

Not for nothing I went to math club in 7-th grade

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

2 hours and 30 minutes... then I have to go to bed later

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

    Wait, when does it turn to 2 hours

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

      They removed one of the problems that's why

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

        Oh thanks, I didn't notice that

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2 years ago, # |
  Vote: I like it -174 Vote: I do not like it

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

Waiting for tourist Vs jiangly showdown. Will no.1 change in this Chinese round?

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

Very excited!

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

    Don't worry, such a competition becomes too often, and if the rating was taken away from you, then there is something else. You need to believe for the better.

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

Give thanks to MikeMirzayanov, for He is good; For His glorious platforms Codeforces and Polygon are everlasting.

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2 years ago, # |
  Vote: I like it -65 Vote: I do not like it

so if The two divisions will share 4 problems, can i register for both and solve the common problems and get double rating.

I know this might sounds stupid please don't downvote.

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

    If your rating is below CM you won't even be able to register to a div 1 contest and if your rating is equal or above CM you obviously can't register to div 2 contests given that you are a div 1 contestant.

    Edit: Sorry, I think that CM's actually can participate in div 2s, but not when there is a div 1 round going on at the same time.

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

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

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2 years ago, # |
Rev. 2   Vote: I like it +29 Vote: I do not like it

"We have some more testers now, let's thank them!". Thank you, testers

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

waiting for your Codeforces Round #1919

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

My first div1 OuO hope to solve 1 problem :p

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

    Div1A is just a Div2C so I think you can do it easily as you were able to become candidate master. Good luck!

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

    Now I think I am not lucky.If the round is rated,I will never participate in Chinese rounds if I don't know the author.

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

Good luck everyone!!!!

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

Is div1C same as div2E this time? The score looks like it might not.

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

    I don't think so, div2E will probably be div1D

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      2 years ago, # ^ |
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      looks more reasonable. So they removed this problem from div2E but kept it as div1C.

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

Good luck

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

looking forward to your round

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

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

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

The title description is too bad

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2 years ago, # |
  Vote: I like it +145 Vote: I do not like it

What happened to div1 E?

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2 years ago, # |
  Vote: I like it +65 Vote: I do not like it

If jiangly knows problem E like everybody else, it looks like he is going to be the new number one player

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

    Maybe he knows but won't be top1 this way

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2 years ago, # |
  Vote: I like it +35 Vote: I do not like it

ToroidalForces

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

    These toroidals made me reread the statement again and again

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

      Sad because if they didn't resort to that terrible formula with mods-that-match-the-extents, I might've stayed with my first/correct read of it. I've worked on stuff with toroidal maps (as an option), it's clear enough to me to simply say that the map/grid wraps around vertically/horizontally/both!

      But no... had to wonder why mod was there, so burned time considering possibility of multiple 'neighbors' telescoping out in all 4 directions (because we overwrite common sense with synthetic stuff all the time here!).

      It's the same sort of terrible where the setter described rotating an array using mod-of-index... like, it's clear to those who already know what rotating is and are only worried about left/right... but otherwise, weird-pseudocode-in-text-form shouldn't be a substitute for clear/standard definitions.

      Don't mind me, just adding a layer of salt to glaze my throne of bricks and clown makeup.

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

        I think it would be easier to understand with mentioning that it is just normal side-neighbours, only with wraparound over borders.

        And with labeling neighbours on the picture with "up", "down", "left", "right"

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2 years ago, # |
  Vote: I like it +89 Vote: I do not like it

the contribution of this post will be negative.

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2 years ago, # |
  Vote: I like it +162 Vote: I do not like it

unratedforces

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

Great problems!

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

    The topic span is too large. I feel like I'm in prison

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

    Can you please explain the 4th test case in the sample of Div2C? How is the answer yes for that case? I solved B but couldn't even understand this case of problem C.

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

BruhHHH

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

Problem B in div.2 is very hard. Actually ,I didn't like this round(MY OPINION)!!

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    2 years ago, # ^ |
    Rev. 2   Vote: I like it +15 Vote: I do not like it

    I think the statement was made more complicated than it should have been.

    Video Solution for Problem B and Problem C

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

    It is a good problem. The idea is similar to last div4 question in which we think in a manner "what if we terminate at the current point ?"

    we can terminate only after removing one odd degree. so following are options

    option 1: Remove one min odd degree if it exists and terminate
    option 2: Remove one min even degree node and one min odd degree node and terminate
    option 3: Remove 2 least even degree nodes and one min odd degree and terminate
    

    so on.....

    I couldn't solve as I constrained myself that I should remove only neighbors of last removed node and complicated things and lost so much time and ended up in runtime error.

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

      Option 2 is redundant. It’s always more optimal to just remove the minimum odd degree so you never want to use operation 2.

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

        Yes Option 3 is reductant. Considering 2 least unhappiness even degree nodes is not benificial than one least unhappiness even degree nodes.

        Option 2 of mine is flawed. I constrained myself by deleting only minimum unhapiness even degree node as first node. Actually it can be any even node.

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

          The only two options that need to be considered:

          1. Remove 1 odd degree node.
          1. Remove 2 even degree node that are connected to each other.

          My solution: 165558438

          Your solution probably works too, so there are probably many ways to approach this problem.

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

      Assume total degree is odd (otherwise, the answer is trivially 0), there are really only two options:

      1. Remove a single person with odd degree
      2. Remove two people that (a) share an edge with each other, and (b) both have even degree

      As for why this works, removing a person with even degree will not change the total degree parity (odd stay odd, even stays even), but it does flip the parity of the deleted person's friends (odd becomes even, even becomes odd). Therefore, in order to make the total degree even, you need to either remove a single initially odd person only, or you first remove an even person in order to remove a friend of theirs who was initially even but now becomes odd. There is no other benefit to removing an even person.

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

        Actually we want total edges to be even according to question If edges are even ==> ans is 0 (trivial case) If edges are odd ==> we need to make it even by deleting vertices.

        case 1) so if we remove odd degree node, we removed odd no of edges. since total edges is odd. (odd — odd = even) we are done

        case 2) If we remove even degree node V, we remove even no of edges. since total edges is odd (odd -even = odd) we are not yet done since total edges is odd, but neighbor's of deleted even node gets their parity inverted(odd becomes even, even becomes odd).

        But there might be improvement to option 1 answer as we have created some new odd degree nodes.` Now you can remove any odd degree node and check if it gives best answer.

        There is no need check delete even degree neighbors(after deleting initial even degree we choose) As these are odd degree nodes before deleting the even node V

        Because unhappiness of the any even degree node(after deletion of vertex V) connected to vertex 'V' is greater than equal to minimum unhappiness of all odd nodes which is case 1 answer. so there is no point in deleting the even degree(after deletion of vertex V) neighbors. Hence the current even node doesn't give best answer and it is not beneficial.

        we do the above process for every even node and this is the best possible answer if we chosen current even vertex as initial vertex

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

          Yeah, basically, there is no benefit to performing case 3 at all. Ultimately, the objective is to remove one odd-degree person.

          If an even-degree person X has all of their friends with odd-degree, then there is no benefit to removing X at all. Removing X changes their odd friends to become even, but we're trying to find odd people, not even people.

