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:
dario2994 for his marvelous coordination of this round,
wyyxhjth for providing feedback on our early ideas,
flukehn,FairyWinx,JupiterJazz,Vladithur,naman1601,dorijanlendvaj,c337134154,STommydx,LeoRiether,hxu10,Warriors_fsy,idxcalcal,jiaangk_,yakamoto,kiwiHM,Joyemang,forxen for testing our round and giving detailed feedback,
FairyWinx for finding that one of our problems is coincident with a well-known problem (We replaced it),
Vladithur for polishing our statements,
KAN for his great help to our round,
MikeMirzayanov for the amazing Codeforces and Polygon platforms!
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!
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:
We have some more testers now, let's thank them!
We adjusted our score distribution slightly.
UPD3: the Div. 1 part of the round is declared unrated.
Sorry for the late editorial.
Congratulate to winners:
Auto comment: topic has been updated by Rhodoks (previous revision, new revision, compare).
Have a look here before downvoting this blog. I don't think Rhodoks has anything to do with whatever happened with div1 E.
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.
It is sad when someone work with you makes a mistake and drags you down to the water .
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?
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 :)
Still the vitosevski'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).
So are you a kidnapper?
They were just babysitting the problems! stop blaming them for no reason!
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
which we have a very clear example recently
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.
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.
You should have your answer now XD
Just curious, the number of testers is a bit low? How did you accurately estimate the difficulty value of each problem?
It will increase soon. The testing is in progress and the final score distribution will be decided according to their feedback.
sir why are you impersonating human god mejiamejia???
Give contribution please <3
As a loser
Good luck for everyone ❤
Good luck to you too
Changed my mind. Worst round ever
Chinese rounds are always frightening, because of the maths involved.
Can't improve if you shy away from Chinese rounds.
yes , i think you are right.
Well, it may not be that many math problems in our round. Just give it a try! :)
the "may not be" is killer xDDD
I also think taking the Chinese round will be a big boost and I feel that good thinking skills in maths are pretty good
I think you commented on the wrong blog
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
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
Not for nothing I went to math club in 7-th grade
2 hours and 30 minutes... then I have to go to bed later
Wait, when does it turn to 2 hours
They removed one of the problems that's why
Oh thanks, I didn't notice that
are u stupid or something?
He might be new to coding. You don't have to be so rude.
Codeforces is best
Waiting for tourist Vs jiangly showdown. Will no.1 change in this Chinese round?
I'm also interested but tourist won that round and I think tourist will win this time too.
I'm also interested, I think that the tourist will win
Um_nik also has a chance to become number one after this round, don't forget that.
tourist vs jiangly vs Um_nik
I agree, but if he tries hard.
I also agree. who votes for whom
I vote tourist. _____________________________________________________________ and I wish you all great success so that you get more ratings
Obviously , no changed due to some obvious reason ~
This contest is longer than typical Div 2 contests. Tbh this is a good thing bc the probs might be more challenging.
I wish that everyone was given more than -1000 ratings.
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.
Thank you, but I never said I was worried xD
Give thanks to MikeMirzayanov, for He is good; For His glorious platforms Codeforces and Polygon are everlasting.
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.
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.
Auto comment: topic has been updated by Rhodoks (previous revision, new revision, compare).
Great Round Hope i'll solve 1++ problems in this round Good luck everyone
"We have some more testers now, let's thank them!". Thank you, testers
waiting for your Codeforces Round #1919
My first div1 OuO hope to solve 1 problem :p
Div1A is just a Div2C so I think you can do it easily as you were able to become candidate master. Good luck!
Thank you :)
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.
Good luck everyone!!!!
Is div1C same as div2E this time? The score looks like it might not.
I don't think so, div2E will probably be div1D
looks more reasonable. So they removed this problem from div2E but kept it as div1C.
looking forward to your round
Auto comment: topic has been updated by Rhodoks (previous revision, new revision, compare).
