Привет!
VK Cup 2022 - Отборочный раунд (Engine) начнётся уже скоро: 15.01.2023 15:05 (Московское время). Это соревнование предназначено для тех, кто решил хотя бы 6 задач из 8 в квалификационном раунде VK Cup 2022. Раунд будет рейтинговым.
Остальных приглашаем на открытый для всех Codeforces Round 844 (Div. 1 + Div. 2, основан на Отборочном раунде VK Cup 2022), который начнётся в то же время и тоже будет рейтинговым.
Все задачи придуманы и подготовлены мной. Также этот раунд стал лучше благодаря KAN, errorgorn, lperovskaya, dario2994, Monogon, Arpa.
Участникам будет предложено 8 задач и 3 часа на их решение.
64 лучших участника закрытого отборочного раунда получат фирменные футболки VK Cup, а 16 лучших пройдут в финал и смогут побороться за призы 4-5 февраля очно в офисе VK или онлайн:
- 1-е место — 300 000 рублей;
- 2-е место — 250 000 рублей;
- 3-е место — 150 000 рублей;
- 4-е место — 100 000 рублей.
UPD: Разбор
Поздравляем победителей отборочного раунда:
и остальных участников, попавших в число сильнейших.
Также поздравляем победителей открытого для всех раунда:
и отдельно maroonrk как автора единственного принятого решения по задаче H2.
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
我的天啊我打破了你愚蠢的锁链!
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
haha broke your silly chain
haha restarted the silly chain.
Omg tourist round.
You do realise I can just break it once it reaches a significant length again?
You do realize that I can just restart it once you break it again?
if(no_of_chain_makers > chain_breakers) cout<<"Relax"<<endl;
Now for everyone's sake,Im restarting the chain again. Thank's for the support guys.
Omg turist round.
You do realise that I can just break it once again?
I'll start again.
Omg turist round.
Omg turist round.
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round.
omg tourist round
omg tourist round
omg tourist round
omg tourist round
Omg tourist round.
Omg tourist round
omg tourist round
Omg tourist round.
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg round tourist
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
turist lol
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
omg tourist round
We need more!
Notice the unusual time
omg tourist round !!
omg tourist round
OMG tourist round
omg tourist round
omg tourist round
no
If it's possible can you please explain how did you get Specialist Rank?
just be good at programming lol
omg tourist round
Damn, 3 hours for only 8 problems. Scary
omg tourist round
Damn, 3 hours for only 8 problems. Scary
omg tourist round
omg tourist round
I am so interested in your graph, how did you maintain a slope of 45 degrees ?
وانتا مالك؟؟
What language is this, looks Arabic I don't know it. Speak in English. Also, I expressed my interest in speedy_boy graph NOT yours. Actually, your graph isn't interesting at all mr.newbie! The comment directed to you is below : Newbie detected, opinion rejected XD
Respect to newbie
yes sir. all due respect! but he seems talking in a rude manner to me by a foreign language tho I didnt talk to him and replied to ur comment only :-( However, your graph with slope 45 is so cool ✪ω✪ speedy_boy
He didn't maintain it after this contest though, if he did he should be orange by now.
newbie detected.. opinion rejected xD
Надеюсь, будет задача про All Cups
Shortest announcement ever *_*
Yes, hope the problem statements will be short as well. :D
Also the most upvoted one :)
Not even close...
(although this might still not be the most upvoted one)
Tourist can't compete against benq in this round, but he can make some crazy geometry problems to force benq to lose rating and get back to rank #1.
What an wonderful Opportunity. Haha to be crowned again.
This contest is only for Russians ig.
Yes VK Cup is for Russians, but this is a mirror of this contest and everyone can participate.
What a great idea
good
3 hours? How difficult the problems will be
Maybe my "master experience card" will be expired in this contest.
talent, and nice avatar
I don't care about the difficulty lvl of the problems if it's Tourist Round. All i know that it is going to be fun.
Note the unusual timing
omg tourist round ...^,^
what you guys think Benq gonna or not gonna participate this contest?
what you guys think is Benq gonna or not gonna participate this contest?
Benq will not participate, because tourist has intentionally made problems with topics where Benq is weak. This is to make Benq fall to second place in top rating chart, so that tourist can become #1 again.
Oh,man !! Is there any topic on earth in which these legendary guys are weak?
I don't know, maybe some optimization of matrix multiplication algorithm from $$$O(n^{2.43})$$$ to $$$O(n^{2.42})$$$.
To make benq lose rating you have to specifically make it hard for him. If it’s hard for everyone he won’t lose rating.
