Hi, Codeforces!

AIM Tech Codeforces Round 3 will take place on August 24, at 19:35 MSK.

The round is prepared by AIM Tech employees: Kostroma, AlexDmitriev, yarrr, ValenKof, Edvard, bobrdobr, malcolm, NVAL, n_makeenkov, Agul, Extr and zeliboba. Round will take place during Petrozavodsk Summer Camp, which is sponsored by our company.

We made our problems a little easier than at AIM Tech Round 1 and AIM Tech Round 2 but we promise they won’t be less interesting. Scoring system will be static.

Thanks to Mike Mirzayanov(MikeMirzayanov) for brilliant platforms Polygon and Codeforces and problem coordinator Gleb Evstropov (GlebsHP). Many thanks to AlexFetisov and winger for round testing!

Our company specialises in proprietary trading, the key concepts in our work are big data, low latency and high frequency. Our team mainly consists of graduates from the MSU Faculty of Mechanics and Mathematics and Moscow Institute of Physics and Technology (MIPT).

We wish you good luck and high ~~frequency~~ rating!

Scoring in both divisions 500-1000-1500-2000-2500

Now we know what is Edvard new job :)

It would be nice to make combinative d1/d2 round.

I guess one can actually prepare the same problems for both divisions, just with different variable/time/memory constraints for Div1 and Div2 participants. But this sounds more woe than fun :)

Will there be T-Shirts as stated here? http://codeforces.com/blog/entry/23240?#comment-276543 :D

these aimtech guys really do have a very interesting field of working, artificial intelligence management, i never heard of it before.

May be something new & so much interesting in this problem set. Eagerly waiting for it... Thanks to AIM Tech & the employees.

Standard duration (2 hours)?

Yes, standard rated round

AIM Tech Round 1: -40, AIM Tech Round 2: -43. I hope my rating change will be positive this time :"D

What is negative rating change?

It's sort of like a lot of downvotes.

That's good you've never experienced it!

You said that last time, too — and it turned out to be just from "3rd place = 3 problems solved" to "6th place = 3 problems solved" :D

5th actually, but nevertheless that was a progress =)

13th place = 3 problems :"D

I know the AIM Tech Round is harder than the ordinary div2,but I will try my best to solve the first problem or the second problem.

Not this time ! Although Question B had tricky cases to deal. Overall, Questions were easy as compared to regular DIV2s. Happy coding.

You didn't give the contest!

I used my another account to take part in this competition.

Hope another exciting contest, interesting problem set :)

what does frequency mean?

"dum.....dum.....dum.....dum.....dum.....dum.....dum" is low frequency, whereas

"dumdumdumdumdumdumdum" is high frequency =)

After long time Div 1 users is getting rated contest =D

i wish high rating for all :D

that who gets the low rating?

I think in the setting rating system if a person go up by x another person should go down by x too! and because of new users the sum of all rating is going up!

It means when you become happy cause of your rating change, there is some one sad of your happiness?!

which is sad :/

truth always hurts. Just a truth-seeker can believe it.;)

Is it called Law of Conservation of Rating?

Codeforces surely put some alchemy in there! :p

Thanks , also to you

*wish

Thanks for codeforces to provide the chance for me to practice with you that from all over the world whitout money,thank you very much, i mean it :)

The comment is hidden because of too negative feedback, don't click here to view it

may your words come true.

they did.

maybe it is new start.

Jet lag makes me feel a litte tired, prefering an earlier time like the day before yesterday:)

Wish that everyone can get a satisfactory rating.

If the rating change formula is correct you are telling an impossible event :( Anyway, wish the most of the user a satisfactory rating.

Good luck boys !

What does "Scoring system will be static." mean? Usually it is dynamic if I'm correct.

That means basic score for each problem is fixed before the contest (and it decreases during the contest linearly). Dynamic scoring means that basic score for each problem depends on number of accepted solutions.

Actually, I think static scoring simply means that it is not 'dynamic scoring' which is detailed in,

http://codeforces.com/blog/entry/4172

Basically, instead of scores decreasing from a fixed amount, the 'initial scores' themselves would depend on the number of contestants who have managed to solve the problem.

We don't see a lot of dynamic scoring competitions these days but they used to be relatively common about two years ago.

Good luck and have fun! (c)Edvard

So fast Scoring System Update .

