Hello, Codeforces!

I'd like to invite you to Codeforces Round #484 (Div. 2). It'll be held on May/17/2018 19:35 (Moscow time) and Div. 1 participants can join it out of competition. Please pay attention that round starts in the **unusual** time!

Me and Andrey API Pikas prepared the tasks for this round.

Great thanks to Nikolay KAN Kalinin for helping us preparing the contest, to Mikhail MikeMirzayanov Mirzayanov for the great Polygon platform and Codeforces and also to Kirill KittyLover Samelyuk, Nikolay Apollon76 Permyakov, Dmitry Krainov_Dmitry Krainov, Andrey GreenGrape Rayskiy and Alexey Um_nik Danilyuk for writing solutions.

You'll have to solve six tasks in two hours. The scoring distribution will be announced later

Good luck!

**UPD** Score distribution **500-1000-1500-2000-2500-3000**

**UPD2** Congratulations to the winners of the *division 2*:

And also our congratulations to the winners of *both* divisions:

**UPD3** The editorial will be published tomorrow.

**UPD4** Editorial

Is there div 3 and is div 3 rated?

Div2 rounds are rated for people on div3 too

Questions about div3 is getting more annoying than "Is it rated?". Seriously, stop.

"Div 3" means there is the special contests rated only for people under 1600. Nothing more. "Div 3" does not exists anywhere except for the special "div 3" rounds.

Is it

UNRATED?Great :). You could fix timeanddate.com link to make it display time in user's timezone. Take a look on Mike's comment: http://codeforces.com/blog/entry/59322?#comment-429178

Displayed time screenshots of #483 and #484:

Thank you! It looks better now (:

Yeaaa! Round by my favourite problemsetter! I waited it so long. <3 <3 <3

I think, I can become Candidate Master tomorrow.

Hi, bro! It seems that I can upgrade my color too!

Third palindrome round in 2018 !!

What do you mean?

Number of the round is palindrome :)

484 is the number of the round and it's palindrome.

I already know that.Downvoted u

The first square round in 2018!

The first square, whose each digit is a power of two, the second digit is the first time two, the third is the second divided by two, for division two, and each digit of the square root of the round number is two, and dividing the square root of the round number by two gives a number whose digits sum to two.

Damn, I forgot that the sum of the digits is a power of two, more precisely 2^2^2.

In case you didn't have enough...

484 = 22

^{2}never enough ...

22 = 17 + 5 : date of contest ² = div 2

13, 14, 15, 17

I smell a lot of hacks

I smell easy C and ot hard D

Q. Why do you start blushing when a new Codeforces round is announced?

A. It turns me Red. ^_^

How can div. 2 only contest make specialist a gradmaster? Does this make any sence to you?

It is a joke. I am not talking about myself.

Reading comments be like...

is it rated for purples ?

Purple users are rated in div2-only contests since rating system change. See http://codeforces.com/blog/entry/59228 for more details.

Should have a contest on yesterday. 13,14,15,17....why missed 16??

There was one in gym yesterday too.

I didn't notice....my bad..

What about for

Div 3members?Most contests in Codeforces:

1) Div.1 + Div.2: Candidate masters and higher (1900 and higher) will be in the first division while experts and lower (lower than 1900) will be in the second division

2) Div.2 Only: Candidate master and lower (lower than 2100)

3) Div.3: Specialist and lower (lower than 1600)

So please stop commenting the same question every contest.

Hey, sorry that I'm coming here only now in 2 or more month.Irony, div 3 came around 2 weeks ago.

Then please search before asking. BTW welcome again :D

Hii

Wish high ratings to all.

If i remember correctly, now the contest is delayed by 1 hour compared to original announcement, which is not good for people waiting for this contest.

Bad timing for the user all over the world

I hope this round won't be like that

Guess I'm a faithful guy...

Hacking is such a ridiculous thing, maybe good for some guys but unfair... But not always bad, the stress and excitement is cool, agree?

no a good time but i like it!:)

better starting time for Bangladeshi contestant....

