Educational Codeforces Round 37 [Rated for Div. 2]
By BledDest, history, 7 years ago, translation, In English

Hello, Codeforces!

On February 02, 17:35 MSK Educational Codeforces Round 37 will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

As usual, the round will be rated for Div. 2. It will be held on extented ACM ICPC rules. After the end of the contest you will have one day to hack any solution you want. You will have access to copy any solution and test it locally. You will be given 7 problems and 2 hours to solve them.

The problems were prepared by Mikhail awoo Piklyaev, Vladimir vovuh Petrov, Adilbek adedalic Dalabaev and me. One of the problems was proposed by unbelievable02.

Good luck to all participants!

I also have a message from our partners, Harbour.Space University:

We are happy to announce the addition of two new coaches to our Hello India x Russia Programming Bootcamp: Gleb GlebsHP Evstropov, Artem VArtem Vasilyev and Filipp DPR-pavlin Rukhovich will all be attending at our India location, Amrita School of Engineering.

We look forward to seeing you all there this spring! For those of you who haven't registered, there's still time.

The Early Bird discount will be set at 20% for those who register before February 12th, 2018.

If you have any questions we can help you with, please connect with us:

Phone number: +34 674 291 422 Email Address: [email protected]

UPD. Editorial is here.

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Comments Comments (187)
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7 years ago, # |
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:)

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    7 years ago, # ^ |
      Vote: I like it -44 Vote: I do not like it

    Hello every gays! Please say**"China No.1"**, or I will kill you by my assist in the PBUG!

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7 years ago, # |
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I like BledDest's contests.

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7 years ago, # |
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please make sure that the contest is rated for div-2 , sometimes it becomes unrated for some statement/data system faults , please check them again

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7 years ago, # |
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3rd rated in this week <3

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7 years ago, # |
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Sadly, it coincides with Hackerrank's HourRank 26 :(

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    7 years ago, # ^ |
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    Hackerrank's HourRank 26 coincides with this round(there is a difference)

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7 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Once upon a time, educational rounds used to be unrated.

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7 years ago, # |
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guess who forgot to thank Mike

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7 years ago, # |
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Is it rated?

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    7 years ago, # ^ |
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    Are u retarded??

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      7 years ago, # ^ |
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      It doesn't take L to figure out that he is indeed retarded lol.

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    7 years ago, # ^ |
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    Of course. It is rated for div.2.

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7 years ago, # |
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System Tester halyavin haven't registered yet.:)

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    7 years ago, # ^ |
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    He has :) Already 100+ hacks

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    7 years ago, # ^ |
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    halyavin has 187 successful hacks and 15 unsuccessful hacks, and he has solved all problems in this contest.

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7 years ago, # |
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clash with HourRank on HackerRank .

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7 years ago, # |
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Woooow!!!

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    7 years ago, # ^ |
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    VK Cub 2018... VK Cub 2018 everywhere

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    7 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    I hope they add contests between Feb 7 and March 10 lol. One month is too long a wait for a contest.

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      7 years ago, # ^ |
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      Got a bit more contests than I expected :P. Love CF <3

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    7 years ago, # ^ |
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    Wow shit! VK Cup 2018 Wild_card round 2 has length is a week.

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +6 Vote: I do not like it

    There are so many contests. I like VK Cup very much, but they are too late for me(23:35,UTC+8).

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7 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Since CF made Educational round rated ,"That 24 hour long waiting for standings and staying at your toes" makes it so exhausting to participate in a round :'( .

Anyway that's the beauty :D .

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7 years ago, # |
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Hello Codeforces, how can I suggest new challenges for Educational Round !!!

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7 years ago, # |
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It will be the best Educational Round!!! Because, in list of authors have unbelievable02!!!

;)

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7 years ago, # |
  Vote: I like it -14 Vote: I do not like it

নিরাপদ সড়ক চাই

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7 years ago, # |
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The contest will be rated for me if I only do hacking?

