Привет, Codeforces!
Рад пригласить вас на Codeforces Round 637 (Div. 1) - Thanks, Ivan Belonogov! и Codeforces Round 637 (Div. 2) - Thanks, Ivan Belonogov!, которые пройдут в 23.04.2020 17:45 (Московское время). Раунд будет рейтинговым для обоих дивизионов (я надеюсь).
Все задачи были придуманы и подготовлены Алексеем Aleks5d Упирвицким, Алексеем alexX512 Перевышиным и мной, Денисом I_love_myself Сапожниковым.
Большое спасибо нашим многочисленным тестерам: 300iq, voidmax, ashmelev, Akulyat, okwedook, Minnakhmetov, divanik, Zakoden, Jostic11, Nakinamo, 4qqqq и allisyonok за тестирование задач и ценные советы, isaf27 за координирование и помощь в подготовке раунда и MikeMirzayanov за отличные системы Codeforces и Polygon.
Вам будет дано 6 задач в обоих дивизионах и 2.5 часа на их решение. Советую прочитать все задачи. Удачи, высокого рейтинга и удовольствия от решения задач!
Разбаловка будет объявлена ближе к началу раунда.
Уже заметили необычное в названии раунда? Вот короткое сообщение от MikeMirzayanov:
Этим раундом мы хотим передать пламенный привет и еще раз сказать спасибо за поддержку Ивану Belonogov Белоногову. И дело не только в существенном подарке по случаю 10-летия платформы. Начав участвовать в 2011-м, Иван прошел долгий путь от участника раундов второго дивизиона до международного гроссмейстера, стал чемпионом мира ICPC в составе команды ИТМО. Такой вот яркий мотивационный пример success story! Спасибо, Иван!
UPD1: Разбаловка будет следующая:
Div1: 500-750-1250-1750-2250-3000
Div2: 500-1000-1500-1750-2250-2750
UPD2: Из-за технических проблем раунд переносится на 10 минут. Извините за неудобства!
UPD3: Разбор опубликован!
UPD4: Это был наш первый раунд на codeforces и мы сделали ошибки везде где только можно, приношу свои огромные извинения. Мы обсудили всё и приняли решение: Div1 раунд будет не рейтинговым, Div2 останется рейтинговым. Более подробно вы можете прочитать в этом посте. Надеемся, что вы насладились решением интересных задач и не стали обращать внимание на наши ошибки.
Just go away! Don't read this....
Let's take the initiative to never Comment.Lol!
Yeah! This happens every time when I comment!
And just do it one more time!
"This happens every time"
maybe because your comments are no help or use for the community.
Good luck and have fun, everyone! :D
Some division rounds are of 2 hours while some of 2.5 hours. I am curious as they both have same number of questions(on average). What is the reason then? Is it that the 2.5 hour rounds are slightly tougher/ implementation heavy?
Ahaha such a funny meme)))) YES BUT ACTUALLY NO)))) It's like "YES", but actually "NO" :)))))
Disturbing joke!
Aha.. figured it! They contain questions similar to 2hr round contests but in a difficult language!
Well, I took Virtual participation last night and spent 0.5h understanding the meaning of Problem B.
The statement is too difficult for people who do not use English as mother tongue to understand quickly.
As one of the testers of this round, I can say that tasks are too interesting. Hope that the authors ' efforts will not be in vain :_)
Wow!
Bait
It definitely is.
orz
yep
You were right. Really good problems.
I hope it'll be easy.
How many problems will there be?
5 or 6 or 7
or
6 or 7 or 8?
"You will be given 6 problems in both divisions"
I wish this blog was posted at codeforces top page so I don't have to find this blog.
https://mirror.codeforces.com/blog/entry/68665
i have some questions...
WOW !! Short announcement!
Hope the problem statements will be also short and we will enjoy the problems. Thanks.
Hope the problem statements will be short.
My first Div. 1 contest! Hoping I can still participate in Div. 1 after this.
My first Div. 1 contest! Hoping I can still participate in Div. 1 after this.
what is ivan's codeforces handel?
Belonogov
It has been mentioned that $$$Please,\,read\,\, all\,\, the\,\, problems.$$$ Does this mean that problems might not be in increasing order of difficulty?
nope there is story of Fricking Nastya and Simp Denis.
.
it seems like creators dont like short problems
agreed
The only question about the contest
Will the No. 1 spot in the rating change ?
tourist or jqdai0815 or Anyone else
chill bro,LGM dont need to worry
Deleted
But still we can see your deleted comment :p
"I love myself" is such a good user name :D
Hope this round will have strong pretests.
Wish everyone high rating and I hope that most of you go to the next rank in this contest.
hopefully the problem statements would be short..
Добрый день, MikeMirzayanov!
И дело не только в существенном подарке по случаю 10-летия платформы. Начав участвовать в 2011-м, Иван прошел долгий путь от участника раундов второго дивизиона до международного гроссмейстера, стал чемпионом мира ICPC в составе команды ИТМО. Такой вот яркий мотивационный пример success story!
Раз дело не только в пожертвовании, хочу уточнить: я начал учавствовать в 2012-м году, начав со второго дивизона; до революции цветов какое-то время был красным, а в этом сезоне прошел в финал acm icpc. Достаточно ли моя success story ярка и мотивационна, чтобы в мою честь тоже провести раунд?
Что непонятно в словах "не только"???
Недостаточно — за всё это время так и не научился правильно писать слово "участвовать"!
Much more frequent contests
К сожалению, не смогу принять участие в раунде, Денис. Но я надеюсь он пройдёт успешно. Удачи!
Спасибо)
I_love_myself will there be any sub-part problems?
Good！
Thanks, Ivan Belonogov.
Good luck, guys!
We will be live upsolving the round when it ends:
https://youtu.be/0kr6YSroclg
hey!! I watched your videos. Nice explanation. Good job done :) I have a doubt in problem E. You figured out that cyclic dp will create a problem as the recursion will catch itself in loop. So, we should use dijikstra, to find minimum distance from source(0) to destination(n), by considering dp states. So, the solution Time complexity will be : m*(g+r)*log(m*(g+r)), which will cause TLE. So, how to optimise it?
Thank you, Ivan!
Interesting nickname
I love my nickname as much as myself
Thank you mister Ivan Belonogov Belonogov
Hope this round has strong pretests...
What will be the score distribution?
Good luck and have fun
Can we please have the scoring distribution now I_love_myself ?
I'm curious to know why people want to know the scoring distribution? Does scoring distribution impact how you are going to attempt the questions? Or is there anything else? (For me it is sufficient to know that the later questions have more marks then the previous ones)
for Mental preparation
Done!
Сейчас бы учавствовать в раундах от упирвицкого
Верните мои апвоты, ироды!
Good luck everyone!
[Deleted]
It has been published. Please check the UPD1
Programmers are so bored in quarantine, they are giving downvote to maximum comments O_o
and yours as well
and mine as well
poor me
orz you
Thanks Ivan for his support ! Hope that one day I could be like him <3
Why has it been extended by 15 mins?
yes
delayed 10 min , not 15
why it is delayed?
Small problems on Codeforces and Polygon. Nothing serious.
But the Problem statements were not so small. :)
Due to technical difficulties :(
I was checking my phone for a few minutes, then saw the contest page again and thought I had slept and woke up for the next contest. /s
next time try not being drunk while attending a contest
haha okay!
When it is delayed ...
delay 10m :((
Yet another long round, 2 hours 30 minutes (+10 minutes delay?)
no electricity in my place...So its good..hope electricity comes back in 10 minutes
codeforces knew it , they did it for you
Electricity came just after writing the comment...
:'(
10 minutes delay after a long time.
Round hasn't even started, but codeforces already start to lag :/
your pic is a love.. <3
when I went to have my dinner, I had 12 minutes left.. When I came back I still had 12 left
The problem of being faster that light, is that you can only live in the darkness.jpg
IT took me one minute to realize my network is still working well and the time on my clock is still right...