Can't open problem C and D. Showing Error !!

Have you tried restarting your computer or opening Chodeforces in a different browser?

How to solve F?

If we look at the dfs tree of the graph:

Submission as a reference: 38382900

How to solve problem D?

You can brute force for all posible k using sets in O(nlogn).

what is the test 5 ?

I used co-ordinate compression. I got WA TC #5 when I was printing K in terms of the compressed value instead of the real value.

I got WA on test 5 for a different reason.

My answer: 57209037

Judge: 45567032

My algo iterated a values as candidates for k. When a is sorted by size, these values are consecutive in the input:

57209037 (found best solution at this value)

45567031 (found worse solution at this value)

Obviously, the best solution can also be found at 45567031+1.

If anyone needs complete test 5 input, I've put it here: https://pastebin.com/vHuv6LE4

What is stored in the sets? Also... Brute force requires to iterate through the whole array right?

We can store lengths of locations and when they are reached, with a array or bst like std::set or std::map. To do this we can sort the sequence and enumerate k from 1 to n, using disjoint sets.

Can you elaborate please?

based on this we can simply use an array instead of std::set, thus we can do this step in O(n).

And of course k can only be equal to a[i] + 1

I got the k=ai+1 part during the contest but can you tell me about how to brute force ??

I have one set to store segments of elements less than k and one multiset to store lengths of all segments in first set. Then i add all a[i] from the smallest to the biggest, where current k = a[i] + 1. When you add one there're three scenarios: it forms a new segment, it unites with one segment, it merges two segments together. I think it's clear how to do updates and how to keep multiset of length.

Oh thanks I get it.

Alternately, start with the segment [1...n], and add a[i] from largest to smallest. This way, it's guaranteed that the only operation happening is breaking up an existing segment, which makes implementation easier.

So when at some step segment set size is begger than it's sizes at all previous steps and when all elements in multiset are equal I update the answer with a[i] + 1.

I thought of this, forgot to implement it because the samples didn't need it, and managed to implement it with about 10 seconds to spare in the contest.

...Needless to say, I wasn't able to submit in time :(

Ah Codeforces, thou art a cruel mistress...

Maintain a dsu of active indices and process the indices with increasing order of values. At each step, let

`cur_k`

be`value + 1`

. Check if neighboring indices are active and accordingly merge them. Also maintain the count of total active indices for a given size of set. Update the answer if the current size is equal to total active indices.Problems are very hackable !

the solution to problem c was available at https://www.geeksforgeeks.org/maximum-edge-removal-tree-make-even-forest/

I took 30mins to solve it. Authors need to be more cautious before preparing a problem. :(

striver_79 try to solve problems rather than searching for problems

haha, if I was to search for the problem, I would have got all the pretests passed in 5 minutes. I tried for an hour, but when I did fail for it, then I was searching for the DFS part when I came across it. Learn to observe before saying something. :)

Looking for the dfs part and then came across that nice explanation btw what are you looking in DFS. "WHAT TO OBSERVE". I also type DFS in google but I didn't came across that question maybe your browser is a legendary one.

I do not owe an explanation to you that is for sure buddy. And stop blaming someone for pointing out a link which seems similar. :) I guess you learn everything without googling. You got so much of inner talent :)

But learning is not googling in between the contest. In between try to apply your own mind.

grab popcorn :P

*grabs

Problem E can be solved by the fact that this rectangle can be mirrored adjacently on the coordinate plane.

No, I didn't solve it.

I've done a similar problem in Google Foobar. If I had been given 40 mins I could've solved it. :(

What is the 6th pretest in C?

Problem C is available at Hacker Rank: Your text to link here... . Check this out..

I read "connected components with equal size" instead of "even size" in C. Yikes

Feedback:

C: nice

D: A bit ambiguous, I needed two clarifications to understand it.

E: The mirroring trick is very standard. I think I've seen a very similar (maybe the same) problem in uva.