#GLHFI am a newbie since last febrauary. I hope to be a pupil in this round. I hate my gray color xD

Yes, I know that feeling bro, just as much as I'm hating my cyan colour, good luck today :D

I study in the Faculty of Mechanics, perhaps my study will help me solve one of the problems today :P

I will be back

Take your time :D

tourist is registered!

9 legendary grandmasters participating :O

Makes no difference to your life though ;)

Div1B was super evil..

Got hacked with 0 0 0 0 I think. I'm so silly!!!

I think e.g. 01 and 011111 are also tricky.

Had it handled.

Funny thing is, if I got hacked just 1 minute earlier(I got hacked with approx 10 seconds to go) I'd still not have answered it correctly. Indeed tricky.

how to solve it?

i did a long stupid algorithm

zero*one=a[1][0]+a[0][1].

Then, it doesn't matter how you arrange them as long as the above holds true.

You, sir, you lied. You're not dumb

thank you <3

and tell this guy CantGetAnyACWhyAmISoWeak :P

I believe I've already told him once proof

Is that your real account, by the way? ;)

I didn't find a simple way to handle special cases and just used brute force to handle all the special case, hope there is no bug :D

Oh， my D failed...

Good problems, but... After that contest, I hate the number 7 :(

I see you passed it,i had same problem,do you have any clue what the test could be ?

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Or just "a" :)

And what should have been the output?

z

last letter z

Oh ok, I didn't know that you had to shift at least one substring once...

No. One guy got hacked for that testcase.

exactly one shift

I will defeat tourist one day... Yes I will...

0 0 0 0 for Div2 D/Div1 B :(

Why did I put "Impossible" for that? :(

I missed the case, and I would've put Impossible if I considered it in time, lol.

My hacker is EEEEEEEEEEEEVIL

Answer would be 0 or 1

Yeah, I read that just now in a (now deleted) comment.

which hacker? this case is in the pretests

Oh, yeah! I accidentally printed '1' for it! Lol, where did it fail then @_@

edit : Here's my sad story. I changed the order of two if statements and got AC

But I thought string "0" satisfy it.

I think you're right both 0 and 1 satisfy it.

Made B 5 sec. after end of round

Via 9fag.com?

No, 9GAG.com

Since "Round will take place during Petrozavodsk Summer Camp", I guess system test will start late?

what was the hack test for div2C

aaaaaa?

damn that was a silly mistake

but it wasn't a hack it was the 7th pretest

I wish aaa was not a pretest in Div 1A.

It wasn't

but it was definitely string consists of all a.

Yep, there was

`a`

but some people still failed longer`aaa`

sWhat is the reason for failing longer

`aaa`

sThey may have added in a special case in their code if the input is "a" but forgot to check if there is more than one 'a'.

hehe, I got a little problem there too.

The difficulty was perfect and problems were interesting. A very nice round (I didn't read div1-d though). I hope there will be AIM Tech Round 4, one day.

translation: give me upvote!

No, translation: I solved E, oh yeah!

Xellos was closer, sorry.

Also, I think that many setters prepare problems mostly to make us think "wow, I enjoyed that a lot". If someone made their job so well, I'm sure he/she deserves some applause.

I didn't solve D and E and I'm not even mad because the problems were so nice.

Different from when you see those innovative query on tree heavy-light decomposition problems and you're like "I don't even care if my rating goes down for not solving this, it's so boring"

Oh, sorry I can't give you more than one upvote. Exactly same feeling about these problems =)

Most likely it will be during the next Petrozavodsk Camp

Tasks were nice! (Translation 2 bits away from AC in E). Either way, tasks were really nice :P.

I think that the description of Div2A was a bit misleading. "... When it happens Kolya empties the waste section (even if there are no more oranges) and

continuesto squeeze the juice"To me that implies that Kolya stops squeezing the current orange, empties the waste section and continues to squeeze the same orange until there's no more juice. But magically, Kolya actually squeezes the entire orange and then cleans everything, resetting the waste bucket.

even I thought of that, my whole contest got wasted because of this ......

Very slow codeforces, but tasks were pretty interesting :)

I suppose it will be a lot of fail submissions after end of testing.

That is the first and only feeling I had after the end of today's contest.. Lets hope for the best

Div 2 C was easier than Div2 B.