Congo bro! Round delayed for *9 minutes
Antipleasant tasks, I wish authors to end third grade faster
Can't agree more. I just give up this round.
Extreme long statements, to reach the actual question its taking time.
I did register for the Div. 1, the website even did notice me about the contest started and I clicked Enter button
But then after I finished the 1st problem, the website says I never register!!!!!
Do I really need to register for both Div. 1 and Div. 2 to join the contest??
u cant register to div. 1 ur green
The problem statements are too complicated!
IMO problem C is so cool!
Recently, the div2 rounds are not being scheduled on weekends like it used to happen earlier. Been missing a lot of recent rounds because of this. Looking forward to more weekend contests :)
Что курил автор, когда составлял условие для задачи про двери и горы?))
Div2 D weak pretests :C
$$$My\,\,Dad:-\,$$$ Don't run behind the girls. They are too complicated. you can be ruined.
$$$Me\,to\,\,my\, \,Dad$$$ (after this round) $$$:-$$$ They can complicate a problem statement too if they appear.
Nice story, awesome DAD
I haven't seen a cancer like this before.
I would say div2 C is too complicated
Tasks description was so long, I had to print it on paper because text didn't fit my screen resoltion.
I mean, task should not test our reading skills. It took me way too long to understand and at the end I ended up not understanding the problem anyway.
UGH !!!!
StatementForces !
Denis(in the statement) is such a poor guy.
I hate those div1 coders who make fake profiles just to take part in div2, pls stop doing this! Are you guys afraid to take part in div1, afraid to lose your rating? If that's the case i feel sorry for you. I might have a lot less rating than you guys but i take part in contests to learn new things, not just to improve my rating!
statements are too long to be read...
VeryLongProblemStatementForces
I don't know about E and F, but the rest of the problems from Div2 were absolute trash: no creative thinking, pure implementation. And the length of problem statements is another story.
The problems are shreiking and saying comprehension skills are better to be developed than coding skills. Hope codeforces will not make such lengthy problem statements in future which further decomposes just to a for loop only . Those who read fast are fast scorers in this contest.
Haha.. Loved the story line!
Very much complicated problem statements. And 2-3 lines of these complex story is necessary to solve the problems. Rest of them only confused me.
True
How to solve Div-2 D. Will greedy work here ?
Actually soln is greedy but take care of valid string after using exactly k.
I thought of doing dp. But, tge implementation was very difficult for me to do.
You can look at English Commentary of Problem D here I hope it helps!
Thanks, understood the mistake I was making.
How to solve Div.1 C?
My solution which takes
O(m*g*log(m*g))passed the pretests. I just use dijkstra with vertices are (index of safe point, time left to move before red light)I didn't think that would pass though :/
You can just use 0-1 bfs instead of dijkstra to remove log.
how does the 0-1 bfs work?
It's same as regular bfs, but instead of queue use deque and if cost of edge is 0 push to the front of deque else to the back.
sorry, I meant to ask how you use it to solve problem C because my dijkstra had wasn't on a graph with 0/1 edge weights
If we change distance to number of (r + g) time segments costs will be 0\1.
ah that's nice. thanks!
I just did Dijkstra and passed the pretests. However, it passed too tight.
Pretest 5 of Div2D?
What is the test case 5 for DIV.2 D?
How to solve div1C/2E ?
I have no issues with stories in problem statements if they fit in well, but the story of Div2B seems like its just been created for the sake of having a story and has practically no relation to the actual problem. In my opnion the entire 4th and 5th paragraphs are utterly useless and just serve to confuse people.
It confused me a lot. It was very sad :_v
WORST STATEMENTS EVER!!! English was so broken and not understandable at some times that I can't even. And to top it all off, inital version of Div1E was some freaking misunderstanding, a very bad joke. It contained a lot of very weird words, like "power" instead of "degree", firstly "pick" instead of vertex, then "peak" instead of vertex and then some very weird sentence for making a move with some random "top" in the middle of it. Of course the statement was corrected, but SILENTLY AGAIN. Less than two weeks ago, I requested that whenever any changes are made then please make an appropriate announcements https://mirror.codeforces.com/blog/entry/75806?#comment-601901 and no improvement here. Quality of statements and that poor misunderstanding called initial statement of Div1E and no broadcast after its change made me rage quit the contest first time in many years.
EDIT: Sad part is that problems alone seemed really nice. But even though I was pretty pumped up with energy and good hopes before the contest, all I could find in today's contest was confusion and anger.
Yet we still get an annoying announcement for "the checkpoints in C may not be given in sorted order".
Yes, like what the frick was that? It made me angry as well. First embarassing thing was to allow an input format without such natural condition. When searching for bugs in my code I reread the statement like two times in the search of condition that they are sorted, because I could not believe my eyes it is not present. And then, I don't care whether you get 1000 questions about whether this sequence is sorted or not, you went for such poorly or evilly prepared contest, so let's take responsibility for your actions and do not make it unfair for people that spend their time and nerves on this. We do not get an announcement "hey, statement of E was complete shit, you should reload it", but we get "hey, you get WA because you didn't notice that some sequence doesnt have to be sorted"...
Not including non-sorted case in pretests(or even examples) was just stupid.
Yeah, the statements were shit.
If somebody is interested in the original statement of D1E:
Surely you all read the book "Alice in Wonderland". In this task, Nastya got to the country of Three strange Bees. The bees are strange because their honeycombs are pentagonal. Nastya got there illegally, so she wants you not to catch her. Help the bees punish the intruder!
This is an interactive problem.
A beehive is a connected undirected graph along the edges of which bees and Nastya can move. A graph satisfies two properties:
The power of any of its vertex is no more than 3. For each edge, there exists a cycle of length not greater than 5 passing through this edge. There are three bees and Nastya. You play for bees. Firstly, you choose the picks where you put the bees. Then Nastya chooses another peak in which she will initially appear. One move is first moving the bees, then Nastya, in turn: For each of your bees, you can either move each one along some edge of the graph, coming out from the top where this bee is or leave it in place. Then Nastya will necessarily move along some edge of the graph, coming out of the vertex in which she is now located. You win if at least one of the bees and Nastya are in the same peak at any time of the game.
If this situation does not occur after n moves, then you lose.
Several bees can be in the same vertex.
Input
The first line contains two integers n (4≤n≤5000) and m (n≤m≤3n) — the number of vertices and edges in the graph.
Each of the next m lines contains two integers v and u (1≤v,u≤n), which mean that there is an edge between the vertices v and u. It is guaranteed that the graph is connected, does not contain loops that the degree of any vertex does not exceed 3 and a cycle of length no more than 5 passes through each edge. Note that the graph may contain multiple edges.
Interaction
At each turn, you must output exactly three vertices a,b,c (1≤a,b,c≤n). For the first time, 3 vertices displayed will indicate which vertices you originally placed bees on. In response, you will receive the top where the jury placed Nastya. Each next 3 vertices will indicate where the 3 bees are after your turn. Each of the bees can, regardless of other bees, both remain at the current peak and move along the edge. After the next output of 3 vertices, in response, you get the number of the new vertex to which Nastya went.
As soon as one of the bees is at the same peak with Nastya or you have reached the limit on the number of moves, your program should complete the work. That is if you made a move, and one of the bees ended up at the same peak with Nastya, your program must complete the work, or if Nastya made a move and ended up at the same peak with one of the bees, you should not make your move and the program should complete the work.
If the number of spells exceeds limit (n, where n is the number of vertices), you will get the Wrong Answer verdict.
Your solution may receive the verdict Idleness Limit Exceeded if you don't output anything or forget to flush the output buffer.