C was a direct copy from geeks for geeks

Mirror trick? Can you say sth? I solve it using Chinese remainder theorem.

Do you know if there's a name for the mirroring trick? I've seen a solution that used it in the past, but I wasn't able to understand it then. I'd like to find more information about it and learn it now.

just consider each side of the table as a mirror and copy the table based on its picture in the mirror. So anytime the ball hit the side it enters the new table and continues its direction.

I don't understand what algorithm follows this. You calculate the intersetcion point of rectangle and a straight line. If you hit the origin or a corner, you are done. If you hit the wall, you reflect the whole rectangle and keep going.

You repeat the above 10^9 and tle.

The point of the reflection isn't to simulate the process: that can be done without the reflection at all. The point of the reflection is:

You can now travel in a straight line

You get generalized coordinates for the corners: for example, using the labeling in erdos' picture in the other comment, point A will have coordinates (2mk, 2nl) for integers k, l.

You can get similar generalized coordinates for the other 3 corners, and then it's simply checking whether integers k and l exist such that the point given by them lies on the line you have, in the direction you want.

Can you please provide a resource link to learn it? Thanks in advance.

I'm sorry, but I don't have an online resource for this sort of thing.

Given the context, I'm pretty sure that there'll be a few problems of this type discussed on AOPS, so look there maybe?

To be honest, this is my first time seeing the reflection being used in a coding problem. I've only used it in maths problems before, but it seems like a nice Olympiad problem trick, so AOPS seems like the place to look.

I'll look around a bit and see if I can find something more concrete.

Thanks, again. :)

This link might be helpful for the proof and solution of diophantine equations.

There is a thread in below link which explains the similar case

Source: https://math.stackexchange.com/a/2295371

I agree with you and saw a couple of problems like E. (e.g. in Google foobar) I'm wondering why did I read it at the end of contest! :-/

I used tree dp to solve C but now i find out that there's some simple solution -_-

dp solution is pretty simple too

So are you sure what you want us to find in D is what D's statement says?

Shit in bold's true but it it's not true that each location has a corresponding

nonemptyset.Now getting back to what the problem asks for:

And at the same time, what you want us to do is find the

kwhich maximizes the number ofnonemptysets, not the number of locations.Even after the announcement, which was too late anyway, nobody bothered to actually update the statement.

This is because if a shark travels 2 days in a row, he will still be at a new location on the day in between, correct?

After reading the problem statement I thought the answer to this test case 8 1 2 7 3 4 8 5 6 should be 0 as it satisfies all the conditions asked i.e 1) the shark was in each location the same number of days, 2) the number of locations is maximum possible satisfying the first condition, 3) k is smallest possible satisfying the first and second conditions

I have the same confusion regarding this problem (but not anymore). Let me help clarify this. Assume the answer is k, one location is counted as non-empty consecutive segment which every number in it is strictly lower than k.

For first sample case, if the answer is 7, the number of location is 3, that is {1,2}, {3,4}, {5, 6} and valid because each segment size is two. If you choose 0 as the answer, the number of location is zero. So there's a better answer than 0.

Happy Ramadan holiday to all. I wish happiness and health to everyone.

This statement didn't exist at the beginning of the contest. My solution was hacked because of it. and there was no clarification about it!!!!!!! AGrigorii [ ]()

it did exist

The If statement didn't

ajecc No, it didn't exist.

MikeMirzayanov

Actually we missed initially that the note was incorrect for

n= 2. The mistake was fixed a few minutes before the round, as soon as MikeMirzayanov spotted it, and the update was pushed from Polygon to CF. It takes some time, but for me the statement opened a minute before the contest already contained the fix, so I decided that everything was ok and I did't have to make an announcement. It turned out that for some people the old, cached statement opened without the fix. Luckily almost everyone understood what we meant correctly.Till I submitted my solution ,the fix didn't exist. I wrote my solution based on what I read in the statement. It wasn't my mistake.

