Привет, Codeforces!
В 10.10.2021 12:05 (Московское время) состоится Educational Codeforces Round 115 (рейтинговый для Div. 2). Обратите внимание на необычное время старта раунда.
Продолжается серия образовательных раундов в рамках инициативы Harbour.Space University! Подробности о сотрудничестве Harbour.Space University и Codeforces можно прочитать в посте.
Этот раунд будет рейтинговым для участников с рейтингом менее 2100. Соревнование будет проводиться по немного расширенным правилам ICPC. Штраф за каждую неверную посылку до посылки, являющейся полным решением, равен 10 минутам. После окончания раунда будет период времени длительностью в 12 часов, в течение которого вы можете попробовать взломать абсолютно любое решение (в том числе свое). Причем исходный код будет предоставлен не только для чтения, но и для копирования.
Вам будет предложено 6 или 7 задач на 2 часа. Мы надеемся, что вам они покажутся интересными.
Задачи вместе со мной придумывали и готовили Александр fcspartakm Фролов, Михаил awoo Пикляев и Максим Neon Мещеряков. Также большое спасибо Михаилу MikeMirzayanov Мирзаянову за системы Polygon и Codeforces.
Удачи в раунде! Успешных решений!
Также от наших друзей и партнёров из Harbour.Space есть сообщение для вас:
Привет, Codeforces! 
Мы поздравляем одного из наших преподавателей Николая KAN Калинина с его первым местом в финале чемпионата мира ICPC, который проходил в Москве, Россия. Годы тренировок Николая и его команды из Нижегородского государственного университета привели их к вершине турнирной таблицы, победе над командами из 116 других университетов и чемпионству.
Также поздравляем нашего будущего студента Егора 244mhq Дубовика, завоевавшего серебряную медаль в составе команды Белорусского государственного университета. Егор присоединится к нам в магистратуре "Computer Science" в ближайшие недели.
Мы с нетерпением ждем встречи с Николаем снова в январе следующего года, когда он будет вести свой курс Advanced Algorithms and Data Structures вместе с Майком Мирзаяновым. В этом курсе студенты сосредотачиваются на ключевых алгоритмах и структурах данных, которые составляют инструментарий современного специалиста.
Мы всегда рады видеть участников сообщества Codeforces в качестве наших студентов здесь, в Harbour.Space, поэтому мы снова предоставили специальную скидку (до 70%) на участие в одном курсе в Барселоне, Испания (расходы на проезд и проживание не включены).
Желаем удачи и до встречи в следующий раз!
Harbour.Space University
UPD: Разбор опубликован.
is becoming quite usual now.
It is very friendly to East Asian players, I hope it will be more than a few times . We don't need to stay up till 2 a.m. to sleep.QWQ
Don't you have Chinese version of codeforces?
It seems not
cf is wonderful
Google, Youtube, Twitter, Amazon... are also great
china have its wonderful Google，YouTube，Twitter，Amazon
we have a mirror site This but the contests are all sync-ed with This Codeforces
Yes,I'm very happy.
Great time for Chinese players!
Why ?
It's 17:05 pm.
Yes ! very good !!!
Thank you for the contest. I hope that my rating will be increase after doing this contest
Hmm...I think it's not as easy as you think
not as bad as you think too
And your contribution will decrease.
good luck to everyone for the competition!
Why is it so early dudeee, some people have school
Yes of course "School in Sunday"
what's wrong with school on sunday? some schools have weekends on Friday and Saturday instead of Saturday and sunday
for example what school?
These unusually timed contests are not for unusual schools!
Allame helli
Some people also have school during the usual contest time.
For Chinese and Japanese，this tmie is better than usual time, as the usual time is 10：35~01:05 in Peking, China.
Too sad we are not seeing vovuh's handle in writers part of an educational round...
[deleted]
[deleted]
and so do I.
How I wish vinchurkar had written "You know the rules ", it would be a nice rickroll. You know the rules, and so do I...
Same too, good luck.
All the best guys!
Why the participation has reduced a little bit from previous 2-3 contests?
Due to universities mid exams I guess !!
RIP sleep schedule. Gotta get up at 5am for this contest.
