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 05.01.2019 19:35 (Московское время) и продлится 2.5 часа. В каждом дивизионе будет предложено по 6 задач.
Задачи раунда были придуманы и подготовлены 300iq, scanhex, cookiedoth, VeryLonelyRaccoon, ----------, kkarnauk, forestryks, TheWayISteppedOutTheCar, LordVoldebug, romanovsavelij, golub, ismagilov.code,alexey_kuldoshin, LadyPython, Jatana под руководством преподавателей izban, VArtem, meshanya, pashka.
Также спасибо за тестирование раунда isaf27, peltorator, Kurpilyansky.
И, конечно же, спасибо MikeMirzayanov за великолепные системы Codeforces и Polygon.
Всем удачи!
UPD: так как регистрация открывается раньше пересчета рейтинга Hello 2019, в случае изменения дивизона, участники будут перекинуты в другой дивизион.
UPD: Разбор.
Я горжусб тобой сынок
Мне страшно за умственные способности деда...
3iq ;)
То есть за умственные способности отца не так страшно?
А мне страшно за умственные способности сына 300iq
3000iq
Тогда хорошо, что у 300iq нет сына
Беру свои слова назад, задачи говно. Сын(?)очка, ты обосрался
I'd very much like to see rated rounds which are shorter than 2 hours, not longer.
For me, it's much easier to dedicate 1.5 consecutive hours to writing a contest than to have 2.5 or 3 hours available.
.
Your comment was downvoted desperately, therefore we can conclude that CF community is unfavourably disposed towards Gassa making a contest.
sorry, what i replied above means i look forward to Gassa's marathon contest ..
I think that 1.5 hours isn't enough for CF contest with current format (5-6 problems).
We should add a contest that runs for 1.5 hours on 2 or 3 problems of high difficulty with batch scoring. Unrated, of course.
I think maybe the div.3 rounds could be set as 1.5hrs long.
2 consecutive contests to start the year? bring me some of that
Cool! 300iq is now purple!
Edit: Not anymore :(
10 min delay :|
Haha, maybe you're comment in the wrong thread
13 problem setters for 8 problems ...........
And that's excluding testers
Why are you freezing up?
Do you have any idea about ICPC rules ? -_-
Yes, I have! I know that the standings were Frozen(or Paused)
But I think no one got my joke! :(
Then learn How to Joke before Joking -_-
The contest length is 2.5 hours in the blog but it shows 2 hours in the upcoming contests.
I think it is of 2 hour because normally all the contest of this type is of 2 hours... @300iq please make us sure
It will be 2.5 hours length
It is updated now.
I think CODEFORCES hates me! Whatever comment(or reply) I make (I tried many kinds. Contributed, Made Memes, Helped people etc.), always get dozens of downvotes. I'm really sad about it. I don't want to contribute anymore! It's already -24 and going down.
grey-dude >:( it's under -24 now
well, think about the person you helped or people who laughed when they saw your memes,
what i'm saying, is "Care about community!" and not about contribution points. if you helped somebody, that is good enough for you to keep doing it.
The way you comment reminds me of attention girls in instagram
Because people are rankist!
This is so sad alexa play despacito
Lol. isaf27 and
peltoratorpeltorator.Oh no, the magic has gone :( Now I can't play to the chameleon
No, it only changed because of rating update.
очередной раунд от сине-фиолетовых ((
After a bad experience of acm icpc, finally I am going to start this year with a new way and this is the first contest for my new journey for this year..
Good luck dude!
Thanks
Hold up.
Well，some later to Chinese users :P.
Every JBer show me ur hands up
The registration page says the contest is two hours long--is it still intended to run for 2.5 hours?
Maybe, Yes! (2.5 hours) Now it shows the length correctly. (Maybe it's updated)
Yup, looks like they corrected it :)
Another contest with 300iq in problemsetters
Looking forward to good tasks
The time of the match is very unfriendly to Chinese players.
Practically everybody in East Asia too
Yah. But south-east Asins are having a good time I think >_<
Nope :( I'm in the part of South-East Asia where the contest starts at 11:35 PM :(
Yah, it will start at 10:35 PM in our country. I think we are almost at the same area . By the way, think about chinese people. they have to sit for contest at midnight :( Maybe we are a bit lucky :D
Oof.. Red and Orange problemsetters in disguise! XD
Seems like it’s going to be a good round
starts at 01:35 in South Korea... I'm willing to exchange tomorrows day time to a codeforce round XD
Score distribution?