To flush the output buffer, you need to do the following immediately after printing the query and the line end:
fflush(stdout) or cout.flush() in C++; System.out.flush() in Java; flush(output) in Pascal; stdout.flush() in Python; for other languages see documentation. In this problem interactor is adaptive. This means that depending on all your previous moves, Nastya's behavior may change.
Hacks are not available for this problem.
What is Test 3 in 1C ?
Probably this?
This one was useful to me:
It was hard to comprehend the problem statements.
I took more time to understand C than to code the solution
For Div.2 C I am sharing my strategy to solve it which has passed the pretests.
Take an example of n = 7 and 1-indexed array arr and position allotment p:
Step 1: i = 1 arr = [1,1,1,1,1,1,1] Now generator can select any index from 1 to 7. Suppose it selects index-4. Now we can't select position-4 further. p = [?,?,?,1,?,?,?]
Step 2: i = 2 arr = [1,1,1,?,2,1,1] Now generator will have to select will have to select position-5 as it is the
maximum position. p = [?,?,?,1,2,?,?]
Step 3: i = 3 arr = [1,1,1,?,?,3,1] Now position 6. p = [?,?,?,1,2,3,?]
Step 4: i = 4 arr = [1,1,1,?,?,?,4] Now position 7. p = [?,?,?,1,2,3,4]
Step 5: i = 5 arr = [1,1,1,?,?,?,?] Now again generator can select any index from 1 to 3. Suppose it selects index-3. p = [?,?,5,1,2,3,4]
Step 6: i = 6 arr = [1,1,?,?,?,?,?] Now again generator can select any index from 1 to 2. Suppose it selects index-1. p = [6,?,5,1,2,3,4]
Step 7: i = 7 arr = [?,1,?,?,?,?,?] p = [6,7,5,1,2,3,4]
KEY OBSERVATIONS:
1: If the position j is selected for number i, Then the remaining positions with a greater index should be placed with i+1 and so on i.e. position j+1 with i+1,j+2 with i+2... 2: Positions less than j will again depend on selection by the generator as all the values will be 1. The assignment of the remaining positions could be considered
again as the new problem with the unassigned minimum number which is 5 in the above example.
Any permutation follows these observations results "Yes" else "No".
I Read problem C for 40 minutes then only got what it is saying. In B also sometimes setter says that peaks have to be answered and sometimes says parts of doors another dielemma NO comments for A(Which is a first ever Div2 A seen by me!)
How to solve E? dijkstra's gave TLE.
Maybe my English is so poor that I can't even understand the meaning of the problem.I am so sad.
No, it was the English of round's authors.
i feel so smart writing a 100 line function to check the current number in D, only to notice that the strings were given in the problem statement :), and unable to submit it in time because of that.
I dunno why the time limit for div.2 D was so tight. It cost me like 14 submissions to pass the pretests.
Was fun typing all the digits on Div1B before realizing that it's listed...
Good to know I'm not the only person who actually wrote them all out (though I'm probably the only one who wrote one incorrectly and got a WA).
I realized it now, after reading your comment. XD
Since you have to copy manually anyway, how about:
you can also do
I Copied the line from the problem statement and put it in a string vector .
A: So easy. B: So easy. C: I need some time to find the construction method. Oh I make it! D: It's slightly difficult, but I can solve it with more time. E: ??? ...OK, Good night.
That's all.
I think the constraints on problem E (Nastya and Unexpected Guest) were too tight for Java. After my first try the program failed due to time on the third pretest and after implementing some stupid optimizations (not changing the O complexity) I got the program to reach the pretest 8. Here's my program if anyone's wondering: https://pastebin.com/Z3Xjg1rk. I'm fairly positive my program would work if the time limit was 2 seconds instead of 1.
If you had used 0-1 BFS, which is probably the intended idea, even a Java solution would have passed.
Can you explain how to use 0-1 BFS in the problem? How do you interpret the problem as one with 0-1 edges?
Make the graph where vertices are pairs of numbers
representing being at the $$$p$$$th line at time $$$t$$$. There's an edge between $$$(g, p)$$$ and $$$(0, p)$$$ of weight $$$1$$$, which corresponds to waiting at that spot during the red light. Also put edges between $$$(g, p)$$$ and $$$(g+dist(p, p+1), p+1)$$$ and $$$(g+dist(p, p-1), p-1)$$$ (as long as those numbers are in the right ranges) which correspond to moving forward or backward. These edges have weight $$$0$$$, since there's no red light you have to wait for.
The video tutorial of Div-2D.
Great video.
But it was possible to solve using only 2 states. Solution
Just imagine sposoring this round xDDDD
should ask for a refund
I don't want to blame any of the authors, the tasks (at least A,B,C,D) were good in my opinion. However, the staments for B and C are long and hard to read for no reason at all.
Was it just me or were some of the problem statements rather long and complex? I decided to start with Div2 C and feel like I made a big mistake. I spent forever trying to understand what the problem was asking for. Decided not to submit anything as I had lost too much time. Which is unfortunate as some of the other problems were very interesting..
Video Tutorial for today's div2D/div1B
Is there not a maxtest in the pretests of Div2D/1B? My solution takes 70 * n ^ 2 instead of 10 * n * k but still passes pretests in around 200ms even though it should take upto 3e8 operations. Fast simple operations or weak pretests? 77820531
My idea for D, unfortunately I spent a lot of time solving slightly different versions of this problem and understood the statement correctly too late:
This is the comment I wanted to post 3 HOURS AGO
Trust me the test cases are weaker than my eyesight!
I am blind :(
Can someone please explain 2 C ?
incase the input was
7 8 9 4 5 6 1 2 3 then the output will be Yes. 7 8 9 4 5 6 10 1 2 3 then the output will be No. hope this gives you a fair idea.
what i did was : added the first contiguous increasing subarray at the end of the deque, and pushing the 2nd contiguous increasing subarray using push_front in deque.
If I end up with with a sorted deque then the ouput will be yes, else no. refer my code : https://mirror.codeforces.com/contest/1341/submission/77810951
I believe the question statement were a little confusing atleast for me !
Did anybody solve F using algebra? Without signs it would be doable with matrices/permutations for example, but it became harder with the signs.
Can you please share your approach ?
Do you mean a normal solution to F, or the approach with matrices/permutations to the problem without signs?
I meant the approach with matrices/permutations to the problem without signs.
With permutations, for each type of bracket choose some permutation which consists only of cycles of length at most $$$2$$$, possibly different permutations for different types. Now, the composition of permutations of the brackets from the left to the right will be the identity if the string is a correct bracket sequence and won't be and identity with high probability otherwise (if you chose long enough permutations and selected them randomly for example). Build a segment tree ofc.
Wasted most of the contest on this approach. :(
A function on finite sets can't really satisfy fg = 1 and gf != 1, it doesn't seem too fruitful
Yes, I realized this during contest but thought that if we checked all balances independently after that, that would fix it. But it doesnt work on "([)(])".
Yea, I've even rushed for implementing checking all balances independently. I saw this case somewhere else after the contest, it looks hard.
Do you know how to do it faster than $$$O(n \cdot \sqrt{n\cdot \log(n)})$$$ btw? (Assuming $$$q=n$$$).
Decent contest!
Solution notes for A-D (Div 2): A) The range of sum of x is [(a-b)*n,(a+b)*n] and it works iff it intersects with [c-d,c+d]
B) Use two pointers/sliding window type approach, where you add or subtract depending on the point you remove and add each step. Becomes O(n) instead of naive O(n^2)
C) Basically if you put something at a point such that all other points to the right of it are filled, then we don't need to do anything. Otherwise, if we put l at point x, we put l+1 at point x+1, l+2 at point x+2..and so on until we hit a point at the right that we put previously, so we need to check if their position in the permutation is consistent with this.
D) Basically dp[n][k] from n-1 to 0, where dp[i][j] stores the leading digit of the best result you can get with strings [i,n) and exactly j moves available.