You can`t re-test only A problem without test with size = 2? I think it would be fair

So this is over and my rate decreased beacause of a mistake which is not mine.

Stop crying over dumb shit. Mistakes happen and this contest is not the end of the world.Also its just -20 which can be recovered very easily.

As for me, this worsened my result by 45, but I totally agree that this is just a rating, but it's not very pleasant either

A question copied directly from geeksforgeeks, wtf ???? Contest should be unrated.Its a big disgrace.

The practice shows that it won't be unrated. There was a case when absolutely the same problem appeared twice even on codeforces itself but the round was rated.

If I am not wrong it was in an educational round, in which the ideas can be common. And it did not affect many div 2 coders. I might be saying this since I performed bad in the contest. But still C was just google and first link and it shouldn't be the case.

I think there were plenty of "just google and first link" cases too. But still I think we should let coordinators decide whether it's fair or not.

I have seen many times(I guess at least 3) where exactly same problem was in other contest, but mostly all of such contests were rated.

Seriously Mike, people copying code from other sites getting more credit than people working their ass off for more than half an hour. Fantastic work Mike.

Yes. I carefully thought and coded it for 47 minutes only to realize after contest that it was googleable. Shame !!! -_-

Was not able to open all problem statements in one window :( Bug

Does this solve F? https://www.quora.com/Given-a-strongly-connected-component-can-you-determine-in-linear-time-if-the-removal-of-a-single-node-turns-the-SCC-into-a-directed-acyclic-graph

How to solve C?

https://www.geeksforgeeks.org/maximum-edge-removal-tree-make-even-forest/

o.O

I think only his last solution got hacked, and all previous solutions were marked as hacked.

did he cheated using clone account?

What topic should i learn to solve C div2?

You should know basic graph theory(specially dfs)

for problem C, just learn how to Google the statement!

...and bfs too. I solved with bfs

Almost same as today's Problem C Is there any difference conceptually ???? May be it is not what the contestant expect from the problem setter..

In the yesterday contest, solution for C was also in Wikipedia.

Standings

I will leave codeforces if the contest is not made unrated, or semirated for the sake of it. But it is not fair for people to copy codes and press submit button.

How to solve C if it would have been a graph with cycles ?

tree hasn`t got cycles

I know that. I am just asking how would one go about solving the problem if it would have been a graph with cycles as the approach used for solving the tree one will not work here.

Sorry, I did not understand correctly what you mean

I may be wrong but in case of undirected connected graph we have to consider only those edges which if removed results in more than one connected component otherwise clearly if removal of an edge does not divide the graph then answer depends on n is even or odd. Now for the previous one you can have a look at bridge in a graph --> https://www.geeksforgeeks.org/bridge-in-a-graph/ Removing such edges and doing the same what we did in actual C question should do the thing.

But Edges might be of order which does not satisfy the constraints if n is 1e5.

By doing the same thing (which i wrote above) i mean that edge should be removed only if it divides two parts in even number (each).

The approach seems fine but I think finding out Bridges and removing would give "minimum" number of edges required for deletion and not "maximum".

No removing bridges does not govern whether we get minimum or maximum edges. When i say that you remove edge only if it disconnects the graph and both sides nodes should be even. So automatically all the edges which satisfy this property will get counted and that's what we did it in given C we count all those edges which divides the nodes into two even halves. Please correct if i m wrong !!

in case of a connected graph ???

About the choice of wording in condition 1 of A

I like this really much!! XD

You are worthy of being a problem-settter on codeforces.

In Question B why doesn't 3*nlogn pass the test cases. I tried using maps in c++ but it is showing time limit exceeded in test 45 . Can someone point out mistake in my code : Link

This part doesn't looks like O(1) or O(log n). I think, it's O(n) at worst case (because of "cnt2 = cnt1;" six lines higher).

Oh got it. Thanks for helping

Yay, I became master, but sadly only because of div2 rounds.

