Всем привет!
Сейчас проходит зимняя смена ЛКШ (Летней Компьютерной Школы), и мы в составе параллели А*+ с ее преподавателями подготовили полноценный Codeforces Round.
Раунд состоится в 05.01.2020 17:05 (Московское время) и продлится 2 часа. В каждом дивизионе будет предложено по 6 задач.
Задачи раунда были придуманы и подготовлены ismagilov.code, devid, Volkov_Ivan, Jatana, karasek, polinarria, cookiedoth, AlesyaIvanova, KhB, AliceG, D.Pavlenko, VFeafanov, LordVoIdebug, forestryks, Ilistratov, seiko.iwasawa, ibraevdmitriy, Drozd_off под руководством преподавателей PavelKunyavskiy, VArtem, meshanya, budalnik.
Спасибо aneesh2312 MarcosK Stepavly Infinity25 tourist antontrygubO_o isaf27 dyrbulshchyl, Kurpilyansky, vintage_Vlad_Makeev за тестирование!
И, конечно же, спасибо MikeMirzayanov за великолепные системы Codeforces и Polygon, и 300iq за координацию раунда.
Всем удачи!
UPD: В задаче E была допущена ошибка в авторском решении, а именно переполнение типа long long при подсчете ответа. До первого теста, на котором оно случалось, дошли 4 участника (ainta aid Um_nik ecnerwala), из которых 2 прошли претесты, а не должны были, и еще 2 не прошли претесты, хотя были должны. В автоматическом режиме протестировать так, чтобы оба ответа принимались не представляется возможным, т.к. переполнение меняет тесты из-за способа шифрования запросов. Поэтому, мы решили, протестировать два решения прошедших до перетестирования решения на старом наборе тестов, а остальные — на новом. В результате, три из решений прошли тесты, с чем мы подзравляем авторов, а одно получило WA80 на 17-ом ответе внутри теста, что явно не связано с проблемами с переполнением. В дорешивание можно сдать только решение, которое учитывает проблему с переполнением.
UPD: Разбор!
OMG, round by cyans only!!! _overrated_, where are you looking???
*_overrated_
I am looking at my current color. Go go cyans!!!!!
cyan round!
*Fake cyans
*Trans-cyans
It wouldn't be complete without Mr. Rodgers wearing a cyan shirt.
I have changed my color to cyan in order to participate in this round.
Me too. Let's make it all cyan, but I hope I don't lose points and get really cyan :D
Me too
I did not change my color from cyan, but I hope this is my last cyan round
The easiest way is probably falling to pupil -.-
I am in the first step to achieve the goal using the hard way :D
Congrats!
thaanks <3
Me too :/
Round by cyans and for cyans only.
why everyone is specialist?
Будет ли новый Гончар Фёдор? LordVoldebug LordVoIdebug LordVoIdebug ответьте.
Все возможно)))
Читайте условия задач.
why everyone is specialist?
pls show me one person that will want to participate in a round prepared by 15+ cyan shkolniki
How on earth is MikeMirzayanov cyan.
CyanForces!
[deleted]
What happened to the time being displayed in the local time zone in the blogs?
Fixed
Do I need to be cyan in order to take part in this round ?
Yes.
Yes.
Don't be afraid to show your real rating, like we did. Say no to rate-shaming.
Thanks for tolerance. We really appreciate it
It's called ratism.
I thought my laptop was glitching at first.
And I thought I was colourblind.
Oh yeah i will definitely participate as there is Ibraev Dimon.(А вы просто указали всю параллель А0 + А?))
you forgot to mention PoLeena moosena && andrew odincoff
Oh, here we go again. Fakes on codeforces.
(У тебя комментарий на русском?)
Ну по хорошему это называется «smurf account” а не «fake”.
Бтв очень надеюсь что раунд будет лучше чем прошлогодний, потому что тогда было реально ужасно(авторы не бейте).
Тот раунд был топчик алло
Прошу указать хотя бы одну красивую задачу в том раунде
Задача див1б
На мой вкус это не задача а один большой кукарек
На мой вкус это отличная задача на скорость рук
Когда задачи на силу(скорость) рук стали украшать раунд?)
А почему Вы думаете, что задачи на скорость рук не могут быть красивыми?
Задачи на скорость рук такого уровня обычно или решаются за 5 минут или за 2 часа(в обоих случаях без доказательства). Честно не вижу ничего красивого в этом.
В той задаче было несложное доказательство, что очень мне понравилось
Я не про сложность доказательства, а про то что его на туре никто не придумывает прежде чем сдать.
Лично я его придумал прежде чем начать писать
Очень крутой.
You guys made me cyan
Well , I changed my mind.
Cyan apocalypse
их боялись даже 4еченцы.
самая опасная опг и отмороженная группировка за всю историю человечества.
After two consecutive mathforces I hope this contest will not be a mathforces.
Dont worry cyans dont know math
Btw
Mike is the real Loki.
