### feecIe6418's blog

By feecIe6418, history, 16 months ago,

Hello Codeforces!

gyh20 and I are glad to invite you to Codeforces Round #670 (Div. 2) which will start on Sep/12/2020 16:45 (Moscow time). Note the unusual start time of the round.

The contest will last for two hours, and you will have five tasks to solve. The tasks are prepared by me and gyh20. This round is rated for participants whose rating is not higher than 2099. You can see that my current rating is exactly 2099 :)

There might be an interactive problem. You can learn about them here.

We would like to thank:

We tried our best to make the statements short and clear, pretests strong and problems interesting. We hope you like the problems!

Score distribution will be announced shortly before the round.

Good luck and have fun!

Upd: Score distribution is 500-750-1250-1750-2500.

Upd: For problem reasons, the contest is delayed for 10 minutes. We are very sorry to keep you waiting, sorry again.

Upd: Score distribution is changed to 500-1000-1500-2000-2750.

Upd: The round is finished. We're really sorry for B being well-known (none of the testers knew the harder version of this problem in ABC173E). Still, congratulations to the winners!

Div1 (unofficial):

Div2:

Upd: Editorial is out here.

• +772

 » 16 months ago, # | ← Rev. 3 →   +286 As a tester,I want some contribution!
•  » » 16 months ago, # ^ |   +69 Have a strong heart, dear.
•  » » 16 months ago, # ^ |   +60 So why are we downvoting the tester >_<
•  » » 16 months ago, # ^ |   +56 A tester leaves a rated contest for us, Have some decency please
•  » » » 16 months ago, # ^ |   -85 I won't mind leaving a rated contest just to see my name in the blog and then getting easy contributions later.
•  » » » » 16 months ago, # ^ |   +16 You see you're expecting contributions too and so does _zzw4257
•  » » 16 months ago, # ^ |   +30 what we get if we have some contribution.i often see people are asking for contribution ??
•  » » » 16 months ago, # ^ |   +15 nothing,in general but it is an indication that you have been useful to the community.Also there is a list for top contributors
•  » » » » 16 months ago, # ^ |   +7 okay Thanks :P
 » 16 months ago, # |   +263 As a coauthor,I am so glad to hold the contest, and i want you-know-what.
•  » » 16 months ago, # ^ |   0 I want positive contribution instead of negative contribution!
•  » » » 16 months ago, # ^ |   +31 Don't forget to give the tester _zzw4257 you-know-what as well. :)
•  » » » 16 months ago, # ^ |   +260 sad life :(
•  » » » » 16 months ago, # ^ |   +10 This happens to me when contest is delayed it hurts alot as i am preparing my self to encounter the problem and just 10 mins more added....
•  » » » » » 16 months ago, # ^ |   +12 Thanks a lot for replying, I had forgotten about the contest. You reminded me. :)
 » 16 months ago, # |   +59 feecIe6418 Can you highlight the start time of the contest.
•  » » 16 months ago, # ^ |   +5 I think it is already highlighted.
•  » » » 16 months ago, # ^ |   +37 I mean it start one hour earlier than regular time.
•  » » » » 16 months ago, # ^ |   +65 OK, added a sentence
 » 16 months ago, # |   +19 Really looking forward to the contest!I think it will be very good!
•  » » 16 months ago, # ^ |   0 Are you son of Time_tears ? :v
•  » » » 16 months ago, # ^ | ← Rev. 2 →   +24 Definitely not! You can see that he created the account a lot later than I did and there is nothing I can do.Sorry,it's just a typo.
•  » » » » 16 months ago, # ^ |   +16 Sorry for telling that..But when i read your id name , i think the tester fisherking and your id name can be related . Again sorry sir !
•  » » » 16 months ago, # ^ |   -27 Certainly correct
 » 16 months ago, # |   +64 As a problem improver, have fun!
•  » » 16 months ago, # ^ |   +8 What I improved is following:In the task E, there was one more constraint "It is guaranteed, that an integer $p>1$ such that $x$ is divisible by $p^2$ does not exist. You want to find $x$."I noticed that we can erase this constraint. If you couldn't solve E, it's good that thinking this version first!
 » 16 months ago, # | ← Rev. 3 →   -118 Off Topic : What is the cutoff for candidate master.. 1800 ? or 1900 ?
 » 16 months ago, # |   -26 Your rating is exactly 2099 but unfortunately the round is not rated for you.
 » 16 months ago, # |   -83 feecIe6418, Will there be any subtasks? I really think having subtasks makes contest more interesting.
•  » » 16 months ago, # ^ | ← Rev. 2 →   -33 The contest is ICPC style not IOI style,so there will be no subtasks.
•  » » » 16 months ago, # ^ |   +13 First of all thanks Elegy_No_Programmer for letting me know, but sadly I disagree with your explanation why this contest can't have subtasks. You can see many Div-2 contests had subtasks for eg:Codeforces Round #658 (Div. 2). Also each problems are scored equally in ICPC( like we have in CF educational round ) which is not the case for CF Div-2 rounds.
•  » » 16 months ago, # ^ |   +46 We used to have subtasks, but we found it meaningless to have subtasks on our problems. So we deleted them in the end.
 » 16 months ago, # | ← Rev. 2 →   -54 Every Round is rated for me, except div 1 :3
•  » » 16 months ago, # ^ |   +45 Div1 people: Are we a joke to you?
•  » » » 16 months ago, # ^ |   +8 forget to mention div 1, Yes excluding div 1.
•  » » » 16 months ago, # ^ | ← Rev. 2 →   -6 As if there were any div1 only rounds
•  » » » 16 months ago, # ^ |   -20 Changing profile picture?? kdjonty31
 » 16 months ago, # |   +27 Are the authors middle school students?
 » 16 months ago, # |   +42 Special thanks to csani for helping us with writing the editorialsEditorials are already written?
•  » » 16 months ago, # ^ |   +31 They are written, but not published.
•  » » » 16 months ago, # ^ |   0 then why not publish them?
•  » » » » 16 months ago, # ^ |   +7 So you cant look at the solutions in contest