Can i get some downvotes please so we bait people to click on view
Can i get some downvotes on this comment also please!!
really excited, I hope my rating increases.
The title description is too bad
What happened to div1 E?
Hello bro, life is a fu** movie!
but but……The samples are the same
If jiangly knows problem E like everybody else, it looks like he is going to be the new number one player
Maybe he knows but won't be top1 this way
These toroidals made me reread the statement again and again
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.
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"
the contribution of this post will be negative.
The topic span is too large. I feel like I'm in prison
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.
Other axis, you can use two 2x5 bands.
Got it. Thanks!
Problem B in div.2 is very hard. Actually ,I didn't like this round(MY OPINION)!!
I think the statement was made more complicated than it should have been.
Video Solution for Problem B and Problem C
You're right about the statements. Thanks for the video ❤
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
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.
Option 2 is redundant. It’s always more optimal to just remove the minimum odd degree so you never want to use operation 2.
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.
The only two options that need to be considered:
My solution: 165558438
Your solution probably works too, so there are probably many ways to approach this problem.
Assume total degree is odd (otherwise, the answer is trivially 0), there are really only two options:
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.
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
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.
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.
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.
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?
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.
But why do you talk about problems here? The contest is running now anyway.
I'm not talking about the solutions, just the problems themselves and stating some facts.
Well, is stating some facts allowed? I think it isn't.
B isn't a graph problem. Simple maths knowledge (about sum of even/odd numbers) can be used to solve the problem.
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.
But I think I have to build the graph to solve it.......
E for easy, D for difficult I guess.
B is not a graph problem I guess. Hopefully my solution is right and I don't get FST :\
Questions about problems
2022-07-24 18:54:49 Problem E. Two Arrays
So should I copy a solution from the comment and paste to solve this problem?
Is it really fair?
2022-07-24 18:53:22 Problem E. Two Arrays
What happened with this problem? I see this is coincidence with some other problems.
Is this contest unrated?
The contest is rated.
where did you find the solution?
see comments above.
this blog will get negative delta contribution after the contest , another unbalanced round -_-
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 :(
Why big coordinates in 2D/1B ? It only makes the problem (way) more painful, and I'd rather go kill myself than writing it.
Large coordinates prevent you from writing value-based complexity solutions
Do not discuss such things during the round...
Wow greenheadstrange aka amiya solved all problems!!
I'm not Miraclefafa today, but a huge fan of his.
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???
Yes, in clarification.
F**k U.Really shiiiiit round!!!UNRATED!!!!!!!!
fku, it should unrated.
I also agree but they gave me +51 rating units.
Please make this round unrated. Problem E is absolutely unfair.
Always choose justice even when compared to great positive delta!
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.
actually, from a blog
A blog open to public.
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.
Yes, I copied the code from a Chinese online judge. And I believe many people did so.
Available but meaningless
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.
Why not unrated?
Guys, you discuss the round too much when it is not even finished
Though I take my words back, apparently div1 has real problems /:
Petition to make this contest unrated
https://www.change.org/ to rescue.
Please make this round unrated
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.
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.
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
negative delta :(
how to solve B ?
Can't wait to see negative votes on this post!
It seems that since several seconds after the contest, the votes have been negative and kept decreasing.lmao.
Please ban whoever decided it's nice idea to link to the problem before the round has ended.
If this round is rated, then God really eats shit.
Didn't you do good?
What's the relation between God and this round ? :|
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.
fk this round
Div1 China-sided round
Managed to solve D2D @ hope no one bothered to make anti-python-hash-hacks xD
When the hardest problem is a well-known problem in China:
Because many school in China have used this problem for practice.
So it is well-known.
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.
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.
Chinese reminder theorem.
bruh, D is sqrt decomposition. just no time left to write 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
oh yea, i should read limits with more attention. sqrt is not going to work (
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.
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.
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)
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.
