Всем привет!
28 и 30 июля состоятся туры EJOI — личного соревнования по программированию для юниоров, проходящей по правилам международной олимпиады школьников по информатике.
EJOI составляется из самых интересных и сложных задач, которые были предложены многочисленным коллективом наших авторов, поэтому, не желая лишать вас возможности поломать голову над задачами полного проблемсета вместе с конкурсными участниками, мы собираемся провести параллельно со вторым туром рейтинговый раунд Codeforces, основанный на задачах обоих туров олимпиады.
В связи с этим мы просим всех участников сообщества, собирающихся участвовать в соревновании, проявить уважение к себе и другим участникам соревнования и не пытаться читерить никоим образом, в частности, выясняя задачи у участников соревнования в Иннополисе. Если вы узнали какие-либо из задач EJOI (участвуя в ней лично, от кого-то из участников или каким-либо иным образом), пожалуйста, не пишите раунд. Участников олимпиады мы просим воздержаться от публичного обсуждения задач. Любое нарушение правил выше будет являться поводом для дисквалификации.
Раунд состоится в 30.07.2018 11:15 (Московское время) и продлится 2.5 часа. В каждом дивизионе будет предложено по 6 задач.
Задачи раунда были придуманы и подготовлены tourist, PavelKunyavskiy, niyaznigmatul, 300iq, GlebsHP, pashka, qoo2p5, VArtem, demon1999, Flyrise, ifsmirnov, isaf27, yeputons, cdkrot.
Также спасибо за тестирование туров never_giveup, vintage_Vlad_Makeev, izban, GoToCoding, Egor, doreshnikov, Sert, disa, wrg0ababd, craborac, budalnik.
И, конечно же, спасибо MikeMirzayanov за великолепные системы Codeforces и Polygon.
Всем удачи!
UPD: поздравляем победителей!
D1:
1) Um_nik
2) LHiC
3) dacin21
4) ksun48
5) Swistakk
D2:
2) WaldarDoppen
3) IHaveHir
4) cly_none
5) Shadi.Gh
Please translate to English
On July 28th and 30th, EJOI tours will take place — a personal programming competition for juniors, taking place according to the rules of the international schoolchildren's competition in informatics.
EJOI is made up of the most interesting and complex tasks that were proposed by a large team of our authors, therefore, not wanting to deprive you of the opportunity to break your head over the problems of the full problem of the network together with the contest participants, we are going to run a rating round of Codeforces, based on the tasks of both rounds of the Olympiad.
In this regard, we ask all members of the community who are going to participate in the competition, show respect for themselves and other participants of the competition and not try to cheat in any way, in particular, clarifying the tasks of the participants in the competition in Innopolis. If you learned any of the tasks of EJOI (by participating in it personally, from someone from the participants or in some other way), please do not write a round. We ask the participants of the Olympiad to refrain from public discussion of the tasks. Any violation of the rules above will be an excuse for disqualification.
The round will be held on Monday, July 30, 2018 at 11:05 and will last 2.5 hours. In each division, six tasks will be proposed.
at 11:05 o'clock in Moscow time.
Sorry, it is translated now.
Thanks.
Your level is increasing man.
When you click on EJOI, it writes no such blog entry.
Thanks, fixed.
Nice IQ(300).
23 coders are here.
Wish there will be no internal errors, because... well... 500...
#iykwim #okthisisabadjoke #imout
isaf27 is among the names of this round while peltorator is among the names of the last round :D
upd:my fault,isaf27 is also among the names of the last round
isaf27 is also among the names of the last round
Congratulations everyone on the 5th century of contests here on Codeforces :) :D
5th Century of Codeforces round, not Codeforces Contest. :)
As educational rounds are also Codeforces Contest but not Codeforces round. :)
Let's hope for a nice 500th anniversary!!
Happy 500th round, Codeforces!!!
Be scared, be very scared.
ez katka round
Registered.
Young man. Are you eager for force?
(Please downvote this comment)
i am not in !!
very excited to join a contest prepared by all of these legend.
DELETED
DELETED
A not bad time for Asian users :) ?
Hey, maybe the contests based on OI have more unusual time which is not in evening or night? :P (Like Moscow Open Olympiad (Round #469))
HAHA!You are so funny.
I can see by your cf icon you too have refined taste.
Yeah,it is so tiring to take part in contests at midnight,especially when you need to go to school tomorrow
but tomorrow's div 3 will start at night again ...