I submit 2 codes for problem A,my both code is accepted but on rank list time of second code is considered and also 40 points of penalty is also deducted .why it is so?

It takes the lastest submission in case you found bug(ssss), some godlike cases that may not be verified in pretests.

You get -50 points for every resubmission, and only the last submission is taken into account for testing.

your latest submission is considered.

thanx!!

When will the system test start please? So I can decide whether i will sleep before school tomorrow Xp(only 3 hours before I must wake up again) Haha

school in august ? no wonder asians are smart haha :D

Can someone give me a hit for Div 2.D??

Me too, I also get confuse with Div2 D, don't understand the problem anymore

div2d had so many edge cases... The main can be solved purely using math. As for edge cases, there are (0,0,0,0) (0,1,0,0) (0,0,1,0) (X,0,0,0) (0,0,0,X) where X is a triangular number. You can deduce number of 0's and 1's in since a and d must be triangular, and a+b+c+d must be (n0+n1) choose 2, otherwise "Impossible". The last edge case is when a=0 or d=0, since there can be 1 or 0 of them. But if theres 0 then it reduce to X 0 0 0 and 0 0 0 X case Now you can say that when a=0 then n0=1 and d=0 means n1=1. Now prove that anything else left is always possible to build. :)

although I have stuck with the 3rd case, I can say my idea.

first, you can get number of ones and number of zeros in the string.

where:

1) No. of zeros * (No. of zeros — 1) / 2 = a00

2) No. of ones * (No. of ones — 1) / 2 = a11

You can check some validations using these numbers, a01 ans a11.

No to build your string.

then you should validate your string and you can do this in O(Length of strnig) by counting how many "01" and "10" subsequences and then check them with a01 and a10.

I implemented this idea but I couldn't to figure this tricky case "0 0 0 0"

sorry for bad English. :)

can anyone explain the example instead of the solution? :(, I even don't understand the example (and the problem also)

I will reverse the problem statement:

You are given a string S consists of 0's and 1's.

How many times these subsequences { "00", "01", "10", "11" } do occur?

The problem itself gave you how many time each subsequence had occurred and you should get the string.

could you explain more detail?, sequence {x, y} mean what?,

that was my idea but oh god I was so stupid I was confused between substring and subsequence for like 30mins and I was like "wtf they are wrong on example 2" :/

What is the difference between substring and subsequence? could you give me an example?

for instance in abcdef :

"bcd" is a substring and a subsequence because letters are neighbors in the string and are in the same order

"bdf" is a subsequence which is letters in the main string in the same order but not necessarily next to each other. But it is not a substring

so here there exist a subsequence "01" if there is a 0 with a 1 later in the string.

in 11111100000 there is no "01" sequence but in 111100001 there are 4 "01" sequences

Okay, thank you I understood the problem, finally

Let the number of 1's in the string be X.

Let the number of 0's in the string be Y.

Then,

1. the number of 11 subsequences = X choose 2

2. the number of 00 subsequences = Y choose 2

3. the number of 10 subsequences + 01 subsequences = X * Y (one of them is from 1's set other from 0's set)

Assert the above conditions for solution to exist.

Now let the requuired string be B, and let us start from a string A and try to construct B from it, where A =

"11111..

(X times)0000(Y times)" — i.e., all 1's followed by 0's.A has X * Y number of '10' subsequence and 0 number of '01' subsequences. Now, notice if we shift the last 1 (the Xth 1) one position right swapping it with the first 0

=> 1111..(

X-1 times) 01000..(Y-1 times)Now, A has X * Y — 1 number of '10' subsequences and 1 number of '01' subsequences.

Keep doing this sort of shifting, you'll eventually be able to arrive at a01 number of subsequences and a10 iff (a01 + a10 == X * Y).

Beware of corner cases like: "0 0 0 0"

"K 0 0 0"

"0 0 0 K" ...

Thanks for your Idea .. Solved it ... After 3 WA & 7 TLE....

Will there be anti-quick sort case on Div2-B for Java solutions?

Interesting systemtest. Waiting half an hour, then almost instantly 58%, then it stops again :D

Because partly the testing is conducted during the contest

I know, I was a problemsetter :D

That doesn't explain the discontinuity, though.

