vovuh's blog

By vovuh, history, 3 years ago, In English

1328A - Divisibility Problem

Idea: MikeMirzayanov

Tutorial
Solution

1328B - K-th Beautiful String

Idea: MikeMirzayanov

Tutorial
Solution

1328C - Ternary XOR

Idea: vovuh

Tutorial
Solution

1328D - Carousel

Idea: MikeMirzayanov

Tutorial
Solution

1328E - Tree Queries

Idea: MikeMirzayanov and vovuh

Tutorial
Solution

1328F - Make k Equal

Idea: MikeMirzayanov

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

Could someone please elaborate the tutorial for question B? Specifically, from where did we get the equations k<=n-i-1 and n-i-1?

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    3 years ago, # ^ |
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    Starting at the end, for every index, let's imagine some string such that it looks like aaaaaabb.

    Now imagine the left-most b moving to the left. For each move to the left, the right-most b has one more possible location. For example, aaaaaabb has 1 position for the second b, aaaaabab has 2, and so on.

    This remaining k value can also help us calculate the position of the right-most b because of the fact that it's remainder once it is less than n-i-1 hints at where it should go.

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      3 years ago, # ^ |
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      Got your point. Thank you! :)

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      3 years ago, # ^ |
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      I still didnt got your point. How we are forming these equations k<=n-i-1 and n-i-1?

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        3 years ago, # ^ |
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        Consider a string s of length n. First, number the positions of the letters from 1(rightmost letter) to n(leftmost letter).

        Lexicographically, the smallest string will have a 'b' in positions 2 and 1. It is easy to observe that there will be i - 1 strings having leftmost letter in the ith position. So, all you need to do is k -= (i - 1) as long as k > 0 and i goes from 2 to n. Suppose, j is the smallest value of i for which you weren't able to perform this operation. That means the leftmost b will be at position j. Since, this is j from the right, it will be equivalent to n - j + 1 from the left. This will be s[n - j].

        Now, if you would have k = 1 then rightmost b will be at position of 1. In general, for some remaining value of k, the rightmost b would be at position k. So, that would mean n - k + 1 from the left. That would mean s[n - k].

        For eg: n = 7 and k = 5. You can do k -= 1 and k -= 2 to get k = 2. j will be 4. That means s[7 - 4] = 'b'. Also, if k = 2 then s[7 - 2] = 'b'. Thus, s = "aaababa" for k = 5 and n = 7.

        I hope this helps. It will be easier to understand it if you work it out a bit more on paper with examples of your own.

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3 years ago, # |
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nice contest :)

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3 years ago, # |
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I am unable to understand third part of D's code . Any hint .. it seemed very easy ,but still i got wa at tc 2.

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    3 years ago, # ^ |
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    you can always get an answer if you color every odd vertex with 1 and every even with 2 but when n is odd then first and last will have same color.so if you have a consecutive pair with same value then you can color both of them with same value which will change parity of all subsequent numbers.EX: 1,2,2,3,4 can be colored as 1,2,2,1,2 instead of 1,2,1,2,1. and if no such pair exist then you have to use 3rd color.

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    2 months ago, # ^ |
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    if else if else else if if else After 100 if else since 10000 "if elses" can be written by newbie, it was considered to be D

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3 years ago, # |
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I had a different approach for 1st problem but it did not accept And when I compile on my idea it worked good.please help me

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    3 years ago, # ^ |
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    I checked your code, the idea is absolutely correct. However, you need to print a newline after every distinct answer otherwise answers for all test cases would be shown side by side. cout<<count_<<endl; Make this change in your code and it should work fine!

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      3 years ago, # ^ |
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      \n is faster than endl

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        3 years ago, # ^ |
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        Yes, but that's a different topic. Since he wrote the previous terms in CPP manner, I suggested using endl. And for this problem, this is not even a big deal!

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    3 years ago, # ^ |
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    Your solution will not work for large cases.

    Let me expand the solution given, i.e how this equation came up b - a%b — let's derive it.

    We know we can write a as a = b * k + a%b where k is the quotient. k can also be written as k = (a - a%b)/b. We are interested to find the next k for which remainder is zero, so we need k+1.

    So, a + c = (k+1)b + 0 where c is the count/number of steps (+1) to add to a to make a%b zero. Substituting k from above to find c:

    a+c = b * ((a- a%b)/b + 1)) = (a - a%b) + b

    a+c = a - a%b + b

    c = b - a%b. There it is!

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3 years ago, # |
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The constraints for problem B were weak many o(k) solutions passed. Example — https://codeforces.com/contest/1328/submission/74509641. I tried to hack it but can't

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3 years ago, # |
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I did B in O(1) per test case

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    3 years ago, # ^ |
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    Printing the string is O(n)

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3 years ago, # |
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Beautiful approach for E!

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3 years ago, # |
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Is there a name for the technique used in E? Or where can I read more on it (formal proofs, etc.)?

