dhirajfx3's blog

By dhirajfx3, history, 13 months ago, In English,

Hello Codeforces,

Manthan, Codefest'18 will take place on Sep/02/2018 17:35 (Moscow time) with a duration of 2 hours (tentative). The round is rated for both Div1 and Div2 participants and will consist of 8 problems.

The Department of Computer Science and Engineering, IIT (BHU) is conducting Codefest from 31st August-2nd September. Manthan (मंथन in Hindi, meaning Brainstorming), the algorithmic programming contest under the banner of Codefest, is being held as a special Codeforces round. The round follows regular Codeforces rules.

The round has been prepared by hitman623, karansiwach360, GT_18, ezio07, Enigma27, csgocsgo and me (dhirajfx3). Special thanks to praran26 and dark_n8 for their contribution in the preparation of the round.

We express our heartiest thanks to KAN, gritukan, 300iq, isaf27 and _kun_ for their help in preparing the contest and MikeMirzayanov for the awesome Codeforces and Polygon platforms!

Prizes

Overall 1st place: INR 25000, Overall 2nd place: INR 18000, Overall 3rd place: INR 12000

1st place in India: INR 10,000

1st place in IIT(BHU) Varanasi: INR 4,000 1st place in freshman/sophomore year, IIT(BHU) Varanasi: INR 1,000

About Codefest: Codefest is the annual coding festival of the Department of Computer Science and Engineering, IIT (BHU) Varanasi, which is held online and is open to participation by all! Register on the Codefest website now! Total prizes worth ₹500,000/- up for grabs with events covering domains from Math, Machine Learning, Natural Language Processing and Capture The Flag style competitions. Go to the Codefest website to find out more!

As usual, the scoring distribution will be announced just before the round.

UPD1: Scoring 500-750-1000-1500-2250-3000-3500-4000

UPD2: Following are the winners of the contest

1. tourist

2. CongLingDanPaiShang3k5

3. LHiC

Best in India

amit_swami

Good luck and have fun!

UPD3: Link to editorial

 
 
 
 
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13 months ago, # |
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The best part of Codefest <3...

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13 months ago, # |
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I hope it will be easier than last year

Last year problem B was a DP :((

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    13 months ago, # ^ |
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    so it was your 5th contest last year.

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    13 months ago, # ^ |
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    Yeah, it's relatively easy than last time.

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    13 months ago, # ^ |
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    Last year problems were harder than problems of the year before last year :)

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      13 months ago, # ^ |
      Rev. 2   Vote: I like it -15 Vote: I do not like it

      So, the possibility of being the problems harder than the last year's increases. :(

      Wish to have interesting problems. Good Luck everyone! :)

      (It's my first reply)

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        13 months ago, # ^ |
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        Or maybe problem setters were different.

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    13 months ago, # ^ |
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    Dude i don't know how, but you got to div1 in only 1 year. Last year's B was very hard for me too back then. I read it now and solved it instantly. I am sure it will be the same way for you. It's actually just playing with prefixes and sufixes.

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    13 months ago, # ^ |
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    Just solved B from last year, DP was definetely overkill.

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13 months ago, # |
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I hope for better problems and better statements than the last year's contest.

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    13 months ago, # ^ |
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    People remember an year old event -- the scar must have been real.

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13 months ago, # |
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For me, I can only hope that the topic is simple. How poor!

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13 months ago, # |
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Here it comes 2-SAT on splay trees of 3D persistent segment trees of dynamic convex hulls tasks.

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13 months ago, # |
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National day of vietnam <3

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13 months ago, # |
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Is there a place in the CodeForces website, showing your division? I know division boundaries don't change frequently, but it would be nice to see.

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13 months ago, # |
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Email says that duration is 2.5 hours but here I see 2

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    13 months ago, # ^ |
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    lol, mail says 7 problems, here it's written 8 problems!!

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      13 months ago, # ^ |
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      Just confirming: there are 8 problems to be solved in 2 hours.

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13 months ago, # |
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I hope for more problems that can be solved by the amazing SSE/AVX algorithm.