In China we have to stay up until 2am to participate in contests with usual times :(
Hoping for a great contest OvO
The start time of this contest is very friendly to Chinese contestants! Hope I'll perform well in this contest.
In China, you can take a nap, wake up and start playing games
The game I said refers to playing codeforces. Please don't get me wrong
I hope I will be specialist after this round.
I enjoyed this round There is really very interesting problems
How to solve D?
I tried to use inclusion-exclusion principle. The answer is (different a) + (different b) — (different a and b). I have figure out how to calculate "different a" and "different b", but didn't figure out how to calculate "different a and b".
Lol, I did the same. Couldn't pass it during the contest. Later debugged and got accepted. Definitely it was an overkill..
I calculated the total possible solutions which will be nC3 and then subtracted the number of bad combinations(observe that in a bad combination 2 problems will be having the same topic type and 2 problems will be having the same difficulty).
Damn idk why i got stuck in a much harder solution .. thanks
you are not alone!
I understand. Thanks!
Try to calculate the number of bad combinations and then it's trivial.
How to solve E?
with dp. the main point is for each flip you just have to update O(n) dp values which makes it O(nq) that fits in time limit
I was wondering the whole time why's $$$q \leq 10^{4}$$$ and not $$$10^{5}$$$, I knew that solution was somewhat related to it. Unfortunalely, I wasn't able to figure it out in time.
Can you please explain further, my brain is fried up right now and I am unable to see the dp states..:P
yes sure let dp[x][y] be the number of stair cases that start from point x,y also dp1[x][y] : number of stair cases from x,y that second move is right and dp2[x][y] : number of stair cases from x , y that second move is down .
now we can easily update all these dp values (you can see my submission) sorry for bad english
No no the english is quite clear, and thanks. I think I will see ur submission now.
and also for each query update you just have to update these points : (x,y) (x-1,y) (x,y-1)
(x-1,y-1) (x-2,y-1) (x-1,y-2)
...
which is in total at most 3*min(n,m)
Ahh yes I see.. BTW do u think this was solvable by Segment Tree or some similar data structure? I would think so considering the type of problem.
maybe but it is not trivial
Yeahh
Hey, can you please explain why only those points have to be updated? I think points like (x, y-2) also need to be updated
No staircase from x,y-2 will cross x,y
There is a technique know in Brazil as Color Update, it consists in store in a set intervals that we assign a given color, each position can have at most one color. You can use this idea to color an interval, ask the maximal interval of the same color that contains a given point, etc. Using this idea just build maximal staircases, for each position store in which maximal staircases it is, and in which position. Then using Color Update technique the solution is straightforward.
You can see this idea in my code 131497457.
How to solve D ? I found it much harder than E or am i missing something ?
just find cases where the given condition fails and remove it from total cases
Inclusion and Exclusion Principle, the total number we can get without any constrains is (N choose 3), then we subtract the invalid ones.
the invalid ones must have 2 similar elements in both sides so we just count that.
https://mirror.codeforces.com/contest/1598/submission/131438774
could you elaborate how you calculated invalid ones
Invalid ones should have at least two similar elements in both arrays let's say we picked the current element.
A is the topic
B is the difficulty
so
A should appear at least one more time.
B also should appear at least one more time.
so we count the number of other A's multiply by the number of other B's and keep subtracting this from the answer, and this will work nicely with the constrains we have as (A,B) can't repeat again together.
sorry it's hard to explain, but you can go through my code for better understanding.
kind of got it. Thanks for the effort
Use Multiplication Principle.
Preprocess each element's number of occurrences , and mark them as cnta[] and cntb[].Enumerate i from 1 to n , and the invalid ones are sigema((cnta[a[i]] — 1) * (cntb[b[i]] — 1)).Hope you can understand it.(My English isn't good).
thanks man.
understood.
What's more.Because of "It is guaranteed that there are no two problems that have the same topic and difficulty at the same time".
Invalid ones are a group with two identical "elements A" and two identical "elements B". So we just need to select the first pair first, and then select the other two pairs which have one "element A" and one "element B" which are the same as the first pair. The formula is just written in my last comment.