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3 years ago, # |
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I have a different approach for B.
We know that after n*(n-1)/2 permutations, both b's will be together. So we can basically make an array of all such numbers (1,3,6,10,..),and for a given k, we can find an n s.t n*(n-1)/2<=k (by binary search). Let j be the pos of that n. We can clearly see that for this value the no. of a's after the 2 b's will be exactly j (i.e, we can determine that particular permutation).
So, from that permutation, we can lexicographically check next permutations till we reach k and print that kth permutation.

C++ soln:74461692

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3 years ago, # |
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" Write some LCA algorithms and other hard stuff or write about 15 lines of code and solve the problem " — One of the best editorial (E problem)!

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3 years ago, # |
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I believe E can be solved by just maintain an array of vertex while do dfs. When you enter a vertex mark it and all it son in the maintained array, when you leave a vertex remove all it sons from the array. Now we can answer the queries in the order we visit the vertexes in the dfs. When we enter to a vertex v , we know exactly which vertexes appear on the path to it or in dist 1 and we can answer all the queries where u = v.

My mistake, I changed the problem in my head and made U alway the last vertex of the query.

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    3 years ago, # ^ |
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    • when you leave a vertex remove just all it sons from the array but not the vertex you just left.
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3 years ago, # |
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I don't understand the solution for E after it explained how to figure out if vertex u is parent of vertex v. Can somebody explain it please ? (I don't want the explanation for checking if one vertex is ancestor of another, I want explanation for AFTER it).

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    3 years ago, # ^ |
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    1. if Vertex A is ancestor of Vertex B, on the way of reaching B we will cross A

    2. In dfs(), we mark every vertex by its parent. (say u is parent of v, we mark v with u).

    3. On the path to reach the deepest vertex(say vertex x) in each query:

    3a. we mark all vertices with parent vertex 1(because vertex 1 is always on the path , therefore we can reach all of the immdediate children of vertex 1, since they all have distance of '1 unit' from the path )

    3b. we mark all the ancestors of 'x'( cause we will cross them to reach 'x')

    3c. we mark all the immdediate children of these ancestors we found( because we still can reach them as maximum distance from the path to x is '1')

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3 years ago, # |
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My solution for Question D without using Graphs and DP.

https://codeforces.com/contest/1328/submission/74499296

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3 years ago, # |
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The site is too f#cking slow today, it's taking minutes to logout/login, all links are loading really slow, is this normal or is this new??

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    3 years ago, # ^ |
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    because of the quarantine that is in place in a lot of countries a lot of people have more time to practice CP so the site gets a lot more people on it.

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3 years ago, # |
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Anyone else used Quadratic equations to solve the main part of Question B in O(1) for each test case??

My solution :- https://codeforces.com/contest/1328/submission/74439092

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3 years ago, # |
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Can someone suggest some good resources to learn how to make changes in DFS methods amd efficient graph algorithms...like many a times I see dfs1 and dfs2 kimd of functions.... Expecting genuine help!!

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3 years ago, # |
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Dear MikeMirzayanov and vovuh, nice problems. But Someone stole my rating after this round. I was dark blue color, so i should be unrated for this Div 3 !

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3 years ago, # |
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Thank MikeMirzayanov and vovuh for interesting problems and fast editorial!

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3 years ago, # |
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I had used StringBuilders for the problem C. Here is my submission https://codeforces.com/contest/1328/submission/74463191 Can any java coder tell me why did it give a TLE?

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3 years ago, # |
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this is my solution 74533033 to problem C in yesterday's contest. i dont understand what's wrong with it. it seems to work fine. can anyone help?

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    3 years ago, # ^ |
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    I have fixed your solution: 74544016. You need only to resize your strings such as x, a and b.

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      3 years ago, # ^ |
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      thanks:)

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      3 years ago, # ^ |
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      vaaven I generally take whole string as input at once and not by each character. So I have a doubt that while taking character wise input(cin>>s[i]), why resizing the string to n, was not needed for smaller inputs(n~5) as in test cases 1 and 2. Runtime error came in test case 3 for n=250.

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3 years ago, # |
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For problem F, I sorted the array and used the difference between consecutive elements and the number of moves required to make all the numbers before a particular number to make all of them equal to that number. I did this for both prefix and suffix part and only counted the required number of moves. 74541622

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    3 years ago, # ^ |
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    how do you sure that you have " at least k equal elements in the array." in each step?

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      3 years ago, # ^ |
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      for that part i take the sum of steps from both left side(represented by 0) and right side (represented by 1) on a particular element(representing that all n-1 elements are converted to that element )and subtract it by n-k (meaning that these elements are extra element that were converted).
      

      for initialization part i took first k elements that means that first k-1 elements are converted to the the kth element and subtracted the answer with the number of occurrences of the kth element. i also did the same thing for the kth element from the last.

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        3 years ago, # ^ |
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        Why you subtracted the number of occurrences of kth element?