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    13 months ago, # ^ |
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    then a matrix multiplication will suffice

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13 months ago, # |
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Manthan in hindi is " मन + थन " which means " mind + boobs ".

So, I guess it means boobs on my mind :P

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13 months ago, # |
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Is it rated for division 3?

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13 months ago, # |
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Best of luck.

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13 months ago, # |
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So is it 2 or 2.5 hours finally?

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13 months ago, # |
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It's my first Manthan,hope to become blue!

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13 months ago, # |
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Can anyone tell what मंथन (manthan) actually means in Hindi? My Hindi is too bad...

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    13 months ago, # ^ |
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    It actually means Brainstorming, but if you're looking for something funny, then the words can be broken to get an another, very interesting meaning. Click

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      13 months ago, # ^ |
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      lol, this can't have been accidental...

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13 months ago, # |
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Hope the problem statement will be too much interesting and funny. :)

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13 months ago, # |
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I didn't see 4000 problem before this round.

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13 months ago, # |
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.

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    13 months ago, # ^ |
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    It probably pertains to when the height of tree is at least 4, such that there is an erroneous nesting while doing the traversal. For e.g.

    INPUT:

    6
    1 3
    1 2
    2 4
    3 5
    4 6
    1 2 3 4 5 6
    

    EXPECTED OUTPUT: Yes

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13 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For Problem D, what should be the output for following input:-
4
1 2
1 3
2 4
4 2 1 3

Should it be "Yes" because as stated in announcement, a[1] is not necessarily 1?

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    13 months ago, # ^ |
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    Announcement states that a[1] is not guranteed to be 1 in tests.

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    13 months ago, # ^ |
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    I think so

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    13 months ago, # ^ |
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    According to the first step in BFS algorithm, vertex 1 is always chosen as root for the algorithm, but a1 is not guaranteed to be 1.

    In this case, answer is "NO".

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      13 months ago, # ^ |
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      Oh Thanks. Reading the announcement carefully, I too think that the answer should be "No". But why did they explicitly announce that? I was initially checking a1 to be 1, and then removed that part of code after the announcement. :(

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        13 months ago, # ^ |
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        They have probably seen the downvote storm in round 505 because of many FST (failed system test) and learnt the lesson from there =))

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13 months ago, # |
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 4 problems under 10 mins, who said 2 hours isn't enough for 8 problems???

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13 months ago, # |
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Hack for D?

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    13 months ago, # ^ |
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    My hacks were:

    200000
    1 2
    1 3
    1 4
    ...
    1 200000
    1 200000 199999 199998 199997 ... 2
    

    (TLE)

    3
    1 2
    2 3
    2 3 1
    

    (WA)

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    13 months ago, # ^ |
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    If a[1] is not 1 but the bfs is correct. That's how I was hacked, at least.

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13 months ago, # |
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How to solve E?

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    13 months ago, # ^ |
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    Let's first consider slow solution. Let deg[v] be the current degree of vertex v. Add (deg[v], v) into set, delete minumum degree vertex while it's degree < k and update degrees of vertices. Answer will be this set's size. Now to support queries, we can just do everything from backwards, instead of adding you will delete edge, and now it is easy to implement in O(n*log(n)).

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      13 months ago, # ^ |
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      Does your solution still work if the graph consisted of 2 disconnected components?

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      13 months ago, # ^ |
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      update degrees of vertices.

      How fast does this update have to be? Because if I understood right, each node is deleted only once, but you have to get to all node's "neighbours" to decrease their degrees before deleting. Wouldn't it be something like O(n^2logn) or O(nmlogn) in total? I know how to optimise it, but can this program pass without any optimisation?

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        13 months ago, # ^ |
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        "Naive" implementation is fine.

        The number of neighbour accesses is bounded above by the total number of edges (ie. m), so it comes out to something like O(n*logn + m)

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    13 months ago, # ^ |
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    Hint: Reverse the process (instead of adding edges from 1 to m, erase edges from m to 1).

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      13 months ago, # ^ |
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      Would be grateful, if you could help with why this code gives Runtime Error on CF compiler.