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    7 years ago, # ^ |
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    I thought you couldn't hack without lock (and submit)? Has something changed?

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      7 years ago, # ^ |
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      In Educational, after that 2 hours for solving, hacks are open for 24 hours. What I was asking was if I register but not solving anything and just do some hacks in that 24 hours period, will my rating be modified.

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        7 years ago, # ^ |
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        Your rank gets modified indirectly... If you hack a person who is ranked above you, your rank decreases. So the more hacks the better.

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    7 years ago, # ^ |
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    No, hacks gives nothing

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    7 years ago, # ^ |
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    No. If you didn't register, it isn't rated for you.

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7 years ago, # |
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wow more than 7200 registration..:)

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7 years ago, # |
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wish you high rating~

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7 years ago, # |
  Vote: I like it +4 Vote: I do not like it

F is very similar to gss4 on spoj.

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7 years ago, # |
  Vote: I like it +54 Vote: I do not like it

halyavin Please hack my A,B,C.

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

LOL, solved F but cant solve C XD

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    7 years ago, # ^ |
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    how did you solve F ? I tried to use Segment tree but couldn't update or get divisors in short time .

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      7 years ago, # ^ |
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      D(x) decreases very fast. And if x <= 2 we can dont update this position. So i have 2 segment trees and precalced all d[x] in NLoG :

      for(i=1;i<MAX;i++) for(j=i;j<MAX;j+=i) dp[j]++;

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        7 years ago, # ^ |
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        yes but how to update a range using different values . I don't think lazy propagation can work .

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          7 years ago, # ^ |
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          In this problem you can use set instead of range update. It will be very fast because d(x) decreases fast too

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          7 years ago, # ^ |
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          Lazy propagation can be applied by keeping track of the number of time you want to apply D(ai) operation on the range. For each range [l : r] store the numbers who have values > 2 because D(2) = 2 & D(1) = 1 and it's redundant to apply update operation on these 2 numbers. so when you are updating the range of numbers, pass the value of Lazy[range] as argument so that you can apply D(ai) operation on numbers > 2 in the range Lazy[range] of times and then return the updated numbers in the range. i hope this helps.

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        7 years ago, # ^ |
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        I understand that this solution will run fast enough, but how to analyze its running time complexity?

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          7 years ago, # ^ |
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          I think that x has max log3(x) dividers, so complexity is about qloglog3(max(ai))

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            7 years ago, # ^ |
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            , O(k1 / 3) in practice. Then we perform at most operations per index, resulting , is that it?

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              7 years ago, # ^ |
                Vote: I like it +8 Vote: I do not like it

              Yes, you are right

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                7 years ago, # ^ |
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                Actually, the O(k1 / 3) factor should be replaced by the maximum number of times you can make ai <--- d(ai) before ai becomes 1 or 2. For ai ≤ 106, this number is around 4.

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    7 years ago, # ^ |
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    wow lot of coder cant solve F,

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7 years ago, # |
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So, what's the testcase 18 in problem D???

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7 years ago, # |
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is the idea for G to use inclusion exclusion for each query observing fact that there will be atmost 8 distinct prime factors.

Or any better??

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7 years ago, # |
  Vote: I like it +14 Vote: I do not like it

problem E : 190E - Counter Attack

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7 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

In problem F what is maximum number of updates on ith index till value of a[i] becomes equal to 2,if not initially equal to 1.

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7 years ago, # |
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Spend an hour in Question B, wrongly understood the question.

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    7 years ago, # ^ |
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    I Got WA just because of an initialization. Really feels bad :(

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7 years ago, # |
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Idea for E ?