What is the difference? Master is master. If you don't want to participate in div 2, then wait for div1...

well actually its opposite, there are some purples that skip div 1, because they don't want to be beaten up by reds and yellows and join div 2 to compete with blues instead.

They can be beaten up by blues then.

I don't understand why everybody is complaining and thinks that it would be easier to get to 2100. The only true thing is that there will be more opportunities to get there. It doesn't mean it would be easier. If you don't deserve 2100 you won't get it, one way or another.

For example I already managed to lose a lot of points in div 2 contest on Monday. And today I was 122. If there were no violet coders, I would have been 25'th which would have given me much more rating.

Lol You don't understand? Before this shit revolution, to get Master(old) you need to most likely solve 3 probs in Div 1 which is equivalent to all div 2 problems. Now get just good ranks which in most case is possible with just 4 div 2 problems. As you can see in this round, 370 officially solved 4th hardest, you think thats good enough for someone Master?

"I don't understand why everybody is complaining and thinks that it would be easier to get to 2100" Because it actually is faggot. If you don't understand, then compare how many users have got there maximum rating in past div 2 only rounds.

"to get Div 1 you need to most likely solve 3 probs in Div 1" — makes sense. Sorry about my previous post.

TO GET MASTER (OLD VERSION) !

You've became at least, not like me :D (9 points)

Congradulations!

At least you have a chance to participate to next Div2 round, not like me :D (passed to master with 11 points).

Congratulations!

It looks very strange that O(nlogn) solution for D works without any used memory: 38374872.

Problem C is a duplicate problem: https://www.geeksforgeeks.org/maximum-edge-removal-tree-make-even-forest/

Problem F is a duplicate problem: https://sio2.mimuw.edu.pl/c/pa-2015-1/p/kon/, https://www.quora.com/Given-a-strongly-connected-component-can-you-determine-in-linear-time-if-the-removal-of-a-single-node-turns-the-SCC-into-a-directed-acyclic-graph

That claim about F is totally ridiculous

Sure. https://sio2.mimuw.edu.pl/c/pa-2015-1/s/89776/source/ 38363244

The Quora answer is wrong and comments don't seem to give any hint to actual solution.

As for the Polish problem, the translation shows that it is a harder version of Problem F. However, it isn't possible for problem setters to verify that their problems don't exist even on non-English sites.

Yes, it's the 4th time that I came across this problem.

It is not. I didn't realize this and now I checked that my incorrect solution to kon (9/10) is accepted after 2 small modifications (for 2 or more disjoint cycles and for empty list of cities).

Though I am not sure if I would have been able to figure out the first one during the contest...

Anyway, top 5 has gone...

Sorry but I have to say that CF won't care about such a thing!

Sometimes CF copies exact problems from itself though.

It's not copying if the second setter didn't realize that the idea for the problem was already used before.

I mean that they won't care about it after realizing copying.

CF be like: do we have issues with the tests? if yes, then it's unrated else it's rated for sure. I've never seen a round to be unrated for a more logical reason.

speed of light coder !!****Wow !

The guy knows his theory.

I think systests on D are very weak. I found a big mistake in my code after submit but surprisingly gave me an AC. My AC code: http://codeforces.com/contest/982/submission/38384511 A simple test case when my output differs from another AC submissions: 11 7 9 8 100 1 2 3 200 4 5 6 There something invalid in this input?

Problemsetter is the best ❤️.

My AC submission give output 10 and your code output 7. Maybe your code do any undefined behavior?

No, its just wrong...

In Problem B What if the number of extroverts(1) at any point becomes greater than number of introverts(0)? Many solutions will fail then ![problem:982B]

It is mentioned in the problem statement itself that there is always a suitable seat for an extrovert, So no such case will occur.

My bad ! I didn't understood the problem statement properly and wasted much time :/. Anyway Thanks for help !

It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row.©Can someone help me in understanding D ...??

For each different k[i] = a[i]+1, you check the size of all the contiguous days with a distance covered strictly less than k[i].