          Here is a different angle that you can look at it:

          Let's consider whether the optimal solution involves deleting person i. If person i has odd degree, then removing person i is enough (case 1) and we don't need to get rid of any more people.

          But if person i has even degree, then removing person i isn't enough. We need to delete a friend of person i as well, so that person i becomes odd degree. If we delete a friend j with even degree, then delete i and j is enough (case 2). If, instead, we consider deleting a friend j with odd degree, then deleting j alone is enough (and is already covered by case 1), and it's better to delete j alone than to delete both i and j (and maybe even others). So cases 1 and 2 are all we need to consider.

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

Meanwhile Carrot:

image

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

I strongly dislike this round.

  1. B is a graph problem and unsuitably hard. I think it is because they deleted the original problem B and can't come up with a new one.

  2. Unsuitable difficulty. Div.2 D is harder than Div.2 E while Div.1 D is way harder than Div.1 E.

  3. Coincidence. Div.1 E coincides with a well-known problem, which is totally unfair and unexpectable.

I actually hoped that this round will help change the image of Chinese rounds in the community, but in fact, I am wrong.

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    2 years ago, # ^ |
    Rev. 8   Vote: I like it 0 Vote: I do not like it

    As a tester (and first time tester), I agree that div1B/div2D is harder than usual, but I think it's a good problem. I solved div1B/div2D in about 80 minutes, and only 3 of the 13 testers (mostly red and orange testers) solved div1B/div2D in the virtual contest. I advised the problem setter add the 4th example of problem D, to give users more hints. (originally it has only first three examples and time limit is 3s). We also increase the time limit of div2B to make it more friendly to python user.

    UPD: I apologize the problem 1E (I thought this problem is far beyond my ability so I didn't make any suggestions about this problem, I should have googled it). It's OK to complain and down vote this contest, even my comments, but please don't say something bad to the Rhodoks himself. He is a very nice person and take our testers suggestions very seriously, and he didn't mean to make a bad contest (he is not the author of 1E). It's just a mistake.

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

      But what about the coincidence of problem E and the unsuitable difficulty of B? Like, now Div.1 D have only several solves while E have almost the same of C. What's the explantation?

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        2 years ago, # ^ |
        Rev. 9   Vote: I like it +18 Vote: I do not like it

        We apologize for the problem 1E. None of us solved 1E in the virtual test. I thought this problem was far beyond my ability so I didn't make any suggestions about this. Next time as a tester, I should search each problem by google, even if I am not able to solve this problem.

        The problem setter Rhodoks is a very nice person. We made many advices and he took our advices very seriously, we even remove one problem because one of our testers find this problem not "original". I should have suggested him to add one easy problem to make the contest more balanced.

        I think he wants to make the contest better, not make a bad contest deliberately. We tested this round too late to make adjustments. (like create a new problem) It's pretty sad that he received so many downvotes (even if the notorious div1E is not written by him). My suggestion is that do not down vote his solution post or say something bad to the person himself, it's unfair to him.

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

    But why do you talk about problems here? The contest is running now anyway.

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      2 years ago, # ^ |
        Vote: I like it -51 Vote: I do not like it

      I'm not talking about the solutions, just the problems themselves and stating some facts.

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

        Well, is stating some facts allowed? I think it isn't.

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

    B isn't a graph problem. Simple maths knowledge (about sum of even/odd numbers) can be used to solve the problem.

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

    E for easy, D for difficult I guess.

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

    B is not a graph problem I guess. Hopefully my solution is right and I don't get FST :\

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2 years ago, # |
Rev. 6   Vote: I like it +35 Vote: I do not like it

Questions about problems

156039 KroosTheKeenGlint

2022-07-24 18:54:49 Problem E. Two Arrays


(Question)

So should I copy a solution from the comment and paste to solve this problem?

Is it really fair?


(Offical Answer)

Yes


156036 KroosTheKeenGlint

2022-07-24 18:53:22 Problem E. Two Arrays


(Question)

What happened with this problem? I see this is coincidence with some other problems.

Is this contest unrated?


(Offical Answer)

The contest is rated.

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

this blog will get negative delta contribution after the contest , another unbalanced round -_-

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

Unbalancedforces

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

SaikeForces

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

I participated in Div.2. The difficulty gap between Problem A and Problem B is too large. Also, Problem D and E are a difficult problem for most people to solve. The problems are good, but the problemset is disappointing :(

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

Why big coordinates in 2D/1B ? It only makes the problem (way) more painful, and I'd rather go kill myself than writing it.

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

Why did problem E exist before??
What will the authors do regarding this??
Should I copy the code of the problem???

A comment above me asked the authors in clarifications (probably) about this problem, they told him he can copy the code to solve this problem, how is this fair??? and even if it is legal, what about plagiarism check???

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

F**k U.Really shiiiiit round!!!UNRATED!!!!!!!!

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

fku, it should unrated.

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

Unbalanced Unbalancedforces

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2 years ago, # |
Rev. 2   Vote: I like it +23 Vote: I do not like it

...

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2 years ago, # |
  Vote: I like it +428 Vote: I do not like it

Please make this round unrated. Problem E is absolutely unfair.

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

    Second.

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

    Always choose justice even when compared to great positive delta!

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

    Is it available to copy on some secret Chinese server? :) I participated in that OpenCup and managed to remember/find the problem and then the editorial, but it still would take some time to implement.

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

    Is it possible to just make Div 1 unrated (as problem D1E was publicly available), but keeping Div 2 still rated ?

    As for Div 2, the problems were fine, though not balanced.

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2 years ago, # |
  Vote: I like it +163 Vote: I do not like it

Why not unrated?

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

Guys, you discuss the round too much when it is not even finished

Though I take my words back, apparently div1 has real problems /:

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2 years ago, # |
  Vote: I like it +277 Vote: I do not like it

Petition to make this contest unrated

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

Unratedforces

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2 years ago, # |
  Vote: I like it +85 Vote: I do not like it

Please make this round unrated

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2 years ago, # |
  Vote: I like it +113 Vote: I do not like it

Everyone is posting spoilers, and nobody's taking actions for them?

It's very unfair if this gets rated, since just looking at the comments will give you a great hint to solving the problem.

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

    It's acceptable if the same problem already exists, it happens. I can understand that.

    But why do people post that fact here? If you know the problem/solution, just get advantage of it. It's obviously against rules to post them before the round ends.

    Serious actions should have been done to delete them as soon as possible, but why did the organizers just leave them be?

    EDIT: Okay, the problem is even copied, what a shame.

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

Really bad round, D and E are absolutely unsolvable for div2 level. It's obvious by the amount of the submissions, 3000 for C and only 50 for D

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

negative delta :(

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

Can't wait to see negative votes on this post!

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

    It seems that since several seconds after the contest, the votes have been negative and kept decreasing.lmao.

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2 years ago, # |
  Vote: I like it +201 Vote: I do not like it

Please ban whoever decided it's nice idea to link to the problem before the round has ended.

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

If this round is rated, then God really eats shit.

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

I participated in Div.2. There is a dramatical difficulty difference in Div.2 between A/B/C and D/E which made this contest a semi type race. The ranks of people solving three problems are from 60 to 2500 currently so there's the differences between participants are not represented well.