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    7 years ago, # ^ |
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    My idea is, first form a graph with the non edges. Insert every element from 1 to n into the set. For every element from 1 to n if it is in the set perform the dfs. DFS: In dfs we take a vertex and then iterate through all the remaining elements in the set and check if there is no edge between this vertex and the vertex in the set. if yes, continue. else delete this vertex from the set. Now keep track of all the elements you deleted from the set. Main: Now for every element you pushed into the vector or deleted, perform the dfs. Now this is one connected component. Now repeat the procedure for the remaining elements. Here is the solution https://ide.geeksforgeeks.org/0JuIOaiKWO But i dont know if it is correct.

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      7 years ago, # ^ |
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      i tried to do same thing as you told.can you please check my code 34866232 ?

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        7 years ago, # ^ |
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        What i think is 1: You are not erasing from the set. So it will be O(n2); You need to erase in the dfs. And actually it is not dfs. I gave name as dfs cuz i din think of any other name. You are doing it by brute force. As you are calling for each vertex. Just check my code https://ide.geeksforgeeks.org/0JuIOaiKWO.
        2: You are using map. it has O(logn) while vector has O(1).

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    7 years ago, # ^ |
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    i) n < 2000 Just solve using BFS/DFS

    ii) n >= 2000 We can easily know that there is a largest connected component and small size connected components if n >= 2000.(Since eliminated edge is smaller than 200000)

    except largest connected component, let size of connected components = k Then (n-k)*k edge must eliminated to make that cc becomes alone. Since n >= 2000, k is smaller than 150.

    So first we think set of all vertex which has more than n-150 eliminated edges. Then that set's size is smaller than 150.(roughly)

    BFS/DFS on that set.

    It is my idea but i cannot sure it is right. Because I'm failed to implement.

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +4 Vote: I do not like it

    Assuming node X has anti-edge with {u1, u2, ..., uk}. Then, there are two thing we need to handle:

    1. Union X with u1, u2, ..., uk.

    2. Union every node inside each of the following segment: [0, u1 - 1], [u1 + 1, u2 - 1], ..., [uk + 1, $n$].

    Which is reduced to this problem from VK Cup 2015 Final.

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    7 years ago, # ^ |
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    If the component graph is not connected , then the original graph is connected => number of connected component is 1. else , start bfs in the complement graph in a different manner , start from a vertex push in the queue all the unreached vertices and continue like this and for each node we use DSU to count number of sets which in result is the number of connected components.

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7 years ago, # |
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can someone plz tell me why this solution doesnt work ? https://ide.geeksforgeeks.org/eN7gAAmHB0

I gave this solution just 5 min after the contest. but it got rejected on test 4. Althouh i solved using queue and bfs after 15 min, i still cant find what is wrong with this.

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7 years ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

34860361

sorry for my mistake :(

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    7 years ago, # ^ |
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    Sum of all n of all the test cases is <=1000

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7 years ago, # |
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Shit! "cnt x y" I thought that we get water from y then pour to x :(

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7 years ago, # |
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Will unsuccessful hack result in decrease in points ?

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7 years ago, # |
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I know that to solve problem F,we are able to use segment tree and I can't solve this problem.I need someone help me to solve it.Need we use some skill?

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    7 years ago, # ^ |
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    check any accepted code.

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    7 years ago, # ^ |
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    My solution: build 2 segment trees, one to maintain the sum of elements in segments of array a, one to maintain the count of floor numbers in segments of array a.

    Here, let's say that floor numbers are numbers which D(x) = x. (People say only 1 and 2 satisfied that rule, but I'm not so certain about that, so, just check the criterium manually — it won't cost much actually ;) ).

    You can do both REPLACE and SUM queries without lazy propagation — simply traverse throughout the trees. However, to avoid TLEs, you can skip traversing if all elements in a segment are floor numbers.

    For example, if a segment is from the 3rd to 6th element, and the floor numbers count is 4, you can skip the following traverse (since these numbers can't be replaced by any distinct integers anyway). If you're working on a REPLACE query, simply ignore it and break, otherwise, add the sum of elements in the segment represented by the node, then break the current traverse.