For those in US wondering how the hell to wake up at 4 AM, let me give you some advice:
3 AM here in Colombia :)
Well, did you follow this advice? Somehow I actually managed not to sleep, hope I can solve more than one problem :D
haha, no, I went early to bed. Im not sure if not to sleep is a good idea haha
Is it emmm, no no nothing.
Should I skip my school to participate?
If u can, why not?
Well, I just did this in Round #469 which has the same beginning time ┑(￣ _ ￣)┍
Херня какая-то
Yes, I made some mistakes, sorry. Next meme gonna be better ;) ________________________________________________________
Our greatest glory is not in never falling, but in rising every time we fall.
Don't worry, here at codeforce we like to say "practice makes perfect", whether it is segment tree or shitpost.
another round with tourist's problems ... ಠ_ಠ
God bless me ...
Have now made my schedule free for Codeforces #500! <3
The best feeling ever (Despite it being at 1:30 pm here, on a Monday!)
I honestly expect quite a lot of people seeing EJOI Day 1's tasks from contestants, which messes with the standings by having an unfair advantage. Expecting everyone in an online competition to play fair isn't enough.
Either make this contest unrated, or change Day1's tasks with original ones.
Leaving such an obvious method for cheating unchecked certainly shouldn't happen for Round #500, which instead should be celebrated as a milestone in fair competitive programming.
I think that the main purpose of the contest is to solve interesting tasks, not to earn ratings. We will try to do everything to fight with cheaters, but I don't think that cheaters can ruin the contest. Just enjoy the tasks!
The onsite contestants were asked to don't disclose the problems.
Anyway, there are ways to cheat in codeforces rounds even without onsite competition present. So we hope, that most people on codeforces just simply don't cheat.
I really like the tone of this comment, "Do this, do that, it should be". Should we also buy you a Ferrari and a nice house?
I think this team of experienced contest authors together with Mike can figure something out, don't you think?
300iq dude you gotta chill out. You've been included in lots of rounds recently.
I wonder if this contest would be too difficult for Div 2 contestants.... That's because actual EJOI is for pro junior coders..
I think that first 2 or 3 problems won't be from the EJOI, because A, B and sometimes C are very easy and I think there aren't that easy problems on EJOI. Please correct me if I am wrong.
feeling good to see tourist in problem setters...i think this contest is going to be more interesting than other contest........
upvotes for tourist.
UPD1: Now downvote for starboy_jb.
UPD2: don't do anything with this comment.
downvote for sguu
keke
14 writers just for a single contest!
I will miss this contest because I have to take part in HDU 2018 Multi-University Training Contest 3contest is here...
me too,hhhh!
?detaR tI sI
Happy round #500
What controls registration time? We can register in Educational round but not in #501 yet.
Подскажите, зачем ставить контест в 11 утра мск? Не могу найти ни одной часовой зоны, для которой время начала на час-два позже не было бы удобнее.
Полагаю, чтобы начать вместе с основным ивентом
Я понимаю, почему не надо начинать раньше основного ивента, но если начинать позже, то какая разница во сколько?
Возможно, удобно для школьников и студентов, у которых сейчас каникулы.
Чем удобнее? Думаю большинство вставать рано не любит.
So many writers? Pfft. I hope this won't be disappointing. Stupid writers always disappointing. I wish you all minus rating changes. F*** y'all
I hate 10 minutes delay :(
Better than long submission queues during the round.
That's better than 1 hour delay,isn't it?
Not really, if you have time.
10 min delay :'(
Ten minutes for dinner:(
Ну ёптель-моптель. Когда-нибудь КФ вовремя начнётся?
What's the score distribution?
As a result of 10 minutes delay, I will miss my supper.
same here, I just want to participate to contest until end of my lunch time, which means 55 minute of the contest. Now they decreased it to 45 minutes by delaying it :(
After getting a %l64d error, I wasn't able to submit my code on problem A, it always said that I've submitted exactly the same code before, even if I changed things in it. I also wasn't able to ask questions on the problem.
It may be caused because I registered in the 10 minutes delay time.
When you solve A,B,C then leave cuz you don't have anything to do. :')
Same in div1...
I guess I'm back to div2 again...
How to solve B?
Represent each row and column as a node, and connect a row and column if cell (x,y) exists. Answer is number of connected components minus one.