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

fk this round

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

Div1 China-sided round

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

Managed to solve D2D @ hope no one bothered to make anti-python-hash-hacks xD

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2 years ago, # |
Rev. 2   Vote: I like it +215 Vote: I do not like it

When the hardest problem is a well-known problem in China:

Capture

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

    Because many school in China have used this problem for practice.

    So it is well-known.

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

      The 10-th man even wrote a blog for that.

      You could see via this link

      It is under the G.game, you could click the second block under it and see the code.

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

      I've got the info just now.

      Two big mock contest have used this problem, and at least 150 people in China have participate in these two contest.

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    »
    2 years ago, # ^ |
    Rev. 2   Vote: I like it +22 Vote: I do not like it

    Chinese reminder theorem.

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

bruh, D is sqrt decomposition. just no time left to write it

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

    My solution is not sqrt . In fact I'm not sure if it is solvable with sqrt. Large constraints on coordinates seemingly prevent it

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

      oh yea, i should read limits with more attention. sqrt is not going to work (

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

        Could you describe your idea if you think it still holds for small coordinates. Just curious, cannot come up with it even for small ones.

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

So, if a contestant is reading the comments and thinks this contest will be unrated, so he or she gives up solving the problems (especially for the countries that the time is not usual and they probably need a sleep), and that causes the negative rating change, is it fair???

Well, I'm talking about me myself, but I have also learnt about someone else just like me, including both Div.1 contestants and Div.2 contestants.

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2 years ago, # |
  Vote: I like it +81 Vote: I do not like it

哈哈,原题场,down vote!

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

Since many participants may copy E from the Internet, I don't think it fair and rational to make this div.1 round rated.(Though I get +30 to +40)

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2 years ago, # |
  Vote: I like it +183 Vote: I do not like it

Problem E got got 0 solves in the opencup and Suddenly it got 171 AC in the contest.Morever this was informed in the comments as well multiple times.The contest just became finding the right code online in time and should be unrated.

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

Please make this round unrated or I will eat shit.

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2 years ago, # |
  Vote: I like it +1133 Vote: I do not like it

What the fuck?

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

This post will get more negative rating than my negative delta today. B statement was weird. C was more like "You know then ok" Idk if it actually is a good problem. Couldn't solve C this time. :( IDK if it's me or this contest was not as good as usual contests.

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

Unrate this round

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2 years ago, # |
Rev. 2   Vote: I like it +73 Vote: I do not like it

rnm, 退钱!

fxxk, refund!

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

    退钱退钱!!!推rating!!!

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

The author of this contest only know copy?

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

me: Annoying so many 114514 ... => Oh, D1C is amazing, I love it! => WTF D1E, why more than 100AC???

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2 years ago, # |
  Vote: I like it +52 Vote: I do not like it

lmao what the fuck

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

jls is going to be angry for 1E.

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2 years ago, # |
  Vote: I like it +188 Vote: I do not like it

Fun fact: E's Sample and the Format is the same as a "similar" Problem.

Similar Pro:

E:

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

BIG difficulty gap between D1A & D1B

and FUCK D1E

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

Nothing else but shit.

Unrated, plz.

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

OK, I know that Div1E is well known and I was implementing it with the help of the editorial, but I couldn't find the code online. How does it come that 171 people solved it? I wouldn't expect that many even if the statement directly said "It's problem G from GP of Bytedance 2019"

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2 years ago, # |
  Vote: I like it +81 Vote: I do not like it

Opinion: I don't think that the round must be unrated. I would certainly prefer it to be unrated for selfish reasons, and I was angry that E is copy-able or at least googlable and even left the round midway. But I don't think that having a well-known problem is enough to make a round unrated.

Also I don't understand why did people post the link to the old problem in the comments during the round, it feels like a deliberate attempt to make the round unrated.

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

    Facing a well-known problem and knowing it is really well-known and finding that many cheaters are copying the code can make me annoyed during the contest and get a bad performance.

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

    Well, I personally think it should be unrated in the case that the problem was intentionally copied from the other source.

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

    I can understand the reason: they just misunderstood that the round had already been completely ruined (but actually it hadn't until the link was posted). Of course "deliberate attempt to make the round unrated" might be another reason.

    It made the situation even worse that during the contest people posted comments in the article https://codeforces.net/blog/entry/65510 so it appeared in Recent actions. Contestants who try to look for the problem using Codeforces could accidentally notice it.

    IMHO the situation is similar to the case where malicious people would have done DDoS or whatever to ruin the contest. People who gave information about the problem need to be punished to some extent, and some people (including me) insist the round should be unrated.

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2 years ago, # |
  Vote: I like it +86 Vote: I do not like it
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2 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Serious?

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

Ggs, gonna lose about 10000000 points

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

thanks for the fast Tutorial on problem E

I see a lot of solutions for it in comments when it's still running ... xD

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

Trash round and unrate it please.

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

These problems were way too hard, I underperformed by a long shot.

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2 years ago, # |
Rev. 2   Vote: I like it +23 Vote: I do not like it

This is the worst Round I have ever participated in.

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

The contest should be unrated because Div1.E is an old problem and can be found by https://www.acmicpc.net/problem/23679.

The tutorial was public on China before finishing the contest. https://www.cnblogs.com/Flying2018/p/acmicpc2874.html

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

    Well, it is probably a reason for making div1 unrated (though they could just skip copy-solutions). But it should not affect div2

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

Over 100 contestants passed problem E, so this contest must be unrated.

I suggest to ban the discussion of original problems during the contest. With this rule ,contests like this will probably still be rated, because only few people know the original problem, and the result can show the skills of contestants.

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

This contest was very good, my family and I hope you don't play again in the forever!

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

Is Div2 unrated?

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

    Why would it be?

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

      Because Div1 must be unrated,i don't know what will happen on Div2

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

        It would be stupid to unrate div2 because of problems that concern only div1

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

          You have a great result today, would be unnice to unrate it :/ You might get nicer color.

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

            Yeah, if I'm not hacked and the round is rated, I could get back to violet for the first time in years.

            Hacks is a separate story. Since I solved D in python and there is no std::map in python, it can be hacked with some antihash. Knew about it during the round but didn't bother to protect. I just hope that hackers didn't bother about it either

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

I wanted to write normal round and became CM. And what is it? Speedforces and Div 1E is not new tasks. I hate this round.

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2 years ago, # |
  Vote: I like it +137 Vote: I do not like it

Div. 1 E is a total failure. Participating above my average level and then seeing everyone just copying the solution of the hardest problem in the contest from some Chinese server (and thus beating me) completely ruined my day. The round indeed must be unrated. The testing/coordination of the round is done poorly, such coincidences must not happen, it's an abominable situation.

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

In China,SaiKr Round 2 (Div. 1 + Div. 2, Rated)

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

There should be atleast one strong Chinese tester in each round.

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

very balanced div2 lmao

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

I don't know why this happened, I was in jail for an hour, what a fuck!!!

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

You can see a bunch of Chinese at the top of div1。funny wow!!!f**k unratedforce

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

https://blog.csdn.net/qq_35577488/article/details/117813076
Problem 1C also can be found on the internet.
This Blog is published on 2021.8.20,with the problem statements and code.
Well, The restriction differs but doesn't matter

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

Trash round. I think nearly 90% of people who pass div1. E are looking at the solution. The examples are the same! So trash.