•  » » » » 16 months ago, # ^ |   +5 He will publish as soon as you completely read all the problem statements and try to solve it for 2 hours :v
•  » » » » » 16 months ago, # ^ |   +4 What if he did not read all the problems ?
•  » » » » » » 16 months ago, # ^ |   0 He should be :v
•  » » » 16 months ago, # ^ | ← Rev. 2 →   +1 Can we have them now? By the way, thanks for the contest, loved it:)
•  » » » 16 months ago, # ^ |   0 Please Publish Editorials
•  » » » 16 months ago, # ^ |   0 TechNite Can you publish editorial now. Thanks!!
•  » » » » 16 months ago, # ^ |   +1 It has been published.
 » 16 months ago, # |   +64 As a tester, I would recommend you to participate in this. The problems are really interesting. Good Luck and have fun.
•  » » 16 months ago, # ^ | ← Rev. 2 →   +47 .
•  » » » 16 months ago, # ^ |   +4 Not all testers comment on this!!! And everyone has their own opinion on what's interesting.
•  » » » 16 months ago, # ^ |   +56 Not really, it's just that testers who think that the round is bad usually don't say "this round is bad, don't participate" in the comments.
 » 16 months ago, # | ← Rev. 2 →   +80 That day is my birthday. It will be a memorable contest lmao :))
•  » » 16 months ago, # ^ | ← Rev. 3 →   +15 hpbd :3
•  » » » 16 months ago, # ^ |   0 Happy Birthday!
•  » » » 16 months ago, # ^ |   +9 "Has a female profile pic" hpbd bro
•  » » 16 months ago, # ^ |   0 happy Birthday. good luck.
•  » » 16 months ago, # ^ |   +18 Today is my birthday :) I was born on 9/11
•  » » » 16 months ago, # ^ |   0 belated happy birthday dude[orz pro max]
•  » » 16 months ago, # ^ |   0 Many Many Happy Returns!!!! duock28cbg
•  » » 3 months ago, # ^ |   0
 » 16 months ago, # |   +14 $Yoo$ now it's becoming trend of $make \,the\, statements \,short \,and \,clear$
•  » » 16 months ago, # ^ |   0 Its useful too. Isn't it?Trgt_2021_explosion
•  » » » 16 months ago, # ^ |   0 As far as there is no story telling(irrelevant statements i.e. not related to that Task) then even complex scenario is fine because converting some complex statements into Mathematical equations/logic is a skill and one should try to learn it.
•  » » » » 16 months ago, # ^ |   +2 Hi,Trgt_2021_explosion I support your thoughts. But I think a clear statement refers to a clear mention of the requirements. Like "if multiple solutions, print any" or this kind of thing... not making a problem easy or spoon-fed.
 » 16 months ago, # | ← Rev. 2 →   +33 Round #$666$ contains $5$ problems for each division. Round #$668$ contains $5$ problems for each division. Round #$669$ contains $5$ problems. Round #$670$ contains $5$ problems. Is it a trend or...something else?
•  » » 16 months ago, # ^ |   +48 No.In fact most of the CF contests a few years ago had 5 problems.
 » 16 months ago, # | ← Rev. 4 →   -8
 » 16 months ago, # |   +24
 » 16 months ago, # | ← Rev. 2 →   -22 I have a question for the authors. Do you know Jiangly? Are you friends?（Sorry. I now feel that this question is not so appropriate. Thank you very much for the author's answer.）
•  » » 16 months ago, # ^ |   +44 We know Jiangly, because he is the best one in the city,but we are not so good as him so he definitely doesn't know us.
•  » » » 11 months ago, # ^ |   0 so, jiangly is a male ?
•  » » » » 11 months ago, # ^ |   0 Of course!
 » 16 months ago, # |   +22 Excited to take part in my first contest!
 » 16 months ago, # |   +3 I have known that the author is very excellent,bless all and hope everyone get higher rating.
 » 16 months ago, # |   +6 After two years without competing here, it is time to come back! :)
•  » » 16 months ago, # ^ |   +13 your rating graph is quite interesting!
 » 16 months ago, # |   +21 Hope that the interactive one will be nice just like the one in the last contest...
 » 16 months ago, # |   +18 Thank GOD for no scary pictures this time...
 » 16 months ago, # |   +8 Great!! Problems with short and clear statement are really good. Hope this will be a good round.
•  » » 16 months ago, # ^ |   +10 with strong pretest.
•  » » » 16 months ago, # ^ | ← Rev. 2 →   +8 Yeah,this is also very good since my problem will not hack after pretest passed.This hacked really sad me.
 » 16 months ago, # |   +15 Why do all the Div.2 rounds consist of 5 problems nowadays? Why not 6/7?
•  » » 16 months ago, # ^ |   +46 How many times have you solved more than 3-4 questions lol?On a serious note, you and I can't even imagine the amount of work it goes into setting a problem and preparing the statements, test cases, etc. So, simply saying things is very easy, when you don't know what is going on behind the scenes
•  » » » 16 months ago, # ^ | ← Rev. 2 →   +10 emmm, recent div.2 rounds are really easier than that before. just like me, before Round 658 , I only solved 2-3 questions. But after Round 658, I always solve 4 questions in Round which had 5 problems.To some degree, it can be named speedforces.
•  » » » » 16 months ago, # ^ |   +55 Maybe you just progress
•  » » » » » 16 months ago, # ^ | ← Rev. 2 →   -36 .
•  » » » » 16 months ago, # ^ | ← Rev. 2 →   +9 That means you are improving! Do you want to stay at the level of only being able to solve 2-3 problems?And, let's just assume for a minute, that the problems have become easier. Do you still know how much effort goes into making those problems, setting up everything, co-ordinating with so many people, getting people to test your problems etc.?To be very honest, I have tried to sit down for hours by myself to think about making a problem, but I never made a single problem which wasn't already popularly known.
•  » » » » 16 months ago, # ^ | ← Rev. 4 →   +27 You should have played Round 669.
•  » » » 16 months ago, # ^ |   +77 is this really necessary to reply someone with this first line!? is it really matter how many problems he solved to tell his opinion and ask his question!? these kind of replies will force them to create alt acc(and probably ruin the cf community). dont bully anyone because of their rating PLS!