Thanks.
Make a bipartite graph (the left side of the graph is for ri and the right side is for ci, edge (ri, ci) exists if we have element (ri, ci)), then count the number of connected component.
This solution works because if you will be able to make element (ri, ci) if you can reach node ci from node ri.
DSU on rows and columns. When you're given a cell, you merge the x and y coordinates. The answer is the number of components in total — 1.
szawinis Why are you doing merge(x,n+y)? A detailed explanation of your approach would be really helpful..
Well, that's just a way of separating the x and y coordinates while keeping them in the same dsu array. Since the first n vertices are taken up by the x coordinates, we can just append all the y coordinates by doing n+y.
For div1A you can prove it is always optimal to make all the x's a contiguous subarray when the ai is sorted right
Yeah, so sliding window will work.
Note that when you're doing a proof you may get stuck at one point while trying to induce a contradiction, but you should also pay attention to whether the y's form a contiguous subarray as well.
If you assume that neither the x's nor the y's make a contiguous subarray, you should be able to easily induce a contradiction of the minimality.
Yes, because you can make the x's contiguous without affecting difference between the highest and lowest y (can even decrease it), resulting in difference between the highest and lowest x being decreased (or not changed at least).
How many cases are there for Div1D?
I think 2. Update: I don't know.
Div 1:
I tried C with a DP approach but got WA on test 8: 40959527.
Can anyone point out my mistake?
Please, explain what is d[i][j][0] and d[i][j][1] in your solution (we are not telepaths)
Alright. Should have some clarifications.
In this, I declared dp[i][j] as the minimum time used to build j houses, using hills with 0-based-index not higher than i.
dp[i][j][0] is the minimum time to build j houses without building a house at the i-th hill, while dp[i][j][1] is the minimum time to build j houses, including one at that hill.
You need to update dp[i][j][1] with minimum of dp[ [0,i-2] ][j-1][1]
I haven't found the mistake yet, but I met difficulty when I was trying to solve it in your way which uses dp[i][j][0] and dp[i][j][1]. So I tried to use dp[i][j][0] ,dp[i][j][1] and dp[i][j][2]. dp[i][j][0] is the minimum time to build j houses without building a house at the i-th hill or i-1-th hill. dp[i][j][1] is the minimum time to build j houses, including one at that hill. dp[i][j][2] is the minimum time to build j houses, including one at i-1-th hill.
Could you please explain the transitions in your solution?
I omitted a dimension of the array, maybe this cause my solution hard to understand. f[i][j][0]=min(f[i-1][j][0],f[i-1][j][2])is obivious. f[i][j][1]=min(f[i-1][j-1][0]+max(0,a[i-1]-a[i]+1),f[i-1][j-1][2]+max(0,min(a[i-1],a[i-2]-1)-a[i]+1)) Transition from f[i-1][j-1][0] means that you only have to lower the height of i-1-th hill from its origin height, which costs max(0,a[i-1]-a[i]+1) hours. Transition from f[i-1][j-1][2] means that you have to lower the height of i-1-th hill from the height which had been lowered when calculating f[i-1][j-1][2] and is already min(a[i-1],a[i-2]-1), so this costs max(0,min(a[i-1],a[i-2]-1)-a[i]+1) hours. f[i][j][2]=f[i-1][j][1]+max(0,a[i]-a[i-1]+1), because there is no other choice. I finally realised my poor English when I was trying to explain my solution... Hope you can understand it.
Thank you very much!
The first bug I see: while counting d[1][1][1] you don't consider a case when the first (0-th) hill is lowered. Can you please determine if d[i][j][1] means that you must take the i-th hill or that you can take the i-th hill?
Looks like I do have trouble clarifying stuffs... My apologies.
dp[i][j][1] means i-th hill must be taken.
I think the bug is this: when you take the-th hill you have to lower the next hill. But then you forget about it and don't put it in consideration when you take the d[i-2] value.
You have to consider the first i hills in isolation. For example see my solution (but here 0 means that we don't count the i-th hill, and 1 means that we can count or not count it).
Solution
Finally, mistake found — a stupid one I must say.
i64 dp[5001][2500][2];(It should be 2501 for the second dimension).Like, I've neglected Carrays for a very long time (only using it today since nested vectors didn't suit in 1s TL), and then this happened...
Still thanks for all the support! rembocoder Tutis ouuan
I always use dp[5010][2510][2] in this case.
