ADJA's blog

By ADJA, 9 years ago, translation, In English

Hello, Codeforces!

We are glad to announce that on 4th of June at 19:30 MSK Codeforces Round #306 will be held. Authors of this contest are me (Adilet Zhaxybay) and Timur Sitdikov (Timur_Sitdikov). The round will be rated for the second division, however, participants from the first division can, as usually, participate in it unofficially.

We want to thank all the people, who helped us to prepare the contest: Max Akhmedov (Zlobober), who helped us with the problems, Bekzhan Kassenov (BekzhanKassenov) and Sergey Lazarev (SergeyLazarev), who tested the round, and Maria Belova (Delinur), who translated problem statements. Also we want to say great thanks to Mike Mirzayanov (MikeMirzayanov) for creating Codeforces and Polygon.

By the way, as far as we know, Timur_Sitdikov is the first participant from Uzbekistan, who took part in the preparation of Codeforces round. We hope that everything will go well :)

Good luck to all!

UPD The scoring will be dynamic

UPD2 Editorial can be found here

UPD3 Congratulations to winners!

  1. mff

  2. I_Love_Nodir.Daminov

  3. tun

  4. I_love_Ngoc_cmn_Thuy

  5. goodhope

Round is over, thanks to everybody, who took part in it!

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9 years ago, # |
Rev. 2   Vote: I like it +30 Vote: I do not like it

This will be my first Kazakh author contest :D I wish find interesting problems and many hacks ;)

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    9 years ago, # ^ |
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    This will be my first Kazakh author content :D I wish find interesting problems and many hacks ;) hahahahahaha pls aka.Sohieb downvote pls me upvote

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9 years ago, # |
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excuse me! maybe is not good place to ask but I don't understand: http://mirror.codeforces.com/problemset/problem/545/C

look at test case 7 why answer is 5? tree x=1 to left, tree x=41 to left, tree x=55 to left, tree x=59 to right, tree x=68 to left, and last tree with x=105 to right so we can cut down 6 tree and correct answer is 5 what's wrong?

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9 years ago, # |
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Your previous round was easy ( Never mind. I love easy round :D ). Hopefully this round will be easy too ;) Wish you two good luck and prepare more rounds for us :)

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    9 years ago, # ^ |
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    I instead, hope I learn something new from this round, participating and finishing a round without progression is not much of a good idea.

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    9 years ago, # ^ |
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    It's weird that recently ratings don't update as soon as past and editorials (except PrinceOfPersia) come late! what are doing there??

    UPD: 2 hours passed and we are stared at monitor waiting for raitings and nothing happened!

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      9 years ago, # ^ |
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      May be they are too busy to find the cheaters and eliminate them. It's an important task.

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        9 years ago, # ^ |
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        What methods they use to find cheaters ?

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          9 years ago, # ^ |
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          I don't know how they exactly find cheaters. But I guess, they may have some software which can find the partial matching percentage of pair of codes. Then they check the suspected codes manually to be more precise. But those are my thoughts.

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        9 years ago, # ^ |
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        even if they eliminate some cheaters, many cheaters are left (like bell-sama in previous contest)!

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9 years ago, # |
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Wish a standard problem set with strong data set :P

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9 years ago, # |
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It's maybe second Kazak contest?

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    9 years ago, # ^ |
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    It isn't Kazak contest, it's Kazakh contest

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    9 years ago, # ^ |
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    Also this contest is the third one)

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      9 years ago, # ^ |
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      Also, it is not only Kazakh contest :)

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9 years ago, # |
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Scoring?????????

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9 years ago, # |
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This time they even not write that "scoring system will be announced later" ;)

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9 years ago, # |
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Hope will be in div1 after this contest.

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9 years ago, # |
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As usual many fake div2 participants. So unfair.

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9 years ago, # |
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Loved your previous contest! I hope I can do better in this one. :)

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9 years ago, # |
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Problem D is nice, I like it.