      Works fine on Local Compiler as well as Ideone.

      https://ideone.com/B8dFH9

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        13 months ago, # ^ |
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        You shouldn't erase while iterating:

        for(auto c:gr[x])
        {
            gr[x].erase(c);
            ...
        }
        

        See this.

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          13 months ago, # ^ |
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          Yup! Thanks a lot mate!

          After all this time, you still keep learning something new :)

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    13 months ago, # ^ |
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    We can answer the queries backwards.

    We build a graph based on the relation, and maintain the degree of each vertices in a set.

    When we answer a query, we remove the vertices with degree < k, and remove their corresponding edges recursively. Then, all vertices has atleast k neighbours. The answer is the number of remaining vertices. After computing the m-th query, we need to remove the m-th edges given in the input.

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13 months ago, # |
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Pretest 4 for D ?

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    13 months ago, # ^ |
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    I got caught there too, I think you have to check (when iterating over guys with depth d-1 as parents) to skip over leaves.

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    13 months ago, # ^ |
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    I think something like traversing nodes in level i in some order but their children in level i+1 are traversed in a different order (node k appeared before node l in level i but children of node l appeared before children of node k).

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    13 months ago, # ^ |
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    Just looking at the levels will not be enough. For example if 4 is the child of 3, and 5 is the child of 2, then 1 3 2 5 4 also comes out to be correct whereas the correct order should be 1 3 2 4 5.

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      13 months ago, # ^ |
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      Basically what this is saying is, if parent(a) is visited before parent(b) then a must be visited before b.

      Am I right ?

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    13 months ago, # ^ |
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    I guess that test might be checking for naive solutions which only check all nodes in one level, which is not sufficient. It is important to check nodes in each depth in groups based on the node on depth — 1. Compare these two cases:

    7
    1 2
    2 3
    2 4
    1 5
    5 6
    5 7
    1 2 5 3 6 4 7
    

    ANS: No , because 3 and 4 have to be next to each other in the traversal path, whereas

    7
    1 2
    2 3
    2 4
    1 5
    5 6
    5 7
    1 2 5 3 4 6 7
    

    the answer to this test is Yes. Unfortunately, I didn't finish my solution in time...

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13 months ago, # |
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Can someone please tell me what is wrong with this solution: 42389932

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I'm trying to submit now, but it redirects me to this blog. Why?

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    13 months ago, # ^ |
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    You'll have to wait until system testing is done. Only then the problem will be added for practice.

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https://pastebin.com/71wMF7nP Could someone help me figure out what's wrong> failing pretest 4.

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    13 months ago, # ^ |
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    7

    1 2

    1 3

    2 4

    2 5

    3 6

    3 7

    1 2 3 4 6 5 7

    Your code says "yes" to this, while it should be "no". Children of a node in BFS would be printed together. So, correct order is 4 5 6 7. or 6 7 4 5. Depending on 2 3 or 3 2.

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Can anyone help why does this code give R.E on CF? For Problem E.

Works fine on Local Compiler as well as Ideone.

https://ideone.com/B8dFH9

[Resolved]. I was iterating and deleting the set simultaneously.

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13 months ago, # |
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Great problem statements! No fluff and extreme clarity.

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How to solve F?

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    13 months ago, # ^ |
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    The problem is equivalent to counting for every vi = 1 + i * (k - 1) the sum .

    Assume that with two equal numbers, the one with higher index is larger.

    To solve this, find for every i the values prev[i] and next[i], where prev[i] is the minimum index i such that maximum value in [prev[i], i] = val[i], and next[i] the maximum index such that the maximum value in [i, next[i]] = val[i]. Now:

    Let cou[i] =  the number of intervals [a, b] such that len([a, b]) = vj for some j and prev[i] ≤ a ≤ i ≤ b ≤ next[i]. We can calculate this with inclusion-exclusion: let count(len) be the number of intervals [a, b] such that len([a, b]) = vj for some j, and 1 ≤ a ≤ b ≤ len. Now cou[i] = count(next[i] - prev[i] + 1) - count(i - prev[i]) - count(next[i] - i).