Heck, I should spend sometimes revising Carrays again, just to have the flow in my head. Such issues are straight intolerable :D
I can finally see my rating change before going to sleep as a Chinese!
And how beautiful when it is going to be a great rise.
Have been monitoring your progress, and I can only say congrats! ;)
Codeforces's progress proves the revolution successful and makes itself better!^_^
Maybe I won't be able to? The rating change hasn't been shown yet.
Now I can see the rating change.
Which problems were from EJOI. The ones from Div.1?
D1 B+ were from EJOI
Cool. Thank for the quick answear.
Can tell us tasks by days ?
https://pcms.university.innopolis.ru/results/org/ejoi/2018/day2-salty.html
I thought, which tasks was A on first day, which one was B etc. I do not see that option on the link :)
Move your cursor on A,B,C,D,E,F.
Thanks, I am stupid :D
It is interesting that there is more AC on C div 1 than on B div 1.
¯\_(ツ)_/¯
Wow, had E coded for last 8 minutes but wifi went out I guess, and iPhone took too long to recharge so not submission :'(
EDIT: :'(
Cheer up T_T..
It seems that you would have won this round if this didn't happen. So sorry.
But still, congratulations on becoming nutella :)
Ehm, I spent some 20min going in wrong directions on E, then wasted a lot of time on D. About an hour after the contest, I decided to revisit E and figured out I can take Euler cycles in about a minute... and the rest is straightforward.
My Modest Solutions
A-Piles with stones
http://coolforces.blogspot.com/2018/07/1013a-codeforces-piles-with-stones-solution.html
B-And
http://coolforces.blogspot.com/2018/07/1013B-codeforces-And-solution.html
Why cannot solve AB-Strings by greedy T_T...
Was I missed something?
How was your greedy, exactly?
Also, please keep in mind that the commands with string are linear in time complexity, in case you implemented them directly on your solution.
Yeah, I also did not understand why does not it work to just break an "ab" in one string and a "ba" in another. Though I just saw the number of OKs and the number of WAs and decided not to think further =/
For a test like:
It's better to make the number of cuts in the two strings more equal. For example, we balance them on the very first move:
Oh.. I think this was it! Thanks!!
so the first time pre-processing was that I was missing
You have 6 total letters in the input data and 8 in the output, seems like you've meant:
Thanks, fixed!
Uhm, are you sure? If you consider it correct, I don't understand the format you've chosen for it.
Argh... sorry! Another try above. Yeah, now yours and mine look equal.
Okay, seems alright now, thanks for the test
Everyone solved D as DFS!!! I never thought of it
Can someone explain it to me??
Let's present a table as bipartite graph. N vertices for rows and M for columns. Then an element means an edge in this graph. If we have a path of three edges, we can connect the ends of this path with an edge. This means that if we have a longer path a_1, a_2, a_3, ..., a_k, we can at first connect a_1 and a_4, so the path becomes shorter, and so afterwards we can connect a_1 and a_k. So if two vertices are connected, then we can draw an edge between them already. Thus we only need to buy edges to make all the graph connected, so the answer is the number of components — 1.
that moment when you spend 1 Hour try solve D then you realize that you misunderstand problem because you are shit in English
How to solve Div2 C? tried a lot but could not get it :(
I approached it with greedy solution — minimize one dimension of the desired rectangle as much as possible.
To achieve that, we will sort 2 * n coordinates, and choose n consecutive coordinates in the new sorted array to be coordinates of the same dimension (either x or y, doesn't really matter). Technically, we'll keep a sliding window and try all possible n-length consecutive subarrays.
I solved Div. 2 B in 16 min. but then accidently resent it after one hour and a half. Both of them are correct Which solution will be rated?
last submission will count
Are there ways to recover smh?
your first solution will not be judged. It will get skipped
It hurts. Thanks.
18 and 43 tests in B kills all submissions
THIS COMMENT WAS A WONG SENT
DELETED
For the problem D in Div.1, Both ko_osaga (submission) and Konijntje (submission) got accepted. But for the input below, ko_osaga and Konijntje's outputs are different.
<< INPUT >>
<< ko_osaga's output >>
<< Konijntje's output >>
hmm, I think it should be 6 for this case.
you should join codeforces team
My answer also has 6 actions.