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

Div 2B looked like a graph problem, not being good at graph found some way to solve it without graphs, passed pretest might fail main test. Wish i had gone for c first, overall i feel this contest was really tough

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2 years ago, # |
  Vote: I like it +37 Vote: I do not like it

I'm not expecting this result since you've already removed one problem. :(

PLZ make this round unrated.

By the way, the previous problems (1A, 1B) seems to be good, but problem.E really broke our contest experience.

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

    1B was good? seriously?

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

      Maybe it's a little harder than normal, but since this round is a 5-problem round, I believe that it's at least acceptable.

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

If the round didn't get unrated then atleast delete problem E from calculations of score, Its unfair for a well-known problem to exist as a new problem!

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

    Just wanna throw in, seeing so many people solve div1E (more than div1B) made me invest all my time in E instead of B. Removing E from calculations won't give this time back and I guess others did so too. Just wanted to point this out.

    Either way I agree, I think this round should be unrated [or at least E should be taken out].

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

I hate this round

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

Imagine running plagiarism checker on tens of thousands of submissions for hours, when the real plagiarist turns out to be the problemsetter :(

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

This is the best round I have ever seen, I can hardly imagine a round with a perfect balance and difficulty, the level of the authors is quite high, all levels of coders were able to get a perfect round, I felt physically and mentally happy when I played this game.

The questions in this cf were very interesting and I learned very many meaningful tricks from them, the difficulty slope was very reasonable, the sample coverage was very wide, and I even got a pass on the sample that only made the code pass.

What I admire about the author is that he has the courage to submit this kind of contest for review. If I had come up with such a topic, I would have been ashamed, but the author is open and honest, a real gentleman, he is the best courageous person I have ever met, bar none.

When I clicked on the leaderboard of the contest, I even wondered if I had clicked on the rating list. other low quality contests had purple and grey in the leaderboard, but in this contest, purple, blue, cyan and green were clearly defined, which made me admire the author from the bottom of my heart.

Finally, I wish the problem setter a long life, a happy family, good health and a speedy recovery from the loss of his mother.

by sora1336 https://codeforces.net/blog/entry/93538?#comment-827037

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

    You should not say anything, about someone's mother, RESPECT EVERYONE

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

o-o I don't think I've ever seen a contest get downvoted so quickly.

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

CN Round qwq

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

Div2B, I actually do not get it. Looking at the codes it seems simple, but I do not understand why they work. What observation do I miss?

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

    if number of pairs is even u can call everyone right? if it is odd u will have to remove 1 pair, so it becomes a greedy as for each pair we check whether to not call first one,second one or not call both.sorry for my bad english.

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

      Why does removing 1 or 2 persons garantee that the remaining number of pairs is even?

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        2 years ago, # ^ |
        Rev. 2   Vote: I like it +5 Vote: I do not like it

        If the person appears odd number of times, then it is obvious why it works

        Now what if there are no such people. It means that if you take a pair, both participants appears in pairs even number of times (but one of them is shared!) so that means that if you remove both, you will remove odd number of pairs

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

        If a person has an odd number of friendships, then removing that person removes an odd number of cakes. Therefore, if we started with an odd number of cakes, we now have an even number of cakes.

        If you remove a person with an even number of friendships, you will still remain with the same parity of cakes HOWEVER each friend that had an even number of friendships will now have an odd number of friendships for which you still have cakes to make. And then you're back to the first case (one person with an odd number of friendships).

        There is no other potential optimal move if the initial number m is odd. If you were to remove one that is initially even and then remove one that was already odd before the removal of the first one, then why did you not remove the odd one and leave it at that? Makes no sense either to remove two that don't have a friendship and are both even-friended.

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

          really appreciate!

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

          Why can't it ever be the case that we remove 3 people?

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

            Because you can always either just remove 1 of those 3 people(if there is one with odd degree) or 2 of those 3 people(if all have even degree and there is an edge between those 2), otherwise you wouldn't be able to remove those 3 people, and removing only some of the 3 people is obviously better than removing all 3.

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

    If M is even, you don't need to delete any node. Else, you either delete a single node having odd degree or 2 nodes having even degree which are neighbours.. Can you explain for Div2c?

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

    If edges count is even, answer is 0. Otherwise, if the optimal solution has an uninvited person with odd adjacency count, it is enough to not invite that person alone. While if all the uninvited persons in a solution have even adjacency count, then at least 2 of them must be adjacent because otherwise the deleted edges count would have been still odd, so it is enough to not invite such 2 persons alone. So the is answer is minimum(minimum unhappiness of a person with odd adjacency count, minimum unhappiness sum of 2 adjacent persons with even adjacency counts).

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

    Hey!

    If m is even, the answer is 0, because we can invite everyone and we will have an even number of eaten cakes.

    If m is odd, then let's try to invite everyone except some guys. We have two options:

    Option 1. Skip one guy:

    Let's take a look at the number of friends that each club member has. If we decide to not invite someone who has an odd number of friends, then we can try to not invite him and get an even number of eaten cakes.

    Option 2. Skip two guys:

    Here we want to remove 2 guys in such a way that after removal there will last an even number of friends invited. Let's consider two friends, such that both of them have an even number of friends. Because both of them have each other as a friend, after the removal of these two guys we'll have an even number of eaten cakes.

    Let's choose the best answer.

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

ABC is solvable, but DE... I think there is a big difference in complexity between C and D.

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

i understand and agree on the fact that contest was not good because of many reasons,but still the kind of comments that we are posting is not at all healthy from a community point of view,so lets give feedback's but in a constructive way as everyone is a human here.Hope u understand my point:).

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

Sparky_Master_WCH1226 is about to lose their bottom contributor spot for a totally unexpected reason (for me at least).

Update: Sparky_Master_WCH1226's bottom contributor spot has been overtaken by the round author.

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

    I want to take it back!!! It used to be me!!!

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

    Why is Sparky_Master_WCH1226 contributing so low and what are the reasons?what did he do wrong?

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

Shame

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2 years ago, # |
  Vote: I like it -35 Vote: I do not like it

I attribute this failed attempt to reaching Master to Div1E. So close yet so far.

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

what's wrong with my solution for C ? i thought that it's always optimal to color the grid either horizontally or vertically but getting WA on test 2 , Here is my code
165577141

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    TEST1
    TEST2
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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You need fill at least two adjacent lines with same color, otherwise you would get only two toroidal neighboors.

    Example:

    1121
    1121
    1121
    

    Is not nice picture

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

      isn't filling rows or columns greater than 3 enough to produce a beautiful grid ? or did i misunderstand what toroidal means ?

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

        I dont understand question
        I meant that your code would print yes on that example:

        1
        3 4 2
        9 3
        

        Though answer must be no

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2 years ago, # |
Rev. 2   Vote: I like it +66 Vote: I do not like it

 Is it unrated?

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

can some 1 now plz explain what were the toroid neighbours ?? i couldn't really fully understand what for a cell (x,y) which cells were supposed to be its neighbours?

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    2 years ago, # ^ |
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    If some neighbour is out of the matrix, in that case we pick the neighbour from opposite side, like for n = 5, m = 6, for (1,1) the toroid neighbours will be (1,2),(2,1),(1,6),(5,1).

