### Rhodoks's blog

By Rhodoks, 3 weeks ago,

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 24.07.2022 17:35 (Московское время). 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

Div2

• -3405

 » 3 weeks ago, # |   +40 Auto comment: topic has been updated by Rhodoks (previous revision, new revision, compare).
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   +37 Have a look here before downvoting this blog. I don't think Rhodoks has anything to do with whatever happened with div1 E.
•  » » » 3 weeks ago, # ^ |   -24 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.
•  » » » 3 weeks ago, # ^ |   +31 It is sad when someone work with you makes a mistake and drags you down to the water .
 » 3 weeks ago, # | ← Rev. 2 →   +78 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?
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +338 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 :)
•  » » » 3 weeks ago, # ^ |   +60 Still the vito_t'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).
•  » » » 3 weeks ago, # ^ |   +397 So are you a kidnapper?
•  » » » » 3 weeks ago, # ^ |   +47 They were just babysitting the problems! stop blaming them for no reason!
•  » » 3 weeks ago, # ^ |   +29 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
•  » » » 3 weeks ago, # ^ |   +18 instances where for example D < C generally which we have a very clear example recently
•  » » » 3 weeks ago, # ^ | ← Rev. 3 →   +36 It's not only when D
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +11 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.
•  » » 3 weeks ago, # ^ |   +35 You should have your answer now XD
 » 3 weeks ago, # |   -28 Just curious, the number of testers is a bit low? How did you accurately estimate the difficulty value of each problem?
•  » » 3 weeks ago, # ^ |   +90 It will increase soon. The testing is in progress and the final score distribution will be decided according to their feedback.
•  » » 3 weeks ago, # ^ |   -25 sir why are you impersonating human god mejiamejia???
 » 3 weeks ago, # |   +150 As a tester... Give contribution please <3
•  » » 3 weeks ago, # ^ |   -102 As a loser
 » 3 weeks ago, # |   +2 Good luck for everyone ❤
•  » » 3 weeks ago, # ^ |   +17 Good luck to you too
 » 3 weeks ago, # | ← Rev. 2 →   0 Changed my mind. Worst round ever
 » 3 weeks ago, # |   +52 Chinese rounds are always frightening, because of the maths involved.
•  » » 3 weeks ago, # ^ |   +8 Can't improve if you shy away from Chinese rounds.
•  » » » 3 weeks ago, # ^ |   0 yes , i think you are right.
•  » » 3 weeks ago, # ^ |   -114 Well, it may not be that many math problems in our round. Just give it a try! :)
•  » » » 3 weeks ago, # ^ |   +18 the "may not be" is killer xDDD
•  » » » » 3 weeks ago, # ^ |   +3 I also think taking the Chinese round will be a big boost and I feel that good thinking skills in maths are pretty good
 » 3 weeks ago, # | ← Rev. 2 →   -44 w
•  » » 3 weeks ago, # ^ |   +15 I think you commented on the wrong blog
•  » » 3 weeks ago, # ^ |   +5 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
 » 3 weeks ago, # |   +131 Not for nothing I went to math club in 7-th grade
 » 3 weeks ago, # |   +3 2 hours and 30 minutes... then I have to go to bed later
•  » » 3 weeks ago, # ^ |   +14 Wait, when does it turn to 2 hours
•  » » » 3 weeks ago, # ^ |   +19 They removed one of the problems that's why
•  » » » » 3 weeks ago, # ^ |   +3 Oh thanks, I didn't notice that
 » 3 weeks ago, # |   -174
•  » » 3 weeks ago, # ^ |   -30 are u stupid or something?
•  » » » 2 weeks ago, # ^ |   0 He might be new to coding. You don't have to be so rude.
 » 3 weeks ago, # | ← Rev. 3 →   -78 Codeforces is best
 » 3 weeks ago, # |   +40 Waiting for tourist Vs jiangly showdown. Will no.1 change in this Chinese round?
•  » » 3 weeks ago, # ^ |   +4 I'm also interested but tourist won that round and I think tourist will win this time too.