Scoring Distribution?
lol what a contest :D
Wow!!! Problem F of div2 in not present in div1.
wonder!! why it is so??
Nice difficulty-balanced div2D and div2E/F
Is there a straight-forward solution for Div1B that's not a disgusting ton of mindless casework? That problem ruined the contest for me, so I hope there's something at least somewhat neat.
I think its just two cases and then DP
I don't think I have done any caseworking
If my solution is correct:
Good table is either
1) Each row is XYXYXYXY
So after you fix 2 symbols in the first row (just iterate over all possible combinations), symbols in all other rows are known. For each row you should check what is "cheaper" to make — XYXYXYXY... or YXYXYXYX...
2) Same, but with rows instead of columns.
Blah, you're right, I suck. Did the first part of that observation and then did the second still with the mindset of rows, which made stuff blurry.
Either each row or each column consists of two alternating symbols.
A was a little difficult to understand.
what is test 6 in Div2 D ?
n=7 p[i]={1 2 3 1 5 5} s[i]={1 -1 2 -1 -1 5 5} ans = 6 a[v]= {1,1,0,0,4,0,0};
It's not optimal to make the value of node U as s[U]-s[parent of parent of U] if S[U]!=-1
why not
If value of mid node (s[mid] is equal to -1) val then it contributes to answer only val and child of mid will contribute S[child]-s[parent of mid]-val but if val is 0 then they contibute sum of S[child]-S[parent of mid]. First case is sum of S[child] — sum of S[parent of mid]-val * numberofchild which is better
Optimal value of val is minimum of S[child]-S[parent of mid]
@CopeCope, Can you please say, why this code get WA? http://mirror.codeforces.com/contest/1099/submission/48076693
Oh Got The Point .. I messed it up.
If your subtree's root is at an even depth, then making its value 0 is not optimal. Consider the tree with edges 1-2, 2-3 and 2-4. If the values are 0,-1,5,5 respectively, you want the node 2 to take 5 and not pass on the value to the nodes 3 and 4
so what is the optimal answer ?
The total optimal sum is 5, and not 10. Since node 2 is the root, it contributes to the sum for both 3 and 4, hence effectively reducing the total sum.
yes I understood the test
how to get the optimal answer in general
Check minimum sv from even node's sons, substract sv from node's father and you get optimal av (and sv too) for this node. If node is even and hasn't sons, you set its av as 0. For odd nodes you just calculate sv — sfather.
mj
Suppose we calculated optimal sv and av values for every node on path 1 -> Parent of CurrNode. Let SumOnPath be sCurrNodesFather
If CurrNode's depth is odd, we can unambiguously determine its av, the sv is given in input.
Now move to the even depth case with 1+ sons Setting aCurrNode only affects contiguous sons ason value — the sson is constant. Since we can set aCurrNode to any value, we want to choose one that minimizes sum of ason.
ason can be represented as
ason = sson — aCurrNode — SumOnPath
So
Starting from root (where we have av and sv ) we can calculate all the values using dynamic programming.
Try: 4 1 2 2 3 -1 8 5
Is the answer 8 ?
yes
Consider the case : 4 1 2 2 3 -1 7 7
Answer should be 7
Try this: 3 1 2 2 -1 3 Answer is 3
Can someone give a hint for E Div2?
Notice that either all the rows or all the columns must have only 2 letters each. Thus, iterate for all the possibilities.
Why ? I get it, instincts, no one demonstrates it himself during the contest. But why is it so ?
It is clear that no column/row has only 1 letter. Then, we need to show that if some row has 3 or 4 different letters, then all of the columns must have only 2 different letters (and the same switching rows and columns).