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

Sorry, I cannot find the button to downvote this COPY round. Can any one help me? >_<

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2 years ago, # |
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Was div2B a standard problem? I just couldn't get the idea.

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

Grass(cao), I can't finish 2B.

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

Div. 1 contest should be unrated. But why should Div. 2 contest be unrated?

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    2 years ago, # ^ |
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    Yeah, no particular reason (although D2E is a copied problem as well but it didn't influence the overall quality of the problems). I liked the round tbh, just want to know whether my memory-optimized D would pass at this point.

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

copy round !down vote!

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

My grandma could make better round than you

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

It took me almost 1 hour to understand B.

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

Why D2B so hard? Well,maybe it is only hard for me.(

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2 years ago, # |
Rev. 4   Vote: I like it +18 Vote: I do not like it

This guy's gonna break interlude's record on the most downvoted blog soon
Update: He did it guys
Update 2: Well I actually feel bad for him now after finding out that he's not responsible for that particular problem. Remember to give him an upvote.

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

Please make this round unrated.

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

New worst contributor!(

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

Can anyone explain the problem Div2.C? I tried to cover the matrix horizontally and vertically but my solution fails at test case 2.

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    2 years ago, # ^ |
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    maybe consider that each color tile at least covers 2 rows(or columns)?

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    2 years ago, # ^ |
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    Have you check if the number of row/column is odd? if any of the number is odd,you should check if 3 consecutive vaild rows/columns can be filled.

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      2 years ago, # ^ |
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      did all of what you are saying, I think I am missing some cases during implementation somehow

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

        I think there is a problem in this part: // calculated the total number of rows that can be colored in groups of 2 for (ll i = 0; i < k; i++) { rows_colored += 2LL * (can_color[i] / 2LL); if (can_color[i] > 2) { if (can_color[i] % 2) odd = true; else even = true; } } For example,can_color[i]=5,you can color 5 consecutive rows/columns since it is valid instead of 2*(5/2)=4 rows/columns.

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

          that is getting managed with the help of odd=true assignment.

          Update: got the mistake, such a stupid mistake :(

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            2 years ago, # ^ |
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            you can use this testcase to think about the question: 2 3 10 3 6 9 15 4 10 3 8 12 20 you can fill 2/3/5 columns for both of conditions.In your code,odd and even will be labeled "true".However,when using 'bool by_cols = row_wise(m, n, pigment);'rows_colored=8 and returned false;

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2 years ago, # |
  Vote: I like it +219 Vote: I do not like it

Do someone succeed to find div1-E on the internet with your own power?

I want to know the method and searching technique for the future

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

C(div1) and E(div1) are not original, and C appeared in a contest in China last year.

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2 years ago, # |
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I think it's the worst round I've ever played

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2 years ago, # |
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I think the B and C should swap in div2.:(

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    2 years ago, # ^ |
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    then also it won't be a reasonable problem for B.

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

I just discovered new algorithm! It is very efficient and easy to understand. It's name is: Finding solutions online Algorithm.

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

...

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

Although I'm so stupid, but I think questioner is stupid too

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

Extra 5 minutes on A cuz of translation misunderstandings... I only understood the problem after reading it in English... Prolly the worst contest of the recent ones... Plz next time work more on balance + russian translation. You tried your best, but it was not enough. Gl next time!

No offence, btw. Nice lil round, if you consider this being their first round ever. It is already commendable, that you've tried. B was really hard for Div2B, but i was so surprised, that my solution passed, so i liked it :D

Also, was the "We will delete at most 2 vertexes" the observation to be done, in B, i mean? If yes, then i really impressed by my graph knowledge, hahaha

UPD: Can't imagine the author's motivation right now... So much hatred down the comment section... Just summarize all your problems, solve them, and become better problem-setters in the future! Thx for round (I think this will be the only thanks, you recieve for this one...)

UDP1: Sorry for so many UPD's, but is the next observation, that if we delete 2 vertexes, then they will be connected? This makes implementation complexity O(n + m), not O(n * m)

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    2 years ago, # ^ |
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    I think this will be the only thanks, you recieve for this one...

    I don't think so. Thanks for the round!

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2 years ago, # |
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Why?? Just why??

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

Most downvoted blog in Codeforces history?

(btw ugly implementation-heavy div2D)

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    2 years ago, # ^ |
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    Can you give me some hints for Div2D / Div1B

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      2 years ago, # ^ |
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      Build the chart (like in the picture of the statement, can be built for O(n))

      Then examine conditions a point has to satisfy (you don't need to iterate all points or use sophisticated data structures)

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        2 years ago, # ^ |
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        I found out how to build the "chart", but could not find a condition for each point...

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      2 years ago, # ^ |
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      Firstly, notice that you only need to check whether there is a rain point with the height > $$$m$$$, not the whole number line.

      Calculate the initial height (i.e. the height before any rain point removal) of each point. This can be done by sorting the points by their coordinates, then for each point $$$i$$$ update the heights of the neighbouring points which are affected by point $$$i$$$ (I did binary search + difference arrays).

      To see if removing an arbitrary point $$$i$$$ will prevent a flood, you'll need to find the maximum of the heights of the rain points after the removal of point $$$i$$$. Yet again use the sorted array of rain points and calculate how the rain points to the left of $$$i$$$ would be affected and for the ones to the right of $$$i$$$ do the same. I did all of this with some formula trickery, lots of binary search and a data structure allowing for finding the maximum value in a subarray.

      (Maybe I did a bit of an overkill, but I couldn't find an easier solution)

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        2 years ago, # ^ |
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        But isn’t in the last test case

        6 12
        4 5
        1 6
        12 5
        5 5
        9 7
        8 3
        

        On point with x 9, rain value will be 12, which is not more than 12, but removing this point will remove flooding, as answer is

        100110

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

Actually the comments available during the contest may be a bigger issue. Even if C is the same problem, a reasonable amount of people solved them; but E gets significantly many solvers because there is a link to the sol during the contest. I really don't think people will search for E explicitly without noticing the comment that this is a duplicate problem.

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

Out of curiosity, what was the initial placement of the problem that was taken out?

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2 years ago, # |
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Please do us all a favour and make this round unrated MikeMirzayanov

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    2 years ago, # ^ |
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    Why would Div. 2 be unrated? Stolen problems didn't make contest much worse.

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    2 years ago, # ^ |
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    The round is unrated for Div1 as conveyed here.

    You are demanding this favour because you'll get a delta negative.

    Meanwhile, they removed one problem from both the divisions because it was "well-known" to copy-paste another "well-known" problem.

    Nice

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

can't have shit in china

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2 years ago, # |
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When I think the difficulty of div2 D&E is vertically high, I found that the difficulty of div1 E is also a same verticality (but is "vertically low"). I feel balance in psychology:)

I really like the sentence that is "What the fuck ?" spoken 30 minutes ago from jiangly.(that's a joke hh)

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


Seems it'll be unrated :D

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

btw why the comment feature is available during the rounds?

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    2 years ago, # ^ |
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    I think both talks and comments should be banned during an official contest.