When a solution considers a number of cases, there can be quite a few of them (mine, for example, has a total of 18 calls to the main solution function, just to be sure). So, if one of X such cases is solved incorrectly, informally, it is X times harder to catch it with tests. So it is understandable if something like that slipped through.
I made an identical TC in contest to test my code. It's a basic one, and if I was a tester I won't tolerate it. (But still I appreciate all setters effort!)
FYI, smaller countercases :
Answer is 4.
Nice.
NO ITS NOT NICE
Very nice
I read this in Skeletor's voice.
a sudden difficulty rise between div1 C and D .
Means we need code faster again
Guys there was some cheating going on today. Guys like n00bie knew the questions beforehand and hence achieved a good rank. What a disgrace!
I guess I'll just repeat Jatin.
Get a life.
Can anybody tell me how to solve div1D?
Consider all substrings
abandba: the switch positions where the two consecutive letters are not equal. Note that each cut can remove at most one switch position. Now, if we cut...a|b...in one string and...b|a...in the other, and swap the prefixes, we remove two switch positions simultaneously. However, one swap can remove no more than two of them, one from s and one from t.What's left to do when, for example, s has no switch positions and t has X > 0 of them? We can remove one switch position from t with one swap. But we can do better: cut the string t so that approximately half of these positions go into s. Then we can again remove two switch positions per swap, which is twice as fast.
In fact, we can start by counting the number of switch positions in s and in t (let them be X > Y), and if they are not equal, make the first cut so that (X - Y) / 2 of them go from s into t, making their count approximately equal. After that, proceed with the greedy solution which removes one switch position from s and one from t in each swap.
What's left is to consider which string will turn into
aaa...aand which intobbb...b, and formalizing the "approximately equal" above. We can just solve for a few ± 1 cases and take the minimum to be sure.Thanks a lot.
Don't. Solve E instead.
Seriously, if it looks simple and there are few correct submissions, maybe the next problem will be more doable — or if it isn't, at least it won't give you a headache.
I found E pretty easy for 2500 pointer, but D was even easier, but just tedious with implementation. For me doing E was better decision than D, but you can't simply discard such problem a priori, it wasn't that bad.
I agree that D was easier than E idea-wise, but if the implementation is tedious enough that there are only 20 correct submissions (while 180 people attempted) in a 2.5 hour contest, and if the first correct submission came after 80 minutes... that implementation is pretty brutal. I think if D and E were swapped, there would be more correct submissions on that D than on that E.
I'm not talking about discarding it a priori, but about recognising that it's not worth spending a lot of time on — either from experience, from looking at the statistics or trying it and seeing that there are a lot of cases and it's easy to miss some (so easy it happened to the authors, even).
editorial?
Rating should be updated now But it has been 3 hours since ending and rating haven't been updated even
Seems like it's not that easy this time. Like they also rolled back the rating changes for round 499.
a sudden disappear of round499 rating record?
whats wrong with cf 499 round??
see this: https://mirror.codeforces.com/blog/entry/60921
Что происходит с цветом в таблице результатов? Кажется, цвет людей считается будто бы без учета 499 раунда? Можно ли это как-то поправить? (например, я на данный момент фиолетовый в таблице результатов div2 контеста 500, хотя должен быть синим, так как до начала контеста мой рейтинг составлял 1899)
It is temporary
Is there any problem with previous round ( Round # 499 ) rating !
Problem has been fixed.
Will there be an editorial?
Editorial?
http://mirror.codeforces.com/blog/entry/60920
А разбор будет?
For people who came up with D in contest, how did you come up with solution to a problem like this? Did you see similar problems or did you figure it out from writing out the pairs/graphs by hand? I understand solution, but coming up with it seems like total magic to me.
Well, yes, associating rectangular grid with bipartite graph is a known thing.
congratulations man. Only your hatters will downvote this comment.
It makes sense that we need to have all (x,y) values in the same connected component, if all cells that we make for free are (x,y) where x is from set R and y is from set C and R is set of all x's and C is set of al y's in that component.
A
Suddenly noticed that both the Announcement and the Editorial aren't linked to any contest...
And the result is no contest materials link in the contest Dashboard or problemset
UPD: Fixed.
The comment is hidden because of too negative feedback, click here to view it
Where can I find the tests to the output-only problem from EJOI?
when are u going to upload editorial??
The editorial existed all this time, but the link to it wasn't posted on this post. http://mirror.codeforces.com/blog/entry/60920