Then, suppose that row i has 3 different symbols. As no two consecutive letters can be the same, we can guarantee that there is an index j such that si, j - 1, si, j and si, j + 1 are pairwise different. Also, as {si, j - 1, si, j, si + 1, j - 1, si + 1, j} = {si + 1, j - 1, si + 1, j, si + 2, j - 1, si + 2, j} (= {'A', 'C', 'G', 'T'}), it must happen that the letters in si, j - 1, si, j are the same as those in si + 2, j - 1, si + 2, j. Analogously, {si, j, si, j + 1} = {si + 2, j, si + 2, j + 1}. Since si, j - 1 ≠ si, j + 1, from our construction, in order to accomplish these two equalities si, j = si + 2, j, and thus, si, j - 1 = si + 2, j - 1 and si, j + 1 = si + 2, j + 1. You can continue this argument for the whole row, and discover that row i and row i + 2 are the same. Using induction, every row j such that i mod 2 = j mod 2 must be the same.
But then, every column repeats characters every 2 steps, as we wanted to show.
The problems were very good! Fantastic! Thank you for great contest!!!
Nice to hear, thanks
Why did this post get so many upvotes ? I get it, the contest was awesome but anybody could've commented that...
Can anyone share their approach to E? Also, for div 2 F, I think the solution would be dfs-based, at each step finding how many cookies can be eaten if Mitya ends the game at that point. Can someone confirm this?
Yes that's right if we sort all cookies from root to node v, we want maximum x such that the first x cookies cost is smaller equal to T — 2 * (sum of all edge from root to v). We can do it with segment tree and after that in every move (except root) Mitya goes to the second child (by amount of its dp) and we can find the answer.
Hey..could you elaborate further on how to find the maximum x using segment tree? The remaining solution is clear.. Thanks!
For each node save number of cookie and sum of cookie and determine that go to left node or right node plus number of left node cookie.
My code :48007201
Horrible, misleading problem statements!!
But you did very well. Congratulations!!!
In problem Div1D, don't you have to dynamically keep the Huffman tree in order to answer the queries, or is there an easier approach?
You can change it to just insertions and undoes with the dynamic connectivity trick. However I don't see how to do undoes without persistence, which would definitely TLE.
Sort the numbers. Calculate the partial sum. At first, set ans = cnt-1. Then, for each i reduce ans by one if a[i] > partial_sum[i-1]*2. This can be done with a segment tree because the number of such cases is at most lg(1e9).
I passed pretests with this idea:
sort all numbers in ascending order, count how many numbers such that they are bigger than twice the sum of all previous numbers, let this count be K and N is total number of numbers, then the answer is N — K.
There is an observation, that the result is #fishes - #(fishes which are more than 2 timesheavier than all fishes lighter than them (after sorting)).
Next observation: is some fish can decrease the result by one, it is a result of a lower_bound operation for some power of 2. It helps easily find fishes which can decrease the result.
Yeah, once you figure out the observation, it’s easy.
How did you prove 1st observation ??
I don't prove, I code :P
Then, where did you come up with that from?
I guessed/observed that they should eat themselves from the smallest ones. What happens when second smallest fish after eating the smallest fish becomes greater than the third smallest fish? The order is distorted, but the intuition is, that because of that the fishes became very close to each other, so now I will be able to do many dangerous fights (until the situation with much greater fish happens, because then for sure I won't be able to do the dangerous fight).
what do you do when you get WA in this case? I mean how do you tell whether it's a bug or wrong observation? :P
It appears he never has bugs:))
apparently he never make wrong observations too :D
went for overkill on DIV2D with hld and still WA
What a great contest!!
Was O(QlogQlog109) supposed to fail in D1D or only my implementation has a too big constant?
I passed pretests with such complexity.
:(
Maybe the 3000 ms TL was too high then. :/
Your implementation has a very bad constant. We have java solution which uses less than 1,3 seconds in all tests
Why would Div1C be excluded and not being used for problem F of Div.2 version, instead there comes a new problem in place of this? :O
I guess they just had too many problems (because they were doing this as a class work) and wanted to include all of them.
It was no a classwork. We just thought that div2 is too easy for div1 and we didn't want to give our div1 for div1 because it can be too hard for them, and we wanted to give for div1 more problem there you need to write not very simple code, so we decided to put it there
Most contests are imbalanced these days :'(
problem D was very nice :DD , how to solve Div2E ?