You go through all the a[i] in ascending order.

if a[i] does not have a neighbour lower than him, it creates a new contiguous day sequence of length 1. Then, you merge it with his neighbour(s) (if they are lower than a[i]), which destroys their sequence and creates a new sequence of length 1 + total length of the sequences of the neighbours. It's important because you need to keep track of how many different lengths of sequences you have at all time, cause you will need it at each step.

To keep track of how many different lengths of sequences you have after each a[i], you can create a table that tracks how many sequences each length has, and a variable that tracks how many lengths you have (that you increase by one each time you insert a sequence whose length is not represented yet, and decrease whenever you remove the last sequence of a given size). If there is only one length in the table, then you know that the number of sequences for k[i] is (number of distances <=a[i]) /(only length of sequence represented).

The answer is the minimum k[i] that leads to the maximum value of this ratio. You can do the merge with an union/find, making the merging a O(n*log(n)), and computing the ratio each time can be done in constant time, leading to (I think) a solution in O(n*log(n)).

Got it..... thanks a lot !!!

AGrigorii When is the editorial gonna be published? If it's already published in Russian, please do provide the link. Thanks

Sorry. I have a long trip to another city. Few moments later but today

In russian

Thanks

I found a surprisingly simple solution for F. The key observation is that if you have a node with an outdegree of 1, you can delete that node and push all incoming nodes to the successor (with a special case for self-loops). This process can reduce the number of edges if an incoming node was already pointing to the successor node.

As an example, node 3 in the image below can be deleted since it has only one outgoing edge:

Which results in:

Note that this reduced the outdegree of node 1, making it now eligible for deletion.

If you have more than one node remaining after this reduction, then the answer is impossible. I haven't tried to prove this, but it resulted in an AC submission. :)

For each node, it is essential to maintain an unordered_set for incoming and outgoing nodes to detect duplicate edges. This makes it slightly slower (due to the set operations), but it's fast enough to pass in C++ (albeit not in Java, from my experience). My accepted submission: 38409568.

I thought about a proof of your solution, and I think it is correct. If anyone find any wrong with it, just point out the wrong point.

The first part is about the validity of deleting the point with one out-degree. We can think about a point u that satisfies the requirement and we call v is a point that have a direct edge to u. We can easily find that every circle that pass the edge u-v must pass the point that u can only arrive at. So we can think u is useless and delete it from the graph (of course it contains the process of changing some edges).

The second part is about why the answer is impossible when there are more than one point remained. Now, the graph satisfies every point has at least two out-edge, of course we can remove arbitrary out-edge to make every point's out-edge is two. Then we build the dfs tree of the graph, and it is a binary tree. We only need to discuss about the out-edge of the leaves. We assume that this graph is valid for answer. Then we need to construct the out-edge of the leaves by a optimal way to make the graph is valid. So we can first make every leaf link to the root, and the other out-edge link to the next leaf in a way you like. But we will find that one leaf can't make another link because it wouldn't be valid no matter where it connect to. So we can prove this part.

This solution has a drawback is that it can't output all the points that satisfy the problem. But it realy easy to type.

(If there are many grammar errors, please don't blame me. Because my English is realy not good.)

I did not manage to solve div2-C using Python 2. My submission 38369140 got runtime error. I think the issue is recursive calls. Is there a trick in Python to get this code AC?

can someone point out the mistake in my code for 2B? 38450232. This is nlgn for sorting and then a map to get the index. Failing on test 11.

The problem is in your while loop:

it gives n^2 because you parse the whole string each iteration. Parse it once before the loop (http://codeforces.com/contest/982/submission/38451293)

tried your approach but with Scanner. Failed on 11 :(. It feels like this cannot be passed without fast IO. 38486099

EDIT: passes with 1809ms :|

Who can tell me what's the meaning of 'even size' in Problem C. The 'even' means equal? or %2==0？？

it means %2==0

Request: Please post the editorial link with problems for this contest. Had a hard time to find this post to find editorials.