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9 years ago, # |
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Unable to submit :( Codeforces temporarily unavailable, it says :(

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9 years ago, # |
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First experience with dynamic scoring, it was different and I liked it.

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9 years ago, # |
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So many hacks with ABAAB and BABBA!

I'm also wrong.

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    9 years ago, # ^ |
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    Yeah...with that particular test case I got 12 successful hackings :P

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    9 years ago, # ^ |
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    My solution also got hacked . I resubmitted it lets see if it passes system testing .

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    9 years ago, # ^ |
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    I figured the hack 5mins before contest ended. Coded it right in time, but not fast enough to submit

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9 years ago, # |
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Who is stdioH? He or She regestered 12 hours ago

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9 years ago, # |
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Thanks to authors for problem E, it was really nice!

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    9 years ago, # ^ |
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    Could you please give an idea of solution?

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      9 years ago, # ^ |
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      If last symbol is 1, then impossible.

      If sequence ends with 1, 0, then result is always 0. So we will try to make at the end.

      If sequnce consists only from zeros, and number of zeros is greater than 2, whe can place second and third from the end zeros to the brackets:

      If sequence suffix is 1, 0, 0, 0, ..., 0, where number of zeros at the end is greater than 2, we can place brackets in a such way: , and now sequence ends with , so result is 0.

      If sequence suffix is 1, 0, 0, than there are two cases:

      1) If there is no other zeros in the sequence, answer is NO.

      2) If there is at least one zero, which differs from zeros at the end, we can place brackets in a such way in a suffix which starts at the last of such zeros: .

      UPD: It was a mistake in last case, but now it seems ok.

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        9 years ago, # ^ |
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        it's not that complicated you can just do this
        if(lastone == 1) NO calculate ans like this ((1 -> (2 -> (3 -> ...... n — 1))))))) -> n) if its 0 then nice else you can't do it! the proof is that i got my E accepted

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    9 years ago, # ^ |
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    Tried to solve it, but failed on pretest 7, idk where I've got it wrong.

    Am I right so if the sequence doesn't end on 0 its impossible to get 0 with this sequence?

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      9 years ago, # ^ |
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      yes, you are right But that's all I thought up:D

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      9 years ago, # ^ |
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      i also got wrong ans on pretest 7. following are all the conclusions i could make: (assuming array name is a[] and size is n) 1. if(a[n-1]==1) ans will be NO 2. for inputs like: 1(1)*00 ans will be NO 3. for all other inputs ans will be yes

      For YES, output pattern is like this: 1. if a[0]==0 and a[n-1]==0: 0->(...)->0 2. for patterns like 1(1)*0(0|1)*0, output will be: [any arrangement for 1(1)*]->(0->[any arrangement for (0|1)*])->0

      I don't know what other cases I am missing. Any type of help is most welcome.

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9 years ago, # |
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Nice problems!

There were often not available to submit or to see someones code in the last 10 minutes.

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9 years ago, # |
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Why is it that always the contests with dynamic scoring are some kind of weird ? :D

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9 years ago, # |
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how to solve E??

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    9 years ago, # ^ |
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    try figuring out what's gonna happen if: last number is 1 last number is 0 first number is 0 first number is 1

    after some thinking, you'll get the idea :)

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9 years ago, # |
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Problem A is very nice. Short problem statement and a lot of hacks. Really enjoyed it! :D

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Is 0 a valid input on problem C?

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9 years ago, # |
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i will get a wa on problem A

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9 years ago, # |
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i could'nt submit my code in the last 5 min :(

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    9 years ago, # ^ |
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    I cannot hack other either. I guess there were too many submissions at that time so that the server couldn't afford.

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    9 years ago, # ^ |
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    I did 2 mins before the end !

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9 years ago, # |
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I could not even pass the pretests of problem A. I believe I understood the problem wrong the whole time. Can anybody tell me what did the problem actually mean?