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      7 years ago, # ^ |
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      Or you can notice that each element will be processed no more than times and do type one operations explicitly.

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        7 years ago, # ^ |
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        I'm not sure if I get what you meant, can you clarify more?

        I knew the maximum processes possible for an element, this is why I said I don't need lazy propagation for my segment trees, just traversing with worst complexity O(segment_length) (but the average complexity is whole lot lower thanks to the notice of no-longer-processed-elements in a segment).

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

I see that it said Accepted rather than Pretests Passed for this contest. So does that mean there are no Systests? Not quite sure as its my first Educational Round.

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    7 years ago, # ^ |
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    For educational rounds system tests will be run which include the hacks as well.

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7 years ago, # |
  Vote: I like it -14 Vote: I do not like it

the very tough contest... unclarity with the statement

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7 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Submitting a problem was harder than solving it!!

Please repair the server

P.S. posting this comment as well

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7 years ago, # |
  Vote: I like it -17 Vote: I do not like it

Submitting a problem was harder than solving it!!

Please repair the server

P.S. posting this comment as well

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7 years ago, # |
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in E can there be ever more than 100000 components?
I think for that you have to provide something like 100000*(100000-1)/2 forbidden edges.
I actually overlooked the constraints.
If there can be more than that components then there is a free hack for You.
submission link

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7 years ago, # |
  Vote: I like it -18 Vote: I do not like it

Submitting a problem was harder than solving it!!

Please repair the server

P.S. posting this comment as well

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    7 years ago, # ^ |
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    I had no problems. Maybe some latency issues on your side. Check by pinging.

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7 years ago, # |
  Vote: I like it -35 Vote: I do not like it

E and F are copied. Make the round UNRATED.

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7 years ago, # |
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Why do div1 users have predicted rating on cf rating predictor? Will this mess up the prediction?

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7 years ago, # |
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That moment when you have the idea to solve F (failed during contest, but finally got it AC-ed after), and you completely have no ideas in D and E...

Well, with the insanely poor performance in 3 first problems, goodbye rating (again)...

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7 years ago, # |
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Any idea why this is not working ? https://cf-predictor-frontend.herokuapp.com/contestSelector.jsp

It was working the previous educational round even during the hacking phase.

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7 years ago, # |
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That moment when you finish solving E 30 seconds after contest ended :))

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7 years ago, # |
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Try to hack me, please.

This is my submissions page. :)

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7 years ago, # |
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Why is this invalid input for B? I don't see any constraint that is not fulfilled

1
2
5000 5000
5000 5000

edit: Validator 'validator.exe' returns exit code 3 [FAIL Integer parameter [name=l_1] equals to 5000, violates the range [1, 2000] (stdin)]

why 2000???

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7 years ago, # |
← Rev. 3   Vote: I like it +5 Vote: I do not like it

halyavin is coming !

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7 years ago, # |
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System testing will be done by halyavin

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7 years ago, # |
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System Tester — halyavin is on fire !

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7 years ago, # |
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can someone explain the Idea of D , Please?

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    7 years ago, # ^ |
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    given that if you sum all the tanks water in one tank, then you can take whatever you want water in amount multiple of k. now, the problem is to bring the (V%k) amount. make a knapsack from all the tanks on an array of all values under K. The crucial idea is that you maybe need to module over k when you are summing tanks values in knapsack process.

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7 years ago, # |
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I have a silly (post-contest) solution for problem E.

First of all, it is mathematically provable that if G is non-connected OR if diam(G) >= 3, then !G is connected (where !G is the complement of G). So if any of the above hold for !G we draw the conclusion that G is connected. We quickly eliminate these cases.