    The answer is .

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      13 months ago, # ^ |
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      Oh, how can I not figure out how to do the last step.... it is so neat a solution...

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        13 months ago, # ^ |
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        same with me.I don't know why i didnt think of inclusion exclusion after doing the hard part.

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      13 months ago, # ^ |
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      Chic Explanation!

      Also, mango_lassi, your named look quite cool, with yellow/orange :P Now red seems to go a bit awkward. Nevertheless, Congrats on becoming a GrandMaster.

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      13 months ago, # ^ |
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      You have to run this for each i and for each j, right? Then if the sequence is strictly decreasing and k = 2. Wouldn't this TLE? Or do you combine calculation for all j's together?

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13 months ago, # |
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I tried to hack someone but I got "generator crashed". Can someone tell me what that means?

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my D will fail because I forgot about that a[0] condition :(

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    13 months ago, # ^ |
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    Well I explicitly removed the a[0] condition because I misunderstood the announcement, smh

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What's intended solution for H? My O(n log2 n) solution got TLE on test 10. :(

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    13 months ago, # ^ |
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    Could you please share your idea with me? I tried to solve it with Suffix Automaton and Link-cut Trees but didn't manage to finish debugging in time.

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    13 months ago, # ^ |
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    I can suggest the following solution:

    Build suffix automaton for string s. Append symbols to s one by one and answer queries whenever we reach their right border r. To answer the query with string x: feed symbols of x to automaton one by one and iterate over the next (bigger than current in x) symbol. Now we need to check is last occurrence so far of suffix of particular length of current state of automaton  ≥ l. It could be done in the following way: consider the tree of suffix links of automaton, call it T. Then you append new symbol to s, update biggest right border for the state corresponding to the current prefix of s (we must also update right borders for all suffix links of that state, but lets don't do that and ask queries for maximum in subtree of T instead).

    This is in a way I implemented it during the contest (and had wa2 due to some typos). I think it should pass (you will make much less queries than n·A).

    UPD: it passed in 405 ms 42402008

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      13 months ago, # ^ |
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      Thanks.

      In fact my idea is very similar to yours, but I used Link-cut Trees to maintain maximum value in a sub-tree.

      Due to the slow data structure, my solution is .

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13 months ago, # |
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Username russianpig get hacked 6 times in a duration of 13 minutes, which sound suspicious to me.

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How to solve... A?

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    13 months ago, # ^ |
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    solution =

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      13 months ago, # ^ |
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      No, it is not. Your solution would fail if n = 2. log2(2) = 1, while answer = 2.

      real solution = (int)(log2(n)) + 1

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        13 months ago, # ^ |
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        Edited. The wierdest part is that I actually solved the task :P

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          13 months ago, # ^ |
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          Hope it also passes the system tests.

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        13 months ago, # ^ |
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        I think the tricky part is not solving the problem but interpreting how you solved it :D

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    13 months ago, # ^ |
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    Same. I solved B and C, wasn't able to solve A. I feel like an idiot as everyone got the logic within 5 minutes.

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    13 months ago, # ^ |
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    Every number has a primary representation. And never does it happen that you use some power of 2 twice. Thus get all powers of 2 lower than this number! That is (int)log2(n)+1.

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      13 months ago, # ^ |
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      So, again, does this problem actually has something to do with prime numbers? I was dancing around those bastards, and almost got an idea, but it seems that it's just powers of two, right?

      Why tho? Some link from that "acquisition" (Aha!) effect is missing and I don't understand which exactly.

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      13 months ago, # ^ |
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      I really don't get the intuition behind using powers of 2. Why powers of 2 only, Why not other numbers ?

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        13 months ago, # ^ |
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        Because every number can be written as sum of powers of 2.

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          13 months ago, # ^ |
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          But it can be written as powers of 3 too.

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            13 months ago, # ^ |
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            No. Try writing 6 as sum of powers of three using each power of three only once.

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            13 months ago, # ^ |
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            How many power-of-three packets do we need to cover all numbers up to 222(base 3)(It is 26(base 10))?
            