I think Benq only wanna compete with tourist who is well-known as the No.1 in competitive programming. In my opinion, he will not participate contests without tourist such as this contest
Tomorrow i will miss Benq vs tourist but no worries because later they will be facing each other soon
omg 3hr round ><
Omg 3hr Round ><
I think, Benq will not perticipate in this contest. But I will participate and face to the world number one Programme's problems...☝️
Score distribution.......???
I'm crazy.
omg tourist round
omg tourist round
Omg clash with ABC.
.
My IQ has decreased by 20 points while trying to read this. Thanks.
This is the shortest announcement for a round that I have ever seen :))
Legend don't need to say much. Their presence is enough alone sir!
omg tourist round . Probme 1 will be 1200+ Rated now
Nope. It was 800 Geometry. Haha
It is rated ？(⊙﹏⊙)
well, obviously, sorry bother……
TFW you expect tourist to claim back first place then see that tourist himself is the author
.
It was good one . Don't know what's wrong ? Maybe too rude or offensive for some people ?? Atleast you should have blurred the authors above. It's disrespectful for them.
Contest Clashing with atcoder Beginner Contest 285.
Tourist round >>> Any Other round on this planet. As Simple as that .
I might just end up top1 because problems are being designed by tourist.
score distribution when?
omg tourist round
omg tourist round
ok
omg tourist round
ОМГ. Раунд туриста.
Thanks to MikeMirzayanov for amazing platforms Codeforces and Polygon.Contest Collision (CF & ABC):
AtCoder -> 17:30 IST
CF -> 17:35 IST
CF is based off VK cup. It can't ve postponed
omg tourist round
i do bad in tourist rounds, hopefully this will change tomorrow
why problem in this year contests are didn't given any rating till now?
conflict with ABC285... what a pity that i can't participate in both contests...
That's right...
Your profile picture looks great
excited !!
tourist gang
The One Piece is real!!!
Originally I am not planned to participate this contest because I have a date then. But when I see the author, I just postponed the date and register for this contest. Wish I can turn cyan once again.
You guys are having girlfriends!
Reference
omg tourist
I wish the next round will be written by Benq
Obviously tourist round will be full of fantastic problems. Good luck, yeah! ^=^
I will always be a fan of tourist!
omg tourist round
It is briefly before the beginning of the round now, and score distribution hasnt been announced yet...
Where is score distribution?
I think I may not be able to participate because I feel like going to the toilet right now. What a wrong timing. :(
glhf
scared！i hope i can solve A。。
OMG tourist round
I regret I couldnt participate in it as I got stuck with some personal work
why the rating of H2 is higher than the rating of H1
it said that H2 is the hard version,doesn't it?
its not H1 vs H2 but rather H1 vs H1+H2
A solution for H2 always works for H1. So if you solve H2 you get the points for both.
F is an amazing problem! couldn't code it in time though, but mindsolved it
Can somebody please tell the solution?
$$$dp[i][j]$$$ = $$$P$$$(Minimum Prefix $$$\geq$$$ $$$ -j $$$ )
In E, finding maximum total area is not that hard but finding the exact subrectangles seems quite difficult. Stuck on it's implementation. Any approach regarding the same?
Maintain a stack for the rightmost intervals for u, d, ud rectangles. It was very painful implementing this, especially when the contest started at 4am for my local time.
Hey, thanks. I just implemented the logic using your stack idea. Yeah, felt like handling a lot of cases!
My solution
Pretty hard contest.
No idea for D. Have idea for E but got WA. Now I'll return to CM.
how's C done ....?
It's pretty annoying. You need to consider all j that n is multiple of j, calculate how many char will be changed if you make the string to j kinds of different chars. You need consider 2 cases: j<j0 , j>=j0, where j0=the number of different chars in the initial string. If j<j0 you need change some chars with low frequency.
After you decided the optimal j, you need to add all chars you'll put to the final string into a "char pool" (implemented as int pool[26]), first try to make every chars remain the same(if(pool[s[i]]>0) pool[s[i]]--; flag[i]=true;) the assign i where flag[i]=false any char in the pool.
Well consider two indices i and j. now calculate all such x that make both a[i] + x and a[j] + x a perfect square. the rest should be easy
In E we can first shrink rectangles with same type (row1, row2, row1+row2) and then consider for each 2-height rectangle, shrink it if it's penetrated by any 1-height rectangle, and shrink every 1-height rectangles who doesn't penetrate it but cross with it.
However I got WA at pretest 15. Also C was very annoying. I spent about an hour for it.