How do you do E? I think E was some sort of dp/dijkstra shortest paths, but couldn't figure out how to find if two points where reachable from each other in exactly g steps quickly... there could be a lot of alternating involved so there were lots of states.
Was it only me or the statement of DIV2 B was kind of misleading as initially it said the segment end should not be peaks but to pass the pretests they were considering segments ending at peaks also. Waisted a lot of time figuring this. Such an easy problem took 1 hr to get passed because of poor statements. Kindly make statements good as they really affect the rating.
Uhh ending of a segment cannot be a peak.
Nice problem statements. First time seeing problem statements of different problems have continuing/same story.
In addition to the bad and unnecessarily long statements, I was rather surprised by the constraints / time limit on Div1 C — either make it so Dijkstra (with log) fails for all, or make it so Dijkstra passes for the most common programming language. In this case some C++ solution passed and some failed without seemingly any difference. I challenged several people successfully and some unsuccessfully (some passing for 0.966s). Was it that hard to make the constraints on M <= 20,000 or 5,000 (instead of 10,000)?
Agreed--constant factor played way too much of a role here. I'm not sure how they managed to miss such a glaring issue with a dozen or so testers--the Dijkstra solution obviously exists, so they should have made a clearer decision about whether to fail it or not. Relying on hacks to catch a wrong solution is an extremely bad idea because it means that people who are hacked have a chance to fix their solutions and resubmit, while people who aren't just FST.
I thought 1D was a very nice problem, but at this point I'm not sure it makes up for the extremely weak pretests and long statements of the rest of the set.
As a tester, who solved it using Dijkstra I can tell we saw the problem, but couldn't do anything about it. Some slower bfs implementations are working in the same time as fast Dijkstra. So we decided that you either write optimal solution (bfs) or squeeze in Dijkstra.
I used this A*-esque heuristic considering distance from current position to destination which seemed to help: https://mirror.codeforces.com/contest/1340/submission/77862540
And afterwards I re-implemented with my own heap because I've found STL's can be a bit slow: https://mirror.codeforces.com/contest/1340/submission/77863711
My solution (using Dijkstra) uses STL priority_queue and works in around 800ms.
If Dijkstra isn't the intended solution, you should fail it in the pretests...
I am one of those not able to understand what 1341C - Nastya and Strange Generator asks for, tried for an hour. Luckyly was able to solve D afterwards, but that was no fun.
Imagine the level of frustration when you have to throw a door at MoUnTaIn PeAkS to break them.
It was the most frustrating problem statement ever! The whole thing was so misleading!
Even in div2 d he is talking about broken sticks! when something is broken how are you supposed to turn it on?
The array in Div1 C isn't sorted and there is no test for that in pretests...
Can someone help me ? What states of dp should i use for Div2 Problem D?
Two states are required. index of the board and current value of k.
I did Div2 D with greedy approach Nlogn(Hacked)
We've got a wrong answer on test 1 on system tests 77835164!
Yeah, it took me some years of practice to achieve such outcome with deterministic solution :)
Was checker for problem D wrong during the contest? Even when my solution output two equal pairs in the sample, it got WA 3, not WA 1.
Yes, there was a tiny issue with the checker, it was fixed, but some solutions failed (8 solutions).
We have written to all, whose solution failed.
Shitty pretests for D :(
I can understand that the problem statements were a little longer but I seriously think that the story deserves appreciation. Natsaya getting confused seeing a heart and breaking the door of heart on the mountains into pieces, Denis getting distracted about her love for a random permutation generator, Denis not able to stand still while he is after a girl and so even moving in the opposite directions. Man! That's really good writing.
Which door stretches over mountain peaks man? I mean seriously??
I know. It doesn't make sense. But I find it extremely hilarious.
Weak pretest in Div2A. :(
Hacked by Ashish
I too did it without using DP
Hacked you as well :)
It stopped my hearbeat for a second
Wait how can you hack after the contest?
That's uphacking, you won't loose points for that and hence your rank won't be affected, you can read more about it here
If I would be loosing points I would have to wait more to become Blue
Hi, can you try hack my solution? 77797225
5 0100111 1001111 1100111 1010110 0110010 Here is test case he is using
Tnx a lot!
Done :)
You are on fire
are the system tests for d1B are ridiculously weak?
Saw some greedy solutions passing sys tests :-/
........................sad
Note that in the greedy solution one can use prefix sums of min needed sticks and max usable sticks.
Using these we can check from left to right in every position the maximum number of usable sticks, and with that simply build the biggest possible digit.
Given that the limited number of digits and 7-segments the complexity is something like O(n*10*10) and hence runs fairly fast.
Example 77847086
I wouldn't call your solution a greedy. It's more of a backtracking with a lot of early breaking, which is like the dp solution without memoization.
Hacked :)
Ashish you a legend if you can hack 77846297
Done :P
orz
verma_ankit484 Sorry XD
orz
If it not too much to ask please hack this.
It seems correct to me, my solution is somewhat similar to what you are doing.
Thanks
Ashish On Fire !!
Shitty pretests for Div2 D.
This contest was the worst way to thank someone...
I'm a bit worried that Russian speaking participants were a bit privileged (guess why). Statements really seem to be taken from google translate, is it true?
Don't underestimate Google Translate!
"This road is not always can be crossed in one green light."
that sounds worse than google translate. Actually, translating to russian and back again gives: This road cannot always be crossed in one green light.
"From any square, you can get to any other in exactly the same way using these roads and not visiting the same square twice."
I mean even google translate works better.
Funny thing: while reading the Russian statements, I thought the original language is English, and translation is sub-par at times.
Its like bad and worse. :(
So is this round unrated, since ratings are not updated? Kinda sad to have cleared the problems only to find out it. :(
Wait for some time it will get updated
Damn, the first test where points in Div1 C are not sorted in the input is test 70...
Hacking changes the round.
I've solved div1E, but I'm pretty sure the following is a counter example.
My idea is to move your 3 bees to minimize the sum of the distances to the (up to) 3 neighbors of N (where you must match bees to neighbors). I've also seen other AC submissions with the same idea. (But instead of sum of distances they minimize the largest distance, then the 2nd largest distance, then the smallest distance).
For a counterexample, consider the 'ladder graph' made into a cycle: two cycles
AandBof lengthnnext to each other, whereA_iis connected withB_i. Note that each vertex has degree3, and that each edges is included in a cycle of length4(i.e. less than5).After picking initial bee positions close to
A_0andB_0, N can start at e.g.A_{n/4}. Now all my bees will move along the circle1step, and then N can just move one step up toA_{n/4+1}. This can continue forever I think.Yeah, I also got AC with incorrect solution. I think this problem has a lot of incorrect solutions for which is very hard to create counterexamples without knowing an idea of solution. But this cycled ladder graph is counterexample to lots of solutions, and it is very strange that it wasn't in tests.
Sure it's difficult to make good test cases here, but I think the only way to make a cycle of length > 5 without chords is to make it so that the entire graph is one 'double' cycle. I think it's not possible to have 2 'independent' chordless cycles of length > 5.
So given that this is the only special graph, you'd think that this needs extra testing.
I and VladProg (who also got AC with wrong solution) also discussed solution that tries to move bees to different adjacent vertexes to Nastya's vertex without going through another adjacent vertexes to Nastya's vertex. This solution will pass cycled ladder test (because if you remove all adjacent vertexes to Nastya's vertex there will be no long cycle) but there exists another counterexample, like cycled pentagons.
Ah, that's a neat idea, but yeah won't work for even bulkier cycles.
Another idea is to choose the initial positions roughly at 0, 1/3rd, and 2/3rd along the cycle. E.g. by fixing 0 and then choosing
u,vsuch thatd[0][u] + d[u][v] + d[v][0]is maximal.Also found a construction to turn any graph
Ginto a graphHsuitable for this problem such thatGis a minor ofH, so you can definitely get more complicated inputs than just cycles.Anyway I'd love to see a proof that this approach works even in graphs without long cycles. The 'every edge in cycle of length <= 5' condition is useful because now some of the bees will always 'take the long route' to block off potential exits, but I can't yet prove that the max distance or sum of distances goes down by at least 1 or 3 respectively each turn.