"Because partly the testing is conducted during the contest"

"I know, I was a problemsetter :D"

Just red coder things :D

Because the judge alternates between Div 1 and Div 2.

Div 1 B is all red...

My solution got TLE on problem C. Are you serious???

Edit: Sorry, maybe I need to sleep more. >_<

i didn't solve it too !

Isn't it ?

Broom test OP

So even Nutella grandmasters make this mistake sometimes :)

this problem is too difficult.

My O(n) got 1543ms, after my O(nlogn) got TLE on pretests. It's easy to underestimate constant factors.

Could've been a great contest for me, but no..........

I feel the same way lol.

Same here. :(

I solved 4 problems but got WA on 2 of them due to stupid corner cases. Pretests were weak. >_<

One of the most interesting, well developed CF round ever. Thank you so much! :)

I can't believe I got WA on test 163 out of 164. -.-

...

How to solve div1-B?

First, you should notice that a00 and a11 are triangular numbers(and it will give you the values of the number of 0's ans 1's, here you should be careful with a00 = 0, because there could be 0 or 1 zeros, the same for a11). Second, a10 + a01 must be equal to xy, and in fact (you can prove by induction on (x + y) that it's everything you need to construct an string with the conditions you want).

So first of all you have to find how many zeros and ones there will be in total. in order to do that you look at the number of 00 and 11. Say number of 00 is 3 that means that we have 3 zeros in our string because you can choose 3 combinations of size 2 from 3 zeros. So basically if you have n zeros in your string the number of 00 will be n*(n-1)/2 and the same goes with 1's. Though there is something we have to look at, which is when you have 0 of 00's. then maybe there is 1 0 but definitely not 2 because if it was to number of 00 would be 1. So it will be 1 if number of 01 or 10 is more than 0. and same goes for 11's. after you determine the number of 0's and 1's you do the following: you start putting 0's and 1's. if you put a 0 you decrease total 0's. if you put a 1 you decrease total 1's. and if you put a 0 that means you increased number of 01 by the amount of remaining 1's because you will place remaining 1's after that 0 and it will be number of reamining 1's times so same goes for 1's . an algorithm would look like this for that :

and there are still some cases. if a,b,c,d=0 then you print 0 or 1. if a b c= 0 but d is not you print only 1's. and vise versa. and if a or d is not in the form of n*(n-1)/2 it is impossible as well.

How did you figure out that the approach of just writing out 0s and 1s greedily "while the current string doesn't violate the rule" would work? My first reaction was that I had to consider many different combinations, and, frankly, I still don't quite understand why this works.

well it is not that hard to figure it out. lets say we have 5 0's and 5 1's. if you put 0 at the beginning how many 01's are you adding for that 0 ? 5 times right ? because it would look like this: 0xxxxxxxxx where x is 1 or 0 but in total there will be 5 1's in x's so for that 0 they will make 5 01's. and now you have 4 0's. lets say you will put 1 now. 01xxxxxxxx . and now we have 4 o's left. how many 10's can you make with that 1 ? 4 times right ? because you cant use the 0 at the beginning to make 10. and it goes like this, think simple dont make it complicated :)

Thanks.

Why can't I see test cases in practice mode? I want to know where my solutions fail.

Pretty sure you can, at least the start of them... (just go to the submission page 20134499 and scroll down)

Your solution for B fails for "0 1000000000 0 0". The testcase for C is too big for me to see.

Yes, thanks for your reply.

A few moments after posting my comments, I was able to see testcases, but before I could see nothing, as if the contest was still going on.

Was stuck on problem B with test case n = 1 till the final 1 minute...lucky to discover that before end.

on problem B a hacker saved me on case n=1 lol

Hard time today lol!!!

Your profile picture seems appropriate for you :)

Look on the good side, you did not got a WA on main tests.

the statement of problem A was terrible :( so bad

yes, it took me 20min to figure that out

it took me 30 min to understand it and 3 min to code it :(

It took me 1 min to load the page, 1 min to read the problem and another last one to code and submit it :P

Meant no offense, just joking though, but why you found it terrible, I thought it was clear enough.