What I did is calculate the answer firstly for two numbers, let's say a and b. We want to transform a into x^2 and b into (x+k)^2 (because b > a), and that means that $$$b-a = (x+k)^2 - x^2$$$, so $$$x = {(b-a-k^2)}/{(2k)}$$$. You can just try every k from 1 to sqrt(10^9) to see whether that k satisfy this condition, then, for each valid k, calculate the number of perfect squares you'd also obtain from the other numbers in the array.
Let's choose some x. If ai & aj becomes perfect square after adding x, then: $$$a_{i}$$$+x = $$$u^{2}$$$ & $$$a_{j}$$$+x = $$$v^{2}$$$ Substracting , we get $$$a_{i}$$$ — $$$a_{j}$$$ = $$$u^{2}$$$ — $$$v^{2}$$$
=> $$$a_{i}$$$ — $$$a_{j}$$$ = (u-v)*(u+v)
Now split ($$$a_{i}$$$ — $$$a_{j}$$$) into 2 factors f1 & f2 such that f1*f2 = $$$a_{i}$$$ — $$$a_{j}$$$.
So, f1 = u-v & f2=u+v.
Find, u from here. Substitute in $$$a_{i}$$$+x = $$$u^{2}$$$ & you will get x.
Find the count over all possible x in this way & take the minimum one.
I thought trying for every factor of a 10^9 number for 50*(50*49/2) times would cause TLE…
my thinking was similar to you though instead of tle i got wa
Now I upsolved D. My submission:189361440
The proof of that the submission is available:
In the algorithm we consider for all divisor k of (aj-ai) where j>i. The maximun number of divisors we check in one case is:
sum(1<=i<j<=n, sqrt(aj-ai)/2)
The "/2" is because if (aj-ai)%4==0 k must be even, if (aj-ai)%4==1 or 3 then k must be odd.
Notice that sum(a[i+1]-a[i])<=A (A=max(a[i])=10^9). We can see for each d, sum(1<=i<=i+d<=n, sqrt(a[i+d]-a[i])/2) is maximized when all a[i+d]-a[i] are same (can be proved by the mean value inequality) and the maximun value is about (n-d)*sqrt(A*d/n)/2=n*(1-t)*sqrt(A*t), where t=d/n. We sum up it for d=1...n-1 and the sum is about O(n^2*sqrt(A)), which fits the time limit.
It doesn't TLE, because you're factoring the differences of $$$a_i$$$ and $$$a_j$$$, all of which can't be around $$$10^9$$$ simultaneously.
it's around 4e7 operations (which should be doable even with 1s time limit), plus the time limit is 4s.
But for case a[i]=1+20408160*(i-1), it runs for 10622587 operations.
OHHHHH I forgoted there's this line in the statement:
I missed in the contest.... I've thought sum(n) could be 2500 and missed the intended solution.
I just upsolved D with a similar solution. I iterated over all n^2 pairs of numbers. For each number, i iterated over all values of f1 (so sqrt(max(a)) time) and did some math to see if it worked. Then I took all the possible values of x I got, and I simulated the process for each value of x.
Problem A was kind of ugly. C also required (at least in my case) a quite hefty implementation.
problem C is tough :((( Can anybody show me some hint? :((
Definitely, not an easier one!!
I spend almost 3 hours but still couldn't solve it :((( But i see so many people could solve it :((( Pardon me for my bad english
Just brute every i from 1 to 26(i is how many letters will be left in string -> n % i == 0) and then you should calculate how many ops(call it cnti) you needed to make string balance(i used greedy in this part). So your first answer is min among cnti.
Implementation is nightmarish.
Suppose you want to modify the string to have exactly $$$x$$$ distinct characters. How many times will each character appear? Which $$$x$$$ characters should you pick? Can you calculate the minimum number of changes to reach $$$x$$$ distinct characters?
ofc the first question immediately got hit by annoying geometry question lol(don't worry imo it's also interesting)
how's problem C solved ...... nothing struck my brain for like 1.5 hrs i took multiple examples but i couldn't generalize it .
You can keep only 1 <= i <= 26 characters with n/i frequency for each one
Try them all and minimize the answer
Can't believe I clutched C like that.
Seeing the drawing for problem A before reading the problem statement was intimidating lol
Anyone know what pretest 11 was about in problem E ?
I got WA11 and 2 1 3 2 3 2 3 2 4 helped me find the mistake. When I was using only 1 row out of 2-row rectangle I forgot to clear the other row in the segment tree
any hint on D?
How to solve D? :(
See here
The answer can always be 1. Now let's see what will happen if the answer is $$$>1$$$.