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      2 years ago, # ^ |
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      I think it shouldn't be disabled for those non-participants, but, if so, people can create a new account/use another account to do the same thing.

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

Why is systesting still pending?

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

I think sleeping during this round is better than taking parts in this fuck round.

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2 years ago, # |
  Vote: I like it +3 Vote: I do not like it
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    2 years ago, # ^ |
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    Yeah, just a little different on constraints.

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    2 years ago, # ^ |
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    Yes it is nearly the same.

    Compared with div1E,it is ____ XD

    And it is also just a common digitally DP problem,and the number of solvers is not unjust.

    So it is clearly to see how trash the round is that this problem is even not big compared with div1E.

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

I wasted my 2 hours by joining the round instead of sleeping, now I am here to get free upvotes

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

this contest should be unrated!!!!!!

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

CQXYM & interlude (Round #745): Good problems (except B), bad round.

Rhodoks & zxyoi (Round #810): Bad problems, bad round.

The problemsetters suck.

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

Is it another water235?

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    2 years ago, # ^ |
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    Sure.LOL

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    2 years ago, # ^ |
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    what does water235 mean? can you please tell?

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      2 years ago, # ^ |
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      water235 is one of the most famous cheater in china.

      water235 likes copying problem from Codeforces Gym, just the same as this round.

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      2 years ago, # ^ |
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      water235 is a "great" teacher who told the students what problems they would meet in contest before that contest began.

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

If a submission turns out to be plagiarism, it is skipped.

Similar, if not higher, standard applies to problemsetting. If a problem turns out to be plagiarism, the contest should be gone.

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2 years ago, # |
Rev. 2   Vote: I like it -17 Vote: I do not like it
nice round
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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it
DIv. 1 contest is unrated.
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2 years ago, # |
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I would have a good sleep if I wouldn't take part in this round(

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2 years ago, # |
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It would be fair to make div1 unrated,of course.But div2 was a nice round,why make that unrated? I mean people who participated in div2 really deservesc their rating.

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2 years ago, # |
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I said that this div1 was very bad. Although I didn't have a chance to play div1, I can fully understand the harm that this kind of cheating or unfairness will bring to those high-end players! Strongly recommend not rating!

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

I felt that the difficulty of the div2 A~C questions in this one was appropriate, it was the later questions that were too difficult.

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    2 years ago, # ^ |
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    I have an idea why this happened:

    Earlier in the contest announcement it said 3 problems would be shared Division 1 and 2 and that each contest would have 6 problems each. But then after an update 1 of these shared problems had to be removed. I'm pretty sure what happened is that Problems D and E in this contest were supposed to be Problems E and F before, and there used to be a Problem D in between. But it got removed, and it caused Division 2 to go full speedforces mode with the diff spike from C to D.

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      2 years ago, # ^ |
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      Now that I think about it.

      Could it be that the removed problem was also deliberately copied...

      Paranoia mode on

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2 years ago, # |
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thank you problem copiers for letting me waste 2 hrs on your shxtty problems.

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2 years ago, # |
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From now on everyone should try to google problems, not solve them. It's GoogleForces now

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2 years ago, # |
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This post have more negative rating than my highest rating multiplied by -1. XD

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2 years ago, # |
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I think if you had a couple more chinese testers, things might have been different. Maybe having testers from different parts of the world (mostly where CP is rather advanced), e.g. ensuring diversity, would solve such problem for the future.

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2 years ago, # |
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Lets make this most downvoted announcement, and my comment as most downvoted comment.

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2 years ago, # |
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zxyoi !orz

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2 years ago, # |
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xi jing ping paid them to do this.

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2 years ago, # |
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In fact, this is not the only round that I can find the same problem that I have done before. In the last round, Education Codeforce Round 132, I found that I have done problem C last year and problem D just this week. So I just use the code that my friend and I have wrote before the contest and the AC code we have see in a blog,(we even have wrote another blog to talk about the problem before the contest!), and get the system message that said it was a violation of the rules since we use the same code. But that was because we have done it before the contest! The rules in http://codeforces.net/blog/entry/8790 said that use the code that have been wrote before the contest is allowed, I comply with the message to plead a miscarriage of justice against me, but no one reply, There was no one to deal with it. So,

the system message:“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.”

the rule:“Solutions and test generators can only use source code completely written by you, with the following two exceptions: the code was written and published/distributed before the start of the round, the code is generated using tools that were written and published/distributed before the start of the round.”

They look useless!There's no administrator to reverse my misjudgment. no administrator tell me that if I need to Provide more evidence. Even if you insist on misjudging me, please let me know to Provide more evidence to prove that!

http://codeforces.net/blog/entry/8790 http://acm.hdu.edu.cn/showproblem.php?pid=4915 http://t.csdn.cn/5qwDN https://www.acwing.com/file_system/file/content/whole/index/content/6131167/

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

Let's discuss the problems for a change.

How to prove that it is enough to consider columnar/rowwise coloring in D2C/D1A ?

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    2 years ago, # ^ |
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    because u can prove if u want to maintain that toroidal property u need to fill complete row/col and also have atleast 1 more adjacent row/column (respectively)/

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    2 years ago, # ^ |
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    each node can only chose a direction to Switch to other color,if a node use a direction to switch to a different color, then for others, they can only choice the direction that Parallel to the others.therefor, it is a problem that only need to think with row or col

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    2 years ago, # ^ |
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    Either suppose you can fill the whole matrix with one color or there will be two adjacent cells with different colors. If the one is below the other it results in rowwise coloring, else it results in a columnar coloring (you can see this by drawing two adjacent different cells and trying to fill the blanks).

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    2 years ago, # ^ |
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    Sorry if this is rather long, maybe someone can help shorten it :P

    proof
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    2 years ago, # ^ |
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    Make these two observations,

    1. If two cells of the same color share one corner then the other two squares in the $$$2 \times 2$$$ grid must have the same colour.

    2. If you have a rectangle of same colour cells, pick some corner then atleast one of the non-rectangle cells adjacent to it have to be of the same colour.

    Diagrams

    Now you have to start with a tetris T-shape block of cells of same colour, and from 1. and 2. are repeatedly forced to extend it into a larger rectangle, being able to stop only once an entire row/column is covered by the same colour.

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    2 years ago, # ^ |
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    Consider two toroidal-neighbors of different color. They must have exactly 3 neighbors with their color, which is T in some orientation. So, there are two T 'centered' at those cells in opposite orientations. But this leads to two other pairs of toroidal-neighbors of different color, same applies to them. This proves that we should have at least 2 columns (rows) of one color and at least 2 columns (rows) of other color from every two toroidal-neighbors. Then, it's easy to show that rows and columns can not be mixed together using proof by contradiction.

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2 years ago, # |
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When tutorial???

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2 years ago, # |
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Only five seconds of reload this blog

Spoiler
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2 years ago, # |
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can anybody tell me why my code keep getting TLE on test 2 for question C as it is O(k) order so it should pass the test https://codeforces.net/contest/1711/submission/165596272

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    2 years ago, # ^ |
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    The order is not O(k). For example, suppose there are only three rows and m columns. Your algorithm would fill up two columns at a time, which is only six cells, so it would take (a_i / 3) steps just to deplete color i. Since a_i could be as high as 10^9, this is not considered to be in O(k).