HELP,
what is pretest 4 for the ACTG problem?
what is test 14 in Div2 D ?
n=7 p[i]={1 2 3 1 5 5} s[i]={1 -1 2 -1 -1 5 5} ans = 6 a[v]= {1,1,0,0,4,0,0};
my output is 6,is it correct?
can u check whats wrong in my code? link
Very great contest for non Div-1 coders like me. I particularly liked the gradient of submissions and the Div2-D problem. Ideal contest to increase rating for a graph lover like me.
what is test 4 in div2 F?
Mine was reading the statement again "The total time , including the descend and ascend". I don't know if yours failed from the same cause;
It was just a random test
Does anyone know what was pretest 16 for Div2D?
Answer needs to be long long. Explanation : s[i] <= 10^9. Sample case : 5 1 1 2 3 1 -1 -1 10^9 10^9
Isn't it a O(n) solution(Div2 B)? 48005949. If so, why did it pass a test n = 1e9?
May be 1e9 was not part of the pretests
It should be possible to do about 1e9 cpu instructions in less than a second.
So it can fit into a 1e9-for if the constant is small enough.
No
is Div2 F binary search? I was trying to check whether it is possible to eat x cookies, but couldn't succeed.
No. We don't have a solution, using binary search, it is just dp with data structures, you will be able to read it in editorial soon
alexey_kuldoshin fugazi I do have a solution with binary search. BIT+binary search. submission
Oh, ofcourse you can do fenwick + binary search, but u don't need binary search here! U can use binary lifting!) Your complexity will be nlogn but not nlog^2, and fenwick with binary lifting was our main correct solution
I used Binary Search with Segment Trees too, cuz why bother
That moment when you solve E 5 minutes before the end of the round but yiu cant submit it because you have the worst internet connection in the universe :( :( :(
I fucking hate my life :( :( :( Hope its not correct :( :( :(
The problem setter of Div1E should stop creating problems.
Yes, it is a quite idea-less task.
It is a pure technical problem.
first tell who r u
lul he's problem setter of Div1E
My screencast will be available here after it finishes processing: https://youtu.be/WR9rMvE-d9Y
C felt so empty :(
A how could initial height be 1 if we have two rocks.. and calling the second rock "second rock" doesn't that mean the first one is heigher but that wasn't mentioned.
sadly I'm not good in graphs to try D .
if you had examined the sample inputs you would have noticed that wasn't true (for A).
that's called invalid input and they should not do tricks here
Is there a specific reason why integer overflow hacks didn't work on this contest? A person I tried to hack used int everywhere, but somehow his submission in C++11 managed to print 3000000000. Is the int limit bigger than 2^31-1? :O
Classic trick:
Oh, I see now, I fell for that.
Thank you!
Screencast (still processing)
i got the #victoryroyale on this contest
Let me ask. Is F just a HLD on a suffix tree with segment tree with operations like "do x[i]=x[i]+1 on the interval" and "get sum of x[i]*a[i] from the interval"? I think that I was close, I had tree and HLD already but the fact that we have to do queries in a middle of an edge defeated me and I run out of time.
I don't know how to do it with 1D queries, our solution have sweep line at each heavy path...
cool picture in div2F
Is there anyone solve Problem C (div.2) with DP ?
Yes I do; 1 line typo .. BAAM!!
yes, I did but unfortunately wrote 'n' instead of 'k' in the inner loop and not even a single pretest had k>n case.
AC solution after the contest: submission
How to solve Div2 F?
Solution
We can solve this problem by using minimax + greedy and segmentation tree. Assume we are at some node u in the tree, then we have remaining time = T - 2 * sum_l, where sum_l is sum of l over all edges from root to current node u. Now we can use a segmentation tree to greedily eat as many cookies with our remaining time from min. time required per cookie to max. time required per cookie for cookies on path from root to u. This segment tree has a leaf for each possible value of ti (so 1...1000000), and stores per interval total cookies in the interval and total time required to eat all cookies in the interval. When we arrive at node u, we add x[u] cookies to all intervals t[u] lies in. When we backtrack from u, we remove the x[u] cookies. Vasya always cuts the edge that leads to the best result, but in the root Mitya can choose an edge before Vasya is allowed to cut. This solution is O(N log 1000000). Corresponding code: https://mirror.codeforces.com/contest/1099/submission/48030249 .
Just curious, why is div2 F not available in div1?
Was this round unrated?