So it's actually quite easy to embed K_5:
https://imgur.com/a/LC60kW9
If you make all 'edges' 100+ steps long I think it should be possible to escape.
Maybe in practice the condition is/should have been that such long cycles do not exist at all anyway? So formally you can reduce the graph to a point by repeatedly contracting cycles of length <=5 to a point.
Also, congrats on being red now ;)
Thanks ;)
Yeah, in random graph you can replace edges with these "ladders" and you will get graph that satisfies problem statement. (also you can split each vertex with degree > 3 into few vertexes with degree = 3 with length of ladder between them equal to zero). But obviously there exists a graph that has minimum number of bees larger than three. I suppose it's impossible (or possible, but using very big number of queries) to catch Nastya on these graphs.
In div2A/div1B, case where solution can consist of leading zeroes isn't present in the pretests. Thought I'd fst, but didn't. code case
Missed out on 1D because I left out a
++pt;in one of my loops. How fun.Cool problem, though.
In Div1 B I thought k would always be less than or equal to number of sticks which are off.As said in the problem statement, what is the maximum number that can burn on the board if you turn on exactly k sticks (which are off now) Reading the line(specially the sentence in the bracket), it seemed to me that at least k lights are off.Which led to wa in system test.
When you realize tourist solved Div1B in 00:04, while you can't even read the statement in that time
I guess the extra 30 minutes from ingeneral round was for reading the long paragraph in each question...
In div2D/div1B, case where solution can consist of leading zeroes isn't present in the systests. Thought I'd fst, but didn't. code case
I solve A(div1) so l check if we have anybody what a[i] + 1 < a[i+1] answer No else Yes.
hmm, I had the exact same code on all 3 submissions...
Are the data structures on newer versions of C++ faster? 77856421 77857168 77857210
Off the tangent, but what made you try this check?
I opened up the status page and looked at the C++ solutions that passed, and most were in C++ 17 or C++ 17 (64). I switched to C++ from Java pretty recently, so I don't know too much about which version of C++ is best for certain situations. I guess this is one of those situations?
Due to the c++17 (64) being 64 bit, the compiler may emit optimizations that the non (64) variants may not. This will optimize it for the computers that the tests are running on, rather than intentionally handicapping it to keep the performance more similar to other languages. It was discussed when it was added a month or two back
I see. I'll check out that blog. Thanks!
I suggest you make a blog, I would like to know the answer to this too. I mean this is literally double the speed compared to the other 2.
I tried resubmitting some of my code too and its significantly faster
With GNU C++ 17 I think it is slightly faster or there is a little bit more optimization.
I can be wrong, but you have long long (64-bit) type in your solution. That's why GNU C++ 17 got TL while 17(64) got AC — it works 2 times faster with 64-bit numbers.
pretty sure operations with long longs are one instruction each in 64-bit, but multiple in 32-bit, so long longs are a lot faster with 64-bit.
That seems to be it; I switched to integers and got the same execution time, but half the memory usage.
PS: congrats on the red, the color changed right after my refresh on posting
This blog might be related.
)
what in tarnation
GG loosing 700 points for long long . . ., i just changed long long to int and got accept in D1C, why should the constraints be that much though?
I think int should pass. As the distance to n can be at max (r+g)*m.
I know that, but on that point i couldn't see any point to avoid long longs.
I think
long longmight not pass because it might print numbers like this: 3e9+5Problem were nice anyway, thx (especially D1C, E)
Div1E is so beautiful! I still do not fully understand why my solution pass, but it is so stupidly elegant! Unfortunately, made some mistakes in code during contest... Tnx for the wonderful problem :)
Apparently creating variables CAN be your code's bottleneck: 77824459 77859221
This is rather particular to
std::vector;vector::clear()doesn't actually deallocate the vector's internal buffer, so reusing the vector means you actually do many fewer dynamic memory allocations. This is compounded by the fact that vectors dynamically resize as you push_back (doubling when necessary), so you're actually saving a couple allocations per loop. You might be able to pass the problem by just doingvector::reserve(...).Hacked 3 greedy solutions in D1-B using very simple case:
(This corresponds to a digit 1.)
Answer should be 4, but greedy solutions that consider the first compatible digit in descending order would pick 7 first then report impossible. Since there is 1 remaining broken segment, and such solutions only consider "patching" 9 -> 8 in such cases. And solutions which consider the first compatible digit in ascending order would pick 1, and report impossible as well.
Anyone can explain better how to solve Div1C-Div2E with 0-1BFS instead of dijkstra. Please. Thanks in advanced
Construct a graph. Let a vertex in the graph be a pair $$$(i, k)$$$ where $$$i$$$ is the index of some intersection (sorted), and $$$k$$$ is the number of seconds which have passed since the light most recently became green. Every vertex (except when $$$i = 0$$$ or $$$i = m-1$$$) has two outbound edges, corresponding to the intersections before and after it (changing $$$k$$$ as we would expect). Note that we can only transition if we have enough time left before the light changes. In the case where we have to wait for a red light at the destination intersection (i.e. we reach the destination exactly as the light turns red), label the edge with weight $$$1$$$, otherwise $$$0$$$. Now run 0-1 bfs to find the shortest path to all vertices.
Once bfs is finished, iterate over all vertices $$$(i, k)$$$. If the destination is reachable from this vertex in less than $$$g$$$ seconds, the final answer for this vertex is $$$(r+g)\cdot c+k+n-d_i$$$ where $$$c$$$ is the cost to reach this vertex in the graph. Output the minimum final answer for all vertices.
Can you explain the intuition for labelling the edge weight 1 and 0? I understand the dijsktra solution but having a hard time understanding how it can be optimized to 0-1 bfs? I would appreciate it!
I'm not sure which part you don't understand. The total cost of a path is simply the number of times we have to wait for a red light. If we have to wait, the edge has weight $$$1$$$. Otherwise, we don't wait for a light, and the edge has weight $$$0$$$.
I guess the key intuition is that if you tell me, "I waited for a red light $$$c$$$ times, and then walked for $$$k$$$ seconds afterwards, ending up at position $$$d_i$$$," then that is actually enough information for me to deduce the exact number of seconds you took getting from start to finish, and it is given by the formula in my previous comment.
Thank you for the response. After some thinking,
(r + g) * cdoes make sense. because when you wait for the red lightctimes,cgreen light seconds must have occurred as well.But I don't understand the
k + n - d[i]part and iterating over all vertices. Can you explain intuition? I would appreciate answer. I thinkn - d[i]is counter-intuitive because you are subtracting the end from thed[i]. Wouldn't you have to wait additional red lights when considering1tod[i]toN.Edit: https://mirror.codeforces.com/blog/entry/76348?#comment-609660 This is a similar solution, but it iterates all possible ending elapsed time on the last island (i.e. m). It's easier to understand, but of course, I still want to understand your variation
Yes, I agree that some other approaches may be more elegant. The formula I gave above is just what occurred to me during the contest, so I went with it.
The expression $$$(r+g)\cdot c$$$ gives us the amount of seconds from the start until the light last switched to green. The value $$$k$$$ gives us the amount of seconds since the light last switched to green until now. So adding them together gives the number of seconds to reach vertex $$$(i, k)$$$ from the start.
Now, $$$(i, k)$$$ is not necessarily the destination. We would like to know how many additional seconds it will take to reach the destination from the vertex $$$(i, k)$$$. This is simply the distance between us and the destination, i.e. $$$n-d_i$$$. We only consider the cases where we don't have to wait for any additional red lights, so if $$$n-d_i > g-k$$$, we simply don't use this vertex for our answer.