    It would be much faster if you simply performed division to find out how many rows/columns a single color can fill.

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

Please Don't Down vote this blog. As Rhodoks has no involvement with Div 1 E.. We all should stand by him in this situation. ❤️

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2 years ago, # |
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Lol this will have -(tourist rating) downvotes

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2 years ago, # |
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Damn, compare to AOPS community (for math olympiads), the toxicity of this place is just insane

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2 years ago, # |
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Unbalanced round

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2 years ago, # |
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How to check whether the colouring is possible in C? I observed that a particular colour strip should be at least of width two but. I wasn't able to figure out the rest of the solution.

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    2 years ago, # ^ |
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    solve separate for horizontal/vertical painting

    for each color get max width = a_i / {n or m}, sort, go from top to bottom, win if in any moment

    sum — cnt <= {m or n} <= sum

    where cnt — how much can we erase from max width = sum(a_i — 2) — any color can be narrowed down to 2

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      2 years ago, # ^ |
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      Don't have to sort. Just keep a flag for if you are encountering a pigment which can form a strip of width at least 3. This will help in the case you have only one strip of width 1 left uncolored on the board.

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        2 years ago, # ^ |
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        it is not necessary to sort, but it seems to me more intuitive and easier to understand

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    2 years ago, # ^ |
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    That was actually pretty much it. Just see if you can fill up the rows OR columns while ensuring that each color covers at least 2 rows/columns. You do need to be careful when the number of rows or columns is odd though, since you can't have every color filling up only an even number of rows/columns then.

    Given $$$a_i$$$, you can use it to fill rows if $$$a_i \geq 2n$$$, and the number of rows it can cover is $$$\frac{a_i}{n}$$$. You can fill columns similarly. Filling an odd number of rows/columns additionally requires that at least one color can fill an odd number of rows/columns (i.e., that color can fill 3 or more rows/columns).

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

STOP downvoting this blog , Rhodoks doesn't deserve this

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2 years ago, # |
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Screenshot-20220724-200142-Chrome.jpg

ChineseForces

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2 years ago, # |
Rev. 2   Vote: I like it +16 Vote: I do not like it

Hell yeah, back to violet

It's been two years damn

Back then it was a contest with a lot of FST. Now a contest with a notorious coincidence. Lol

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

    You got your color back, and I finally turned purple for the first time) Cool

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2 years ago, # |
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Thank you for the contest, I finally reached CM!

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    2 years ago, # ^ |
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    Congrats !

    You go straight from pupil to CM in less than 5 months after a long time in newbie, may I ask how did you practice ?

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      2 years ago, # ^ |
      Rev. 4   Vote: I like it +27 Vote: I do not like it

      Well, i cant really know what helped me, but i can recommend you:

      • Resources
      1. Read this a couple of times and practice the topics from there
      2. This is really good book
      3. This is good website, I visit it very often
      4. Use this to find out which topics are your weak areas and practice them
      • Practice
      1. Practice. You can use the codeforces problemset for example (I think it doesn't matter where you practice most of the time, the codeforces problemset is the best for most people)
      2. Run some virtual contests to see what level you are and practice solving contests (I only ran a few, dropped half in the middle. That was enough for me to figure out what I was doing wrong)
      • Contests
      1. Participate in ALL contests
      2. Upsolve problems that you can solve
      3. Read all the editorials. If you see new technique — google about it, if there is nothing you dont know, solve problem
      • Some tips
      1. Practice Ad-Hoc problems
      2. Visit the codeforces EDU and some competitive-programming-related youtube channels
      3. If you see that some property was noticed in the editorial, remember it
      4. Always google the best implementation for data structure/algorithm, if there is good implementation in the editorial — use it, practice implementations (just start writing everything from scratch)
      • What I am going to do next
      1. I will practice more
      2. I am going to learn some topics from here
      3. I am going not to give up

      PS: Sorry for bad English
      PPS: Hmmm, i think i should write blog about it when i will reach 2400+ XD
      PPPS: Pls, upvote me if you can

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

Will there be an editorial?

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

First time seeing a top ranked guy cheating.

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

    I don't know if he cheated or not and I don't have telegram, but this guy really sucks at cheating... He is a newbie and only solved A and B in this contest. Also, he used his own (somewhat strange) template, the one he's been using for a long time, so it's unlikely that he "directly copied" anything.

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

a post with -2573? Can we maybe not shoot the messenger? Sucks that the contest was not rated, sucks that some people are dicks, but don't be being angry at the bringer of bad news.

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

You are not a real troll.

You're not a real negative contributor!

Downvoted on Home Page? An announcement blog? What a joke. I worked my ass off to get where I am! And you take these shortcuts and you think suddenly you're my peer? You do what I do because you were framed by the other author and you can make people blame you unjustifiably? I committed my life to being obnoxiously intrusive! You don't slide into it like in a span of 5 hours and reap all the downvotes!

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

CopyForces

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2 years ago, # |
  Vote: I like it -32 Vote: I do not like it

I hate any Chinese Round

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

-99:(

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

In problem Div2C, I cannot understand why I get a WA, can anyone provide a sample for me? https://codeforces.net/contest/1711/submission/165612985

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

div2B is hard, I don't like this round :((

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

is the editorial available? Can anyone plz give me hints for Div2D (Rain).

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

    One hint is that if flooding occurs at all then it must occur at some position where a rainfall was centered.

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

    Same here. Actionally there is editorial (once get there, use ctrl+F '1710B — Rain'). But I didn't get some statements there (and feel like a 120% fool). This one (and like):

    $$$d^{2}_{x_i+1} \leftarrow d^{2}_{x_i+1}-2$$$

    Per my understanding (and following first statement of that original solution):

    $$$d^1_{x_i} = a_i - a_{i-1}$$$

    $$$d^1_{x_{i+1}} = a_{i+1} - a_i$$$

    So

    $$$d^2_{x_{i+1}} = d^1_{x_{i+1}} - d^1_{x_i} = a_{i+1} - 2 * a_i + a_{i-1}$$$

    If $$${a_i}$$$ was set from 0 to well... $$${a_i}$$$, then change would rather be written like:

    $$$d^{2}_{x_i+1} \rightarrow d^{2}_{x_i+1}-2*a_i$$$

    What also confuses me, is why this expression is given for $$$x_{i+1}$$$? Shouldn't it be given for $$$x_i$$$ just like it is given in source code section?

    As long a intensity of $$$a_i$$$ decreases with increasing distance this looks pretty much correct:

    $$$d^{2}_{x_i-p_i+1} \leftarrow d^{2}_{x_i-p_i+1}+1$$$

    ...for $$${x_i-p_i+1}$$$ is a last cell at left which is affected by $$$a_i$$$. But why leftarrow? Shouldn't it be rightarrow ($$$\rightarrow$$$)?

    But. This seems to be wrong again:

    $$$d^{2}_{x_i+p_i+1} \leftarrow d^{2}_{x_i+p_i+1}+1$$$

    It looks like here we should replace $$$+1$$$ with $$$-1$$$:

    $$$d^{2}_{x_i+p_i-1} \leftarrow d^{2}_{x_i+p_i-1}+1$$$

    Once editorial is done with $$$d$$$ expression it proceeds to $$$a_i$$$, m and it doesn't link it with $$$d$$$ formulae at all. So out of narrative is not clean why do we mention that $$$d$$$ stuff??