            The answer is 6. Two 1-coin-packets, two 3-coin-packets, two 9-coin-packets.
            So, if we take only power-of-three packets we will need  2*(int)(log3(n)+1) = 6 packets.
            If we take power-of-two packets we'll need (int)(log2(n)+1) = 6 packets.
            
            As you see answer is same.
            
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        13 months ago, # ^ |
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        Because you have 2 options for each packet, either use it or not.
        Edit:
        The idea can be generalized. For example, Say, the question was to find the number of packets required to represent every number upto n, and each packet with value x had 3 optional values (we could only use one option for each packet) -
        1) 0,
        2) x,
        3) 2x.
        Then, we would require log3(n)⌉ packets to denote every number < n.

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          13 months ago, # ^ |
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          Hmm that makes sense. Maybe that's why we can't use powers of 3 right ? Where exactly would using powers of 3 or some other would fail ? I can't seem to imagine.

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            13 months ago, # ^ |
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            if you want base 3 you need 2 pack each size to get 3 choices. so it would be larger no. of packets.

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            13 months ago, # ^ |
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            I edited my comment for a general case, where using powers of 3 would be applicable.

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              13 months ago, # ^ |
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              Finally understood your update. Misunderstood it at first.

              Now that i got it. It refreshed my number systems knowledge with a different view.

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    13 months ago, # ^ |
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    int(Log2(n))+1. By induction, a packet of 1 gives you 1. and for packets 1, 2, 4, ..., n, you know that this combination gives you from 1 to 2n-1. So adding 2n will give you 2n itself, and 2n+1, .... all the way to 2n+2n-1 = 4n-1 = 2(2n)-1

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      13 months ago, # ^ |
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      How do we know that this combination guarantees minimum number of packets. What's so special about powers of 2 ?

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        13 months ago, # ^ |
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        Every number can be represented as its Binary equivalent, so think of 13 as 1101 in Binary, ie one packet each of 1,4,8. Splitting in powers of 2 ensures that every number can be made using at most one occurrence of each power.

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          13 months ago, # ^ |
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          The question was: can you prove optimality for this construction? What if for some n there is a better split than binary?

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            13 months ago, # ^ |
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            The only other way I could think of splitting (in such a way that every number below x could be produced by using at most one occurrence of each packet) is by splitting into multiple packets of 1. I guess these are the only two ways which satisfy the conditions, and among them binary one would use less packets.

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              13 months ago, # ^ |
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              Well, no, seems like there can be many other optimal answers: for example, 5 units can be split into (1,1,3), and (4,3,1,1) make 9. I hope for someone to publish a formal proof.

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                13 months ago, # ^ |
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                Yeah you're right, there could be multiple ways

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                13 months ago, # ^ |
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                yes we can prove that it is optimaly to use powers of 2 as a base to generate 1 to x, consider you have n buckets of number(1,2,4,..2^(n-1)) those numbers can generate 2^n ****different numbers**** and all this numbers are between 0 and 2^n-1 so if we get the binary representation of x and let d is the highest significant bit so let x=1000010101 if we want to generate all number less than x and the highest significant bit is less than d we want n-1 number and these number generate 2^(n-1) different numbers and finaly to get x we only need one another number wish is 2^d and like this you get all number between 0 and x with least size of bucket

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    13 months ago, # ^ |
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      13 months ago, # ^ |
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      How do we know that this combination guarantees minimum number of packets. What's so special about powers of 2 ?

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    13 months ago, # ^ |
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    In Java, it's one liner: out.println(32 - Integer.numberOfLeadingZeros(n));

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      13 months ago, # ^ |
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      Can you explain please ? How's that working? Seems like some play without counting the bits.

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        13 months ago, # ^ |
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        The answer is basically minimum number of bits to represent n (which is also sufficient to represent any value smaller than n). Integer is 32-bit and you count the number of zeroes in the beginning of the binary representation of n, 32 minus that count gives you the number of bits needed.