Even Sample are Same?
Please make this round unrated or I will eat shit.
What the fuck?
jls, QAQ ,jls!
Okay, I corrected my E 1 minute after the contest, and 171 people solved it?
人都傻了 直接 woc 171people solved
can can need
han jin liang yue shao，han jin liang yue duo！
long ge jiu shi long!
China round, (****)!
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.
Unrate this round
The author of this contest only know copy?
me: Annoying so many 114514 ... => Oh, D1C is amazing, I love it! => WTF D1E, why more than 100AC???
lmao what the fuck
jls is going to be angry for 1E.
Fun fact: E's Sample and the Format is the same as a "similar" Problem.
Mother of coincidence!!! xD
ko_osaga solved it 8 months ago lol
Oh that question B messed up my already messed up rank lol
BIG difficulty gap between D1A & D1B
and FUCK D1E
Ah, a man of culture I see (prof pic)
EDIT: No longer xD
Nothing else but shit.
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"
The post was published while contest is in progress??
But aren't you the problemsetter/idea provider of the original problem?
No, problemset is authored by MSU Red Panda
Oh, I see.
If so, why he asked how to solve it in the comment?
Didn't think of that xD
Edited during the contest most probably
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.
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.
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.
Well, I personally think it should be unrated in the case that the problem was intentionally copied from the other source.
How do you prove that a problem was intentionally copied?
That's a bit hard, but can you explain these comments? https://codeforces.com/blog/entry/105214?#comment-935685
EDIT: update https://codeforces.com/blog/entry/105221
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.
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.
That's might not just a coincidence ~
Ggs, gonna lose about 10000000 points
thanks for the fast Tutorial on problem E
I see a lot of solutions for it in comments when it's still running ... xD
Trash round and unrate it please.
These problems were way too hard, I underperformed by a long shot.
This is the worst Round I have ever participated in.
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
Well, it is probably a reason for making div1 unrated (though they could just skip copy-solutions). But it should not affect div2
Why so Hard B?
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.
This contest was very good, my family and I hope you don't play again in the forever!
Is Div2 unrated?
Why would it be?
Because Div1 must be unrated,i don't know what will happen on Div2
It would be stupid to unrate div2 because of problems that concern only div1
You have a great result today, would be unnice to unrate it :/ You might get nicer color.
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
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.
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.
In China,SaiKr Round 2 (Div. 1 + Div. 2, Rated)
Unbengable for saikr
There should be atleast one strong Chinese tester in each round.
very balanced div2 lmao
I don't know why this happened, I was in jail for an hour, what a fuck!!!
You can see a bunch of Chinese at the top of div1。funny wow！！！f**k unratedforce
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
Trash round. I think nearly 90% of people who pass div1. E are looking at the solution. The examples are the same! So trash.
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
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.
1B was good? seriously?
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.
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!
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].
I hate this round
Imagine running plagiarism checker on tens of thousands of submissions for hours, when the real plagiarist turns out to be the problemsetter :(
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.com/blog/entry/93538?#comment-827037
You should not say anything, about someone's mother, RESPECT EVERYONE
o-o I don't think I've ever seen a contest get downvoted so quickly.
CN Round qwq
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?
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.
Why does removing 1 or 2 persons garantee that the remaining number of pairs is even?
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.
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:
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
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.
Why can't it ever be the case that we remove 3 people?
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.
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?
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.
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).
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.
ABC is solvable, but DE... I think there is a big difference in complexity between C and D.
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:).
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.
I want to take it back!!! It used to be me!!!
Why is Sparky_Master_WCH1226 contributing so low and what are the reasons?what did he do wrong？
Bad Contest, I spent more than 30 minutes to understand B
Red also cheats in contest?
I attribute this failed attempt to reaching Master to Div1E. So close yet so far.
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
You need fill at least two adjacent lines with same color, otherwise you would get only two toroidal neighboors.
Is not nice picture