Now we are left with only cases in which !G is a connected graph with diam(!G) <= 2. If we analyze it, we can draw the conclusion that the original G is almost connected (i.e. it has one very big connected component). We want to brute-force the answer for the original G, but we can't do that yet (it might have a large number of edges). We choose some number (O(M)) of random edges from G and brute-force on this partial graph. We remove all nodes in the largest component and all nodes adjacent to nodes in the largest component (we remember the size of this component). Finally, build the sub-graph of the original G which has only the remaining nodes. Brute force on this sub-graph and output the answer + the size of the big component we discovered before.

This is the submission: 34873106

It seems like this always give the correct answer and the run-time shouldn't be too high provided that we discover a large enough part of the big component during the brute-force on the partial graph. I'm curious if the idea is hackable.

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    7 years ago, # ^ |
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    It seems to work (get AC) without the initial elimination (I guess the condition that one very big component exists holds anyway for larger graphs), but it might be easier to hack with RNG prediction. Maybe.

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7 years ago, # |
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Will a successful hack be considered valid for me in any of these cases ?

  • Hacking a solution which was submitted after the contest time.
  • Hacking a solution for a problem that I didn't solve during the contest.
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7 years ago, # |
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Editorial?

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    7 years ago, # ^ |
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    It will be posted after the hacking phase and the system testing are over

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7 years ago, # |
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This is correct solution for Problem B. But when I use sort function, it gives me wrong answer on Test 5. I am not able to figure out the reason. Please Explain.. (http://codeforces.net/contest/920/submission/34874806)

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    7 years ago, # ^ |
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    The problem in your sort function is that it compares the data only based on the L values, that fails in case of two students with the same L.

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      7 years ago, # ^ |
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      But my 4 test passed in which same value of L exist. The sort function manage the same order if Both L are equal. Still Not get it.

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        7 years ago, # ^ |
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        I'm not sure how exactly it works, but I submitted your solution after changing the sort function and got AC.

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        7 years ago, # ^ |
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        The sort function manage the same order if Both L are equal.

        Not necessarily. See stable_sort

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7 years ago, # |
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How to solve E?

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    7 years ago, # ^ |
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    Read this

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    7 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    search for the node with adjacent nodes fewer than sqrt(n). let it x. there must be at least one , or the overall graph is small. Now, connect all the nodes (which are not disconnected with x) with x. Now, you have a component that has more than or equal (n-sqrt(n)) nodes. brute force on the rest nodes.

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    7 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    using brute force.Now there are (n-1)*n/2-m edges.If n less than 200,it's easy to solve,or we can prove that connected components never exceed 200.so we can sort the edges in order.And using union-find set to merge connected components.This is my submission 34858691.

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +2 Vote: I do not like it

    There is a simple way like this: You use set A to store vetices, set B to store pair of vertices which is not edge. Then we will do dfs, for every vertex u, push all v which are in set A and pair <min(u, v), max(u, v)> doesn't in set B into a vector. Of course all vertices in the vector are adjacent with u. Then you erase all of vertices in vector from A and start dfs from each of them. Complexity: log n for std::set, n for visiting n vertices -> o(n log n)

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      7 years ago, # ^ |
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      Sorry there was a mistake. My solution was accepted so i though it's complexity was o(nlogn). It was o(n^2 logn) for the worst case (for every vertex, we check all the vertices remain in the set, so it should be n^2), but somehow passed pretest (maybe because of the constraint).

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        7 years ago, # ^ |
        ← Rev. 3   Vote: I like it +5 Vote: I do not like it

        No it's not. :)

        Each time, you time you meet a vertex, you either erase from the set ( cost 1), or you don't.

        Notice that you don't only if it belongs to the given E given edges, so you have at most O(E) pairs, and their cost is O(E).

        Totally complexity is linear in number of edges and nodes multiplied by a log factor.

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7 years ago, # |
  Vote: I like it -6 Vote: I do not like it

In CF Educational R 37 in Problem C, I had written the following code, where I have done cumulative sum of 1's and checked if all bits in the string between i and arr[i]-1 are 1 (& if arr[i] < i, then between arr[i] & i-1).