So suppose the answer is 2. This means that there are any two indexes(let's assume i and j) such $$$a[i]+x=c*c$$$ and $$$a[j]+x=d*d$$$. Here c and d can be any valid integer. So now $$$d^2-c^2=(d-c)*(d+c) $$$.This is equal to $$$(a[j]-a[i])=(d-c)*(d+c).$$$
Now if we factorize $$$a[j]-a[i]$$$. We can find the value of $$$d. and .c$$$. Using those values we can find the value of $$$x$$$. Now we will store all the eligible values of x and use the best possible option.
How's the factorization is happening. Suppose $$$a[j]-a[i]$$$ has a factor($$$A$$$).Then there will be also be another factor $$$(a[j]-a[i])/A (assume B)$$$. Now $$$A*B=(d-c)*(d+c)$$$. So $$$d=(A+B)/2$$$ and $$$c=B-d$$$.
I didn't get how you solved to get, d=(A+B)/2 and c=B−d
since $$$d+c=B$$$ and $$$d-c=A$$$. If we add both the equations we get $$$2*d=A+B$$$ which is equal to $$$d=(A+B)/2$$$. For the value of $$$c$$$ it comes from equation 1 by shifting $$$d$$$ from left side to right side.
I swear F is easier than E...
any hint on d?
D is simple but the constraints are way to big. Any idea how to solve it ?
Suppose that $$$b_i^2 = a_i + x, b_j^2 = a_j + x$$$, then we have $$$(b_j - b_i)(b_j + b_i) = a_j - a_i$$$. By enumerating $$$b_j - b_i$$$ we know what $$$b_i$$$ and $$$b_j$$$ are, and also the $$$x$$$.
So ... it is not simple, then?
D was very good
Yeah, it was nice. I initially thought it was some sort of meet-in-the-middle trick, but was pleasantly surprised to see it was a much nicer brute-force.
How did u solve D ?
lets consider i<j and a[i]<a[j] then if both a[i]+x and a[j]+x turn out to be perfect square then
$$$a[i]+x=n_{i}^{2}$$$
$$$a[j]+x=n_{j}^{2}$$$
then from $$$ a[j]-a[i]= (n_{j} - n_{i})*(n_{j} + n_{i}) $$$ you can find all possible x's in $$$O( \sqrt{n} )$$$
then for any x if total number of pair's it makes perfect square simultaneously is y then this $$$ ^{ans} C _{2} = y$$$ should hold
As x can be of the order 1e18 wouldn't O(√n) will be of the order 1e9? which should TLE?
We only need to find the factors of a[j]-a[i] to deduce all the x's and a[j]-a[i] can at max be 1e9.
Can you please Explain About How to Find X using (a[j]-a[i]) what two factors represents of difference and how this term come (factor1 + factor2)/2
i am not getting it :(
see for $$$n_{i}$$$ and $$$n_{j}$$$ to exist $$$n_{j}-n_{i}$$$ should exist and be a factor $$$a[j]-a[i]$$$ because this equation $$$a[j]-a[i] = (n_{j}-n_{i})*(n_{j}+n_{i})$$$ holds.
therefore if we iterate on all the factors of $$$a[j]-a[i]$$$ , (p,q) such that p*q = $$$a[j]-a[i]$$$ then we can treat smaller of p,q & p<q as $$$n_{j}-n_{i}$$$ and the bigger one as $$$n_{j}+n_{i}$$$ and solve these two equations
$$$n_{j}-n_{i}=p$$$
$$$n_{j}+n_{i}=q$$$
find a feasible $$$n_{i}$$$ , $$$n_{j}$$$ if it exists we can get the respective x from equation
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That's what she said
trash problem E, only boring implementation.
I will never think a strong competitor to be a certainly good author any longer.
I thought it has to be obvious after first stage
Implementation is an important aspect of problem solving too; in my opinion, there's nothing wrong with problem E. You need to find a way to systematically reduce the rectangles, and figuring out how to structure your code to do this as efficiently as possible is part of the challenge.
I agree. Compaired with implementation, observation is much too easier in this problem. That makes me feel really bad.
I can't implement it, too. But I think it's not all the author's fault. E is not very good but also not trash.
We should consider to improve ourselves. I have already suffered a lot for my poor implementation ability.
Problem E is a good problem. Please do not send such stupid comment only because you can't solve it. This will make you look like a loser.
I don't solve FGH, but I don't criticize them. You never get to know why I am saying it's trash and others are upvoting me. So you are a fool.