No it was rated, but ratings will be updated later.
My solution for the problem C was shown as "accepted" but then after the contest, they realised I had a wrong answer? Like.. How is that allowed!!?
To give a serious answer:
There are pretests for the contest, where you will be given a particular verdict (in your case, Accepted). Afterwards, the submissions are run on a more thorough set of test cases, where the real verdict is made.
There are a few reasons for this – one is that it lightens the load on the servers to only run the submissions on <20 cases, as opposed to ~100. Another is that it allows for "hacking", where you can look at other people's submissions for a problem that were accepted on pretests but have a bug in them, and create a test case that exposes the bug. All the successful test cases created from hacking are added to the total test case suite that runs after the contest. This is pretty beneficial to problem setters, because it can be difficult to think about the wrong solutions that will be made, and the corner cases that must be created to expose the bugs in the wrong solutions. The hacking system essentially crowdsources the creation of thorough test cases, making a more accurate problem for everybody.
Failing on system tests is, unfortunately, part of the game here. On the bright side, it could be worse: Some contests, like Topcoder and Facebook HackerCup don't run your code on any test cases until after the contest is over.
actually verdict is "pretests passed" that should give one some hint
Новая ава 300iq как бы намекает...
Is it only me or somebody else also who was expecting a higher rating change? Rating Predictor is showing +162 but I got just +73 whereas my friends' ratings changed as per shown in rating predictor.
I saw the same bug previous round
Did it get resolved?
IDK, i think it`s not resolved for this moment...
For me 2.5 hours are better !!
by the f**k, why it TL??? (help me please, I am just don`t understand what wrong???) https://mirror.codeforces.com/contest/1099/submission/48007197
reeWorld We had a solution like yours. Ofcourse it gets to because real complexity of this solution is N^2logN. Look at test, it is a bambook of n/2 vertrexes and in deepest vertex of it there are n/2 vortexes connected with it. X and T are such as you should erase all elements when you can to these children, and when you return you put when again. So you do n^2 operations, and it multiplicates log because these are operations with multiset
ah, realy... Thanks a lot! I even breafly did not have idea it is n^2*log(n) time complexity in worst case
For those who are afraid of precision errors.. Div2-B Solution
Can you explain it,please
We begin with a single square and then add more squares gradually. let a[0] and a[1] be the length and width(so both are 1 initially) Now, each time I'll pick the smaller of the two(which is a[0] since it is sorted) and increment it by 1(i.e. add one more segment to it)..with this extra segment drawn, I'll be able to complete a[1] more squares using this new segment as guide. Stop when at least n squares are drawn(i.e. while cur<n).
always put a space after a comma.
Always start a sentence with a capital letter, grammar.nazi!
Or just notice that solution can only be 2*sqrt(n),sqrt(n)+sqrt(n)+1 or 2*(sqrt(n)+1).
How you proved it?
Optimal solution is to have sides (horizontal and vertical) as much as equal as you can. If you have x horizontal lines and y vertical, then number of squares formed is x*y. So square root minimize number of lines needed.
how did you prove it?*
H ow did you prove it? *
(Every sentence starts with a capital letter.)
Or just check sqrt-20 to sqrt+20 :)))
Code: 47977441
How to solve Div2 D???
Hint: the best S[i] for some unknown node is the minimum S[j] of its children, or the same of its parent if it has no children. After figuring out the optimal S[i] for every node, you can restore the original A[i] for each node.
Figure out the lower-upper bound lu[i] of s[i] . If s[i] is determined then lu[i] = {s[i], s[i]} else lu[i] = {s[p[i]], min(s[u]) with u is a child of i}. Well this is because a[i] >=0 then s[i] >= s[p[i]] and for every single child u of i then s[i] <= s[u].
Obviously a[i] = s[i] — s[p[i]]. Sum it up : S = Sum of a[i] = Sum of s[i] — Sum of s[p[i]]. So now for i = 1 to n you could calculate how many times of s[i] contributes to the sum S above, let call it cnt[i];
To satisfy S is as small as possible:
-if(cnt[i] >=0 ) then s[i] must be the lower bound
-if(cnt[i] < 0) then s[i] must be the upper bound.
So here you now all about s[i] it would be easy to calculate S
когда выйдет разбор?