It is failing for Case #5 and the test case is of length 200,000 !

I have spent close to 5 hours to figure out why it is failing :P. It will be really helpful if someone can help me figure this out (maybe provide a simple test case !)

http://codeforces.net/contest/920/submission/34883063

include

include

include

include

using namespace std;

int main(){ int n; cin>>n; vector arr(n); for(int i=0;i<n;++i){ cin>>arr[i]; --arr[i]; }

string str; cin>>str;

vector<int> cum(n-1);

cum[0]=str[0]-'0';

for(int i=1;i<n-1;++i){
    cum[i]=cum[i-1]+str[i]-'0';
}
int f=0;
for(int i=0;i<n;++i){
    int &val=arr[i];
    if(cum[val-1]-cum[i-1]==val-i) continue;
    else { f=1; break;}
}
if(f) cout<<"NO"<<endl; else cout<<"YES"<<endl;

return 0; }

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

Could anyone provide any small but tricky test case for problem F? I am having trouble understanding the bug with my logic/implementation ( submission link: 34884739 ) and all test cases except #1 seems to be intractable.

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

Can someone explain the DSU solution for E, in details? An example would be really nice. Thanks.

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Looks like it is legal to transfer water from tank to itself in problem D. Is that supposed to be allowed?

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    7 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    The output format doesn't forbid that.

    Unfortunately, the word "other" wasn't supposed to be in the statement.

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7 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Could someone explain why am getting TLE on E with O(m·log(n)) algorithm?Here is my code 34891034 and here is explanation of algorithm and proof of complexity:So I will use set to save nodes that are adjacent to current node,and not marked yet.Note that with BFS I only need one set ,every time node is visited I delete it from set,and when it's turn to traverse adjacent nodes for node A we will delete all nodes that aren't adjacent with A and after that we will add them back.Proof of complexity:Deleting and adding nodes that aren't adjacent to current node will take at most O(m * log(n)) time where constant factor is 4 because we will add and erase those nodes.Note that we will iterate exactly once for every node in set possible so that part is linear.So maximal number of operations is 4 * m * log(n) + n < 15 * 105.What I got wrong?

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7 years ago, # |
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I see that there is a few people on the top 50 got their solution hacked, but the leaderboard still show that their solution is accepted.. is there something wrong?

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7 years ago, # |
← Rev. 2   Vote: I like it +18 Vote: I do not like it

Few hours ago halyavin did 190 hacks.
Now it shows 187 :/

Also, why is eddy1021's rank shown as 1(3) in common standings?

Is there any issue?

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7 years ago, # |
  Vote: I like it +7 Vote: I do not like it

To those who solved E.Connected Components, how do we use DSU??

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

Right now maybe Codeforces is having some issues. It's very slow, I can't submit my solution and (maybe) the leaderboard in Educational Codeforces Round 37 is showing the results of about 10 hours ago.

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7 years ago, # |
  Vote: I like it +24 Vote: I do not like it

And now we have Internal Server Errors on the hack and submission pages.

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

Is there no penalty for wrong submissions in this contest ?

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7 years ago, # |
  Vote: I like it +36 Vote: I do not like it

In statement of problem 920E - Connected Components? it is said "denoting that there is no edge between x and y. Each pair is listed at most once; (x, y) and (y, x) are considered the same (so they are never listed in the same test)". But in test 85 we can found edges 1 2 and 2 1 in one test. Why?

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7 years ago, # |
← Rev. 2   Vote: I like it -43 Vote: I do not like it

Problem E is not new : http://codeforces.net/contest/190/problem/E Also in the statement they say that "Each pair is listed at most once" but in test 85 they are not ! This round should be unrated!

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +7 Vote: I do not like it

    oh i see you're in edge of falling back to cyan. (and didn't solve E though ). I think the problem E should be rejudged not unrated.

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      7 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      I think they should rejudge E regardless.