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        13 months ago, # ^ |
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        2^3=8 now if u want to make any value which is <=8. Than you will need only 2^0,2^1,2^2,2^3. You need each of those 4 value at most 1 time for making each value which is <=8. For example 7 can be made by 2^2+2^1+2^0. We can use this property to solve problem A. Sorry for bad explanation.I can't explain well but I tried my best.

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      13 months ago, # ^ |
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      also, out.println((Integer.toBinaryString(n)).length());

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    13 months ago, # ^ |
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    Here is my intuition. At each step we can generate everything from 0 to x. We start with 0 as we can always generate. The next packet we should add is x+1. Now using all these packets we can generate everything from 0 to 2x+1. If we can't add x+1 as it exceeds n, just add the rest and it will be the last packet.

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13 months ago, # |
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((Sorry for my bad english ㅠ~ㅠ I'm not englisher)) I have two question. 1. I didn't lock during Round. than I can't hack?

  1. What is ?? at page [Standing]? ( http://codeforces.com/contest/1037/standings )

  2. i solve A and B during Round and failed C. than what is in [problems]? ( http://codeforces.com/contest/1037 ) I am now A red-B none-C red

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    13 months ago, # ^ |
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    1. I guess, yes.
    2. Your program is not checked yet.
    3. I didn't understand. I suggest you using a translator at this point.
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    13 months ago, # ^ |
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    it is running system test now(full test cases), your solution will be accepted after pass the system test.

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    13 months ago, # ^ |
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    Just wait a little (15 mins), all the colors will settle and everything will be much clear,

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    13 months ago, # ^ |
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    Thankyou Everyone. I get know about my question!

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13 months ago, # |
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I was wondering why I did well this contest, and then I realized that most of the really good people are at IOI right now.

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13 months ago, # |
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whats wrong with my solution for problem D? I check distances from vertex 1 to a[i] it gives wa4. Who can help? Thanks in advance.

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    13 months ago, # ^ |
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    Check this:

    Input:

    6
    1 2
    1 3
    2 4
    2 5
    3 6
    1 3 2 5 4 6
    

    Output:

    No
    
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    13 months ago, # ^ |
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    I think you maybe do the same thing as me.The childs of a node, will be visited together.
    like this case:
    7
    1 2
    1 3
    2 4
    2 5
    3 6
    3 7
    1 2 3 4 6 5 7
    the answer is No
    and 
    1 2 3 6 7 4 5 
    //the answer is Yes (wrong)
    the answer is No
    and for 1 3 2 6 7 4 5
    the answer is Yes
    test it on your solution.
    
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      13 months ago, # ^ |
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      Output for the second case

      1 2 3 6 7 4 5 , should be "No"

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        13 months ago, # ^ |
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        yes, you are right, maybe second can be : 1 3 2 6 7 45

        thanks for your help

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13 months ago, # |
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In problem D, can the answer be YES if a[1] is not 1 ?

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    13 months ago, # ^ |
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    No. It is told "valid BFS traversal of the given tree starting from vertex 1".

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13 months ago, # |
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can you please tell me what mistake i am making http://codeforces.com/contest/1037/submission/42395202 i am getting WA on test 11.

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13 months ago, # |
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For problem D is it sufficient to check if for the given order, the levels are increasing and the parent indices are increasing?

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    13 months ago, # ^ |
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    Parent indices aren't bound to increase:

    3
    1 3
    3 2
    1 3 2
    
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      13 months ago, # ^ |
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      I'm sorry if there's any other definition of parent indices, but according to what I mean, the parent indices are increasing in this case because 3's parent's index is 1 and 2's parent's index is 2. I am referring to the index of the parent in the given traversal.

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        13 months ago, # ^ |
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        Then you seem to be right, plus you'll have to check if a[1] == 1.

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    13 months ago, # ^ |
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    They need to be level wise, as well as in same order as their parents. So, for input like-
    5
    1 2
    1 3
    2 4
    3 5
    "1 2 3 5 4" is wrong, as 2 came before 3, so children of 2 should also come before children of 3.

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      13 months ago, # ^ |
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      Yes. That is what I mean by parent indices should be increasing.