Jokes on you I became cyan today. ( for a second I thought I broke the system)
guess it's my turn now
That's what she said.
I wonder where is UnstoppableChillMachine...
Is this some kind of anti-ratist movement?
let's remove colors! make them all cyan!
Is this some kind of anti-ratist movement?
let's remove colors! make them all cyan!
Yesterday we participate contest with maximum no of registration and today we are going to participate contest with maximum no of writer.
Round doesn't seem worth participating in. Maybe with 900 more of these problem setters it will equal the quality of an Um_nik round.
Our team rating is 3237.32, which is better than his rating.
I've used this formula: https://mirror.codeforces.com/blog/entry/16986?locale=ru Code and result: https://ideone.com/SGTALm
https://mirror.codeforces.com/blog/entry/62744?#comment-467221
You really can't into jokes, can you?
Well I guess you can cross that rating today :)
Kak zhe tyi slab v postironii
Well, my guess was correct. You have indeed crossed that rating.
Joining the cyan trend!
Hold up.
Specialist round!
.
Happy Birthday to you
Why is it cyan round?
The round is rated for every one with ratings between 1400 and 1600. However, all of you who wish to take part and have not rating between 1400 and 1600 , can register for the round unofficially.
Is it just me, or it's actually looks satisfying to see all cyan like that
cyanity
Me: (Looks at setters) This is incyanity
A creative use of cf new-year magic!
Even if we consider the worst-case scenario, where both divisions have completely different problemsets (i.e., 12 unique questions in total), it is still way less than the number of writers for this round (which is 22). Interesting.
Every problem has been worked on by multiple people. You can easily have one person write both (english and russian) statements and 1-2 others do everything else all at the same time. Yes, every person listed as a writer is actually one, everyone had a part in making the round.
Go team Cyan!
Over 20 writers for 8-9 problems (combined in Div 1,2)... Can we expect each problem to be a mixture of multiple concepts and observations ? And will not be speedforces today ?
P.S. It should be fun today. :)
The round of 1000 cyan .. = the round of Um_nik
Hell yeah, cyan round.
in respect of all the cyans
in respect of all the cyans
Let's make cyan great again everyone!
the number of authors makes me feel good about this round, but I hope it's not a mathforces
When you're not cyan yet:
when you just want to write cyan in comments somehow.
Does it mean that EvenImage would be participating today.
Man, your display picture is disturbing.
Cyan round?
Such an original comment!
I can be cyan after this cyan round!
WOW
I need 5 points to get cyan...
One more cyan! UPD: scores for problems?
How is task complexity calculated?
Seeing a strong cyan community, Hoping to become a Cyan today.
Good Luck everyone!
Why there are some candidate masters registered in div2 and some experts registered in div1?
Because some of them registered for the contest before the Hello 2020 ratings have been recalculated.
It will be rated for them?
I'm pretty sure that yes.
Once I was transferred to Div. 1 several minutes before the start of the contest (I registered when I was still Div. 2), and it was rated.
However I don't know if that's a global thing or not
My rating is 1346, would this div 2 contest be rated for me? I am confused as it is not mentioned in above blog.
Yes div2 is for rating lower than 1900 and div1 is for rating greater or equal to 1900!
Is it Rated?
Yes. It's rated!
Of course NO one really want to do this round unrated .
Did someone can really think that Regular Codeforces round can be unrated?
This is the point where you need to cut down the list of authors on the Contests page to only the most relevant ones because it's tl;dr.
What do you mean by "most relevant ones"? Everyone had a part in making the contest.
Are you asking what the phrase "most relevant ones" means or who should be considered the most relevant ones? In the first case, try a dictionary, in the second case, I can tell you what I'd pick if you describe the preparation process in detail and how everyone contributed.
If, for example, I made a dumb heuristic solution for some problem and a countertest for that got added, I wouldn't consider it relevant enough to call myself one of the authors of the contest.
And everyone can be mentioned for example on this page, with a shorter list posted where it would be an inconvenience to have a very long list — the Contests page, in this case. It's not like everyone who had a part in making a contest should make their own blog post about their part in it, either, right? The logic is the same.
So every person got assigned to a problem, with 2-3 people per problem, and each team then distributed the tasks between its members. There's really a lot to do if you're making a problem in polygon: write a statement and a tutorial in 2 languages, write several solutions (correct and incorrect ones alike), generate tests, make a checker and a validator, and each one of these tasks can pretty much be done independent of others. Of course, the harder the problem, the harder each of these tasks are, but not significantly: everyone involved knows the statements and the solutions well, so preparing any problem is pretty much a routine job. So there really are no "most relevant" problemsetters, they are all roughly on the same level.
Huh, that's... unusual. Most of the time, there's someone who makes sure statements are ok, someone (possibly the same person) who makes sure translations are ok, testers who write the correct+incorrect solutions and comment on what's missing, and for each problem one, very rarely two people who do the main, initial brunt of the work — basic statement, correct solution, basic tests.