Can you please tell me why I got TLE in problem C !! that's so weird!! 20125382

use printf instead of cout. and in order to make it a little bit faster you can do this int size = s.size(); instead of looking at the size every time in the for loop.

printf should make nearly zero difference, cout is only used once

you are completely right lol I thought it was in for loop

String comparison is

O(n), and you do itntimes.s < prev inside loop, comparing strings is O(len)

String comparison

s.length() is calculated after every condition check...which makes your code square complexity....I made this mistake once

Nope, s.length() is constant. (In C++11 at least) C++ Reference

Thanks for letting me know about it!

you are mistaking strlen(s) for s.length() possibly

String comparison is costly time-wise, this line to be specific:

`s < prev`

. It should be replaced with`s[i] < prev[i]`

as you don't need to compare the whole string everytime as you pass on each letter and can compare them individually.Here is your code modified, it got accepted. 20135520

thanks a lot dude, it's so tricky I think.

It is, best way to avoid these sneaky little mistakes in the future is by making them in the first place!

No it's not. In the killer testcase your code is comparing a very long string for each character it has, leading to a

O(n^{2}) behaviour.How to solve Div2B ?

just sort the data....now you need to visit n-1 points. So you have to check for 2 choices — either leave first point or the last point(in the sorted data set) which has 4 ways of doing so. Check my code — 20124041

Can you elaborate more please ? I am not that good programmer you can see :(

Sort the Points. Assume D as the given Point where you are standing presently ...Now,

Now calculate the minimum of those result .

Make sure you are checking all other Corner Cases like ... if the point D is in between Arr[N-1] & Arr[N] . ( there are 3 more Corner case according to my approach . I am sure your are smart enough to find those out )

We have a position x. and an array a. sort array.

the first case x <= A [0] then the answer (a[n-2] — x)

second case x> = A [n-1] if the answer (x — a[1])

in other cases the answer: a minimum of

( (2*(x-a[0]) + (a[n-2]-x)),

(2*(x-a[1]) + (a[n-1]-x)),

((x-a[0]) + 2*(a[n-1]-x)),

((x-a[0]) + 2*(a[n-1]-x)).

Multiply by 2 because they have to pass through this segment 2x. early in one direction and then in another. Choose from 4 variants the most profitable

We do not forget about the option when n = 1; the answer is 0

Div. 2 Problem C has a complexity of O(n) but still I am getting Time Limit Exceed on test 27. My solution link is : http://codeforces.com/contest/709/submission/20121635

it is O(n^2) because string concatenation is O(n)

Maybe you shouldn't calculate s.lenght() every time because it's probably a linear opeartion. Just calculate it after reading and use the result in the "for". (at least that's how c++ works)

In "for (int i=0; i<s.length(); i++) {" You called s.length() every time in the loop, and the time of each time you call "s.length()" is O(n). So your submission is O(n*n). You had better use "int len=s.length(); for (int i=0; i<len; i++) {...

it is O(1) in java. reason for TLE is string concatenation

Think before you speak Arrays.length() is always O(1). See my accepted submission for the same problem.

Java string addition copies every time both the string and then gives new String.

This will eventually turn out to be O(N

^{2}) for large operations. You should use StringBuffer for appending the result.waiting for rating change

Wish I can turn blue now! Waiting for editorial and how to solve DIV 2D?

I have explained it above

What is this about cheaters?

?

The main post has been updated to say that there were cheaters.

Maybe the cheating is in div2C... How some people got it Accepted in few minutes.... !! Maybe it is a old problem......

No, its actually very easy. Problem setters should not assign higher points to easy problems. It is really unfair.

Guys thank you so much for such an awesome contest, really enjoyed it, can't wait for more AIM Tech rounds

only Div1 Rating changes have been updated , why Div2 haven't been updated until now ?

UPD: it's now updatedThis is the first time I solved problem D in a contest.....I was having some bad previous contests..I can expect a good rating change now :)

and you got it, congratulations +114 :D

Missed specialist by just 5 points :(

after all ... contest was easier than previous ones ...

what was the hack of problem c div2

Maybe it was "aaaaaaaa" or something like this....

can someone explain why aaaz is lexicographically < aaa ?

It's not, but you must change exactly one substring.

:'( I can't believe this is the reason I got WA

oh and thank you :)

I think you mean aaaz is lexicographilly < aaaa? It's mentioned in the problem statement that "You have to pick exactly one non-empty substring of s and shift all its letters". so you cannot just leave "aaaa" not changed then the least change that could be done is to change the last 'a' to 'z'.