Among all vertices which do satisfy that condition, we want to find the shortest resulting path. Any vertex with $$$n-d_i \leq g-k$$$ constitutes a valid path, so we minimize over all such choices.
Let d[u][t] the minimal amount of times you had to wait r + g seconds to get to vertex u and the green light has been on for the last t seconds. Transitions are possible if the amount of seconds needed to get to adjacent vertex does not exceed g. If it is possible to get to and adjacent vertex v and the green light had been on for g seconds then you have to wait r more seconds so the light turns green again. Don't know if I made myself clear
Not sure if I'm missing something. But isn't that observation just enough to figure out the dijkstra solution? I'm still unable to understand the intuition how the edges can become 0-1. Because in each transition, the cost if the difference between the two safe islands.
Let the state $$$(u, t)$$$ be that you are currently at island $$$u$$$ and have elapsed $$$t$$$ seconds since green light started. Then transitions will be:
$$$(u, t) \rightarrow (v, t + dist(u, v))$$$
$$$(u, G) \rightarrow (u, 0)$$$
If you fix the time elapsed when you reach the final island to be $$$f$$$ the answer will be: $$$cost(m, f) + f$$$. Minimum time to reach $$$(m, f)$$$ plus the last $$$f$$$ seconds elapsed on the current green light.
Iterate through all final $$$f$$$ and profit.
Notice that the graph built only has two cost, (and one is $$$0$$$) so BFS0/1 can be applied.
mnaeraxr That's a really good explanation! Thank you!
Some examples. Not all of them are incorrect ofc., some just mean what they should mean in the worst manner:
"This road is not always can be crossed in one green light." <3
"...what is the maximum number that can burn on the board..."
"...there is a road of $$$n$$$ lines."
"From any square, you can get to any other in exactly the same way using these roads..."(exactly the same way as what?)
"...and will return to the first and at the same time the maximum time in which he visited will be minimal."
"If the number of spells exceeds limit..."
" power of each vertex is at most $$$3$$$"
"power" instead of "degree"; "peak", "pick", "top" instead of "vertex" — hard to find anything in E, as it's fixed now... Probably I'm not the only one who opens all the statements at the beginning of the contest, so believe me, fixing it during the round doesn't give much.
After the contest, I've read the first paragraphs, I like the one from E:
"The bees are strange because their honeycombs are pentagonal." Huh? What does it mean? Just a story, or should I care about it? Maybe the graphs are planar? Inequality from the input section is close to the planarity one: $$$m \leq 3n$$$. So? No, "powers" are at most $$$3$$$, so so somebody just wrote $$$m \leq 3n$$$ instead of $$$m \leq \frac{3n}{2}$$$...
About A: I don't see anything incorrect, but who let this statement appear as A?
An explanation for those who don't understand "pick", "peak" and "top". For people from Poland it's rather easy: vertex has some translation to Polish (wierzchołek) which also means the same as the top (of the hill for example) or peak, but of course, the meaning is very different. This also suggests that somebody has used google translate here. Russian is a bit similar to Polish, so...
"peak" and "vertex" indeed both translate to "wierzchołek", but it never crossed my mind to associate these two with each other, so it by far was not clear to me, that "peak" means "vertex" :P. And "pick" does not mean "wierzchołek" in any way. P___ pasted initial statement of E (thanks!) above, and "peak", "top", "pick" and "vertex" have all been used alternately. There is no way that a person decides to keep using four different words for one thing. It is clearly produced by Google Translate. Or by a very very poor writer. And has not been proof-read by anybody. isaf27 any comment on this? And please remember, many people open statements at the beginning of each contest to ensure they can access them in case of any CF outage, so silent corrections during the round don't fix the problem.
And raise your hand if after first reading it was clear to you what "there is a road of n lines." means xDD. And to top it off, followed by a beautiful "This road is not always can be crossed in one green light."
I make a petition to make this round unrated for English speakers (ok, this is not serious suggestion, but you get the point).
Man, antontrygubO_o wasn't the coordinator for this round.
Whoops... Thanks, I will correct this.
Yes, I'm sorry about the bad statements.
Actually, they were translated by one friend of authors. The bad thing was, that he doesn't good at mathematical terminology. So, that strange things could be in the statements.
There were more such thinks, I tried to correct them all, but some of them I missed (I don't know how I missed "power" and etc).
My steps to avoid such things later:
1) I will spend more time checking the statements.
2) I will try to make them shorter and more clear
3) I will ask for help from the English-speaking people.
Sorry again.
"Mathematical terminology" was definitely not the only problem with the statements, as "This road is not always can be crossed in one green light" illustrates.
Those are some good suggestions; I notice your own English is not very good, which is perfectly normal, but in this case it's especially important that you follow 3) and get someone whose English is good to review the English version of the statements.
While we are at it, the Russian statements could also use some fixing. Usually the errors don't make a statement ambiguous, but still, grammar and spelling errors are a continuous distraction from solving problems, and a bad example for the community.
A few examples from today's problems:
A: "...могла ли она получится в результате работы генератора." It's even correct in the next paragraph, "...могла ли получиться..."
B, flavor text: "чтобы предложить ей стать парой.", "предлагает быть им вместе". Doesn't sound like Russian. Looks more like a too direct translation from English. And in the latter phrase, the order of words in the English version is also wrong.
B: "если i-я палочка не горит и 1", "если включить ровно k палочек или −1": such complex sentences tend to lack a comma before "и" and "или", transforming the formal statement into something funny.
All problems today: the em-dash signs look like they are written in TEX source as
word ~--- word, which makes the left space longer than the right space. In that case,word~--- wordor perhapsword "--- wordwould be better.I will be happy to contribute/help with the translation to English. How can I sign up for it?
I suspect that friend of the authors used Google Translate.
Testers can also help fix English. It's what I do sometimes when testing. Even for those who aren't very good at English, noting some mistakes helps and if there are many testers, the incremental improvements add up.
By the way, do you have any idea if this scary creature was intended to speak broken English, or it was created by someone's friend who doesn't good at entomology? :)
This is a reference to a well known meme from some very old game, where the translation was very bad and one character said "all your base are belong to us".
https://en.wikipedia.org/wiki/All_your_base_are_belong_to_us
https://www.youtube.com/watch?v=8fvTxv46ano
It's the same in most Slav languages. In Russian, vertex is вершина, and in Slovak it's vrchol — both have the same basic meanings as your example.
I have tried problem A during the contest but i didn't pass even the pretests. Here is the submission i got WA with during the contest: First try Here is the submision i got Ac after contest: Second try
The idea of the first try is to check if some i inside the interval [a-b,a+b] times n is inside the interval [c-d,c+d].
The idea of the second try is actually just the reverse. Check if some i inside [c-d,c+d] divide by n is inside the interval [a-b,a+b]. Why the second try works and the first one does not?
Check if
[(a - b) * n, (a + b) * n]intersects with[c - d, c + d].But this solution is actually very similar to me first try. Thank you.
very similar != correctThe intuition is that the minimum possible answer is n * ( a — b ) and the maximum possible answer is n * (a + b). And you can make any answer between min, max by adjusting. In your code, I'm not sure what the logic is.
I see what you mean. In my code i tried to bruteforce every number in the interval [a-b,a+b] , but know i see that sometimes we my need 13+13+13+12 to make the sum fit into the interval [c-d,c+d], but my code only takes into consideration cases where we need only: 13+13+13+13 ( only equal numbers ) . Thank you.
Lol mate, your solution is actually wrong and passes pretests by sheer luck. Sorry to ask, but how can I hack a solution in Codeforces (new member)?
It actually did not even pass the pretests.
That's because all weights do not have to be the same. When you try to find an integer from a-b to a+b, you assume all the grains to be identical. You are ruling out the possibility of some having a different weight, and hence a floating point average weight.