•  » » » 3 weeks ago, # ^ |   +5 I'm also interested, I think that the tourist will win
•  » » 3 weeks ago, # ^ |   -31 Um_nik also has a chance to become number one after this round, don't forget that.
•  » » » 3 weeks ago, # ^ |   -30
•  » » » 3 weeks ago, # ^ |   0 I agree, but if he tries hard.
•  » » » 3 weeks ago, # ^ |   0 I also agree. who votes for whom
•  » » » 3 weeks ago, # ^ |   +3 I vote tourist. _____________________________________________________________ and I wish you all great success so that you get more ratings
•  » » 3 weeks ago, # ^ |   0 Obviously , no changed due to some obvious reason ~
 » 3 weeks ago, # |   -10 This contest is longer than typical Div 2 contests. Tbh this is a good thing bc the probs might be more challenging.
 » 3 weeks ago, # | ← Rev. 2 →   -64 I wish that everyone was given more than -1000 ratings.
 » 3 weeks ago, # |   0 Very excited!
•  » » 3 weeks ago, # ^ |   0 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.
•  » » » 3 weeks ago, # ^ |   +16 Thank you, but I never said I was worried xD
 » 3 weeks ago, # |   +39 Give thanks to MikeMirzayanov, for He is good; For His glorious platforms Codeforces and Polygon are everlasting.
 » 3 weeks ago, # |   -65 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.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +11 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.
 » 3 weeks ago, # |   -101 有中国人
 » 3 weeks ago, # |   +11 Auto comment: topic has been updated by Rhodoks (previous revision, new revision, compare).
 » 3 weeks ago, # |   0 Great Round Hope i'll solve 1++ problems in this round Good luck everyone
 » 3 weeks ago, # | ← Rev. 2 →   +29 "We have some more testers now, let's thank them!". Thank you, testers
 » 3 weeks ago, # |   +30 waiting for your Codeforces Round #1919
 » 3 weeks ago, # |   +9 My first div1 OuO hope to solve 1 problem :p
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +14 Div1A is just a Div2C so I think you can do it easily as you were able to become candidate master. Good luck!
•  » » » 3 weeks ago, # ^ |   +5 Thank you :)
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +4 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.
 » 3 weeks ago, # |   0 Good luck everyone!!!!
 » 3 weeks ago, # | ← Rev. 2 →   +32 Is div1C same as div2E this time? The score looks like it might not.
•  » » 3 weeks ago, # ^ |   0 I don't think so, div2E will probably be div1D
•  » » » 3 weeks ago, # ^ |   0 looks more reasonable. So they removed this problem from div2E but kept it as div1C.
 » 3 weeks ago, # |   +1 Good luck
 » 3 weeks ago, # |   +1 looking forward to your round
 » 3 weeks ago, # |   +7 Auto comment: topic has been updated by Rhodoks (previous revision, new revision, compare).
 » 3 weeks ago, # |   -138 Can i get some downvotes please so we bait people to click on view
•  » » 3 weeks ago, # ^ |   -55 Can i get some downvotes on this comment also please!!
 » 3 weeks ago, # |   0 really excited, I hope my rating increases.
 » 3 weeks ago, # |   0 The title description is too bad
 » 3 weeks ago, # |   +145 What happened to div1 E?
•  » » 3 weeks ago, # ^ |   +166 Notorious Coincidence.
•  » » » 3 weeks ago, # ^ |   0 oh~MiFaFa, Right
•  » » » » 3 weeks ago, # ^ |   -10 Hello bro, life is a fu** movie!
•  » » » 3 weeks ago, # ^ |   +2 but but……The samples are the same
 » 3 weeks ago, # |   +65 If jiangly knows problem E like everybody else, it looks like he is going to be the new number one player
•  » » 3 weeks ago, # ^ |   +27 Maybe he knows but won't be top1 this way
 » 3 weeks ago, # |   +35 ToroidalForces
•  » » 3 weeks ago, # ^ |   +10 These toroidals made me reread the statement again and again
•  » » » 3 weeks ago, # ^ |   0 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.
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +3 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"
 » 3 weeks ago, # |   +89 the contribution of this post will be negative.