Such misunderstandings ruined my contests too many times during past. Apart from this, I do a lot of silly mistakes like, not noticing constraints properly, forgetting long long, giving less size to arrays or vectors (even did this today at problem B :D ) and many others.

I am just getting accustomed to be a freaking block-head-dull-brained person all the time.

Happy! ^_^

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    9 years ago, # ^ |
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    I found that my solution of A is completely wrong after locking the problem! -_- howeever, the problem says to check whether the given string contains both AB and BA as a substring. and AB and BA should not be overlapped (like ABA )

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      9 years ago, # ^ |
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      I have seen the pretest #4. It was ABABAB

      Yep I thought that NO AB should overlap with any BA and vice-versa.

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9 years ago, # |
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Lots of construction algorithms! Nice round!

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9 years ago, # |
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How can I solve problem D?

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    9 years ago, # ^ |
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    for even k does solution exist ?? if anybody could tell it will be helpful

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      9 years ago, # ^ |
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      Nope, for even K answer will always be NO

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      9 years ago, # ^ |
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      I read a homework question and answer pdf of MIT, and it was mentioned that there does not exist any solution for even degree of all vertices, during the contest.

      Currently could not find that link, will post as soon as I find it again.

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9 years ago, # |
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Are there any solution exists for even N in Problem D?

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    9 years ago, # ^ |
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    I also could not find any ;(.

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    9 years ago, # ^ |
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    It can be proved that there are no solutions possible for even N.

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    9 years ago, # ^ |
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    nope. if every vertex in a simple connected graph has even degree, it can be proven that the graph has an Euilerian circuit. so all of the edges of the graph are at least in a cycle. so there are no bridges in this graph.

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      9 years ago, # ^ |
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      I proved it without the help of Euilerian circuit.

      Assume the graph has n vertices. Then n*k must be even. If we have a bridge break the graph into two parts(one has a vertices and the other has (n-a) vertices), then a*k-1 and (n-a)*k-1 must be even too. So the only possibility is that n is even, a and k is odd.

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        9 years ago, # ^ |
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        Thanks. I just solved it. I was so close. Tricky problem :)

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      9 years ago, # ^ |
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      Another way to think about it is like this: Look at the right side of the bridge, without considering the bridge. The degrees of that subgraph, if it existed, would be k-1,k,k,k,k,k,k,k... Since k is even, the sum of those degrees is odd. In every graph, the sum of degrees must be even, so we win :). Edit: Practically the same as boleyn.su, didnt update fast 'nuff.

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        9 years ago, # ^ |
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        I also thought that way. But I took too much time to find this.

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    9 years ago, # ^ |
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    Can't find it on paper, but also couldn't prove that its impossible, so just give up and moved to E.

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    9 years ago, # ^ |
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    Assume that such a solution exists. Consider its "left" part from the bridge and including the bridge vertex. k vertices in this subgraph have degree n and one (the bridge vertex) has degree n - 1. So, the total number of edges in this subgraph must be

    But since n is even, the numerator is not divisible by 2.

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9 years ago, # |
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i locked one of my solution to hack others :D and bam... my solution got hacked :'(

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9 years ago, # |
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With new feature of highlight code did impossible to hack for me, it was very slow, sometimes did not show the code :S. ¿ Anybody had the same problem? I use Firefox.

I love the new feature :D, just I have that observation.

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9 years ago, # |
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Problem A is so amazing, until in System testing :D

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Editorial ? :D

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9 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

!!!!! my A got WA !!!!! what's wrong with test 8 ? The answer is obviously YES why do they expect NO?!?!?!?!?!

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    9 years ago, # ^ |
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    As I see, Test#8: ABX repeaten many-many times. It contains AB but doesn't contains BA

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    9 years ago, # ^ |
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    Is there any "BA" in testcase 8? If not, then it's NO.

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      9 years ago, # ^ |
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      I think there is: ABX A B X AB

      the first two ones and the 5th and 7th one make : ABBA

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        9 years ago, # ^ |
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        They must be beside each other.

        So you can't take the fifth and the seventh one.