I don't have that experience. When I'm preparing a problem, in order to make strong tests, I want to understand intricacies of the problem and the solution well enough that I might as well write it down, so making the statement, tutorial, one solution and tests is all interconnected. The same goes for the checker if a non-trivial one is required. If I do just part of it, I can miss something important, and if I divide it up with someone else, it happens too often that one of us ends up doing too much or there's just a general lack of communication, and it costs more than it offers. Then, some tasks are independent — validator, simple checker, simple bruteforce etc, and some, like good translations, then require specialised roles.
In your case, it's more like when there's a prioritisation meeting at work where some bigger structured task was already explained and divided into also explained smaller parts before, it gets moved from the todo list and team members are assigned to parts of that task. Since you mentioned that each assigned person knows their problem+solution well enough.
Anyway, regarding the long list: you mention 18 people + 4 teachers/coaches + some testers. Since you mention 2-3 people per problem, I assume that's the 18. If each person is mentioned just for 1 problem and only for the division where that problem appears (C is one problem), that would cut it down. Another, probably better alternative, would be to just keep the list of authors on the Contests page empty. Whoever wants to read it can read it here.
What is scoring distribution!!
Let's all convert to Specialist
Come in as a fake cyan, come out as a real cyan.
i didnt registered because i didnt see starting time how can i now register there is 50 minutes remaininig?
me at the contest cyan round was very difficult for me
Being the legendary grandmaster I was able to make TLE at B!
good contest. specialist here i come!
stuck TLE on div2B for about 1h until i found i used map changed to unordered_map and pp lol
I passed pretests with map in 654ms. Was it intended?
guess i implement it poorly
Nice Impossibleforces !!
Radewoosh RIP
orz Radewoosh
Is this correct for D2E1/D1C1 —
We ask (1,1) + (1,n) + (2,n) By (1,1) we get first character, by (1,n) and (2,n) we can get only prefix substrings. By keeping count of frequencies, we can place characters by reducing the first character one by one.
That worked for me, but asked only (1, n) and (2, n) — you get the first character for free as it is the only string of length 1 in their difference.
On problem C, I spent 20 minutes debugging my code, only to find out that all I was missing was the edge case where N = 1. :(
Overall, great contest in the D2 at least.
https://mirror.codeforces.com/contest/1287/submission/68274741
Please explain why this isn't working. Time complexity is O(n*n*(k+log(n))) Can it be done in better???
String concatenation is not always O(1). Better to use vector of characters. 68258848
Thanks ..... This has made me sad now, didn't knew such a thing :(
worked for me by concatenating only chars to the main string...
in your example in the last case, you concatenate string w to string t:
should be just :
my AC solution : https://mirror.codeforces.com/contest/1287/submission/68262300
That too is just on the boundary!!
My solution (pending systests) is pretty much the same idea. I only used a set instead of a map. Also I locally generated a
n=1500, k=30input locally to make sure my code does not TLE before sending itYou should also note that the log(n) factor is unnecessary, and due to the strict time limit can by itself cause a TLE, assuming a badly implemented code.
this line t=t+v[i][m];
It's inefficient, because basically you create third string and get data from the first string and second string, the total t construction after k iteration will be O(k^2) http://www.cplusplus.com/reference/string/string/operator+/
It will be faster, if you change it to t += v[i][m] It will only add the second string to the first string, The total t construction after k iteration will be O(k) http://www.cplusplus.com/reference/string/string/operator+=/
my code with push_back fails..any reasons why?
thanks gegewepe ........ it worked
How to solve C?
.
i think he asked about div1c
never mind
Why constraints for DIV2 C are so low? It can be solved in O(nlogn) using greedy.
Update: my solution passed system tests
Whats the greedy solution? I could only think of a dp solution with states involving (index, oddRemaining, evenRemaining, lastOddOrEven)
Consider the parities of fixed points in order as they appear in the string. We can see that if two consecutive parities differ, then the the complexity of the substring between them is 1. So, it remains to consider the segments with same parity, sort them according to length, and give the required numbers with the same parity till we run out.
The greedy which I implemented has some annoying edge cases.
Basically, you have to separately consider substrings bordered by the edge, and substrings bordered by two numbers. Plus, you have to deal with the all-zero case and the N=1 case.
Only 4 types of gaps are possible: even-even, odd-odd, even-odd, odd-even
For even-odd and odd-even, the minimum complexity generated is always equal to 1 (doesn't matter if you use all even or all odd or combination of both)
So just take all even-even and odd-odd types of gaps in increasing order of gap value and greedily fill them. E.g. for even-even if enough even numbers are available to fill the whole gap then the minimum complexity generated is 0 otherwise 2.
Gaps in start and end of the array are needed to be handled separately.
I thought of similar solution but couldn't implement it cause B took all the time.
I started writing this as well, but I wasn't sure if it was correct so I just deleted everything and wrote DP, since I knew for sure that would pass.
How to solve using DP ?