 » 3 weeks ago, # |   +162 unratedforces
 » 3 weeks ago, # | ← Rev. 3 →   +7 Great problems!
•  » » 3 weeks ago, # ^ |   +6 The topic span is too large. I feel like I'm in prison
•  » » 3 weeks ago, # ^ |   0 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.
•  » » » 3 weeks ago, # ^ |   +1 Other axis, you can use two 2x5 bands.
•  » » » » 3 weeks ago, # ^ |   0 It's amazing
•  » » » » 3 weeks ago, # ^ |   -10 Got it. Thanks!
 » 3 weeks ago, # |   +4 BruhHHH
 » 3 weeks ago, # |   +29 Problem B in div.2 is very hard. Actually ,I didn't like this round(MY OPINION)!!
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +15 I think the statement was made more complicated than it should have been.Video Solution for Problem B and Problem C
•  » » » 3 weeks ago, # ^ |   0 You're right about the statements. Thanks for the video ❤
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +3 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.
•  » » » 3 weeks ago, # ^ |   0 Option 2 is redundant. It’s always more optimal to just remove the minimum odd degree so you never want to use operation 2.
•  » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 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.
•  » » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 The only two options that need to be considered: Remove 1 odd degree node. Remove 2 even degree node that are connected to each other. My solution: 165558438Your solution probably works too, so there are probably many ways to approach this problem.
•  » » » 3 weeks ago, # ^ |   0 Assume total degree is odd (otherwise, the answer is trivially 0), there are really only two options: Remove a single person with odd degree 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.
•  » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 MB
•  » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 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 donecase 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 VBecause 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
•  » » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 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.
 » 3 weeks ago, # |   -6 done A,B,C
 » 3 weeks ago, # | ← Rev. 2 →   +8 Meanwhile Carrot:
 » 3 weeks ago, # |   +3 I strongly dislike this round. 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. Unsuitable difficulty. Div.2 D is harder than Div.2 E while Div.1 D is way harder than Div.1 E. 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.
•  » » » 3 weeks ago, # ^ |   +9 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?
•  » » 3 weeks ago, # ^ |   +45 But why do you talk about problems here? The contest is running now anyway.
•  » » » 3 weeks ago, # ^ |   -51 I'm not talking about the solutions, just the problems themselves and stating some facts.
•  » » » » 3 weeks ago, # ^ |   +6 Well, is stating some facts allowed? I think it isn't.
•  » » 3 weeks ago, # ^ |   +2 B isn't a graph problem. Simple maths knowledge (about sum of even/odd numbers) can be used to solve the problem.
•  » » » 3 weeks ago, # ^ |   0 Yeah, I agree. I also used simple maths knowledge to solve it. However, it is just not suitable and is hard for anyone who isn't familiar to graph to approach, that's the point.
•  » » » 3 weeks ago, # ^ |   0 But I think I have to build the graph to solve it.......
•  » » 3 weeks ago, # ^ |   +15 E for easy, D for difficult I guess.
•  » » 3 weeks ago, # ^ |   0 B is not a graph problem I guess. Hopefully my solution is right and I don't get FST :\
 » 3 weeks ago, # | ← Rev. 6 →   +35 Questions about problems156039 KroosTheKeenGlint2022-07-24 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 KroosTheKeenGlint2022-07-24 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.
•  » » 3 weeks ago, # ^ |   0 where did you find the solution?
•  » » » 3 weeks ago, # ^ |   0 see comments above.
 » 3 weeks ago, # |   +9 this blog will get negative delta contribution after the contest , another unbalanced round -_-
 » 3 weeks ago, # |   +14 Unbalancedforces
•  » » 3 weeks ago, # ^ |   0 UnvotedforcesUnratedforcesMathforces
 » 3 weeks ago, # |   +31 SaikeForces
 » 3 weeks ago, # |   0 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 :(
 » 3 weeks ago, # | ← Rev. 3 →   +8 Why big coordinates in 2D/1B ? It only makes the problem (way) more painful, and I'd rather go kill myself than writing it.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Large coordinates prevent you from writing value-based complexity solutions
•  » » » 3 weeks ago, # ^ |   +19 Do not discuss such things during the round...