That's because you can solve E in usual, while you have never been solved problem FGH in the past. You just arrogantly make silly comment on problems you can't solve, and try to infect others with your trash emotion. You never get to know why I am saying your comment so brainless, you will just like a mouse hide in the hole and won't listen to others opinions.
OK, next time at div4 A I will provide you with a math problem that has 10,000 corner cases and requires 100KB code and you cannot solve it and I say to you that you should improve your coding skills and not complain about the problem itself
Actually, E was not that difficult to implement.
For me, implementation is like an art: if you do it the right way, it feels godly.
Yes, of Course he will not complain.
I really hate a homeless dog barking at me when I'm walking in the street. It's really annoying. But I can't have too many requirements to a dog, right? They just want a bone.
The same to you, fucked bitch
Angry fool, LOL. I don't mention anyone. If you are angry, that's only because you are shame on yourself<3
I know.
But why are you grey?
More often than not I cannot solve problem E even D in div1+2 but never have I criticized them if it's my fault that causes me to fail to solve it
You are being offensive. You are not offering any explanation for your words. You are attacking when attacked. I really can't stand this behavior by entitled contestants. Many comments by you in the past have a similar attitude, that makes you a bad member of the community. Please be more respectful.
Problem E was rather easy to mind-solve in quadratic time (and still interesting). Rather hard to implement properly in pseudo-linear time. That is one of the main points of the problem. The implementation itself (at least mine) is short, but I wasted more than one hour because of wrong assumptions, wrong ideas, not having the whole picture clear in mind, and blah blah. That does not make the problem trash, it makes me weak. I was very satisfied when I finally solved it.
You are the fool if you assume that codeforces upvotes are a valid support for the content of your messages.
pls shup up until you can solve E
What make you mention that i can't solve E? Please shut up until you can prove it.
so how can you know he can't solve E
He agree it himself bro, please don't ask such stupid question next time.
Arisu_Sakayanagi can't you?
they ACed A-D 1:17 into the contest, then proceeded to WA on pretest 1 on E for two hours.
Now, perhaps I'm too low rated, too unexperienced to judge a master, so take my word with a grain of salt all you like, but that doesn't sound like they can solve "only boring implementation". If they only had a hour or so sure it might have been a time issue, but two hours for ~150 lines of implementation that only involves greedy and sortings? Please.
When I get stuck on implementation for two hours in a contest, I don't say the problem is trash. I say that I'm too bad at programming and that this problem exposes an issue in me. I should be grateful to this problem even though it gave me -delta because it shows me how to improve my ability.
That's very offensive to say Sir. Either problem is Good or it is not meant for you. Tourist has put his hardWork and time into that problem, atleast he deserve a valid constructive criticism sir. Codeforces upvotes are merely a number to me. You are still wrong to me. Also Someone hardwork isn't trash at all.
thanks for this great round
How to solve problem C?
My idea was to enumerate the number of occurrences of each character.
It took me about 2 hours. But I got TLE on pretest2.
(Forgive my poor English.)
Edit:
Why many people downvote this comment?
My idea was to first, implement a function that optimally changes the string to be balanced with $$$x$$$ different characters ($$$O(n\log(n)$$$) in my implementation). And then go to all the divisors of $$$n$$$ less than 27, because the number of characters cannot be greater than 26. Then in total is something like $$$O(26 n\log(n))$$$.
Thank you for helping me to solve this problem.
How to solve F?
I turn the problem into this:
There's an array S indexed from -n to n, the number on index 0 is 1, rest is 0.
Do n operations, on each operation there's S[i]/sum(S) possibility to choose index i, add 1 to S[i], then P possibility to add S[i+1] 1, 1-P possibility to add S[i-1] 1.
Output the possibility to make S[-n...-1] both 0 after n operations.
But I have no idea how to solve this :(
perhaps we should add s[i] 1 too?
Oh I forgot to mention it. Now it’s fixed. Thanks!
If D doesn't use
long longtype to save $$$x$$$, will it be WA on #5? I got WA on #5 and I think it was because I didn't uselong longtype to save $$$x$$$, but I didn't have enough time to change. :(Yes.
Thank you. I will resubmit my code later. :(
F was a lovely problem, absolutely loved it !
How to solve Problem B
More generally, how to approach these questions based on sorting?
Check if you can take k people. You can take k people iff there are exactly k people for whom a<k and there is no person for whom a=k
when upsolving will be open?