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13 months ago, # |
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Aside from D's pretests, this year's problemset is much better compared to the last year.

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13 months ago, # |
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Easy solution for D: for each vertex sort the neighbour list by the order they appear in the given sequence.

Then perform a standard BFS from node 1 on the tree and check if your BFS sequence is equal to the sequence in the input.

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    13 months ago, # ^ |
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    Used the same approach keeping priority_queue as adjacent list but got hacked. Can you please explain this case, 2 1 2 2 1

    why this should give "No"? mine gives "Yes"

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      13 months ago, # ^ |
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      It is told "valid BFS traversal of the given tree starting from vertex 1". In your example BFS is starting from vertex 2.

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13 months ago, # |
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I was using binary search without sorting the vectors in D...corrected the bug in last 10 seconds,,,couldn't submit it...The world doesn't want to see me purple :(

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13 months ago, # |
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There should be auto refresh of problem statement whenever it is updated.

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13 months ago, # |
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First time I managed to solve 4 problems, feels amazing :D

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13 months ago, # |
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Today I encountered an extremely strange behavior of std::stack. I managed to fail today's F problem because of the vector of stacks (42392892). What is strange: I've got an MLE verdict. After the contest I changed the vector of stacks to the vector of vectors and got AC with 20 MB (42402332). Does anyone know what's wrong. Don't I know something about std::stack? Why does it consume so much memory. Or maybe it's just stupid me and I don't know how to use it?

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    13 months ago, # ^ |
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    Cppreference on deque (stack uses deque as underlying container)

    deques typically have large minimal memory cost; a deque holding just one element has to allocate its full internal array (e.g. 8 times the object size on 64-bit libstdc++; 16 times the object size or 4096 bytes, whichever is larger, on 64-bit libc++)

    Hope that helps

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      13 months ago, # ^ |
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      Wow, at least I've got a valuable experience. Thanks!

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    13 months ago, # ^ |
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    Well...

    "The standard container classes vector, deque and list fulfill these requirements. By default, if no container class is specified for a particular stack class instantiation, the standard container deque is used."

    Why tho?

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13 months ago, # |
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Any hint for my solution 42402189 of problem D ? I got Wrong Answer on testcase 11 .UPD: Accepted

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    13 months ago, # ^ |
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    Your code gives "No" on following input-
    4
    1 2
    1 3
    3 4
    1 2 3 4

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13 months ago, # |
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13 months ago, # |
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Who is CongLingDanPaiShang3k5? LGM in 8 contests. Seems like a fake account, but I wonder whose.

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    13 months ago, # ^ |
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    There's no fake account when it comes to being LGM :)

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    13 months ago, # ^ |
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    Also who is amit_swami ? I have never seen anybody else's CP profile who became red as quickly as this guy. (comparison only among Indians).

    I remember some people showing hate on one of his quora answers which he later deleted.

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13 months ago, # |
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it was a good contest,i could have become candidate master today if i had solved D more fastly

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13 months ago, # |
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Seems like there are three groups of problems in this contest. [A-D], [E,F], and [G,H]

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13 months ago, # |
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1 2 1 5 2 3 2 4 5 6 1 5 2 3 4 6

How is this not a BFS Traversal of the given tree?

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    13 months ago, # ^ |
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    You start at 1, visit the neighbors 5 and 2 and put them in the queue. Then you are at node 5, visit the neighbor 6 and put it in the queue -> Error

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    13 months ago, # ^ |
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    Following the BFS,

    • You arrived at 1
    • You push 1's neighbors into the queue using your order
    • Queue is now 5 2
    • You pop queue's head and thus arrive at 5
    • You push 5's neighbors into the queue
    • Queue is now 2 6
    • This means that after processing 2, you will definitely have to process 6, while your sample order says 3.
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    13 months ago, # ^ |
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    6 1 2 1 5 2 3 2 4 5 6 1 5 2 3 4 6 Its hard to understand your given input. Now It is not a BFS traversal. Because 5 is called before 2 so child of 5 should be called before child of 2. So if you call child 5 first than it should be 1 5 2 6 3/4 4/3

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13 months ago, # |
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.