I think this comment is more clear than the last one.By the way , wish you good luck. :)
yup , just got the code accepted.
this comment really summed it up for me ^_^
Definitely missing something. $$$D$$$ was a nice exercise of complement counting. I solved it by adding edges $$$a[i], b[i]$$$ to empty bipartite graph with $$$2$$$ partite sets of size $$$n$$$ and then the problem becomes equivalent to counting the number of paths with $$$3$$$ edges in this graph.(at this point the problem is quite easy by storing frequency array of both $$$a$$$ and $$$b$$$) Of course this is only for better visualization. In reality you never actually do a graph algorithm to solve the problem. But I personally found this as a nice way to see the problem.
Can you please elaborate your visualization?
Uhh just add vertices in the graph and do simple case work for counting the complement. You will find every valid element in the complement corresponds to a path of length $$$3$$$ in the graph u made...
You are not alone.
Can E be solved with link cut tree ?
Definitely moving up from Pupil today!! How was the contest for y`all guys?
There are so many comments in your code :-\
Removing the comments in your submission of problem d, its remarkably similar to the solution in this image
I don't know if you will be caught by plag check but, What's more shocking is the audacity of your comment " Definitely moving up from Pupil today!! How was the contest for y`all guys? "
its remarkably similarit's equal.Very suspicious submission.
lmao adding random comments to avoid plag
You seem to be a Cheater
Ironical to spam comments just to avoid plag check. Where is your dignity lol
you are not going to see any rating change.
You piece of shit. Look into the mirror and ask yourself what have you gained from cheating. You are a disgrace. Your parents will be ashamed of you.
Someone, Please hack this cheater's solution The same copied codes by others are hacked by many people
solved D but tough Java constraints :( https://mirror.codeforces.com/contest/1598/submission/131447806
https://mirror.codeforces.com/contest/1598/submission/131456342 it worked with c++
I used the exact same approach as you did, and got AC in java. 131438128 I am pretty sure that your issue is at the line where you create a new ArrayList (computeIfAbsent). Try to change that to putIfAbsent and it should work fine.
thanks, it worked! do you know that makes
computeIfAbsentso slow?No idea, but my guess is that for some reason it generates a new Arraylist, even if some element is not absent resulting in a new copy every time.
sorry for necroposting but I found out what was the problem with computeIfAbsent.
https://docs.oracle.com/javase/8/docs/api/java/util/Map.html#computeIfAbsent-K-java.util.function.Function-
the reason of MLE here is that computeIfAbsent passes key as argument for ArrayList constructor so ArrayList is instantiated with given capacity which can lead to MLE 166786739 AC with computeIfAbsent
A round a day keeps my rating away.
It seems that I will get about -50 rating after this round. I hope I can remain an Expert.
By the way , Problem D is such an excellent problem although I didn't solve it in this round. It really gives me a lesson.
Talk about the shortcomings of this game (this does not mean that I do not support the author, I support every author of CF), the data range of the C question is very intriguing (why is there a data range that cannot be covered by int64?), I am useless Over int128, so I spent nearly 20 minutes on compilation errors. Of course, the other topics are very good, thank you very much to the authors!
I guess you checked for $$$sum \cdot (n-2)=(sum-a_1-a_2)\cdot n$$$?
Notice, that you can shorten this check to $$$2 \cdot sum = a_1 n + a_2 n$$$ for which both sides do fit in a long long.
int64 works fine for C. You just need to reduce the equation with algebra.
$$$\frac{\sum v_{i}}{n} = \frac{\sum v_{i}}{n-2} -y$$$
$$$(n-2) \sum v_{i} = n \sum v_{i} - ny $$$
$$$ny = 2 \sum v_{i}$$$
$$$y = \frac{2 \sum v_{i}}{n} $$$
How To solve D ?
Let be ai = task, notice, that for a we can take (n * (n — 1) / 2) other tasks and lets just remove pairs of tasks that we can not use with a. This is the amount of the same difficulty * the amount of the same topic
If we consider pairs as coordinates on 2D plane and since all points are distinct, only way three points selected at random dis-satisfy the given conditions is when they form the right angle triangle. Hence subtract number of right angle triangles from NC3.