Define $$$dp[i][j][k]$$$ as the minimum obtainable complexity upto position $$$i$$$ such that exactly $$$j$$$ even numbers have been used so far and the last one used has remainder modulo $$$2$$$ equal to $$$k$$$.
I tried that too, I immediately knew it was DP but for some reason, I went too far with the DP tbh it felt like I was drunk XD
I have submitted this idea but since I was in a hurry (Badly written) there might be a bug in the code, If anyone gets accepted with this idea please link it.
https://mirror.codeforces.com/contest/1287/submission/68272946
I followed the exact same greedy approach but cannot get AC on 12th test case. I am unable to find the issue with my code. Can you help? 68277187
You should try this
7
0 3 0 5 1 4 2
The correct answer is 2, not 3.
Perhaps because the greedy (if it is indeed correct) is error-prone while not very interesting. I started with greedy and after 20mins and 2 WAs just wrote the DP which passed immediately. I actually respect their decision not to force the greedy approach.
No offense, but your solution is pretty ugly. DP produces a shorter and cleaner code.
My point was about time complexity not implementation complexity
Finding the best possible complexity is nice, but to me it sounds like an upsolving excercise. During a contest, the goal is to get points ASAP, so I don't see why waste time implementing it.
No offense,but when there is a solution with better time complexity,it is always better to learn it.
Same question,could have been asked with larger constraints,and dp wouldn't work.
I don't think you read my comment carefully. It is good to learn the fastest solution, but it's better to do so without time pressure during contest. That's what upsolving means.
I get it,it is always better to implement a dp solution than a greedy,if it works under constraints,but I believe the intent to post the greedy solution to help others.
I never said his comment had a bad intent. He asked why constraints were so low, and I answered: DP was simpler than greedy.
Can you please explain your solution to $$$D$$$ ?
That feel when you read that test 20 is incorrect and then get "Pretests passed".
How to solve B?
lets choose two strings i , j any two are valid ( try to prove it ) then you can know the third string in this way loop in those two strings if current two chars are equal then the third string must have the same char else the third string must have a char nethier equal those chars in that position
Terrible
I have a problem with live contests usually I can solve div2A in < 15 min and div2B in < 30 min (in training)
But I need to see the test cases to see why I get WA... if someone know how to get over such a problem..
Yes and i will be newbie after this round
Learn to debug and stress test.
Try Virtual Contests, in that case, you won't be able to see the test cases.
Do you really think putting these problems together in one round was reasonable choice? Already C2 was a tough one and A+B+C1+D (solving only one out of 4 hardest problems) gives me 11th place before systests.
I prefer contests like these to contests where all the top participants solve everything. It gives freedom to choose which problem to solve. Pretests could be stronger though. :(
Why did i go in thinking a cyan round should be like taking candy from a baby :sobs
Does this mean I have successfully hacked the judge?
No
I thought that in problem called "implement a lot of stuff" implementing BigInt was just another part.
Wow! Now I see why Codeforces doesn't support 128bit integers.
+1
Ok, I think results of this round prove that Um_nik is really worth more than 21 cyans.
What's the final resolution to this? It looks like there are no tests that overflow 64-bits now? I think that decoding via $$$ans \% 2^{64} \% 26$$$ and $$$ans \% 26$$$ both pass?
Any gueses what is in test 3 for Div2C?
n == 1
My code gives answer 0. But I still get WA 3
n == 1 && p[1] == 0
Yeah! In this case my code answers 1. Thanks
some ideas:
6
0 0 1 0 0 0
(expects 1)
5
0 2 0 4 0
(expects 4, pretest 4 is like that)
5
0 1 0 3 0
(expects 2)
1
0
(expects 0)
6
0 0 1 3 0 0
(expects 2)
looking for pretest 5 if somebody got it
Pretests are now visible.
Indeed, and I realized my mistake. I had something like
and my intention was to do
because the
else ifin the first version acted as an else for theif(condition2), and not for theif(condition3)like I wanted... What a mistake xdn=1
Anyone solved C2 by asking $$$(1, n)$$$, $$$(1, \frac{n + 1}{2})$$$ and $$$(\frac{n + 1}{2}, n)$$$?
how do you recover answer?
I didn't have time to implement it, which is why I'm asking others who actually solved it here. My rough idea is we should know $$$a_\frac{n+1}{2}$$$ from the beginning, then just build up the answer symmetrically from those 3 parts.
That's my solution but I kept getting WA on 12. I think I have a bug because the implementation is tough.
I think query same half would be work, thus might use one half to recover another half. waiting for editorial..
Sort of, but I have 6 cases: $$$n = 1, n = 2$$$, and $$$n \mod 4$$$. Maybe some are unnecessary.
I'm not sure, but I assume that it recovers the string up to reversal, isn't it?