In problem D, suppose any two consecutive square numbers are p^2 and (p+1)^2 . Then their difference will be (p+1)^2 — p^2 = 2p+1. Now given the maximum element is 1e9. so after choosing some x and changes a[i] to a[i]+x. The maximum number of all a[i]+x will be at most 2e9 or even less than 2e9. Isn't it? Or I am wrong somewhere? If this is the final solution we can brute force all the perfect squares of a[i]. But this solution gives me WA at test 5. Where is this solution wrong then? My solution . TIA :)
Hint: 1e9*1e9=1e18
no, difference is at most
5e8, but actual squares can be up to3e17,and number of squares is still1e9"so after choosing some x and changes a[i] to a[i]+x. The maximum number of all a[i]+x will be at most 2e9 or even less than 2e9"
How you came to this fact?, like x can be upto 10^18.
Got it. Found the mistake. Thanks a lot.
8
6 0 3 3 6 7 2 7
For the third testcase in problem B (reproduced above), why couldn't 2 people (persons 2 and 7) come? They'd both be happy (2 because at least 0 come and 7 because at least 2 come) and the rest because they're each happy if and only if 3 or more come; less than 3 come, so the rest should be happy as well. Why isn't this a valid result?
a person is happy if at least ai persons OTHER THAN i are coming
The problem statement says a[i] excludes the i-th person themselves. So 7 would not be happy, since there aren't two other people who went (just one).
How can you participate in 234 contests consistently for 4 years and still be a newbie?
When will we be able to upsolve?
Wrote solution for 1782D - Many Perfect Squares few minutes after the contest and got WA on test 6. Can anybody tell me what's wrong? (189360636) It passed more than 50000 random tests with $$$n\leq{7}$$$.
You shouldn't stop iterating over $$$k$$$ when you found an appropriate one. If you try out all valid $$$k$$$'s, your solution will pass the tests (check out this submission: link)
Thank you so much!
A pretty hard contest tbh especially C, so hefty implementation
I liked C. Good problem where you think how to implement rationally.
...Cofeforcer codefofced coresorces!...
The tasks were more interesting than usual. Liked B and D!
But C was a bit difficult to implement.
The ratings are updated preliminarily. Tomorrow I will remove cheaters and update the ratings again!
Thank for kicking me back to div2. Maybe I need more practice before being a genuine master.
Welcome to the club!
Now I've upsolved E. My submission:189365023
Basic idea:
First the maximum total area is the initial area. Proof: for any 2 overlapping rectangles, if they are same type (row1, row2 or row1+row2; we'll call them type1, type2 and type3 later), we can shrink one of them to remove the overlapped area and remain the same total area,like this:
[-------[+++++]||||||] -> [-------] [|||||||||||]
('-': left '|':right '+':overlap)
So we can remove all overlaps caused by same type of rectangles. If they are still 2 rectangles who are overlapping, then one of them is type3, the other is type1 or type2 (since type1 cannot overlap with type2, and we've removed all overlaps of same type in previous step). WLOG assume they are type 3 and 1. If type3 is "penetrated" by type1, which means type1.L<=type3.L && type1.R>=type3.R, we need to shrink type3 to row2 (and it becomes type2), like this:
[---[+++]---] --> [---------]
.....[| | |]...............[| | |]
('-': type1 '|':type3 '+':overlap)
Otherwise, we can shrink type1 to remain the area (since type1 can only get out of type3 from at most 1 side). Now we could solve the whole problem.
The naive approach is O(n^2), but by classifing rectangles by different types and sort them by the left borders, we can get an O(n*log(n)) solution. In the first step, for each type of rectangles, we keep prevR=the largest right border of rectangles we've checked, and for every rectangles with left border <= prevR, we move it’s left border to prevR+1 (and if it's left border is greater than the right border, we mark it as removed). In the second step, we maintain 3 pointers p1, p2, p3. For each type3 pointed by p3, we check for every type1 and type2 until there's no more that type of rectangles, or it's left border is greater than type3.R. If there's any type1 penetrates the type3, we mark flag1 as true, similar for flag2. After checked type1 and type2, if(flag1 && flag2) we mark type3 as removed, elif (flag1) we change it's type to 2, elif(flag2) we change it's type to 1. (However, the code implementation is very very very annoying)
Update: Now I've kicked back to div2. Maybe I need more practice to make sure that I can solve all solvable problems in contest.
Worst problem C ever
Cheating case Codeforces Round #844 (Div. 1 + Div. 2, based on VK Cup 2022 — Elimination Round) problem C
Zaoldyeck code :- 189324513
HAKUNAA_MATATA code :- 189349431
Thanks for the fast editorial
I suddenly know why 19000+ people registered but there were only 9000+ people submit.