What an excellent and unique understanding!
what is the meaning of open hacking phase is running, can i get points if i solve problems or what it is ?
You can hack other's submissions. Hacking doesn't earn or loss points in this contest. Feel free to hack if you find any bugs in other's code.
I misunderstood D and thought that both the given conditions should be satisfied. Later obtain the actual answer easily from the answer I calculated. Anyone did the same?
D was combinations problem though looks like graph
break into smaller subproblems simplify using pnc and generalize the formula to compute in O(nlogn)
why u didnt check examples?
I'm lazy and I wanted to solve the problem faster.
In C why (sum*(n-2))==(sum-a1-s2))*n is wrong?
overflow.
sum can be at max 2e5 * 1e9 == 2e14
And then sum*n==2e14*2e5 == 4e19 which is more than the limit of the long data type
I don't think this is wrong, maybe its an overflow issue
bro power will reach to something 10^20 and long long has range 10^18
Can't we find mean using this method ?
double k = 0; //k is the mean int t = 1; for (double x : arr) { k += (x - k) / t; ++t; }avoid doing such operations bro.
it won't give you the correct ans.
there are always chances of floating pint error,
performing such operation is not advisable
The fact is that there are many Chinese top coders like Miracle03 and QAQAutoMaton are busy in teaching some new coders.So they don't have time to write more contests.The rounds you see are written by some students who love coding , and I think they're good enough for these younger coders.I think they need more encouragement but not prejudice.
I do hope you could post comments with less discrimination and prejudice.And I think Chinese writers can get progress very soon.Best wish.
Too good questions (:)
Where is the solution of this contest ？ I think I need it！
there is a solution for problem E
notice that each staircase's length at most 2000
and each point have 4 staircases at most
so U can use brute force to solve it
the complexity will be $$$O(Q(N+M))$$$
Oh, i mean that you can start only 4 staircases at each free cell
when will publish editorial?
May be after hacking phase is finished
Awesome round. Love problem F
F has too many details :( Is it only for me?
solved E just several seconds after the round ends, what a pity! :(
Awesome round with much less extra hackings (comparing with other Edu rounds) ,excepting problem D.
Can anyone help me with the approach to solve B?
There are only 5 days. We can enumerate all pairs of days and find if it is possible to seperate students into two groups. Let's say we choose day A and day B. If there are some students (possibly one) who can't attend day A and day B, then {A, B} isn't the answer. We count the students who can only attend at day A, students who can only attend at day B, and students who can attend at both day A and B.
Nice round. Totally enjoyed it!!
Please explain how to solve D. >_<
Substract the illegal ways from all the ways.
There are in total $$$\dbinom n 3$$$ ways to choose $$$3$$$ elements among $$$n$$$ of them. Now considering how many of them are illegal, that is, none of the two conditions in the statement is satisfied.
To calculate it, we first define
cnta[x]as how many times topic x has occured andcntb[x]as how many times difficulties y has occured.Then assume that we choose the $$$i$$$ th problem, since no two problems share the same $$$a_i$$$ and $$$b_i$$$, we can assume the second problem we choose shares the same $$$a$$$ with $$$a_i$$$ and the third problem shares the same $$$b$$$ with $$$b_i$$$. So we just need to calculate
1ll * (cnta[a[i]] - 1) * (cntb[b[i] - 1]).You can view my code 131440598 for more details.
Sorry for my poor English(
Can you provide a more rigorous proof?
It seems that a lot of the solutions for D are getting hacked (mine included) and they all count the answer the same way you are describing. I don't know why though.
They are hack just because they didn't use
long longIn your code, when it comes to
nA += (ca[a[i]]-1) * (cb[b[i]]-1);, I suppose that it overflows theintlimitationthank you
The penalty for each incorrect submission until the submission with a full solution is 10 minutes What is the meaning of this?
Penalties are for the leader board. More time used means lower ranking.
Oh okay. I am a little confused, will I get penalities for multiple wrong answer submissions for a problem even If i don't solve it?
No, penalty will be added only for those problems which are accepted.