•  » » » 16 months ago, # ^ | ← Rev. 2 →   0 Your point is valid, but he did ask in a curious way(Not implying or demanding anything). In no way was it appropriate to reply with that first line.
 » 16 months ago, # |   +7 I hope it have strong pretest!
 » 16 months ago, # |   -9 I don't want to FST any more.So do you have the strong pretests?
 » 16 months ago, # | ← Rev. 2 →   +44 This is gonna be my first ever contest. So excited about it. Not very experienced in Competitive Programmer but my journey starts from now. I hope this be a hell of a journey!UPD : For those who Upvoted me and wished me luck, thank you so much! I think your wishes came true and i got Rank : 652 and Rating got +671
•  » » 16 months ago, # ^ |   +45 T̶u̶r̶n̶ ̶b̶a̶c̶k̶ ̶w̶h̶i̶l̶e̶ ̶y̶o̶u̶ ̶s̶t̶i̶l̶l̶ ̶c̶a̶nGood luck on your first contest!
 » 16 months ago, # |   -33 Is there any Chinese description
•  » » 16 months ago, # ^ |   +23 No.
 » 16 months ago, # |   -8 Happy Programmers' Day!!
 » 16 months ago, # |   -17 Whats Polygon guys?
•  » » 16 months ago, # ^ |   +8 It's a platform that helps make problems, created by Mike
 » 16 months ago, # |   +10 Socre distribution <- typo
•  » » 16 months ago, # ^ |   +7 fixed, thanks for pointing out
•  » » » 16 months ago, # ^ |   0 Still, congrarulations <- another one
 » 16 months ago, # |   -93 The comment is hidden because of too negative feedback, click here to view it
 » 16 months ago, # |   +17 As far as I've seen, this is your first contest arrangement in CF. Best of luck to you too.(Hope everything will be fine)
•  » » 16 months ago, # ^ |   +30 Thanks!
•  » » 16 months ago, # ^ |   +8 you just raised a flag, 10 mins delayed
 » 16 months ago, # | ← Rev. 2 →   0 Interactive problem will be there or not ?
•  » » 16 months ago, # ^ |   +9 YES
•  » » » 16 months ago, # ^ | ← Rev. 2 →   0 Thanks :)
 » 16 months ago, # |   +17 There might be an interactive problem.I guess that codeforces is trying to make interactive problems a part of the contests, which according to me, is a good thing to do.
•  » » 16 months ago, # ^ |   +3 haa its really good
 » 16 months ago, # |   -131 start the contest!! im feeling horny
•  » » 16 months ago, # ^ |   +10 wtf dude, this is codeforces not a nrop website.
•  » » » 16 months ago, # ^ |   -90 STFU you unrated GAY!!
•  » » » » 16 months ago, # ^ |   +9 Ok, you idiot. Look what you wrote. Now, try to find any comments like this on this website. See there is no one. I think a better place for you is a nrop website. Also, where did you got enough information to deduce I am gay? Because I'm not.
•  » » 16 months ago, # ^ |   0 umm... okay
•  » » » 16 months ago, # ^ |   -34 STFU you bald ass!!
•  » » » » 16 months ago, # ^ |   0 It is not nice to talk to yourself like that.
•  » » » » » 16 months ago, # ^ |   0 Hey, we are not same. Thanks.
•  » » » » » » 16 months ago, # ^ |   0 I know that, but now find a way to prove it. People will start thinking that we are the same.
•  » » 16 months ago, # ^ |   +6 Go for virtual . :p
•  » » » 16 months ago, # ^ |   -34 Yeah sure with your mother.
•  » » » » 16 months ago, # ^ |   0 Don't you know about virtual contest.? It seems you've started painting your depressions here.
 » 16 months ago, # |   +18 Delayed by 10 min :(
•  » » 16 months ago, # ^ |   0 yup!!!
•  » » 16 months ago, # ^ |   +5 umm , we actually know..
•  » » » 16 months ago, # ^ |   +4 That's why_you_are_here :)
•  » » » » 16 months ago, # ^ |   0 that's enlightening,thanks!
 » 16 months ago, # |   +6 Auto comment: topic has been updated by feecIe6418 (previous revision, new revision, compare).
 » 16 months ago, # |   +66 Very sorry,something we didn't expect happened just now ,the contest is delayed by 10 minutes, hope there won't be another delay.
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 Thanks for the info :)
•  » » » 16 months ago, # ^ |   +29 I think no longer , it should be good this time.
•  » » 16 months ago, # ^ |   +8 Just curious to know, what was the last minute change because of which score distribution was changed? I think present scoring distribution was accurate.
 » 16 months ago, # |   -21 wish me good luck.
•  » » 16 months ago, # ^ | ← Rev. 2 →   -19 why though? why to you specifically?
•  » » » 16 months ago, # ^ |   -18 because I am loosing my rating. can't you see?
•  » » » » 16 months ago, # ^ |   +3 Then what you need is practice and not good luck!
•  » » » » » 16 months ago, # ^ |   -8 yes bro I do.. But CP is fun and exciting. Why you taking it as a fight? if someone ask wish me good luck if you can just wish them.. What is harm in it? Spread love and have fun bro. This is what life is.
•  » » » » » » 16 months ago, # ^ |   +3 Fighting? Hehe chill. good luck
•  » » » » 16 months ago, # ^ | ← Rev. 2 →   0 more practice and then make progress. good luck and high rating!
•  » » » » » 16 months ago, # ^ |   0 yes. I do.
•  » » » » 16 months ago, # ^ |   0 goodluck fam
•  » » » » » 16 months ago, # ^ |   0 thank you:)
•  » » » » 16 months ago, # ^ |   0 now contribution too, good luck though!
•  » » » » » 16 months ago, # ^ |   0 exactly. It is -17. LOl. I think people should not behave like this.
•  » » 16 months ago, # ^ |   0 Good Luck Native!!
•  » » 16 months ago, # ^ |   0 good luck!
 » 16 months ago, # |   +23 Delay it as much as you want and solve the problem PLEASE DON'T MAKE IT UNRATED due to those issues
•  » » 16 months ago, # ^ |   0 you want some rating so badly.
 » 16 months ago, # | ← Rev. 2 →   -12 delayforces again lol, but still great thanks for the preparation!
•  » » 16 months ago, # ^ |   +28 We're very sorry. But please, still enjoy the contest.
•  » » » 16 months ago, # ^ |   +6 Thank you for your effort！！
•  » » » 16 months ago, # ^ |   0 always thankful. Thanks for the preparation anyway