5 4 SETT TEST EEET ESTE STES Participant's output 0 Checker comment wrong answer 1st numbers differ — expected: '2', found: '0'
But my dev c++ runs out 2 Can anybody explain that?=_=
You are likely dealing with some kind of undefined behavior (such as indexing an array with an invalid index or using ununitialized variables)
change
char kk[k];tochar kk[31] = "";Interactive problems...
Find out a solution at last but keeping 'Idleness limit exceeded on pretest 1'
How to solve B ?
Choose 2 strings and try to make the third one. You can see that if you have 2 strings, there can be only 1 third substring. All you have to do is to create a map to store the frequence of every string and, after you build the third string, add its frequency to the answer. Be careful, you count every triplet 3 times, so, at the end you should divide then asnwer by 3. :)
I got WA on D and E1 on first submission because i was missing the case N = 1... Has it happened to someone else? :)
Thanks for including $$$n = 1$$$ in pretests for C1 -_-
Really sad. I found where I made a mistake in my prob D code after 2 minutes the contest had over.
How to solve Div2-D?
Start placing integers in nodes in decreasing order — When subtree size = c-value, this is max in subtree, select the one with lowest depth and decrease subtree size of all its ancestor by 1.
Can you explain what you mean by "start placing integers in nodes in decreasing order"? Should you place bigger integers before smaller ones? Or should you place integers from nodes n to 1? And what do you mean by "subtree size = c-value"? What is max in subtree of what? And what do you mean by decreasing the subtree size of all its ancestors?
Should you place bigger integers before smaller ones? — Yes
And what do you mean by "subtree size = c-value — c_values given in problem
What is max in subtree of what? — Maximum a_i among nodes in its subtree
Thanks for the nice Div1C? How to solve Div1D
How to solve Div1D ?
can someone please explain why this solution 68252477 for PROBLEM B of DIV 2 is getting TLE
Your solution is (n * n * k * k * logn) which is quite big
You are using std::set, which has an awful constant on nearly all the operations. Using three ifs instead of set gave AC.
Thanks for replying. I don't think this is the problem but still i will give it a try once submissions are allowed and will update you
update: you are correct. Just by using three loops it passed. Thanks man
You did s += x to add a single character to a string, which will lead to $$$O(k^2)$$$ for forming the string. Instead do s.push_back(x), which will be in $$$O(k)$$$.
s+=x has complexity linear in term of length of x. s=s+x has linear complexity in length of resulting string.
so s+=x will lead to O(k) only
can you explain why s+= x is slower than s.pushback(x) in general ?
Sigh...
I liked problem C, but it can be hard to spot the solution. It is kind of guessing which approach to take.
C is probably my favourite interactive problem now, finally something that is not binary search :Dd
Ask [1, n] and [2, n], sort strings in the answers and remove strings in the second answer from the first. Then only one string of each length remains, and those strings are all the prefixes of our string. By counting which character frequency increased from i-1th prefix to the ith, we find the ith character.
Use C1 to find the first half of the string, then query [1, n]. The total length of the answers is $$$(k+1)(k+2) + \frac{(2k+1)(2k+2)}{2} = 3k^{2} + 6k + 3 = \frac{3}{4} (2k + 2)^{2} \leq 0.777 (n + 1)^{2}$$$ when $$$n = 2k + 1$$$.
Let $$$cou[c][k]$$$ be the total number of occurrences of character $$$c$$$ in substrings of length $$$k$$$ in the last query. Then the number of occurrences of character $$$c$$$ in the first $$$k$$$ and last $$$k$$$ characters of the string is $$$cou[c][1] - (cou[c][k+1] - cou[c][k])$$$, as if the character at position $$$i$$$ is $$$c$$$, it adds $$$\min(i, n+1-i, k)$$$ to $$$cou[c][k]$$$.
We know the first half of the string, now loop $$$k$$$ from $$$1$$$ to $$$\lfloor \frac{n}{2} \rfloor$$$ and find how many times character $$$c$$$ occurs in the first $$$k$$$ and last $$$k$$$ positions. We know all but one of the characters in the first $$$k$$$ and last $$$k$$$ positions, so we can this way check if the $$$k$$$th last character is $$$c$$$.
I did the same thing in C1 div1 / E1 div2, but damn, that 0.777 was scary
You can actually restore string uniquely up to reversal just by asking [1, n]. And then you can ask any [i, i] such that $$$s_i \neq s_{n - i + 1}$$$. So it can be solved by 2 queries with sum of substrings $$$n(n+1)/2+1$$$.
I don't think so, what about this:
xyaxybxyxybxyaxyThese are not reverse from each other, but they have exactly the same output (up to randomness, this is, before returning sort characters on each substring and then sort all substring).
You hacked me. :)
I don't think that binary search interactive problems are in trend now, wow. Last such problem was on Codeforces Round 569 (Div. 1) (in summer).
Last few interactives: 1270D - Strange Device, 1282D - Enchanted Artifact, 1254C - Point Ordering, 1207E - XOR Guessing, 1205C - Palindromic Paths, 1201E2 - Knightmare (hard) — no binary search!