    So I feel confused, and beg for something smart to explain, haha :-)

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

      Even I don't understand it. Here is how I solved. I calculated the amount of rain at every point first. Then for checking whether removing a rain will remove the flood or not, I have taken the idea of calculating equivalent points from this problem. Then I am checking whether removing the rain will cover that range or not. You can check my solve function in this submission, It'll be more clear.

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

A question: Why div2C 5 5 2 12 13 output "No"

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

    Each color must cover at least 2 rows/columns. But if the total number of rows or columns is odd, you need at least one color to cover 3 rows/columns (since it's impossible to cover all rows/columns if each color only covers an even number of rows/columns).

    So for m = n = 5, you need at least one color to fill 3 rows/columns, which requires a value of 15. No color has a value of 15, so the result is NO.

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

Comparing to the mistake of this contest, I only care about the solutions of these problems.

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

    Can the editorial be available as soon as possible? Thanks.

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2 years ago, # |
Rev. 3   Vote: I like it -23 Vote: I do not like it

I am stuck at problem Div 2 B and kind of have a solution but having runtime error, cannot find the bug in the solution, can anyone take a look?

Link to the submission

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

    This part of code:

      if (m % 2 == 0) {
        cout << 0 << "\n";
        return;
      }
    

    should be after

      for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
      }
    

    because you need to input all the data in a test case, or you wouldn't input the right thing in the next case.

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

      Ohh man, I messed up the whole thing...had 10-15mins yesterday to find this one, but didn't find the bug. Thanks buddy!

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

Hello sir Rhodoks. Here is Botswanna SWAT Team(codename Sparky_Master_WCH1226). Your account has been detected of exceeding the lowest contrib from Sparky_Master_WCH1226. Please get some upvote now, or your account may be blocked from accessing codeforces.com. Thanks for support sir.

Yours sincerely, AccountDeleted

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

May someone explain why this submission: 165562980 exceeded the memory limit when hacked?

I see they've defined an int array with length 3e5+10 and a pair<int, int> array with length 3e5+10. With 4 bytes per integer, That's 900,030 * 4 = 2,700,120 bytes (not even 3 MB). There is a map<pair<int, int>> which takes only 3 key-value pairs (m = 3 in the hack).

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

    Znb-Jfrian I would be thankful if you explain how it worked.

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

      Oh, So sorry for not replying sooner. I hope you got your answer :)

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

    The size of the map <pair <int, int>> is not limited to 3. It's true that when you're reading edges, only three entries are added to this map. However, whenever you try to access a key that was not yet added to the map, this key will become added to the map, with a default value (0 for int).

    The nested for loops examine every pair of $$$i$$$ and $$$j$$$. Out of $$${40000 \choose 2} = 799980000$$$ pairs to be checked, only the last three form edges (which will update ans in order to break out). Before these last three pairs, all of the other ~$$$8 \times 10^8$$$ pairs will add new map entries when accessing mp[{a[i].second,a[j].second}] in the second if statement. So this map is gonna get really huge.

    In general, you can avoid this type of map blowup by checking if (mp.count(KEY)) first, but it requires typing more, and the added complication of a new if-else can also sometimes slow down your thought process as well. In this case, however, since the map only takes values of 1, then you could certainly just use mp.count({a[i].second, a[j].second}) directly (without having to actually check anything). I'm not sure if this change will suffice for this particular submission (due to runtime concerns), but it should be helpful to know in the future.

    On that note, I find that programs with high runtime would sometimes be rejected as memory limit exceeded. Sometimes it's obvious as to why the high runtime results in high memory, but often it's due to some low-level design details that I admittedly don't know too much about. But in general, if you see that the memory limit exceeded, it often implies that the runtime limit would be exceeded as well, which is usually easier to identify. In this case, with nested for loops running ~$$$8 \times 10^8$$$ iterations, with each iteration accessing a map whose size increments up to that high as well, it should signal that the runtime would likely break, even if the rejection message is about memory limit.

    I would guess that the hacker was also intending to try breaking the runtime, as opposed to specifically targeting the memory limit. Of course, either outcome would be considered an equally successful hack, so the hacker doesn't have to worry about such distinctions.

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

      Thanks a lot, this was helpful. It still looks weird how it cleared the third pretest (n = 10e5).

      Edit: perhaps it's due to a lucky escape with an early break.

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

        I thought you designed the nested for loop to break after it sees one edge, in which case, it's quite reasonable for it to pass pretests. The logic of checking only a single edge, specifically the edge with the least unhappiness sum over the two endpoints, would actually be correct for this problem (but idk if you did that, since I don't wanna parse that second line of the nested for loop).

        Given how easy it is to check all edges directly (especially since the input format provides an edge list), I'm guessing the pretests weren't expecting submissions that would try to loop through every vertex pair instead of every edge. There might've been a system test to punish that, but the hacker broke it first.

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

[HELP][DIV2 B] Can someone help me explain for problem B in DIV2 we are removing only a single pair (X,Y) such that both X and Y are present in even no. of pairs and getting the minimum of all such pairs.

How can we prove that if we choose two different pairs (X1,Y1) and (X2, Y2) and remove only either X1 & Y1 OR X1 & Y2 OR X2 & Y1 OR X2 & Y2, we won't get the optimal answers ?

Basically why only choose a single pair and use that to get the minimum value ? Why not choose two different pairs and use them ?

PLEASE HELP !

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

What went wrong was Div1E and only a thousand people have participated in Div1, but why does there are more than three thousand downvotes? Please do not follow the crowd.

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

    FYI, I believe the vote weight depends on rank. So for example, if a red coder downvotes, it counts as more than 1 downvote (somebody said it was counted as 10, but I don't know if that's true). Given that the participants of Div 1 usually have higher ranks, we can expect that the downvote count would get quite inflated.

    That being said, I do think there are also many people who weren't affected and probably didn't even bother to read the updates on the full story (that Rhodoks was completely innocent), and they just downvoted because that was the popular trend, so I second your request of not following the cloud.

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

When will upload editorial?

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2 years ago, # |
  Vote: I like it +59 Vote: I do not like it

Where is the Editorial?

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

Personally I am still new to this platform since I am still only a newbie, but when this kind of error happens ,it surely is demotivating to see that people can get answers like from other sources and create unfair competition.

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

    I understand that this can be very demotivating, but please understand that this is actually a very rare occurrence, and that Codeforces strongly condemns this behavior: https://codeforces.net/blog/entry/105221

    Note that making the round unrated means it will have no impact on anyone's ratings. While this might give the impression that the effort exerted into the contest is wasted, please don't forget that the general objective in these contests is to practice and gain more experience into attempting new problems. The effort you put in and the experience you gained should have hopefully contributed to your personal growth, even if unfortunate circumstances prevented it from being reflected on your rating.

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

Participants in Div1: ~1000. Number of down-votes to the post: ~3500.

WTF? why are div2 people, unaffected by div1E controversy, down-voting it?

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

Although there is a copied problem, I like this contest. Problems are intresting and a bit hard for me.

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

Thanks for the problems(in Div 2).