•  » » » 16 months ago, # ^ |   0 So you people make quality problems for us, make quality test cases so that only good solutions pass and respond to our queries during the contest.I don't think you need to apologize for anything. You're already doing more than we deserve.
 » 16 months ago, # |   +3 delay making me more anxious.
 » 16 months ago, # |   +21 Score distribution changes.
 » 16 months ago, # |   -7 Hoping for strong pretests!
 » 16 months ago, # |   +11 How to solve D?
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 Didn't passed system test yet but my solution idea is :-Hints:-ans is tot = a[0]+d where d = sigma( max( 0, a[i+1]-a[i] ) );ans = (tot+(tot>=0))/2;so only edges L and R in queries can change the value of d.I guess now you can figure out the solution
•  » » 16 months ago, # ^ | ← Rev. 3 →   +26 My solution has not yet passed sys tests but here is how I solved it. Suppose we let initial value of non decreasing array be c so initial value of non increasing array will be a[1]-c.Now for each next element if is greater than previous value we can keep the value in non increasing array constant therefore the other array remains non decreasing. We can similarly keep the value in non decreasing array constant if the next element is less than the previous one.Therefore we get the final value of non decreasing array as c+(sum of all positive a[i]-a[i-1]). Now we just have to minimise max(a[1]-c, c+sum of +ve dif). We can see that it will be minimum if both elements are equal.After each query only 2 differences are changed at max so we can store values of each element by Fenwick Tree and recalculate the optimal c. Upd: passed sys tests.
•  » » 16 months ago, # ^ | ← Rev. 7 →   +4 Observe that since $b_{i}$ is non-decreasing and $c_{i}$ is non-increasing, the max is just $max$$(c_{0}, b_{n - 1})$.We can notice that $b_{i} \gt b_{i - 1}$ will be required when $a_{i} \gt a_{i - 1}$ and $c_{i} \lt c_{i - 1}$ will be required when $a_{i} \lt a_{i - 1}$. Clearly $b_{0} + c_{0} = a_{0}$ is a required condition. Let $b_{0} = u$ and $c_{0} = v$.If $sum_{inc}$ is the sum of $a_{i} - a_{i - 1}$ whenever $a_{i} \gt a_{i - 1}$, $b_{n - 1}$ is clearly $u + sum_{inc}$. So we want to choose $u$ such that $max$($a_{0} - u$, $u + sum_{inc}$) is the minimum possible. This is clearly $\frac{a_{0} - sum_{inc}}{2}$ which gives an answer of $\lceil \frac{a_{0} + sum_{inc}}{2} \rceil$.As for the queries, we can notice that only the contributions from $a_{l} - a_{l - 1}$ and $a_{r + 1} - a_{r}$ to $sum_{inc}$ can change. Maybe there is some elegant observation that makes this easy to calculate, but without that also there is an easy way to figure out the change. Subtract the contribution of those two endpoints, update the range (standard lazy prop) and re-add the contribution of them.BE CAREFUL that you perform ceil correctly for negative values.
•  » » » 16 months ago, # ^ |   0 BE CAREFUL that you perform ceil correctly for negative values. Damn. That's why my 4th TC got WA.
•  » » » » 16 months ago, # ^ |   +1 Wait how, if you ceil incorrectly for negative values it should fail sample 3 as you will do $\frac{-2 + 1}{2} = 0$ instead of $\frac{-2}{2} = -1$.
•  » » » » » 16 months ago, # ^ | ← Rev. 2 →   0 Hmmm. You've got a point. I didn't mention that I thought that, that might be the issue.Now that I can check the test cases, I'll check what exactly went down.EDIT: TL;DR: I'm dumb. So I just updated the l and r values of the array, and somehow thought that it would work, without updating the middle values. Will implement the lazy propagation part now.
•  » » » 16 months ago, # ^ |   0 It should be (a0-sum)/2
•  » » » » 16 months ago, # ^ |   0 Fixed, thanks. I accidentally wrote the minimum obtained instead of the value to be value which when substituted gives the minimum.
•  » » » 16 months ago, # ^ |   +8 We can also do away with the Segment/Fenwick Tree. Since only the contribution of $a_l - a_{l-1}$ and $a_{r+1} - a_r$ affect our answer for the current query, keep an array of all differences between consecutive elements and change these two values in that array. Change $sum_{inc}$ accordingly. P.S. Don't forget to change the value of $a_0$ if the current query starts at index 1. Code
•  » » » » 16 months ago, # ^ |   0 Oh yeah since the Fenwick Tree effectively acts like a dynamic difference array here which is sufficient for our needs. Thanks, somehow that didn't strike at that moment during the contest.
 » 16 months ago, # |   +23 How to solve C See CF blog in recent actions Understand you can only have one or two centroids For one centroid print any edge twice For the second remove an edge going from the first centroid to any vertex except the second centroid(e) then make an edge going from the second centroid to e. Easy AK!
•  » » » 16 months ago, # ^ |   0 Yup!
 » 16 months ago, # |   +11 How to solve E?
•  » » 16 months ago, # ^ |   0 I think ask for all prime numbers ( ~9600 ) and make some Inclusion–exclusion principle (ex if you have n = 6 ask for x=2 => s = {1,3,5,6} and for 3 you have to see that 6 is still there (because after the first operation just 1 element %3=0 had to remain in the set) ) .
•  » » 16 months ago, # ^ |   +20 First we use the second operation to all primes < 1000, all the remaining numbers can either be a prime or $x$.We can query A 1 and check if $x$ is a prime.If $x$ is not a prime, we query every prime number $\leq n$, and obtain x by finding all prime divisors of $x$.If $x$ is a prime, we can do the following: delete S prime numbers, and query A 1 to see if $x$ is in the $S$ primes numbers you just deleted. When you find the $S$ numbers that contain $x$ you do $S$ extra queries to find $x$. Let the number of prime numbers be $N$, The number of operations is $N + S + N/S$, and we can set $S$ to $\sqrt n$code: https://codeforces.com/contest/1406/submission/92637463. It passed the pretests anyway.