People either got scared of problem A, or they accidentally slept in. (I started the contest 30 minutes late bc I slept in XD)
Unusual timing also didnt helped the cause. I couldnt participate due to contest getting started ealier than usual
This is the hardest race I've ever done
Can someone plz tell where this code goes TLE. Solution code for problem C 189395773
Hi guys! If you're having troubles with B and C problem you can check the video tutorial at this channel- www.youtube.com/@grindcoding
omg tourist round
Hi there!
If you have any doubts in problem B and C you can checkout the tutorials here- www.youtube.com/@grindcoding . Happy coding!
Anyone tried to solve the problem D using Binary sesrch?
H is a problem designed cleverly (although most haven't ever clicked the link of it). For case k=0 it's a trivial div2C problem (consider the smallest "bounding-box" of a pattern and use inclusive-exclusion principle to count for each bounding-box). For case k=1 it's a normal div2E problem (consider every possible position of the broken light relative to the bounding-box, they'll form a rectangle, if it's contained in the bounding-box, subtract the number of patterns where all lights in this rectangle are turned on). For case k=2 we get 2 rectangles, and we need to guarantee that there must be at least one pair of corresponding positions of rectangles where lights are both off. We consider how many patterns will violate this condition: If for every pair of corresponding positions, there must be some position with light on, we construct a graph whose vertexs are positions in the 2 rectangles, and add an edge between each pair of corresponding vertexs, we need to find the number of ways to color these vertexs where every edge has a vertex with a light on. Since the graph can be disposed to several chains, and the number of such coloring for a chain is Fibbonaci numbers, we need to find the length of each chain (if any chain has 2 adjacent vertexs out of the bounding-box, the answer is 0, because they can't be light on).
I'd rather say it is too classical and easy (to come up with a solution, implementation not included) for most who did click the link of it, and totally not worth higher points than G. And I bet many solved G but could not solve H1 only because of a lack of time. (For me, an additional 5 minutes would suffice.)
It's a waste that H is not a problem with case of k=0,1,2 as easy, medium, hard subtasks, where much more people would try to think about it instead of ignore it completely.
Can anybody teach me that how to solve F? many thanks.
tround
really good round i think
When editorial will be published?
when will we get editorial of this contest?
I feel like D was a lot more easier than C, I still am not sure how to do it. I guess D and C should have swapped places
Solution sketch for F:
Let us say that two characters are paired if they were inserted in the same operation. Note that there are always an even number of characters between any character and its pair. Let $$$g(n, k)$$$ be the probability that after $$$n$$$ operations, there are $$$2k$$$ characters between the leftmost character and its pair. Note that $$$g$$$ does not consider the type of pairs inserted (
()vs)().Compute $$$g$$$ by DP. $$$g(1, 0) = 1$$$ and $$$g(1, k>0) = 0$$$. From any state there are three possible transitions. A new pair may be inserted left of the leftmost character, producing state $$$(n+1, 0)$$$. A new pair may be inserted between the leftmost character and its pair, producing $$$(n+1, k+1)$$$. Finally, a new pair may be inserted entirely to the right of the leftmost character's pair, producing $$$(n+1, k)$$$. $$$g$$$ has a quadratic number of states each with a constant number of transitions.
Suppose that
(has value $$$1$$$ and)has value $$$-1$$$. In a regular bracket sequence, all prefix sums must be non-negative. Let $$$f(n, s)$$$ be the probability that in a string produced from $$$n$$$ operations on an empty string, all prefix sums are $$$\geq s$$$.Compute $$$f$$$ by DP. $$$f(0, s \leq 0) = 1$$$ and $$$f(0, s > 0) = 0$$$. To compute $$$f(n, s)$$$, suppose there are $$$2k$$$ characters between the leftmost character in the resulting string and its pair, and suppose that pair is of type
)(. Inside the pair, we have a sub-problem $$$f(k, s+1)$$$. To the right of the pair, we have a sub-problem $$$f(n-1-k, s)$$$. Each $$$k$$$ contributes a term $$$(1 - \frac{q}{10^4}) \cdot g(n, k) \cdot f(k, s+1) \cdot f(n-1-k, s)$$$ to $$$f(n, s)$$$. Pairs of type()are handled similarly. $$$f$$$ has a quadratic number of states each with a linear number of transitions.Finally, output $$$f(N, 0)$$$ for $$$N$$$ in the input.
Thank you!! I've been waiting for someone to make an editorial on F the whole day.
In computing of f, if all prefix sums at the left are not less than s, then at the right we should make all prefix sums not less than zero? Because if all prefix sums on the right are not less than x, then by adding to each of them s from the left we get that they are not less than x+s?