 » 3 weeks ago, # |   -33 Wow greenheadstrange aka MiracleFaFa solved all problems!!
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +8 I'm not Miraclefafa today, but a huge fan of his.
 » 3 weeks ago, # |   +94 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???
•  » » 3 weeks ago, # ^ |   +4 Yes, in clarification.
 » 3 weeks ago, # |   +67 F**k U.Really shiiiiit round!!!UNRATED!!!!!!!!
 » 3 weeks ago, # |   +99 fku, it should unrated.
 » 3 weeks ago, # |   +20 Unbalanced Unbalancedforces
 » 3 weeks ago, # | ← Rev. 2 →   +23 ...
•  » » 3 weeks ago, # ^ |   0 I also agree but they gave me +51 rating units.
 » 3 weeks ago, # |   +428 Please make this round unrated. Problem E is absolutely unfair.
•  » » 3 weeks ago, # ^ |   +3 Second.
•  » » 3 weeks ago, # ^ |   +12 Always choose justice even when compared to great positive delta!
•  » » 3 weeks ago, # ^ |   +82 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.
•  » » » 3 weeks ago, # ^ |   +28 actually, from a blog
•  » » » » 3 weeks ago, # ^ |   0
•  » » » 3 weeks ago, # ^ |   +3 Someone post the link in the comment during the contest for a long time, so many people may see it, but it is deleted now.
•  » » » 3 weeks ago, # ^ |   0 Yes, I copied the code from a Chinese online judge. And I believe many people did so.
•  » » » 3 weeks ago, # ^ |   0 Available but meaningless
•  » » 3 weeks ago, # ^ |   0 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.
 » 3 weeks ago, # |   +163 Why not unrated?
 » 3 weeks ago, # | ← Rev. 2 →   0 Guys, you discuss the round too much when it is not even finishedThough I take my words back, apparently div1 has real problems /:
 » 3 weeks ago, # |   +277 Petition to make this contest unrated
•  » » 3 weeks ago, # ^ |   0 https://www.change.org/ to rescue.
 » 3 weeks ago, # |   +78 Unratedforces
 » 3 weeks ago, # |   0 lol
 » 3 weeks ago, # |   +85 Please make this round unrated
 » 3 weeks ago, # |   +113 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.
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   +82 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.
 » 3 weeks ago, # |   0 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
 » 3 weeks ago, # |   +8 negative delta :(
•  » » 3 weeks ago, # ^ |   0 how to solve B ?
 » 3 weeks ago, # |   +14 Can't wait to see negative votes on this post!
•  » » 3 weeks ago, # ^ |   0 It seems that since several seconds after the contest, the votes have been negative and kept decreasing.lmao.
 » 3 weeks ago, # |   +201 Please ban whoever decided it's nice idea to link to the problem before the round has ended.
 » 3 weeks ago, # |   +29 If this round is rated, then God really eats shit.
•  » » 3 weeks ago, # ^ |   0 Didn't you do good?
•  » » 3 weeks ago, # ^ |   +16 What's the relation between God and this round ? :|
•  » » » 3 weeks ago, # ^ |   -13 有原题
 » 3 weeks ago, # |   0 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.
 » 3 weeks ago, # |   +2 fk this round
 » 3 weeks ago, # |   0 Div1 China-sided round
 » 3 weeks ago, # |   0 Managed to solve D2D @ hope no one bothered to make anti-python-hash-hacks xD
 » 3 weeks ago, # | ← Rev. 2 →   +215 When the hardest problem is a well-known problem in China:
•  » » 3 weeks ago, # ^ |   +10 Because many school in China have used this problem for practice.So it is well-known.
•  » » » 3 weeks ago, # ^ |   +16 The 10-th man even wrote a blog for that.You could see via this linkIt is under the G.game, you could click the second block under it and see the code.
•  » » » 3 weeks ago, # ^ |   0 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.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +22 Chinese reminder theorem.