•  » » 16 months ago, # ^ |   +7 перебрать все простые делители до N начиная с таких, что их квадрат больше N, проверять можно по группам, например первые 100 такие делители с помощью функции 2 типа, и проверить есть ли среди данных 100 простых чисел нужное с помощью типа 1 и так далее по сто, а в последнем оставшиеся. Затем перебрать все меньшие простые делители и его степени. Например, k — простое, k * k < n, тогда если н делиться на к, с помощью бинпоиска найдем степень вхождения это числа.(iterate through all Prime divisors up to N starting with such that their square is greater than N, you can check by group, for example, the first 100 such divisors using a function of type 2, and check whether among the data 100 primes necessary using type 1, and so on one hundred, and in the last remaining. Then iterate over all the smaller Prime divisors and its powers. For example, k is a Prime, k * k < n, then if n is divisible by K, we use bin search to find the degree of occurrence of this number.)
 » 16 months ago, # | ← Rev. 2 →   -8 Could anyone help me to find why my code for problem C is TLE.92617935My solution is linear,as far as i know.
•  » » 16 months ago, # ^ |   +3 memset is the problem. If you have 10000 tests of n=1 then you will do 10000 (number of test) * 1e5 (maxN) operations
•  » » 16 months ago, # ^ |   +3 You should not clear array sz for each testcase since T can be sufficiently large to give TLE for your implementation. Don't use memset(sz,0,sizeof(sz)) each time. Instead, clear first n cells.
 » 16 months ago, # |   -9 Great Contest!! B was really original problem!!
•  » » 16 months ago, # ^ |   0 how to solve B I was getting WA in test 3
•  » » » 16 months ago, # ^ |   -7 Sort the array, took the first and the last 5 elements, now you have an array of 10 elements, just brute force for every 5 pairs and take maximum value.
•  » » » » 16 months ago, # ^ |   0 Can you help me what's wrong here, i tried a similar approach link
•  » » » » 16 months ago, # ^ |   0 Is this approach correct? Sort the array then take i elements from front(i goes from [0, 5]) and take [n-4+i to n-1] elements from the back. Just compute the max ans from all i. I get WA on test 3, any clues anyone ?
•  » » » » » 16 months ago, # ^ |   +3 Aah, of course it is correct, got my mistake i use using INT_MIN instead of LONG_LONG_MIN ;-)
•  » » » 16 months ago, # ^ | ← Rev. 2 →   0 Notice that there can only be 3 possible answers in B, 1) all largest positive numbers 2) 2 smallest negative and 3 largest positive numbers 3) 4 smallest negative and 1 largest postive numbers. Find the max of all three possible cases
•  » » » 16 months ago, # ^ |   0 sort the array,the last 5 elements are definitely the biggest now so their product can be the answer but apart from that if the product of first and second number is greater than fourth last and fifth last ,or product of first four numbers is greater than product of second last ,third last,fourth last and fifth last (this will happen when starting numbers are negative ) they can be the answer as well. So, only three cases arise,pick the best.
•  » » 16 months ago, # ^ | ← Rev. 2 →   +1 Actually.. problem B is a special case of this one ABC173_e where K=5
•  » » » 16 months ago, # ^ |   0 My comment was meant to be sarcastic
 » 16 months ago, # |   -8 E looked really easy to me, but I guess I was wrong.
 » 16 months ago, # |   0 That's a nice contest!Although I'm not able to solve E, I think the questions are very interesting and I really hope I can have the ability to make such greate problems :)
 » 16 months ago, # |   +1 How to solve C? it seems it requires some observations, please anyone give hint.
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 Basically there can atmax exist only 2 centroids and this happens when there exists an edge u-v such size of subtree of u is equal to size of subtree of v (in opposite directions) or basically s[u]=n-s[u].Now all you have to do is remove an edge from u and connect it to v and only 1 centroid will remain
 » 16 months ago, # |   0 How to solve idleness limit exceeded for problem E? Please help in this solution — https://codeforces.com/contest/1406/submission/92639473
•  » » 16 months ago, # ^ |   +4 It seems that cin>>t; isn't needed
•  » » » 16 months ago, # ^ | ← Rev. 2 →   0 Oh Thanks :)
 » 16 months ago, # |   +1 Did anybody else did o(10^5) solution for b?
•  » » 16 months ago, # ^ |   +1 Yeah but after 4 wrong submissions ;(
•  » » 16 months ago, # ^ |   +1 My solution has $O(n\cdot k)$ time complexity; in this case $k = 5$, so final time complexity is $O(n)$.
 » 16 months ago, # |   +1 How to solve E? I got to finding out all prime factors (and their powers) except one. There can only be primes with power more than 1 until 317. Until 317, we check all powers of each primes < 10^5. For primes higher than this, it can be a factor with at most one power. We can iterate through all primes from here and apply operation B. We also keep track of how many more multiples must be remaining at this point. If it differs, we know that this must be a factor (and something else). If we continue iterating all of them, we get all the prime factors (and their power is 1 as explained) except the first prime number. I couldn't figure out how to get this one. Can anyone solve the puzzle? :)
•  » » 16 months ago, # ^ |   0 I tried an approach similar to yours where I calculate how much should a number appear if it is divisible by a prime p,and I kept on using operation B for every prime while updating the amount of numbers divisible by all other primes bigger than p and if sometime they did not match up or if the answer was 1 I check it again and then find x or otherwise x=1.I don't know what I am missing but it did not work.