So, when you do the reverse, that is division, you are also considering the fractions.
Example: In one of the samples, you can see there is 2, 2, 3. You can't take one integer value for that!
Exactly the second is correct because it also takes floating point into consideration. Thank you. This silly mistake made me lose a lot of rating today.
Can anybody help to understand the problem Div2 C, i still cant understand it :( Specially these paras:
2.For all t from 1 to n, we calculate countt — the number of positions 1≤j≤n such that rj is defined and rj=t.
Consider the positions that are still not taken up by permutation and among those we consider the positions for which the value in the count array is maximum.
The generator selects one of these positions for the number i. The generator can choose any position.
The same happened to me. Understood after reading it for maybe 15 minutes...
PROBLEM STATEMENT :
Let me explain you
1 > We have to check if the given permutation array is possible to generate by using the generator algorithm or not.
2 > Understanding the algorithm :
Step-0> Initially let perm be an array which we are creating to match it with input. Initially perm = [#,#,#......#] (# denotes free space)
Step-1> For every position x [1...n] calculate an array called r which will have next immediate position available in our perm considering x itself.
Let's say perm=[6,#,#,4,2,#] so our r = [2,2,3,6,6,6]
Another example perm=[#,#,#,4,2,1] so our r=[1,2,3,#,#,#] because for last three positions we don't have any valid position to allocate.
Step-2> Generate an array call count where for every i [1..n] it stores the frequency of i in array r.
For above r = [1,2,3,#,#,#] , count = [1,1,1,0,0]
For another example, let perm = [#,2,#,1], so r = [1,3,3,#], based on which we will make count=[1,0,2,0]
Step-3> Now allocate an number to position who have maximum count in array count, if there are more than one maximum in count then choose any position you want(Here we will choose position according to input array).
As for example let perm = [#,2,#,1], so r = [1,3,3,#], based on which we will make count=[1,0,2,0], here position 3 has maximum count in array count having count as 2. So we have to forcibily choose position 3.
All those above step-1 to step-3 are just a process for only one iteration.
So whole algorithm would be,
initialize perm with free space
for every number in [1 ... n] {
generate an array r (our step-1)
generate an array count (our step-2)
allocate number to position in perm who has maximum count (our step-3)
}
check if the generated permutation is same as our input array
EXAMPLE
ip_arr = [2,3,4,5,1]
perm = [#,#,#,#,#,#]
for number=1:
r = [1,2,3,4,5], count=[1,2,3,4,5], perm=[#,#,#,#,#]
We can choose any postion we want to allocate 1 in perm bcoz every count is maximum,
we will choose here 5 as 1 is in the position 5 in ip_arr
for number=2:
r = [1,2,3,4,#], count=[1,1,1,1,0], perm=[#,#,#,#,1]
For allocating 2 we can choose any position from 1 to 4 but not 5 as it has lowest count,
we choose position 1
for number=3:
r = [2,2,3,4,#], count=[0,2,1,1,0], perm=[2,#,#,#,1]
Now here we are forced to choose position 2 as it has maximum count,
so 3 is allocated to perm[2]
for number=4:
r = [3,3,3,4,#], count=[0,0,2,1,0], perm=[2,3,#,#,1]
Again we are forced to choose perm[3]
for number=5:
r = [4,4,4,4,#], count=[0,0,0,1,0], perm=[2,3,4,#,1]
We are now forced to choose perm[4] for number
SOLUTION
Observation to make is that when an lower number occurs in array it has to lead consecutive numbers,
In test case 1: [2,3,4,5,1] 2 has to lead with 3, 3 has to lead with 4, 4 has to lead with 5, 1 has no one to lead
In test case 3: [4,2,3,1] 1 does have any one to lead, 2 has to lead 3, 4 does not have to lead
Eg. [4,5,6,1,2,3] 4->5->6, 1->2->3
Wrong test case : [1,5,2,4,3] here 1 should have 2 next ot it
Wrong test case : [1,3,2] here 1 should have 2 next ot it
Wrong test case : [4,6,5,1,2,3] here 4 should have 5 next to it
Implement it and you will get Accepted...
I apolozize if i made it hard for you...have a good codeforces journey :)
I was thinking about r and count array all the time, i did not think about permutation array that is why i could not able to build/relate r array. Now, it is clear.
Thanks buddy!!Always available for help buddy. Your welcome
You explained it way better than the actual problem statement I thank you for that
It was a nightmare reading that statement at the time of the contest... You're welcome
split the background story with the actual question with a meme, pls. It makes no sense reading long story during contest.
Div1 C, Div2 E problem. I don't know reason that I get WA. Are there anyone who help me?
I used dynamic programming.
I used -2 to block from cycle.
I used 1000000000000ll to block from overflow.
Sorry, my pool English. But if you help me, I would really appreciate it.
https://mirror.codeforces.com/contest/1341/submission/77862656
Interesting solution. I do not believe it is wrong, so I write some random tests and find a case:
20 15
0 1 2 3 4 6 9 10 11 14 15 16 18 19 20
8 0
Now I am debugging it. If you find out why, please let me know.
I realize that dynamic programming can't solve cycle problem perfectly.
Let assume that below sentence
There are problems(p1, p2, p3, p4, p5).
p1 has two subproblem (p2 and p3).
p2 has two subproblem (p4 and p5).
p3 is simple problem and has valid answer(cache value > 0).
p4 has two subproblem (p5 and p1). Here is a cycle.
p5 is simple problem and has invalid answer(cache value = -2, can't arrive target position)
If a problem has a subproblem which has valid answer, this problem has valid answer. If a problem has only subproblems which has invalid answer, this problem has invalid answer.
When we try solve p1, recursive function try to solving p2, p3.
- When we try solve p2, recursive function try to solving p4, p5.
-- When we try solve p4, recursive function try to solving p5, p1
- now p4's and p5's answer is -2. So p2's answer become -2. And this value will be saved in cache.
p2 has invalid answer. But p3 has valid answer. So p1 has valid answer.
Now solving process of p1 is over. p1 has valid answer. But now assume we have to solve p2.
PS : Sorry for my poor English. I am practicing English speaking and writing. I want to share many ideas of problem solving with other people. Good luck to me :)
Good explanation! I think this is the problem that memorized dfs (dp) can not solve but bfs can.
Though problems were nice but statements were unnecessarily twisted!
I just downvoted your contest.
FAQ
What does this mean? The amount of contribution on your blog and maybe codeforces account has decreased by one.
Why did you do this? There are several reasons I may deem a blog to be unworthy of positive contribution. These include, but are not limited to:
Rudeness towards me,
Making a bad contest,
Sarcasm not correctly flagged with italic font,
Giving me negative delta.
Am I banned from the Codeforces? No — not yet. But you should refrain from making contests like this in the future. Otherwise I will be forced to issue an additional downvote, which may put your commenting and posting privileges in jeopardy.
I don't believe my comment deserved a downvote. Can you un-downvote it? Sure, mistakes happen. But only in exceedingly rare circumstances will I undo a downvote. If you would like to issue an appeal, shoot me a private message explaining what I got wrong. I tend to respond to Codeforces PMs within several minutes. Do note, however, that over 99.9% of downvote appeals are rejected, and yours is likely no exception.
How can I prevent this from happening in the future? Accept the downvote and move on. But learn from this mistake: your behavior will not be tolerated on CodeForces.com. I will continue to issue downvotes until you improve your conduct. Remember: Codeforces is privilege, not a right.
Stop whining and enjoy the contest already, it has some beautiful problems.
It's my first codeforces round (but I did several problems for other Olympiads). I made a lot of conclusions and next time I hope that there will be no such problems. Sorry(
No problem man!! It was a good contest and high quality of problems were there...I appreciate you!
When will the editorial come out? It hasn't come out yet.
And there are up to 11 testers for this round smh.
I was the one who deregistered ~30 mins before the start of round.