 » 3 weeks ago, # |   0 bruh, D is sqrt decomposition. just no time left to write it
•  » » 3 weeks ago, # ^ |   +3 My solution is not sqrt . In fact I'm not sure if it is solvable with sqrt. Large constraints on coordinates seemingly prevent it
•  » » » 3 weeks ago, # ^ |   0 oh yea, i should read limits with more attention. sqrt is not going to work (
•  » » » » 3 weeks ago, # ^ |   0 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.
 » 3 weeks ago, # |   +11 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.
 » 3 weeks ago, # |   +81 哈哈，原题场，down vote!
 » 3 weeks ago, # |   +42 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)
 » 3 weeks ago, # |   +183 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.
•  » » 3 weeks ago, # ^ |   +1 Even Sample are Same?
 » 3 weeks ago, # |   +9 Please make this round unrated or I will eat shit.
 » 3 weeks ago, # |   +1133 What the fuck?
•  » » 3 weeks ago, # ^ |   -9 jls, QAQ ,jls!
•  » » 3 weeks ago, # ^ |   +244 Okay, I corrected my E 1 minute after the contest, and 171 people solved it?
•  » » » 3 weeks ago, # ^ |   0 人都傻了 直接 woc 171people solved
•  » » 3 weeks ago, # ^ |   -8 jls,QAQ
•  » » 3 weeks ago, # ^ |   -10 can can need
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   -17 wochao xuangou
•  » » » » 3 weeks ago, # ^ |   -11 haoduoyinliugou
•  » » » » » 3 weeks ago, # ^ |   -10 cancanneed
•  » » » » » » 3 weeks ago, # ^ |   -13 wochao die
•  » » » 3 weeks ago, # ^ |   -13 han jin liang yue shao，han jin liang yue duo！
•  » » » » 3 weeks ago, # ^ |   -15 long ge jiu shi long!
•  » » 3 weeks ago, # ^ |   -19 China round, (****)!
•  » » 3 weeks ago, # ^ |   +5 +
 » 3 weeks ago, # |   +6 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.
 » 3 weeks ago, # |   +26 Unrate this round
 » 3 weeks ago, # | ← Rev. 2 →   +73 rnm, 退钱！fxxk, refund!
•  » » 3 weeks ago, # ^ |   0 退钱退钱！！！推rating！！！
 » 3 weeks ago, # |   +78 The author of this contest only know copy?
 » 3 weeks ago, # |   0 me: Annoying so many 114514 ... => Oh, D1C is amazing, I love it! => WTF D1E, why more than 100AC???
 » 3 weeks ago, # |   +52 lmao what the fuck
 » 3 weeks ago, # |   +5 jls is going to be angry for 1E.
 » 3 weeks ago, # |   +188 Fun fact: E's Sample and the Format is the same as a "similar" Problem.Similar Pro:E:
•  » » 3 weeks ago, # ^ |   0 Mother of coincidence!!! xD
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 ko_osaga solved it 8 months ago lol
 » 3 weeks ago, # |   0 Oh that question B messed up my already messed up rank lol
 » 3 weeks ago, # |   +27 BIG difficulty gap between D1A & D1B and FUCK D1E
•  » » 3 weeks ago, # ^ |   +37 Ah, a man of culture I see (prof pic)
 » 3 weeks ago, # |   +22 Nothing else but shit.Unrated, plz.
 » 3 weeks ago, # |   +71 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"
•  » » 3 weeks ago, # ^ |   +2
•  » » » 3 weeks ago, # ^ |   +113 posted @ 2022-07-24 22:52 Flying2018 阅读(1209) 评论(0) 编辑 收藏 举报 The post was published while contest is in progress??
•  » » » » 3 weeks ago, # ^ |   0 But aren't you the problemsetter/idea provider of the original problem?
•  » » » » » 3 weeks ago, # ^ |   +9 No, problemset is authored by MSU Red Panda
•  » » » » » » 3 weeks ago, # ^ |   +3 Oh, I see.
•  » » » » » 3 weeks ago, # ^ |   0 If so, why he asked how to solve it in the comment?