Any hack tc for D ?
I think the hacks exploit this.
I believe you can have all numbers from $$$1$$$ to $$$100000$$$ and create the pairs $$$(1, 100000)\ (2, 100000)\ \dots\ (100000, 100000)\ (1, 1)\ (1, 2)\ \dots\ (1, 99999)$$$.
In that way there are $$$2\cdot 10^5 - 1$$$ pairs (possible because $$$n\leq2\cdot 10^5$$$) but the product of counts will overflow if they are initialized as
int. (for example considering the first pair the counts will both be $$$99999$$$)I used size_t( vector.size() ) in my code, but it still be hacked.Do you know why?
You've been overflown. Thought your ans is 'long long', you are not yet familiar how expression is evaluated. It's in this line:
Try to output: long long ans = 100'000 * 100'000;
Thanks for your reply!
Could give me one testcase? I test my code with input:
n = 200000
a = [1, ..., 1, 100001, 100002, ..., 200000]
b = [1, 2, ..., 100000, 1, ..., 1 ]
When x.fi = 1, d = 1, that line is:
ans += (100000 — 1) * (100001 — 1). It seems that my code is correct on this testcase with the ouput of 1333303333500000.
Try for this one. But with very large sequence, (1, 1), (1, 2), (1, 3), ... , (2, 1), (3, 1), .....
Nice problems :)))
Nothing to say!XD
Why O(Q*(N+M) + N*M) is failing for problem E (TLE on testcase 41 (not included in pretests))?
My Code : https://mirror.codeforces.com/contest/1598/submission/131480763
Edit : Finally my code was accepted. Replaced vector with pair which brought a major change in execution time (approx 7 times faster). Always try not to use small vectors (if possible).
AC Code : https://mirror.codeforces.com/contest/1598/submission/131498204
Assuming that you are doing a bfs for each query : giving the fact the number of situations is at most (nb of x * nb of y * nb dir = N*M*2), the total complexity of your algorithm is O(N*M*2*Q + N*M) = O(N*M*Q), which is too slow. Sorry for my bad English.
No, the time complexity is fine actually because for changing a particular cell you just need to update max O(n+m) cells, not the whole dp table. I had run test case 41 on codeforces and it's taking 3.3 seconds to give the output. The runtime is slow. Can anyone please optimize this code?
Sorry, my bad
To make your code cleaner, you can use array<ll, 4> instead of those pairs, here is the AC code with that modification 131514905
I just came to know that Um_nik was in the 1st place.. I thought that he finished University and I was watching the closing ceremony and thought that this guy behind the mask has a familiar hairstyle and then after googling I got to know that he was our Um_nik. I tagged you because I wanted to congratulate you publicly but without a blog.. Congratulations and wow you are a hardcore person from specialist to red and then ICPC Champion... way to go man I couldn't be a great coder winning contests like you do till now but it's ok coz everyone's life is different.. but it feels good to see an inspiration of hardwork and straight forward man(this i am saying by your blogs and comments) like you at the top.
Can somebody give a testcase where this is going wrong
try this
For problem C I used double still I think it is overflowing somewhere How can double overflow here.Please help My submission
just use (long long) instead of (int) here: cout<<int(s)<<'\n'; int has been overflowed, not double
When will the solution be available
Does anyone know why the
unordered_mapsolutions to C are resulting in TLE? Is it because of some crafty input which is causing everything in the hash table to be put into single bucket?yeah its because of anti hash test case which is causing the unordered map to have a bad hashing
That's really interesting. Do you know where I could read more on this or how I could generate such cases?
yeah you can read about it here, I have also used these techniques to hack 20+ submissions of C
Thank you!!
my question is that is it valid to hack solution on the basis of short coming of unordered hash , should this be concern of the problem . I think its a useless ground to discard out solution specially in problem C
What about hash map implementations in other languages (say Python dict)? Can they be hacked like this?
Can someone please let me know the problem with my Problem-C solution. It is giving TLE on Test-17. I have used a standard unordered_map based solution, with O(n) complexity as per my understanding.
Any help will be appreciated. Thank you !