Yeah I missed that part :/. I Guess I have to pay more attention to the problem statement.

How to solve div2E ?

I want to know as well, but here's what I think. We have to calculate subtree size of each node(in a rooted tree) and if the subtree<n/2 then we have to find a subtree in the upward-parent-subtree whose size is h, such that removing this subtree makes the upward-parent-subtree <=n/2 . We can check the number of children a node has, and their subtree sizes to get a range of values for h for a particular node.

I have no clear idea how to code this, or even , if its correct. But that's all I came up with.

Yes, it's possible using this method.

http://codeforces.com/contest/708/submission/20136869

You can refer to this approach. Its similar to mine, but by finding the centroid first, we can do this easily in time. If we don't find the centroid first, then we can't do this in time limit.

For Problem C i have coded a solution of Time Complexity O(n) but still i'm getting a TLE .

Can someone point out why this is happening?

https://ideone.com/oPOD8j

You have an O(n^2) solution, because of the way how you add the charachters in

`be=be+s[i];`

. Effectively, each such call copies the whole string that is build thus far.No,i think it's O(n). Suppose you have an array : A{1,2,3,4} and you want to calculate the sum of it.

This addition of integers and addition of character's to a string takes the same complexity i.e. O(n)

Yeah, string s += char is NOT the same as s = s + char. The first appends char, the second computes s + char, then sets s equal to that. I hacked someone on Div 1 B based on this.

You just blew my mind *_* Got an AC by replacing s=s+char with s+=char.

Doesn't seem like a very efficient way to handle things. Is there any need for this separation? They could've overloaded the operators to act similarly, unless , there is some use.

edit: yeah, t=s+a. I'm silly lol

`+`

has to make a copy of the string, otherwise stuff like x = s+char wouldn't work (and the compiler doesn't know s=s+char and s+=char mean the same thing).I literally figured it out while you were writing this comment lol

Could you explain E's approach?

Denote the centroid of tree as X, then reassign root to X (You can easily find out X). Now X is the tree's root, it means that every sub-child node of X has no more than n/2 children.

If vertex U is the new centroid, only n — numChild[u] is possibly greater than n/2. Now you need to find the vertex V (V is not sub-child of U) then break it from tree, re-attach to U. Checking new tree

But how can we do this without iterating the tree back and forth multiple times?

edit : I think I understand how.

1. We find the centroid

Cof the tree and root the tree at this vertex.The answer for this vertex is obviously yes.2. Then we move down one of the subtrees. We are now at vertex

V3. Then, we must remove a subtree from the other child of

Cand attach it to subtree ofV, such that size of that removed subtree isn/2-subtreesize[V].4. We can achieve 3. by using euler array of the tree rooted at

C. This is because the removed subtree will always come from one of the children ofCwhich we are not currently traversing.Correct me if I was wrong anywhere.

`**from the other child of C**`

=> It should be other children of C, which are not child ofV.We can find this vertex by using segment tree

Thanks man :) I think now I've fully understood how its to be done.

Got it thanks!

How to solve div2e?

Remove largest tree from eccentric child if the tree is rooted at i. The largest tree's size should be of course <= N/2 as we will attach this tree directly to i. I used timestamping and some arrays to find the largest tree. You can refer to my last AC code.

It was very hard to debug it during the contest sadly :( .

Anyone Pls tell me how to solve the problem B in Div 2 ?

consider this case 5 8 ........ 3 1 5 10 9 we have to go to n-1 points , so it is clear that we need to sort the points first [1,3,5,9,10] , so we dont go back and forth (which will increase distance), now to go to n-1 points , we either go to first n-1 points or last n-1 points , that is we either go to points [1,3,5,9] or points [3,5,9,10] to make distance minimum , now we simply check in both the cases if we should first go the first point or last point and add it to the ans , something like this..

a1 = min(abs(a[0]-p),abs(a[-2]-p)) + abs(a[0]-a[-2])

a2 = min(abs(a[1]-p),abs(a[-1]-p)) + abs(a[1]-a[-1])

print(min(a1,a2))

What happened to the rating changes? Are cheaters being dealt with right now?

so...... what happened to rating ?