How to solve Div1D ?
I hate subtasks so much!!!
So go to D after solving C1?
I thought D would be way harder than it actually was..
What's wrong with this solution for div2B? I am getting TLE on pretest 10. https://mirror.codeforces.com/contest/1287/submission/68262344
I think in the nested loop, you can reduce the use of set. due to too much of find, erase and insert. Maybe thats why you are getting TLE,
You can check my code here, it does the same as your code using map.
https://mirror.codeforces.com/contest/1287/submission/68278675
Same here https://mirror.codeforces.com/contest/1287/submission/68271441. I am seeing exactly the same solutions with AC ...
same, i changed map to unordered_map and pp
is Div2-C solvable with DP ?
Yes
i am not good at table method but memoization so i was affraid of getting TLE because of Stack Time + N ^ 4
I wrote dp[last index][count of odds][count of evens][is last even or odd(0/1)] : dp[100][100][100][2].
So it is N ^ 3, that easy passed all tests by TLE.
See my submission
Maybe greedy?
68256404
i did that :D
When you are just on time.
So what's the best ratio anyone's gotten in Div1 C2? My solution takes about $$$0.75N^2$$$ and that's likely the intended one. It doesn't feel like this would be the optimal solution.
Edit:
According to aid a query restores the string uniquely up to reversal, which makes the problem much simpler. I've no clue how one proves that, though.
I think you can solve it with one query for the whole string and then one more question for a single string. The algorithm is rather messy so I didn't get time to implement :(
Basic idea: Looking at strings of length $$$N-1$$$, you can get the first and the last symbol. Then, with some considerations, from the strings of length $$$N-2$$$ you can extract 2nd and $$$(N-2)$$$-nd symbol, and so on.
There is one more consideration: the first time you encounter different characters with the above algorithm, you can ask for one of them directly. After that, you can use this difference to figure out where each character is for each different pair in the next iterations.
Look at all strings of length $$$n-1$$$. There are 2 of them. From them you can restore first and last elements, but you don't know which one is first and which one is last. If they are different, ask about the first one. Then if we restored the first and last $$$k$$$ elements, let's look at the strings of length $$$n - k - 1$$$, we know all but 2 of them. Find those 2 strings, from them we can find $$$k+1$$$-st elements. There are some cases like when we already found 2 different characters and not.
How do you handle something like
aabbaa?We have strings "aabba" and "abbaa", from them you know that first and last elements are 'a'.
But those strings can also come from
bbaaab. Can you elaborate your approach?But you know the number of 'a' and 'b' in the whole string. Each of the strings of length $$$n-1$$$ either has 1 less 'a' or 1 less 'b', from that you know 2 border characters.
I had the same approach, but during upsolving I found out that during restoring there can be an issue, where prefix and suffix are the same (w.r.t character frequencies), we have two different letters, and we don't know which letter goes to prefix and which goes to suffix. That's why I got WA23 :/ How do you deal with such situation?
You know how many A's and B's there are because of the 1-length strings, so you can check which character is absent in both substrings by counting
My solution uses around $$$\frac{2}{3} N^2$$$:
C1: $$$\textrm{query}(1, N) - \textrm{query}(2, N)$$$ gives you the set of prefixes of the hidden string, which is enough to construct it. It uses around $$$N^2$$$ strings.
C2: With the same idea from C1 we can learn the first $$$\frac{2}{3}$$$ of the hidden string using $$$\frac{4}{9}N^2$$$ strings. Knowing the middle third of the hidden string and querying its last $$$\frac{2}{3}$$$ (an additional $$$\frac{2}{9}N^2$$$ strings) is enough to construct the final third.
Can you explain how do we get the last third of the string?
We want to construct a string of length $$$2T$$$. We know its first $$$T$$$ characters and we know the list of responses from querying the entire string.
From the unique response with length $$$2T$$$ we know the unordered contents of the full string.
Suppose that we have already figured out the last $$$k$$$ characters of the string. Initially $$$k = 0$$$ and when $$$k = T$$$ we are done. We consider all substrings of length $$$2T - 1 - k$$$. We already know enough characters at the beginning and end of the string to construct all of them which don't start at the very beginning of the string (subtract excluded letters at the beginning and end from the contents of the full string). By elimination we find the one which does start at the beginning: it gives us the $$$(k+1)^\textrm{th}$$$ character from the end.
Thanks
What's with pretests in Div2B not even having the stress cases? It sucks to get TLE on main-tests for some constant optimisation.
CAN YOU MAKE PRETESTS STRONGER?
Very very very weak pretests :(
Attempt 1 : TLE
Attempt 2 : TLE
Attempt 3 : TLE
Attempt 4 : TLE
Attempt 5 : Pretests passed at 1:59:32
Reason for TLE : use of set to find the character not present at a given index rather than simple if else.