The condition that prefix sums of the right part are not less than zero is sufficient but not necessary. Note that $$$s$$$ is non-positive.
Because we always insert
()or)(pairs, the prefix ending at the leftmost character's pair has sum 0.Oh, that's it, got it! Thanks!
my cat looks like yours
She agreed
()satisfies this.)(case if $$$s \leq -1$$$ because the first character's prefix sum is already $$$-1$$$ in this case.Also, this problem is quite tight on time, and $$$\text{mod}$$$ operation is quite costly, so try to minimize the use of it by using long long and only modding a state after the third loop has finished.
AC submission with a few comments: 189490185
you can use dfs with memory array to save time,dfs code can solve this problem in 400ms.189513211
I think f(1,s<=0) is not 1.
There is p possibility that f(1,0) is 1 and (1-p) posssibility that f(1,0) is 0.
Thanks, it should be $$$f(0, s \leq 0) = 1$$$ and $$$f(0, s>0) = 0$$$.
First time seen that there is no thanks to MikeMirzayanov for Codeforces and Polygon platforms.
based tourist
When will editorial be released
Never because tourist doesnt give a f
Can someone share a solution to C?
Here is the code : 189565190
Basically what i have done here, is start iterating from 1 to 26, and for every integer when n%i == 0, called a function solve which gives the most optimal string when the no of distinct characters present in the string are i.
Now, In solve function, first I calculated the no of times a character should be present in the string, that is n/i, then replaced all the characters whose frequency in the string is greater than n/i to of those whose frequency is less than n/i, till all the i characters have n/i frequency.
So I have a solution for F which requires O(N^3) memory and time but the memory is way too much it is something like this : dp[i][j][k] = the probablity of it being a balanced string after i moves and having exactly j sub balanced strings (any balanced string is a number of balanced strings next to each other) having k positions in the whole string were if string )( is added the number of balanced substrings go up by one (in other words if you write the prefix sums of +1 and -1 the prefix sum is +1 at that point). now updating each cell of this dp is O(1) because all you have to consider is where the string () or )( is added. in case of string () if it is added between 2 balanced substrings it adds one to the number of them and also adds one to k value. if it is added in the k positions where the prefix sum is +1 it doesnt add the number of balanced substrings but adds one to k and in other positions it doesnt change k or j. for string )( , it cant be added between the balanced substrings, in the chosen k positions it adds one to the number of balanced substrings and also adds one to k and in other positions it doesnt change anything.
is this approach wrong ? and if it is not can i somehow iterate on something to change memory to O(n^2) ?
I've thought this kind method would work, but consider for (.(*).) where k=2 (represented by '.'), if we add )( at *, it would become (.().().) where k=3.
ahh i see. thanks !
No editorial?
Please Publish Editorial.
where is the tutorial？？？
omg tourist round
Internet explorer tutorial
in C after fixing the number of distinct characters in the final string how to decide which characters i have to keep and which to delete so as to minimise operations ... mathematical proof is appreciated .
Sort the frequency of the initial string and draw a histogram, then draw a rectangle represent the target frequency, see what's the difference between them.
For example, if the frequency of the initial string is 6,4,4,1 ans you need to change all frequency to 5 (and there will be 3 different chars), the histogram will be like this:
-----!
----*
----*
!
('-':the character will remain the same '*':the character will be added '!':the character will be removed) in this case the number of chars will be changed is 2.
internet explorer tutorial
where is tutorial ? we want to upsolve..
Upsolve by yourself
who asked
L Bozo
Who's that?
That's slang
Oh ok. Btw, have u ever seen ligma?
No. And I know what you're up to
Fine, have a nice day
Read editorial != upsolve
Please give the editorial before the next contest starts !!
The slowest editorial ever. Oh, wait, there's no editorial yet!
Spent a day looking at B. Still can't solve. tourist Please make tutorial easy and if possible early.
Thanks.
It is the first announcement to have those triples of no:
1- No editorial
2- No score distribution
3- No thanks to mike mirzayanov
There are thousands of problems in Codeforces, you guys could solve them instead of waiting for editorial and you guys can upsolve the problems anytime later after getting the editorial, Mike won't delete any problems. So instead of worrying about it, solve some other problems:)
Can someone share a solution for problem H?
omg tourist round
lets go
Извините, будет ли, и если да, то когда примерно объяснения задач этого этапа? Или оно уже есть, но не в блоге Туриста? Подскажите, пожалуйста, кто-нибудь
TUTORIAL IS OUT!!!
TUTORIAL IS OUT!!!
TUTORIAL IS OUT!!!
TUTORIAL IS OUT!!!