•  » » » » » » 3 weeks ago, # ^ |   0 Didn't think of that xD
•  » » » » 3 weeks ago, # ^ |   +13 Edited during the contest most probably
•  » » » » 3 weeks ago, # ^ |   0 Probably, this guy who published this blog reviewed his own code at the edit page, this behavior will also cause the posted time to change.
 » 3 weeks ago, # |   +81 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.
•  » » 3 weeks ago, # ^ |   0 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.
•  » » 3 weeks ago, # ^ |   +28 Well, I personally think it should be unrated in the case that the problem was intentionally copied from the other source.
•  » » » 3 weeks ago, # ^ |   +18 How do you prove that a problem was intentionally copied?
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +13 That's a bit hard, but can you explain these comments? https://codeforces.com/blog/entry/105214?#comment-935685EDIT: update https://codeforces.com/blog/entry/105221
•  » » » » » 3 weeks ago, # ^ |   +44 The person is having a mental breakdown because the whole community is very angry at them. They are really sorry that the problem was well-known and that their co-author got heavily downvoted. Also I don't know Chinese traditions for apologizing, it might be just a custom to be self-deprecating in such a situation.
•  » » 3 weeks ago, # ^ |   +5 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.com/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.
 » 3 weeks ago, # |   +86
•  » » 3 weeks ago, # ^ |   +3 That's might not just a coincidence ~
 » 3 weeks ago, # |   -8 Serious?
 » 3 weeks ago, # |   -14 Ggs, gonna lose about 10000000 points
 » 3 weeks ago, # |   +28 thanks for the fast Tutorial on problem E I see a lot of solutions for it in comments when it's still running ... xD
 » 3 weeks ago, # |   +44 Trash round and unrate it please.
 » 3 weeks ago, # |   -13 These problems were way too hard, I underperformed by a long shot.
 » 3 weeks ago, # | ← Rev. 2 →   +23 This is the worst Round I have ever participated in.
 » 3 weeks ago, # |   +7 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
•  » » 3 weeks ago, # ^ |   +15 Well, it is probably a reason for making div1 unrated (though they could just skip copy-solutions). But it should not affect div2
 » 3 weeks ago, # |   0 Why so Hard B?
 » 3 weeks ago, # |   0 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.
 » 3 weeks ago, # |   0 This contest was very good, my family and I hope you don't play again in the forever!
•  » » 3 weeks ago, # ^ |   0 me too!!!
 » 3 weeks ago, # |   +1 Is Div2 unrated?
•  » » 3 weeks ago, # ^ |   0 Why would it be?
•  » » » 3 weeks ago, # ^ |   +17 Because Div1 must be unrated,i don't know what will happen on Div2
•  » » » » 3 weeks ago, # ^ |   +3 It would be stupid to unrate div2 because of problems that concern only div1
•  » » » » » 3 weeks ago, # ^ |   0 You have a great result today, would be unnice to unrate it :/ You might get nicer color.
•  » » » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 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
 » 3 weeks ago, # |   0 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.
 » 3 weeks ago, # |   +137 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.
 » 3 weeks ago, # |   +1 In China,SaiKr Round 2 (Div. 1 + Div. 2, Rated)
•  » » 3 weeks ago, # ^ |   0 Unbengable for saikr
 » 3 weeks ago, # |   +9 There should be atleast one strong Chinese tester in each round.
 » 3 weeks ago, # |   +8 very balanced div2 lmao
 » 3 weeks ago, # |   0 I don't know why this happened, I was in jail for an hour, what a fuck!!!
 » 3 weeks ago, # |   0 You can see a bunch of Chinese at the top of div1。funny wow！！！f**k unratedforce
 » 3 weeks ago, # | ← Rev. 2 →   +8 https://blog.csdn.net/qq_35577488/article/details/117813076Problem 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
 » 3 weeks ago, # |   +21 Trash round. I think nearly 90% of people who pass div1. E are looking at the solution. The examples are the same! So trash.
•  » » 3 weeks ago, # ^ |   +13 More precisely,99%.
 » 3 weeks ago, # |   0 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
 » 3 weeks ago, # |   +37 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.
•  » » 3 weeks ago, # ^ |   -10 1B was good? seriously?