I used policy based data structure so my solution got accepted at last
please can someone tell me why using set give TLE
set.insert() and set.erase() are not O(1) operation but O(logn) operation, although n here is equal to 3, i didnt expected that to be huge issue
https://mirror.codeforces.com/contest/1287/submission/68275963 this solution give output 0 for n=7,p={0,0,0,7,0,0,0}, but actual output is 1
That's not the way on how to write problem B for Div 2.
Duh, getting $$$O(3^n)$$$ is not that hard in F, definitely more doable than anything like $$$O(2^n n^{10})$$$ and indeed as I expected both accepted solutions are $$$O(3^n)$$$ -.-
never give up :))))
unordered or ordered map ? In some problems unordered passes while ordered fails, while in others exact opposite happens.
Always use map,unordered map's time complexity depends on nature of input,if a solution passes for unordered map but fails for map,then it's either bad implementation or bad question
Unordered map is faster, provided that you have a good hash function. The reason why unordered_map is sometimes slower than regular map is that there exist special techniques of generating worst-case tests for unordered map, where standard hash functions give a lot of collisions.
See this blog for more details.
The contest was raely nice but I had a better idea on it
It was realy nice to add a simple problem at the begining of the Div. 2 then you change the n of the problem Div 2. D to 1e5(solveable with easy dfs).
so the contest would be better to solve.
And also D was easier than C and swaping them was a good idea.
Thanks CyanForces!
It was good at all.
can you tell your approach for d, i find dp solution of C much easier than d.
as a dfs return a vector and make this like:
insert all of the vector of kids of v together
the insert v as c[v]_th element (if impossable "NO")
then get the vector of root
then for all i, ans[a[i]] = i
then output the ans[i] for each i
And I think it is the best problem fo teaching dfs to someone :)
Can you explain why that works?
Because every vertex is the c[ i ]_th it it's kid's set.
Because we inserted it as c[ i ]_th element in the vector :)
Do you agree? so now we came to find out what numbers to assign vertex to get the order v1, v2, ..., vn.
so we assign 1 to v1 and 2 to v2 and ... .
now we have some numbers that are the c[ i ]_th in their SubTree. :)
Yes, thank you!
How would you do div. 2 D with n = 1e5?
Do like the comment on top but with a little diffrent.
First with an easy dfs check if every vertex's (SubTreeSize[v] <= c[v]); If not output("NO");
Here we go:
take an ordered set Insert all 1 to n in the set
For each vertex in the dfs erase the c[ v ]_th element and it would be ans[ v ] and it's ans sequence is the same as This (try to prove it by Mathematical Induction and the delete the root and it is ok for the tree's made by deleting it and so on ...
And output ans[i] for every i.
Good one! Actually, just implicit key cartesian tree gives nlogn if vector is replaced with it. Code for n^2, which can be modified with tree instead of inserting into vector
Yes bro got it right!
it was the idea behind it.
What is the time complexity of writer's solution of problem F? I think coming up with O(3^N) solution is not so hard, yet I was able to fit it to TL by pruning: 68281697
$$$O((1 + \sqrt{2})^n + \log{n} \cdot n^2 \cdot 2^n)$$$
Could someone please, tell me, where I went wrong. 68273158
How to solve DIV1B/2D numbers on tree.
I solved in this way: Keep the number of nodes in the subtree for each node(let it S[]). If there exists i such that S[i] <= C[i], then the construction is impossible. Otherwise, the construction is always possible. Keep 1, 2, 3, ... n originally in a set data structure(e.g., treap?); let the the set B. From the root, run DFS and make each A[i] value to C[i]+1th smallest value of B, and erase the value from B. Then, the number of smaller nodes in the siblings are exactly same to C[i].
Do you mean number of smaller nodes in the subtree?
Yes. As we wish to be. :) 68267213 This is my submission. I feel ashamed to be in public(cause the code is dirty..).. But maybe it would be helpful under the function predfs. I used treap data structure to keep the set and find kth element from it. But there would be many alternative solutions like indexed tree + binary search, or etc.
Cleaner submission with Fenwick Tree and binary search(to find kth element): 68289460. (O(nlog^2n))
Никто не знал, но на самом деле эта задача изначально содержала эту картинку в условии (я серьезно):
Достойно уважения на самом деле.
А что, MikeMirzayanov не пропустил в продакшн?
Sorry, but it was a coincidence.
My solution for C was 100 lines in python (making lists of gaps with different ends then sort etc) there should be some shorter approach to avoid this spaghetti mess.
When tutorial will be published???
in problem B, TLE code in contest time 68259187. Ac code after contest 68286650.
just modified a simple function named f(). I think it should not matter. please have a look.
Well, you were using map in the function that gave you overhead
It was a good contest.
Тестиролвали, тестиролвали, да не дотестиролвали
Can someone tell me why this code gives WA on div2A?
APAPP, add i-- after while loop and it should be correct
How to solve Div.2 C/Div.1 A?
Think dp! The state is dp(index, odd remaining, even remaining, parity of last element). See this:https://mirror.codeforces.com/contest/1287/submission/68272946
Thank you!