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      7 years ago, # ^ |
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      did you read the problem "Counter attack" ? They are the same !

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        7 years ago, # ^ |
          Vote: I like it +19 Vote: I do not like it

        Well, did you ever try just solving problems instead of trying to google them during the contest? :D

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          7 years ago, # ^ |
            Vote: I like it +2 Vote: I do not like it

          Come on! if a problem is copied from another, does this not affect the rating?

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            7 years ago, # ^ |
              Vote: I like it +13 Vote: I do not like it

            You're partially right, ya. But this claim is correct only presuming many people have seen this problem beforehand.

            It would be kinda tough to persuade me that all div2 participants remember the problem that took part ~350 rounds ago :D

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

            Its an educational, its literally written in the description that problem might look similar to some in the past..

            p.side.note: this drafts button needs to be placed somewhere else.

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              7 years ago, # ^ |
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              I'am against that for a rated contest the problems could look similar to some in the past.

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      7 years ago, # ^ |
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      Agree with you. Rejudge the problem E and not unrated.

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      7 years ago, # ^ |
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      Now he solved E ;)

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7 years ago, # |
  Vote: I like it +13 Vote: I do not like it

the 85th test case of E is wrong. 7 20 4 6 6 7 4 5 1 2 2 4 1 7 3 5 2 1 6 2 6 1 7 3 3 2 3 6 3 1 3 4 2 5 1 6 7 4 6 3 7 5

(6,3) and (3,6),(6,1) and (1,6),(1,2) and (2,1) are all listed but (x, y) and (y, x) are considered the same (so they are never listed in the same test). will rejudge?

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7 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

Repeated later.

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

Here, in the latest hack for problem E, the verdict to this hack had to be "invalid input" instead of processing it.

And since this is the last test, it's pretty easy to accepte all the solutions which failed on this test and not make it unrated.

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7 years ago, # |
  Vote: I like it +35 Vote: I do not like it

Unfortunately, test 85 in problem E was invalid.

I don't have any permission to rejudge the whole problem, so I tried to find all solutions that failed on this test and rejudge them manually. If your solution still fails on test 85, please tell me so I can rejudge it.

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    7 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Filter the status page by setting the problem section to that problem, and the verdict to "WA", and the test to ">= 84", it may help.

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      7 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      I did exactly this, but maybe I missed something?

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        7 years ago, # ^ |
        ← Rev. 2   Vote: I like it +1 Vote: I do not like it

        True.

        And I am wondering, was that hack state the only one which was processed instead of telling that it had "invalid input"?

        If there are other ones it will be harder to rejudge manually.

        Is it efficient to rejudge only the hack states that successfully passed? so it will detect the wrong tests.

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        7 years ago, # ^ |
        ← Rev. 3   Vote: I like it +1 Vote: I do not like it
        Sorry, repeated.
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7 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

When will the system tests start?

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7 years ago, # |
  Vote: I like it +16 Vote: I do not like it

When will halyavin finish testing us?

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7 years ago, # |
  Vote: I like it -15 Vote: I do not like it

Why the FUCK are the ratings not updated still??

I could have asked politely but the system testing has been over for over 4 MOTHERFUCKING HOURS and i am going FUCKING CRAZY !!!!

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

E — Connected Components?

is it possible to use dfs instead of bfs here ? Solution Accepted

if possible then why i get WA when I use dfs here in the following code WA using dfs

shortly my idea: i insert all node from 1 to n in a set/vector then remove it one by one when they are visited by using dfs/bfs.

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

I think the data of problem E in Test#85 is wrong. In the description of problem E, you said " Each pair is listed at most once; (x, y) and (y, x) are considered the same (so they are never listed in the same test). ", but it seems both (3, 6) and (6, 3) are in Test#85.

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7 years ago, # |
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For problem E, My Submission I am getting Time Limit exceeded. Can anyone please help me where I am going wrong?

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

wrong thread i am sorry ):

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