•  » » 16 months ago, # ^ |   0 I think we can group the primes greater than 317 into like groups of 100. After every 100 operations of the form "B p" we can do an "A 1" to check if the number of elements in the set has decreased by less than 100, in that case we know one of these primes is a factor of x and we go back over the group and perform an "A p" operation for each prime p in that group. I realized this in the last 5 minutes but didn't have time to fix my code :(
 » 16 months ago, # | ← Rev. 2 →   +16 Hard version of problem B: https://atcoder.jp/contests/abc173/tasks/abc173_e
•  » » 16 months ago, # ^ |   0 Sorry for the problem being well-known. None of the testers knew this problem before. (In fact B used to be exactly the same as abc173e, but to make it easier we chose k=5). We're really sorry for it.
•  » » » 16 months ago, # ^ |   0 thank you for this task, I didn't know how to solve it, and now I know
 » 16 months ago, # | ← Rev. 4 →   0 For D is this correct — $result = max(a_1 - x, tot + x)$ where $x = (a_1 - tot)/2$ and $tot = sum(a_i - a_{i-1})$ $if$ $positive$ ?
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 My idea that passed pretests was $sum = a_{1} + tot$ and $result =\frac{ (sum + (sum \ge 0)) }{2}$ (basically ceil division of $sum$ by $2$. What is $t$ in your notation?
•  » » » 16 months ago, # ^ |   0 Sorry t is tot only
•  » » 16 months ago, # ^ |   0 I didn't solve it in-contest but I think you can compute the minimum gain of sequence B and minimum decrease of sequence C by computing the sum of all rising edges (for B) and the sum of all falling edges (for C).The max of the two sequences is the start point of C or the end point of B. So we can shift up or down the entirety of B and C to find the point that equalizes those two points as closely as possible, which should be the optimum.
 » 16 months ago, # |   -7 CopyPasteForces
•  » » 16 months ago, # ^ |   +8 Really sorry for problem B. Still we hope you have enjoyed this contest.
 » 16 months ago, # |   0 Out of curiosity — is there some nice observation about the changes on both ends that allows D to be solved without using lazy prop for the range updates in the queries?
 » 16 months ago, # |   +3 Can anyone tell what is wrong in this code. for B. Link:- https://codeforces.com/contest/1406/submission/92640232
•  » » 16 months ago, # ^ |   0 almost same logic, but wrong answer at pretest2... got no idea
 » 16 months ago, # |   +7 Can anyone tell me what these numbers mean? I have never seen them before.
•  » » 16 months ago, # ^ |   +16 Probably number of tests
•  » » » 16 months ago, # ^ |   0 Is this a new feature?
•  » » » » 16 months ago, # ^ |   +9 Probably YES, but it was previously available on m1.codeforces.com(or m2,m3).
•  » » 16 months ago, # ^ |   +4 This number tells us the number of pretests in this contest
•  » » 16 months ago, # ^ |   +4 The number of pretests. I think you could use it to measure if your solution was good enough. A good feature to possibly prevent system test failing imo.
 » 16 months ago, # |   0 Am I correct in my logic for E?I first find all primes upto n, then print each of them with a 'B'. After this round of operations only two integers, 1 and x if x != 1 should remain or only 1 if x = 1.Then I print out each of the primes with 'A'. If I get 1, then i try its powers till i exceed n or get a 0. At each of these steps, if i get 1, I multiply the ans with x. I think that this should be enough but this fails pretest 2. Can someone please point out why logic falters?
•  » » 16 months ago, # ^ |   0 Number of request is about 2*PrimesUptoN which exceeds the given limit for N = 10^5.
•  » » » 16 months ago, # ^ |   0 My bad, I googled number of primes less than 10000 instead of 100000. I feel so silly right now
•  » » 16 months ago, # ^ |   0 This solution will fail since we can't make more than 10000 operations. This will work only if we are allowed to make ~20000 operations.
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 There are 9592 primes <= 100000. Thus, doing two questions per each one exceeds the limit.
 » 16 months ago, # |   0 When E seems to be easy than D, but that constraint You can perform the following operations no more than 10000 times made it quite tough to solve.
 » 16 months ago, # | ← Rev. 2 →   -37 https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/ just put 5 instead of K
•  » » 16 months ago, # ^ |   +1 Bruh...
•  » » 16 months ago, # ^ |   -7 YUP. This was insane.
•  » » 16 months ago, # ^ |   +65 Sorry for this, we had $k=3$ and find this in the last hour and we don't have other problems, so we just changed $k$ to $5$ so you can't google it, and that's the reason for delaying. Sorry again.
•  » » » 16 months ago, # ^ |   +1 And you guys still managed to create strong pretests for B!
 » 16 months ago, # |   0 For problem A, did the text change in between the contest? I think initially the problem was to split the set into two equal halves!
•  » » 16 months ago, # ^ |   0 The problem was worded very oddly, but no, there was B = Ø in one of the examples.
 » 16 months ago, # | ← Rev. 2 →   0 Can anyone please tell why this solution for problem B fails on pretest 3? link
 » 16 months ago, # |   +6 Thanks to KokiYmgch for "The way to find the centroids of a tree"https://codeforces.com/blog/entry/57593
 » 16 months ago, # |   +3 For Problem E I think we can group the primes greater than 317 into like groups of 100. After every 100 operations of the form "B p" we can do an "A 1" to check if the number of elements in the set has decreased by less than 100, in that case we know one of these primes is a factor of x and we go back over the group and perform an "A p" operation for each prime p in that group. I realized this in the last 5 minutes but didn't have time to fix my code :(
 » 16 months ago, # |   0 Is problem B solved by sorting the array and then using it like circular array by doubling its size by copying first half to 2nd half and use a sliding window of width 5 and find the max product? I Couldn't submit in time.