I m newbie can someone tell me where is the editorial?
Here
https://mirror.codeforces.com/contest/1287/submission/68290995 wrong answer on case 12 which doesn't have a value at index 0.
https://mirror.codeforces.com/contest/1287/submission/68290894 wrong case on case 8 which also doesn't have a value at index 0.
https://mirror.codeforces.com/contest/1287/submission/68290844 wrong on case 12 which also doesn't have a value at index 0.
https://mirror.codeforces.com/contest/1287/submission/68290780 wrong on case 19 which also doesn't have a value at index 0.
please help me with this error .
My approach
i am not able to manage what should be done first .
My approach
in some cases assigning to first works in others after assigning to all gaps first and last are checked. Only 4 types of gaps are possible: even-even, odd-odd, even-odd, odd-even For even-odd and odd-even, the minimum complexity generated is always equal to 1 (doesn't matter if you use all even or all odd or combination of both) So just take all even-even and odd-odd types of gaps in increasing order of gap value and greedily fill them. E.g. for even-even if enough even numbers are available to fill the whole gap then the minimum complexity generated is 0 otherwise 2. Gaps in start and end of the array are needed to be handled separately.
7
0 3 0 5 1 4 2
The correct answer is 2, not 3 , my code is giving 3.
Please help me fix it.
should i sort the gaps in increasing order of length before filling?
yes sort the gaps,before filling numbers My code: 68283299
(https://mirror.codeforces.com/contest/1287/submission/68307501)
Please help me i have done everything still this is not working on test 12 but working fine till 18 tests other than 12th one.
Thanks in Advance.
uncommented code is very difficult to read,try random test cases,compare the output with that of an accepted code
Editorial???
Here
div1 F
O(3^n) can spend only 300ms
my code
Impressive
can someone tell me why i am getting tle on problem(B. Hyperset) here is my solution https://mirror.codeforces.com/contest/1287/submission/68274322
Isn't string concat in C++ $$$O(n)$$$? In which case your approach is $$$O(n^2k^2)$$$.
here is one solution(https://mirror.codeforces.com/contest/1287/submission/68268204) in which string concat is performed but it got accepted.
you should change st=st+s[i] to st += s[i] read my comment above for a little explanation
Grammar mistake in the description of div. 1 F.
"The array is considered destroyed is all its elements are zeroes."
Thanks, fixed!
Waiting for editorial...a bit par contests for green coders...
Here
How to solve div1D?
Let $$$dp[i][j][0/1]$$$ = probability of proton $$$1 \dots i$$$ 's first collision time is $$$j$$$ , $$$i$$$-th proton flies to right/left.
There is at most $$$2n$$$ different values of $$$j$$$.
Use segment tree to optimize，time complexity is $$$O(nlogn)$$$ .
respected sir, i have no clue how my code matches with the other 2 persons you mentioned in the e-mail. this is absolutely a coincidence. i neither share my code with anyone nor do i copy. i used flag method to solve the question which is quite common, and hence may be it matches with these 2 other persons. thank you anirudh agarwal
(https://mirror.codeforces.com/contest/1287/submission/68307501)
Please help me i have done everything still this is not working on test 12 but working fine till 18 tests other than 12th one.
Thanks in Advance.
In D1C, any solution to query two times, [1,n] and another point, is impossible.
Consider the following 4 strings: abaabbab abbabaab baababba babbaaba
If someone only queries [1,8], he will got exactly the same results.
I'm not sure if it's possible for [1,n] and queries of O(1) additional length.
how to solve div 2 C ?
Where I can find editorial?
Here
Where is the Tutorial???
Here
Why is this 68314242 passing and not 68261685 for Div2B?
The only difference is the way in which i find the third string to be searched! Both are constant time!
Can someone explain the DP approach for Div2 C, it is not there in the editorial released.
You can read here
there is O(n) solution in question Div2-C by greedy
1-count number of different parity between number which aren't zero
2-sort contiguous segments of zero by counting sort by their length in O(n), we can use from vector structure like vector p[100] which index is our length of segments and we store start and end of segments, we can iterate vector's elements in O(n)
3-chose smallest contiguous segments of zero which don't start from beginning and don't ends at the last element, these segments are between two numbers like x and y (for example x 0 0 0 ... 0 y)
4-if x % 2 != y % 2 then we have one different parity in this segment always so don't do anything
5-if x % 2 == y % 2 then if you have enough numbers with parity of x % 2 then by adding these number in this segment we have no different parity but if you don't have enough numbers then we have two different parity in this segment
6-after visiting of all segments which aren't beginning of numbers and aren't end of numbers, if there is contiguous segment in beginning of numbers or in the end of numbers then our segment is like ... x 0 0 0 0 ... 0 0 or 0 0 0 0 0 ... x ... then if you have enough numbers with parity of x then we don't have any different parity and if you don't have enough numbers with parity of x then we have different parity with 1.