Unordered map does not work as efficient as hashing. Technique
Is not unordered map == Hashing ?
unordered map worst time complexity is O(n) , so anti hash test cases has been used to hack unordered map without custom hashing
Notice that unordered_map is a hash table and the complexity is average O(n). I believe that is some input that is causing it to get the worst case complexity which is O(n^2)..
So should I prefer ordered_map in future ?
I don't know to be honest. I myself used unordered_map in the contest.. I think that if the log n operations are acceptable we should prefer map since.. Though I think someone more experienced would be able to answer better.
you can still use an unordered map but with custom hashing ,you can copy paste the template from this blog,because in edu round people have enough time to make anti hash test cases.
Thanks KingRayuga, will keep in mind.
I think we can also go with ordered_map as it seems that the time limits are flexible enough to accommodate O(n*logn) solution even when best one is O(n). Please, correct me if I am wrong.
yeah you are right
Just read someone else's comment here and tried with ordered map. Solution Accepted !
So what should I learn/conclude from here regarding the use of ordered/unordered_map. Like when to prefer ordered_map over unordered (because here I thought using unordered should work, but id didn't eventually).
Read this blog by neal https://mirror.codeforces.com/blog/entry/62393
The test cases used for successful hacks will be included in the main tests?
Yeah they are
Probably not relevant to the blog but I have a solution which I think would work for non-distinct pairs for D. Would anyone like to test it out or perhaps reply with a solution of their own which also works for non-distinct pairs.
Here's my submission
https://mirror.codeforces.com/contest/1598/submission/131482336
same solution using unoreded_map is tle (after hack) but using map is accepted.
I used unordered_map in contest
Unordered map implementation- 131448620
Ordered Map implementation- 131510925
I think it can help you to solve your problems: https://mirror.codeforces.com/blog/entry/62393
Had I known this earlier...
When will the ratings change?
How to solve problem F? I have been thinking for a long time but still have no ideas. Can anyone tell me the idea of this problem? Thx.
It's a standard bitmask dp. The most difficult part is speeding up the transitions.
Thx.
Why ratings change taking so long?
I guess because the time of the contest, when the hacking phase finished the timing was too late for Mike (Just a guess) :D
yes editorial was also late
Rating change is taking longer than I thought!!!
I think it's because the unusual time of this round, but also, I have to say that the rating change of this round is the longest I have waited...
when does system testing starts?
I remember it has been finished already in the past few hours.
No, System testing is not finished yet.
You are right.It is still running now.
But it is the second time for system testing.I wonder why it takes.
Why does the system testing run twice?
But the rating didn't change yet.Maybe the technicians met some problems.
MikeMirzayanov please remove the cheaters and update the ratings thanks !
why the systetm testing is caried out multiple times. is this for rejudging with additional testcases.?
Video Editorial of C
why system testing is done so many times ?
same doubt
unordered_map's TLE is about hash, but how about multiset?
I will become pupil but why the delay in giving ratings.
how so orz
How so orz?
How so orz !?
I take back my words, sorry
And you, don't be useless do something.
Don't take them seriously they had a discussion on discord and then just came randomly , just ignore them .
Congrats on pupil btw
how so orz?
What does "how so orz" mean?
congrats on pupil
congrats on pupil
And now you've become pupil. Congratulations!
I find 2 solutions are copied. But even why were they not skipped
Please upload editorial as soon as possible. Thank you for the contest.
Video Solution for E (Staircases)
Video Solution for C (Delete Two Elements)
-1
I agree w you. I found many cheaters in Educational round 115 but their rating still increase
Hello everyone, I have written a blog on improving codeforces rating that I used to follow and it is a really good routine. Do read the blog here.
link broken
Thank you for informing me. I have updated the same.
when will cheaters be remove. I found many cheaters but they were not removed
Can anyone tell me why my rating was decreased by 29 even when my verdict was OK on a single submission. No other submissions were done .Just one and that too with OK verdict
Rating change depends on your rank
If your rating had been increasing whenever you were making a single correct submission in a round you would have been a legendary grandmaster by now.
-1
Yes I also think this. How can I contact admins about this problem
Have the cheaters been removed? As there are many remaining..