        (They are looking for substring, not subsequence)

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        9 years ago, # ^ |
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        I'm afraid you misunderstood problem

        It says to find substrings "AB" and "BA" that not overlape. There isn't substring BA in test#8

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      9 years ago, # ^ |
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      pnnb1997 — How did your submission 11416298 with worst case complexity O(n^2) for question A passed System Tests? :O . Also i can see that there is a test like "ABXABXABX....ABX" just to make sure that O(n^2) with pruning like yours gets time limit exceeded .

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        9 years ago, # ^ |
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        It didn't pass system tests and it's O(n)

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          9 years ago, # ^ |
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          which submission r u seeing? Maybe i made some mistake in posting the submission link which still seems fine to me. I can see in green letters "Accepted" verdict there!

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9 years ago, # |
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on problem A

ABABAB => YES ?!

Are you serious? you have only overlaps!!

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    9 years ago, # ^ |
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    A B A B A B

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    9 years ago, # ^ |
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    I think you misunderstand the problem, the two substrings that you find should not be overlap.

    For example, you can find such answer: AB*BA*

    They don't overlap.

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9 years ago, # |
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What time does regexp require? 11433753 Is there any way to make regexp works faster in Python?

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    9 years ago, # ^ |
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    It was supposed to be O(n*s*_a_) where n is the length of the string, s is the number of states of the NDFA and a is the average number of state transfer edges. However, it seems that Python implemented it another way.

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    9 years ago, # ^ |
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    Your regexp has very big time complexity. Try to make it more efficient. Later you can see my submission with regexp in 31 ms (that's because of compiler's optimization).

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    9 years ago, # ^ |
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    The problem is backtracking, I modified your solution to prevent it. You can check it out here:

    http://mirror.codeforces.com/contest/550/submission/11443577

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Can somebody please explain dynamic scoring to me ? Thanks

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9 years ago, # |
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how to solve b and c?

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    9 years ago, # ^ |
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    For B, you need to try all possibilities which are 2^N. For C, what you need to know is that a number is divisible by 8 if the number formed from its last 3 digits is divisible by 8 which means that if there is a number with K>3 digits divisible by 8, then there is also a 3-digit number divisible by 8. So you need to try all possibilities for 3-digits numbers, 2-digits numbers and 1-digit numbers.

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    9 years ago, # ^ |
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    im going to give you a hint on problem c to not spoil it.
    c-> you just need to know that a number n is divisible by 8 if and only if its 3 rightmost digits together are divisible by 8 For example for the divisibility of 3213123123213888 you just have to check the divisibility of 888 which is divisible.

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9 years ago, # |
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Why the stoi function not working in GNU G++ 11 4.9.2 :( Due to this I'm unable to submit at last minute. And when I uses stringstream my solution accepted.

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    9 years ago, # ^ |
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    Codeforces doesn't use a 'real' g++, but rather MinGW (a Windows port which has several bugs, including missing stoi and to_string functions).

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      9 years ago, # ^ |
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      yeah, MinGW bugs, which annoyed me when I want to use stoi and to_string()....How to fix these bugs, I tried but still not work

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My worst contest YET!!!!
my A, B got WA!!!!!!!!!!!!!!!!!!!!!!!!!
but glad C, D are AC!

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    9 years ago, # ^ |
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    How it is worst? You have been able to solve D first time I think, I engrossed so much in Hacking, I read problem D little late and could not generate regular graph in time.

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      9 years ago, # ^ |
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      problem D was our homework some months ago!!

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        The exact problem D? Can you provide some link of some kind of resource? That would be helpful indeed. :)

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Hey guys, editorial was posted here

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    9 years ago, # ^ |
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    I think it has a typo. It is showing dp[i][ai mod, 2015 - 06 - 04 8] = 1 instead of dp[i][ai mod 8] in the 550C.

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Update the ratings man, so I can go to sleep.

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Does anyone solved A using regular expression, I tried to solve A in this way instead of simple one. Forgot that regular expression takes exponential running time.