•  » » 16 months ago, # ^ |   0
 » 16 months ago, # |   0 Please help me to figure it out why this solution 92602531 for problem B is giving the wrong answer for case 1 and exactly the same solution runs correctly on other IDE (CodeChef and my own local IDE)I thought there may be issues with my current compiler version but after trying it with different versions, it stills gives wrong here but it runs correctly in other IDE's
 » 16 months ago, # |   -59 Div 2 B was from GeeksforGeekshttps://www.geeksforgeeks.org/maximum-product-subsequence-size-k/#:~:text=Answer%20%3D%200.&text=CASE%20III%3A%20if%20maximum%20element,positive%20integer%20in%20the%20subsequence.please make it unrated
•  » » 16 months ago, # ^ |   +4 bruh it is not a reason
•  » » » 16 months ago, # ^ |   -13 Many of my friends and other participants solved B in 3 minutes as it is a very famous interview problem.It was kind of unfair with the participants who haven't solved the question before.
•  » » » » 16 months ago, # ^ |   +10 Even if you know, there is a solution on gfg, you should try it yourself during the contest. Forget about those who copied from gfg.
•  » » » » 16 months ago, # ^ |   0 That is the case for all problems, and the reason why it is a good idea to practice a lot.
•  » » 16 months ago, # ^ |   +5 "pLeAsE MaKE iT uNrAtED"
 » 16 months ago, # |   +16 E was cool! ThanksBut i am actually not understand why we need answer on type B query...
•  » » 16 months ago, # ^ |   0 How to solve it?
•  » » » 16 months ago, # ^ | ← Rev. 4 →   +6 There are abut 9600 primes <= n, and about 70 numbers <= sqrt(n).Let me name prime number which <= sqrt(n) as small and >sqrt(n) as bigAnd actually there is 0 or 1 big number in X factorization.We can understand small primes in X factorization using about 200 queries (asking A queries about p, p*p, p*p*p, ...)And how to understand is there a big prime? We can arrange all big numbers on 100 groups of 100 numbers. Lets check every group separately using B queries and after checking it ask A(1) to understand was bg prime found or not. If it was found check this group again.It will be like 200+9550+200=9950 queriesSorry for my bad English ;(
 » 16 months ago, # |   -10 Best channel for editorial of this contest . https://www.youtube.com/channel/UCBStHvqSDEF751f0CWd3-Pg/ Do view and subscribe
 » 16 months ago, # |   +1 How to run interactor & solution for an interactive problem in terminal/cmd?
 » 16 months ago, # |   0 I tried solving B with multiset can any one tell me why am I facing TLE. Worst complexity is O(n * 2logN). Submission
•  » » 16 months ago, # ^ | ← Rev. 2 →   +1 Why? You iterate from 2 to $n-1$ and inside you iterate over all positive numbers. It's $O(n^2)$
•  » » » 16 months ago, # ^ |   0 My approach was to for every i from 2 to n — 2, find the max, min product of two numbers before ith and max, min product after ith of two number. Then multiply them accordingly based on the value of element i. I did not understand how it is O(n^2).
•  » » » » 16 months ago, # ^ | ← Rev. 2 →   +1 Because at $i = 2$ you iterate all positive numbers ($solvemin(muls)$). At $i = 3$ you iterate all positive numbers except maybe one. At $i = 4$ you iterate all positive number except maybe two and so on.And besides you pass multisets by value. So you at each step copy multisets.P.S. Yes I see now that you iterate at max over two numbers. So copying multisets is you main problem.
•  » » » » » 16 months ago, # ^ |   0 In solvemin(muls) I am breaking after I get two numbers from the multiset.
 » 16 months ago, # |   0 For problem C — if there are two centroids then we disconnect the leaf of 1 centroid and connect it directly to the other centroid. How does this not works ?
•  » » 16 months ago, # ^ | ← Rev. 2 →   +4 you always print 1 2 if there is just one centroid and that s the mistake. The edge 1 2 may not exist
•  » » » 16 months ago, # ^ |   -12 there might be three centroids.
•  » » » » 16 months ago, # ^ |   0 nope , always are 1 or 2
•  » » » » » 16 months ago, # ^ |   0 why though ? (im a beginner)
•  » » » » » » 16 months ago, # ^ |   0 i don't know exactly how to explain. Tree doesn't contains any cycle so you can't have the 1-2 2-3 3-1 case when all nodes are centroids. Wait for the tutorial i m sure that will be a proof.
•  » » » 16 months ago, # ^ |   0 Whhhhhaaattttttttttt, whyyyyy,come on codeforces (ノ｀Д´)ノ彡┻━┻
•  » » » » 16 months ago, # ^ |   0 yeah..i made the same mistake but i fixed it after 30 minutes and 2 WA :|
•  » » » » 16 months ago, # ^ |   0 I'm so mad at my stupidity.
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 Deleted.
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 I did same and it worked for me.
•  » » 16 months ago, # ^ |   0 I did the same and it worked for me. https://codeforces.com/contest/1406/submission/92632511
 » 16 months ago, # |   -47
 » 16 months ago, # | ← Rev. 3 →   +40 My approach to problem E:$Ans=p_1^{k_1}\cdot p_2^{k_2} \cdots$.For $p \le \sqrt{n}$, just ask their exponents directly. This costs about $180$ queries.For $p \ge \sqrt{n}$, there is at most one prime factor. So remove $\sqrt{cnt_{primes}}$ primes in one time and check if the number of remaining numbers is wrong. If wrong, the factor is in the last $\sqrt{cnt_{primes}}$ numbers you remove.There are $cnt_{prime}+2\sqrt{cnt_{prime}}+41+66\times 2$ queries at most. The time complexity is $O(n \ln n)$. Since the system test isn't finished, I don't know if it's correct.edit: passed system test.
•  » » 16 months ago, # ^ |   0 So you would ask the exponents in a descending manner? Like for example $2^{19}$, $2^{18}$, ..., $2^1$, and see if one of them matches?