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   +10 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.
 » 3 weeks ago, # |   +24 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!
•  » » 3 weeks ago, # ^ |   +24 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].
 » 3 weeks ago, # |   +12 I hate this round
 » 3 weeks ago, # |   +50 Imagine running plagiarism checker on tens of thousands of submissions for hours, when the real plagiarist turns out to be the problemsetter :(
 » 3 weeks ago, # |   +23 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.
•  » » 3 weeks ago, # ^ |   0 You should not say anything, about someone's mother, RESPECT EVERYONE
 » 3 weeks ago, # |   +17 o-o I don't think I've ever seen a contest get downvoted so quickly.
 » 3 weeks ago, # |   0 CN Round qwq
 » 3 weeks ago, # |   0 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?
•  » » 3 weeks ago, # ^ |   0 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.
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Why does removing 1 or 2 persons garantee that the remaining number of pairs is even?
•  » » » » 3 weeks ago, # ^ |   0 It doesn't, you have to greedily find a person (or two) such that removing them makes the remaining number of pairs even. Atleast that's what I came up with during the contest.
•  » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 it does not guarantee u will have to check,just make a map and keep a count, which number appears in how many pairs for example:1 22 32 45 6 two appears in 3 pairs so i cant remove 2 coz that will leave an odd number of pairs,but removing 1 and 5 is feasible coz that leaves 2 pairs.
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +5 If the person appears odd number of times, then it is obvious why it worksNow 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
•  » » » » » 3 weeks ago, # ^ |   0 Thanks
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 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.
•  » » » » » 3 weeks ago, # ^ |   0 really appreciate!
•  » » » » » 3 weeks ago, # ^ |   0 Why can't it ever be the case that we remove 3 people?
•  » » » » » » 3 weeks ago, # ^ |   +1 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.
•  » » 3 weeks ago, # ^ |   0 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?
•  » » 3 weeks ago, # ^ |   0 Observe that If the number of pairs is already even, you don't need to do anything. On the other hand, if the number of pairs is odd, the answer can be either of the following:A. The min unhappiness is achieved by not inviting a person with an odd number of pairs.B. Don't invite a person with an even number of pairs. If the number of pairs for any of his friends becomes odd after this, the unhappiness of removing these two people can be unhappiness. The minimum overall will be the answer.
•  » » 3 weeks ago, # ^ |   0 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).
•  » » 3 weeks ago, # ^ | ← Rev. 4 →   0 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.
 » 3 weeks ago, # |   +12 ABC is solvable, but DE... I think there is a big difference in complexity between C and D.
 » 3 weeks ago, # |   +17 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:).
 » 3 weeks ago, # | ← Rev. 2 →   +43 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.
•  » » 3 weeks ago, # ^ |   0 I want to take it back!!! It used to be me!!!
•  » » 3 weeks ago, # ^ |   0 Why is Sparky_Master_WCH1226 contributing so low and what are the reasons?what did he do wrong？
 » 3 weeks ago, # |   0 Bad Contest, I spent more than 30 minutes to understand B
 » 3 weeks ago, # |   +5 Shame
 » 3 weeks ago, # |   0 Red also cheats in contest?
 » 3 weeks ago, # |   -35 I attribute this failed attempt to reaching Master to Div1E. So close yet so far.
 » 3 weeks ago, # |   +1 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 code165577141
•  » » 3 weeks ago, # ^ |   0 TEST14 6 6 4 4 4 4 4 4 Answer: NO  TEST24 3 2 4 8 Answer: NO 
•  » » 3 weeks ago, # ^ |   0 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
•  » » » 3 weeks ago, # ^ |   0 isn't filling rows or columns greater than 3 enough to produce a beautiful grid ? or did i misunderstand what toroidal means ?
•  » » » » 3 weeks ago, # ^ |   0 I dont understand questionI meant that your code would print yes on that example: 1 3 4 2 9 3 `Though answer must be no
 » 3 weeks ago, # | ← Rev. 2 →   +66 Is it unrated?
 » 3 weeks ago, # |   0 qwq
 » 3 weeks ago, # |