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13 months ago, # |
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I solved D by comparing the precedence of parents of ai and ai + 1.

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13 months ago, # |
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Got TLE in F because cin ( even with sync_with_stdio ) I'll never use cin again in my life.

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    13 months ago, # ^ |
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    You didn't add cin.tie(0), cout.tie(0);, with which your code ACs in 1653 ms. Click

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      13 months ago, # ^ |
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      His code would have got AC if he used >= C++14, probably because of the compiler version.

      I always used cin.tie(0); but I used to think that it made difference only when reading/writing alternately. Omitting it seems to hurt his solution's performance by no more than 100ms on >= C++14, but it hurts just enough on C++11.

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        13 months ago, # ^ |
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        You are right. In C++14 it passed with 1500 ms

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          13 months ago, # ^ |
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          By the way, nice idea of iterating through the smallest side to avoid the math work :)

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13 months ago, # |
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Problem D. a [1] must be equal to 1. I missed it( I'm crying (

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13 months ago, # |
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D problem : Don't know why it is giving wrong answer on 11th test case. My idea is I will keep on matching parent of an array element with front element and when all children of front element get finished I will pop first element of queue and start the whole process with new first element. Someone please look into it. My submission : http://codeforces.com/contest/1037/submission/42406584

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13 months ago, # |
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How two learn trees and graph implementation and algorithms for competitive programming ? Any help would be highly appreciated !

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13 months ago, # |
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Can someone tell me why i am getting this error in my code it is working fine in my compiler 42407188

please help me i am not getting what is wrong

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    13 months ago, # ^ |
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    Didn't test or really check the code but i can see right away that your MAX is wrong, it should be twice as that, and since you got an overflow error that might be the cause

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13 months ago, # |
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why this testcase for D should be NO? Can someone explain?

6

1 2

1 5

2 3

2 4

5 6

1 5 2 3 4 6

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13 months ago, # |
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Can somebody tell me what is wrong with my code (42403836) for 4th problem.Actually I am getting WA on test case no.11 which is a large test case that is why I am not able to debug it.

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    13 months ago, # ^ |
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    Your code gives "No" on following input-
    4
    1 2
    2 3
    3 4
    1 2 3 4

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    13 months ago, # ^ |
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    You overwrite k and that's why check correctness only on even levels. I'd rewrite your code with a simpler single counter loop, sets are an overkill. Here's another hack:

    7
    1 2
    2 3
    2 4
    3 5
    3 6
    4 7
    1 2 3 4 5 6 7
    
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      13 months ago, # ^ |
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      HI,shrubb i improve my code now i am not overwriting k,now your test case is also working fine but still getting WA on test case no. 11 , 42416659

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13 months ago, # |
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In problem E Trips , If we had to find the largest group of friends which could go for a walk, Given the condition that at most 1 group can go for walk. (Extra condition added to the question is that all people going on walk should be connected by some links (possibly greater than one link) , the at least k friends of each person condition also remains there).

Then how can we solve this question?

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    13 months ago, # ^ |
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    You said that at most one group can go for a walk but think about it. If you had 2 groups that matched the criteria and you combine them, isn't the new group still correct? :) And about how to solve it. Think of the problem as reversed. If you know the group after all the m edges (friendships) are made then you can substract one edge at every step and calculate the new group. If you want the detailed solution you can look at the editorial but I suggest trying it yourself before.

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      13 months ago, # ^ |
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      It is not necessary that we can combine two groups. The intersection of the two groups can be a null set. And they both independently would be satisfying these conditions.

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        13 months ago, # ^ |
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        You can always combine 2 groups.

        The rule is that if someone is in a group he has to have at least k friends in the same group.

        Now if we have 2 valid groups and we combine them, every person will still have at least k friends because he already had those k friends in the small group.

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          13 months ago, # ^ |
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          Yaa right but if the question was to find the largest connected friends group. (The largest among all small group) then how can we find it?

          We cannot combine 2 groups if they dont have any common person.

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13 months ago, # |
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But Is It Rated?