I solved Problem B after reading through comment section and after numerous WA's and edge cases. I want to know how did people think about the strategy to solve it ? I was able to solve A and C quickly but could not even figure out how to do B during the contest. Can someone give me some pointers ???

I struggled with it as well, but I can give you some tips.

Sorting the points is never wrong in a task like this.

Also, you'll always either choose not to visit the leftmost point or the rightmost point. For each of these, you'll either go like start -> left -> start -> right or start -> right -> start -> left, you take the min out of those 2 posibilities for not taking pos[0] or pos[n — 1].

The edge cases are n = 1, n = 2 and if you're between pos[0] and pos[1] or between pos[n — 2] and pos[n — 1].

I agree that the task is really ugly, but taking it step by step it's solvable.

Not so ugly :D

Source

Thanks.Although I sorted the values during the competition, I had no idea how to proceed thereafter. But yeah, taking it step by step makes it solvable.

I'm a little late but maybe my answer will be relevant because it's fast to write and doesn't deal with many edge cases. It's a little overkill though.

It was basically, store all elements and the initial position and sort them all. Then find where the initial position is in the sorted array, if there is a repetition take the smaller position.

Now just think that you must visit

ipositions on left of initial position, 0 <=i<= n-1, andn-1-ipositions on the right of the initial position. Then you just need to iterator oniand check if these left and right positions are inside the array.Take care that if you visit left first the result may be different of visiting right first so you need to consider both.

That's my submission: 20114511. I found this way very fast to write and easy to cover all edge cases, took me 13 minutes and 0 WAs.

Wow, that's fantastic. Nice logic. No you are not late. Thanks for the idea.

when i will get editorial of this round ?? :)

I am not understand problem C .Plz any one explain it...

Can anyone point out whats wrong with codeforces for my submission for B? http://codeforces.com/contest/709/submission/20144340

I can run the 1st case successfully locally and on http://ideone.com/D2a7Nu.

I want to know the input test cases to practice the problem "after the contest" "Problem A" wrong answer at test 5 and the input is very large and not complete to test it and find the problem. Thanks in advance.

Can anyone please help me with the logic of div 2 E-Centroids?

After finding centroid, I find lca of all nodes. Then, I find the node

vfrom each nodeusuch thatn-subtree[v]<=n/2.

Then, I can make parent of

va child ofu.Before finding

v, I make sure that none of the children of the root(centroid found initially) have a subtree such that it can be directly attached tou//code removed

hi, the link you provided does not show anything. make sure you make it public. i checked your last submission it doesn't seem to use lca in the solution.

I'm sorry! last submission

I realized that LCA is not needed(to my understanding). From the output it seems that I'm printing 1's instead of 0's for many nodes. A possible reason can be that my

Citself is incorrect.C.C, which have the maximum subtree sizes. This is so that when I am traversing inside the subtree with the maximum size, I can cut the other subtree, as stored in ma.If cutting one or the other subtree we have in ma[] doesn't help, meaning that

`n - subtree[ma[0 or 1]] - subtree[current node under consideration] > n/2`

then I cut the edge joining C with the subtree that I'm currently traversing. Clearly, such an operation only needs to check the subgraph containing

Cbecause the other subgraphs will definitely have size<=n/2.But I think I'm missing some test cases. Do you know which cases?

you are just computing the centroid in a wrong way.

reread your code i believe it's a sleepy mistake.

here is an ac submission 20357644. i just changed your centroid function.

BTW, i think this is exactly the same idea as described in the editorial. i don't know if this is intended but check it.

Thank you !!! :)

I suspected something's wrong with the centroid, but I checked the code just 2 minutes ago and still couldn't find the mistake.

I didn't realize it was the editorial explanation. I started with LCA and then realized its not needed. Next time I'll make sure I don't repeat such a thing, and check the editorial beforehand.

and i'm actually glad that it doesn't use LCA :P. as i'm not familiar with this stuff yet. willing to study it in the near future though.

With LCA it was not difficult, although unnecessary. It really is easy.

lca[i][vertex] = the vertex which is at distance 2^i from vertex.(Counting starts at 1. So, a distance of 1 = the node itself)

lca[0][v]=v;

lca[1][v]=parent[v]

lca[k][v]=lca[1] [ lca[k-1][ lca[k-1][v] ] ]

I think you'll understand why we do an extra lca[1] [..].