•  » » » 16 months ago, # ^ | ← Rev. 2 →   +5 Asking in increasing manner is better (i guess). A number $n$ can be written as the product of at most $log(2, n)$ prime factors ($<20$ for $n=100000$), which means you will stop at the first query for most of the primes.
•  » » » 16 months ago, # ^ |   +5 You can also binary search on exponent. But it isn't necessary here.Ascending or descending anything is fine.
•  » » 16 months ago, # ^ |   0 Ive done the exact same thing but ive get WA(i checked it it was because of the number of queries) and cntprime is around 9700 right!?
•  » » » 16 months ago, # ^ |   0 There are 9592 primes exactly. I think 408 queries are enough to do other things.
•  » » 16 months ago, # ^ | ← Rev. 2 →   0 can you please explain with an example . I didn't get for p>=sqrt(n) . What is cnt_primes ?
 » 16 months ago, # |   -66 make this round unrated as B problem's code can be easily found,here are the links-> https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/ https://atcoder.jp/contests/abc173/tasks/abc173_e https://www.codechef.com/problems/MMPRODall those who have copied these code or have taken the logic from there are at an edge ahead of those who are solving problem honestlyUnrate this contest
•  » » 16 months ago, # ^ |   0 Ok. Now shut up
 » 16 months ago, # |   0 Is LONG_MIN not defined in codeforces?
•  » » 16 months ago, # ^ |   0 $#include$
•  » » » 16 months ago, # ^ |   0 I had included this header
•  » » » » 16 months ago, # ^ |   0 So you can use LONG_MIN. Show submission where you cannot use it?
•  » » » » » 16 months ago, # ^ |   0
•  » » » » » » 16 months ago, # ^ |   0 Its compiling, its just WA
•  » » » » » » 16 months ago, # ^ | ← Rev. 4 →   0 $LONG\_MIN$ is minimum for $long$ type$long$ type is signed type that at least same size as $int$. $int$ type is signed type that at least 2 bytes. Sizes of that types are different between compilers. So maybe you want use fixed size types.If you want minimum of 8-byte signed type. You should use i64 and $numeric\_limits::min()$ or $INT64\_MIN$
•  » » » » » » 16 months ago, # ^ |   0 Try LLONG_MIN
•  » » 16 months ago, # ^ |   0 I think LONG_MIN is same as INT_MIN on codeforces, because one time I was getting wrong answer using LONG_MAX and had to change to LONG_LONG_MAX
•  » » » 16 months ago, # ^ |   -19 Yeah because of this I kept getting WA on test 3 for problem B. I think this sould be fixed by codeforces
 » 16 months ago, # | ← Rev. 2 →   +5 am i the only one who printed "1 2\n1 2" if there is only one centroid in C?
•  » » 16 months ago, # ^ | ← Rev. 2 →   +8 that doesn't necessarily work because edge 1 — 2 may not exist in the input treea solution is just to save an edge from the input and output that one instead
•  » » » 16 months ago, # ^ |   0 yeah, that was my only mistake
•  » » 16 months ago, # ^ |   0 same , cost me 4 wrong answers.
 » 16 months ago, # | ← Rev. 3 →   0 Is it possible to solve prob/B with dp ? Thank you for your responses but is there a recurssive dp solution ?
•  » » 16 months ago, # ^ | ← Rev. 3 →   +14 Definitions: $mx[i][r]:$ max product of $r$ numbers in the prefix $1...i$. $mn[i][r]:$ min product of $r$ numbers in the prefix $1...i$. Now, the transitions: $mn[i][r] = \min(mn[i-1][r], \ mn[i-1][r-1]\cdot a[i], \ mx[i-1][r-1]\cdot a[i])$. $mx[i][r] = \max(mx[i-1][r], \ mn[i-1][r-1]\cdot a[i], \ mx[i-1][r-1]\cdot a[i])$. Now, base cases: $mn[0][0] = mx[0][0] = 1$; $mn[0][r] = INF$ and $mx[0][r] = -INF$ for $r \neq 0$;
•  » » » 16 months ago, # ^ |   +3 mn[0][0] and mx[0][0] need to be initialized with 1, not 0
•  » » » » 16 months ago, # ^ |   0 Yes, my bad.
•  » » 16 months ago, # ^ |   +1
 » 16 months ago, # |   0 https://codeforces.com/blog/entry/57593 By just applying this links code to tree C problem can be solved
•  » » 16 months ago, # ^ |   +11 What's next after finding the centroids of the tree using the code in the tree? The problem isn't solved yet!Finding the centroids are just a helping tool and not the full solution
•  » » » 16 months ago, # ^ |   0 If $N$ is odd, is it always the case that the answer is printing a random edge twice (unique centroid)? Can anyone help me in proving that or hacking my submission?I have tried to find a sub-tree of size exactly $N / 2$ if $N$ is even.My submission: 92627903
•  » » » » 16 months ago, # ^ |   +3 Indeed if N is odd, there can be only 1 centroid. Thus printing a random edge twice is valid.Then finding a subtree of size N / 2 if N is even, is also correct. Since those are the only nodes that satisfy sz[node] >= N / 2 and N — sz[node] >= N / 2 -> node is a centroid.Congratulations! Your solution is correct!(I might be wrong though feel free to correct me)
•  » » » 16 months ago, # ^ |   0 IMO the tricky part of the solution is to find the centroid as after finding centroids we can remove any subtree from one centroid and connect it to the another centroid
•  » » » » 16 months ago, # ^ |   +1 And do you have proof of that? Does it always work?You may have found that as intuitive, but many others don'tThe problem tests you if you found that property, not testing you how to find centroids
•  » » 16 months ago, # ^ |   +4 Yeah applying this code gives you AC but only "JUST" applying wont get you anywhere ,this is the basic difference between having a particular tool and knowing where and how to use it.
 » 16 months ago, # |   +1 In problem C, how can we prove that there can at max 2 centroids? I took this as an assumption (based on observation) and fortunately it turned out be correct.
•  » » 16 months ago, # ^ | ← Rev. 2 →   +9 This is my personal proof for this. Correct me if I'm wrong.Proof: assume that there's k > 2 centroids. Then, those k vertices should be connected by (k-1) edges (since they should form a tree). But because by Pigeonhole Principle, at least 1 vertex will certainly get 2 or more subtrees (from the other centroids). Then the centroid would be that vertex. Contradiction.Hence, the number of centroids should be at max 2.