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9 years ago, # |
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Does codeforces not support "to_string" function ? I get compilation error even with C++11 . http://mirror.codeforces.com/contest/550/submission/11434694

Compilation error is

program.cpp: In function 'int main()':

program.cpp:40:25: error: 'to_string' is not a member of 'std'

string temp=std::to_string(i);

                     ^
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my solution gives a weird runtime error for C — problem with error code -1073741510 both in my compiler and on CF but my compiler runs the code and gives this error as warning . http://mirror.codeforces.com/contest/550/submission/11430760 can anyone give any insights?

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    9 years ago, # ^ |
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    Holy mother of unreadable code...

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    Dude, i dont know anything about the runtime but as a friendly advice, i think you should code a little bit cleaner so atleast you can understand your own code. i closed it the moment i opened it.
    sorry for being a little too straight.

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    9 years ago, # ^ |
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    Do you write code in Notepad?

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In div2 A, I wa on the ABAB so sad.. D & E 's constructive algorithms is very ingenious! Have fun in this round!

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^thanks shervin will take that into account in future :)

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9 years ago, # |
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I love this contest because of very short problems without very long story.

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Plz Give Rating.Wait a long time.

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    9 years ago, # ^ |
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    It is 2:45 AM, BST. Still waiting for rating. Tomorrow morning I have an internal practice contest. So, I should sleep now. :-)

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9 years ago, # |
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Still awake for Rating and feel bored.

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Hope I will become #Candidate_Master today. Nice problem set. Enjoyed very much :-)

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    9 years ago, # ^ |
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    Yeh!!! I made it. Now I am #candidate_Master. Hope I will become Master within 2 months. :-)

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9 years ago, # |
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My submission for problem C gave wrong answer on test case #22 but the same code gives different answer on ideone.com and my home compiler which is correct. Please tell me what is the issue. My Submission for contest : http://mirror.codeforces.com/contest/550/submission/11436093 Same code with test case 22 on ideone : http://ideone.com/Y82K0r

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    9 years ago, # ^ |
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    Most probably precision issues due to pow method, try resubmitting after writing your own exponentiation method using fast exponentiation in O(logn)

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9 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Don't get inspired by codechef. Please update the ratings ASAP.

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9 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Problems A, B, C very easy.

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    9 years ago, # ^ |
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    maybe you forget that this is div2 round o_o

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    9 years ago, # ^ |
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    Problem E was also easy. I couldn't solve it during the contest due to my own silly mistakes. Solved it after the contest

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9 years ago, # |
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Two and a half hours after contest and rating...........

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9 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

.

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9 years ago, # |
  Vote: I like it 0 Vote: I do not like it

1700!

I Love This rate

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9 years ago, # |
Rev. 2   Vote: I like it +21 Vote: I do not like it

A, C and E solutions with regexes: 11433416, 11435544, 11438339. Update: obsolescence's solution E — 11442038

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9 years ago, # |
Rev. 2   Vote: I like it +8 Vote: I do not like it

For D, I constructed the graph like described here:

For k=99, it produces 965448 edges, so it is still within the constraints. Being correct, nevetheless, my submission 11429811 timed out. Guess it is just a problem of optimizing output.

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9 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Congratulations to the winners!

Div.2 winners:

  1. mff

  2. I_love_Ngoc_cmn_Thuy

  3. goodhope

  4. kiana810

  5. xwind

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9 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I am getting an error on problem D "wrong output format Unexpected end of file — int32 expected" what does this mean , can anyone help (http://mirror.codeforces.com/contest/550/submission/11450063)

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9 years ago, # |
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D: my experience is low , if I can master more about graph , I believe I will enjoy more,

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9 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Sorry to take part in it late. I feel this is one of the top competitions I took part in. Requires no complex knowledge of algorithms and still the problems were tough. Thoroughly enjoyed the problems. Specially surprised to find red coders getting problem A wrong.