Thank you for the amazing round. This round is very special for me. For the first time ever I can solve almost 5 problems and I can feel being in top 10 before system test.
I got a strange issue, My same code failed when I used 64 bit C++ but passed when submitted using normal c++ 17. Can anyone explain why? Fortunately, this did not happen during the contest. Failed Passed
Perhap that's due to an undefined behavior on Ln 24 & Ln 35 :
if(arr[i] > arr[i-1] && arr[i] > arr[i+1])When $$$i=n-1$$$ , this can slightly cause a UB.
Please try to enlarge the size of
arrand resubmit it. I think then it will be all right.Editorials! Please.
so long problem statements dude, work on it. It was not at all fun or clear.
How to implement the dp of div2D. It looked really hard for me to implement.
I'm feeling real bad :(. How can I debug my code when it's just one digit away from the answer :)
can anyone plz explain how we can solve div 2 D/div 1B using hashing as this guys didhttps://mirror.codeforces.com/contest/1340/submission/77788156
I would like to point out few points regarding this contest please read this carefully whoever has come across similar problems thanks for reading it :-
I wrote $$$n^2$$$ * $$$t$$$ solution it passed in Div2A and failed main cases weakest pre cases testers even failed to check these type of solutions.
I struggled in problem Div2B for 40 min and was super confused in the language with poor correlation of story and what actually the problem was asking for it seemed I was solving an April Fool's Problem.
Finally I chocked at Div2C for about 1 — 1.5 hours just to understand the problem while never being able to code the problem actually.
I was having doubt on my comprehensive skills at that point of time. Meanwhile I saw many of Div2D problems failing main cases since brute force solution was allowed to pass pretests which is horrible in itself.
I respect Codeforces because it is great for learning but I would request my fellow competitve coders to discourage this type of contests making with some random red and yellow by heavily downvoting, guys. Please make standard question statements which is grammatically correct not English mixed with Russian. P.S not hate for Russian language though but mixing doesn't seems pleasing.
Please don't do that again don't make contests at high frequency. It would be just nice if contests are of quality for which Codeforces is well known I saw some of previous 300+ rounds I literally had tears in my eyes that is what quality is. Please don't ruin this reputation with some nonsense rounds with absolute absurdity
If you find this offensive or bad feel free to downvote also if you find it relevant don't forget to hit the upvote button. Have a Good Day
Bro, You didn't even submit a single solution. Also, a problem comprises of two parts- the ability to understand the actual problem and then finding a solution for that. Not to boast but I found the solutions for Div2 B and C quite simple so I guess the author I_love_myself had to make the problem statements interesting.
We have not tested the English statements.... It was a huge mistake. It will not happen in the following times.
Yeah The problem statements were too long but don't judge the problems by that ... Problems were really good with a balanced contest ... Mistakes happen by everyone !!
It seems the tests for div2 A problem are not correct. My solution could not pass these tests but I consider it as correct. Most contestants use solution based on algorithm like checking just n*(a-b) and n*(a+b) against c-d and c+d. And the system considers these solutions like successful ones. For example, p=n*(a-b) q=n*(a+b) r=c-d s=c+d f=0 if(p>s or q<r): f=1 if(f==1): print("NO") else: print("YES") Check this algorithm on such data like n = 10, a = 5, b = 1, c = 55, d = 1 and you find this algorithm incorrect. (p=40, q=60, r=54, s=56 => f=0 => Yes). I ask testers to verify their tests or to request author for problem reformulation.
Some of people wrong in here. for all $$$i$$$ and $$$j$$$, $$$1\leq i,j\leq n$$$ $$$i$$$ and $$$j$$$ different, it shouldn't be $$$x_i = x_j$$$. In other word all $$$x_i$$$ shouldn't be same
Problem Div1D was very cool. Thanks for a nice contest).
can some explain what problem Div2 D wants to say ? I do not get the problem statement .
What i have understood till now : example case 1 :
We can get maximum 8 because we can turn on 7 sticks .
Why answer is -1 ? for example , first string itself has 5 zero , and we can fill them .
Also i did not got few more things in problem statement , like "Number that can burn" , "k sticks break .........turn on exactly k sticks" .
Please some one explain the problem.
You have to change exactly k zeros to ones, such that all strings correspond to valid numbers from 0 to 9. If it is possible, you have to maximise final number.
yeah , so in string "0100001" , i can change 5 zeros to one and i get "1111111" but -1 in statement ?
After all the sticks are used, all the strings should be a digit (0-9). Here, the second and third strings don't form a digit, we have to add sticks there also.
when will the editorial be published??
I think the time limit in Div2D was too tight. I used a recursive dp solution with memoization, which had time complexity of O(n*k), and the time limit was exceeded. Had to convert it to an iterative solution to get it accepted.
I had it accepted using memoization, then carry the string in function was giving tle.
I don't understand the mistake in my code (Div2 Prob D). Can anyone help me out here. Even the test case for which its failing looks similar. It don't show for which digit its failing. My Submission 77853885
The stories were very weird. I liked it!
Div2 D. I get it! Nastya is a cyborg, and Denis punched her in her seven-segment display (засветил ей в табло) trying to persuade her to date him.
I tried to submit a few hacks (problem B) after the contest just now, but they all get "Unexpected Verdict". I'm pretty sure the test is perfectly valid. Anyone know why this is?
EDIT lol, some searching around suggests this means the checker crashes?
pinging I_love_myself, could you take a look?
e.g. hacks 636949, 636950, 636951, 636952, 636965
So, my solution worked 998ms out of 1 second TL?
ಠ_ಠ
What was the meaning of that TL? I really don't understand how it can be lower than 2 seconds in such problems. So that non-C++ users hate you, and C++ users gamble? :) Even in C++, my solution TLs on the older compilers.
To be clear, I know that priority queue was not required in this problem, and without it my solution passes in ~150ms. But I still think TL is unnecessarily tight, and pretests also don't fail you on that.
Anyways, problems were pretty good, thanks to the authors, and looking forward to your next contests!
That's close.:)It's even during the contest!
Can anyone help why my bottom-up dp approach for Div2-D is giving TLE.
77895150
Hey Actually you are using map(creating extra log factor) Just done some changes in your code (change maps to vectors) 77955330
Thanks got it!
How do I know if my Div2D submission is correct (as system tests were weak based on some comments above)? 77847563
EDIT: Sorry, just realized I can view uphacking testcases as well as expected output.
You can use 2d array instead of map for precompute
Perhaps you replied to the wrong comment.
It worked thanks!
thanks MikeMirzayanov for such amazing contest I am very like question C he is awesome
RIP English.
Story of Nastya and a Simp.
fucking network :(
If codeforces cannot afford to hire someone to translate problem statements into decent English, I can agree to do it for a contest named in my honor.
To all of you rushing for the downvote button — lighten up, the above was a joke. :)
When are we going to get editorial.
https://mirror.codeforces.com/blog/entry/76479
Thanks. It is still not added in the contest material section of contest.
Wish for a better contest next time.
I will rather prefer a short and formal statements over reading a novel in the contest.
A novel of questionable artistic value, I might add. But, on the other hand, those novels come with an extra half hour. :)
Это был наш первый раунд на codeforcrs и мы сделали ошибки везде где только можно
Да уж... :)
What is the problem with my solution code, I am getting TLE. My complexity is O(10nk). Can anyone please help? Thanks
I am trying to solve Div2D/Div1B question. I have a DP solution that is [n+1][k+1][10] for O(10*n*k). I TLE on test case #7, but it would seem that this solution (as others have also done it this way) should work within the time limit. Is this not efficient enough, or am I doing something wrong? Additionally, I am not great with PD and I don't understand the [n][k] DP solution at all, so any information on that would also be great!
EDIT: I changed DP solution to [n][k] but TLE on #22. 78075008
Thanks for the nice round <3