как решить E DIV2? суди по всему очень длинная решение но буду очень признателен если объясните)
Editorial?
editorial?
It's here
UPD: Oh... It's not translated yet.
yeah you know what about writing it in English in the first place like a normal person not in your language that no one cares about
Considering Codeforces owners, authors of this contest and a huge part of this community are russians, I think your trolling sucks.
I did'nt get the problem statement of div2-B, can anyone pls elaborate it.
here -> Visualized solution. Solution : -find the first number square num_sqr , which is bigger then n. -if n > (num_sqr — sqrt(num_sqr), print 2 * sqrt(num_sqr) — else print 2 * sqrt(num_sqr) — 1
Why is it true?
Where is the editorial?
It's here!
It's not Complete/Translated fully yet.
when editorial will come ??
when will the editorial come?*
When will you shut the fuck up dumb nazi reeeeeeeeeeee
Хочу выяснить причину незачета задач. Писал все сам никому не скидывал решения
include
using namespace std;
int main() { int wStone, hStone; cin >> wStone >> hStone; int w1, h1; cin >> w1 >> h1; int w2, h2; cin >> w2 >> h2; for(int i = hStone; i>-1; i--) { wStone += i; if(i == h1) wStone -= w1; else if(i == h2) wStone -= w2;
}
В коде всё вроде правильно.
damn man no editorial till now!!
As the editorial is not published till now can somebody tell how to solve div2E and div2F?
I have explained how to solve Div.2 F here: https://mirror.codeforces.com/blog/entry/64296?#comment-482494 .
is it rated?
мэ нэ нраицца данный контекст
I love that contests are being held so often now :) I am learning a lot of things thanks to that , I have high hopes that it will keep going this way !!!!
i don't think so lul
you do not need to put a space between the last word in your question/sentence and the question mark/exclamation mark.
Every sentence starts with capital letter.
I appreciate your work, grammar.nazi. But I think you're in a wrong place. It's a community of Programming/Programmers. People type their comment here to express their idea. So, grammar doesn't matter here that much. People may learn many sides of grammar and be aware of grammatical mistakes but it's kinda disturbing. I think you'll get a bunch of downvotes every time you try to correct people's comments (I upvoted your reply). Don't mind.
[Sorry that my reply may also have some grammatical mistakes. (I'm not a native speaker of English and I think many people aren't too.)]
The account was created for the sole porpuse of trolling so please do not encourage him.
is there a DP approach to solve C?
When is the editorial translation going to be here?
Is there some one know the link of this problem: given a string, and some query [l,r] find the maximum i such that s[l,i]==s[r-i,r]... I'm pretty sure it is in codeforces gym thanks
https://mirror.codeforces.com/gym/100962 D?
I have a sol to solve problem F:
first we can solve the lcp[i,l] l<=i<=r not consider the bound of r restriction.
then we can try the minus old and add new value of point i s[l,l+i]==s[r-i,r].
we can solve first question by sweepline the increasing order
of rank[l], and build 2D segment tree, it can be solved
by O(n*lg(n)^2)
then we consider to find point s[l,l+i]==s[r-i,i].
there is a problem in gym find the maximum i, if we
got maximum i, then we can got sec maximum j that
s[l,l+j]==s[r-j,r] so we can get it is a repetition of s[r-i,r-j-1]==s[r-j,r-j+i-j-1].....it is a arithmetic progression
we can find the next bound of this arithmetic progression by O(1) and this number of segment is not bigger than log(n) with small constant factor
happy new year!! brute force Accepeted I think my code can be hacked,welcome to hack
https://mirror.codeforces.com/contest/1098/submission/48087000
What a great contest!!
Such shameless authors , nearly 25 people involved in the round and still no translation for the editorial. I think this affects not only the authors but also codeforces's reputation.
We sincerely apologize for the delay with the English version of the editorial. It's now available (link).
However, before writing offensive & demanding comments, please understand that yesterday was the last day of the camp, and all of us needed to travel really long way home. Quite obviously English is not a mother tongue for any of the high-school students who prepared the round, and it takes a lot of time to write something readable.
I apologise for being so unthoughtful.
I AK IOI. I AK ACM World Final. I AK Universe OI. I hangbeat tourist. I'm txdy.
.