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          3 years ago, # ^ |
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          Suppose

          k = 5,

          Ar = 2 4 5 6 6 6 7 8

          m[6] = 3

          dp[0] = 0 cost for all element to become 2

          dp[1] = (4-2 * 1) + 0 = 2 cost for all element to become 4

          dp[2] = (5-4 * 2) + 2 = 4 cost for all element to become 5

          dp[3] = (6-5 * 3) + 4 = 7 cost for all element to become 6

          dp[4] = (6-6 * 4) + 7 = 7 cost for all element to become 6

          Now m[6] = 3-2 = 1

          our answer is dp[4] — m[6] = 7 — (1) = 6

          How answer is 6 ?

          Keep in mind that we already have three 6's so we need not to convert 5 -> 6

          Final answer = 6 , 6, 5, 6, 6, 6, 7, 8

          Total cost = 6-2, 6-4, 5, 6, 6, 6, 7, 8

          We only need to convert those which are required so if our Kth element is x and we are suxh more x already present ? we can use that too.

          Similarly

          k = 4

          for ar = 1,1,1,2,3,3

          3,3,2,2,3,3

          answer = 5

          We can do this for both sides and select the minimum one

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3 years ago, # |
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Please explain what exactly is happening in the B problem?how are we knowing the position of b by decreasing by n — i — 1?

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3 years ago, # |
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Can someone please help me figure out why am I getting timeout even if my complexity is O(N + sum(K)) for problem E? Here's a link to my solution. https://pastebin.com/y9Fx5day. Thanks in advance!!

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    3 years ago, # ^ |
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    Maxi can be equal n in worst case. So your solution is O(n^2)

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3 years ago, # |
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can somebody please explain how the scoring works in codeforces? start off by explaining what's does it mean by +,+1,+2.

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3 years ago, # |
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Anyone who has done D using DP ? Please share your solution

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3 years ago, # |
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This was my solution for B.

Solution- 74469154

i used the concept of "Sum of first n terms" So if the string size is n then the the total lexicographical strings produced will be sum of first n-1 terms.

for ex- n=5 so total number of lexicographical strings produced will be sum of first 4 numbers 1+2+3+4=10

Then I made a vector pair consisting of lower range and the upper range of k.
1 contains (1,1)  leftmost b will in the position n-2
2 contains (2,3)  leftmost b will in the position n-3
3 contains (4,6)  leftmost b will in the position n-4
4 contains (7,10) leftmost b will in the position n-5

(The concept is For every movement of leftmost b , Following will be the movement of rightmost b till the its position one more than the position of the leftmost b.

For ex-string-abaab the rightmost b will have positions 3 and 4.

)

Rightmost b's position will depend on the distance between k and its lower range
for ex n=5,k=8
5 will lie in the range(4,6)
leftmost b will in the position 1 (1 based indexing)
rightmost b will in the position n- (difference(n-lower range))  ie 5-(5-4)=4
so the string is 'baaba'

Complexity — O(n)

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    3 years ago, # ^ |
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    Me too

    My simple Implementation for problem B

    My Submission

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3 years ago, # |
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Hello and thanks for contest. I was EXPERT before the contest DIV3 but my rate change -20, WHY? Thanks again.

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    3 years ago, # ^ |
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    Similarly I've seen an EXPERT coder get his rating increase for about 20 in the last div.3 round.Maybe it is a bug?

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    3 years ago, # ^ |
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    probably because you had registered yourself for this contest before rating change of last round (when you were<1600)

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3 years ago, # |
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It says time limit exceeded but is the logic correct?

void solve()

{ int n, k;

cin >> n >> k;

int l = n * (n - 1) / 2;

string s(n, 'a');

int b1 = n - 2, b2 = n - 1;

    for (int i = 1; i < k; i++)
    {
       if ((b2 - b1) == 1)
       {
         b1--;
         b2 = n - 1;
       }
       else
       {
         b2--;
       }
    }



s[b1] = 'b';
s[b2] = 'b';
cout << s << endl;

}

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    3 years ago, # ^ |
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    Yes, logic is correct but it works very slow because you have O(k) where k can be equel 1e9.

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3 years ago, # |
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Can anyone tell me what's the reason of getting TLE for the following solution of problem E? https://codeforces.com/contest/1328/submission/74550098

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    3 years ago, # ^ |
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    Max depth can be n. So you will make n iterations to check only 1 node. Your solution work in O(nm)

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    3 years ago, # ^ |
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    Because you iterate over the whole longest path every query, so the complexity may be O(mn) instead of O($$$\sum\limits_{k = 1}^mk_i$$$)
    The testcase may be a linear chain tree with every query contains a far node from the root

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Problem : E My submission : 74555362 Verdict: WA on TC 56 Can anyone tell what is the mistake? I cant figure it out. Used the same method as told in Editorial. My code is very short. Thanks a lot :)

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    3 years ago, # ^ |
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    I fixed your solution:74558555

    I just add this:

    reverse(temp.begin(), temp.end());
    

    and edit this:

    if(temp[i].F>st or temp[i].S<en)
    
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Anyone who has done D using DP ? Please share your solution

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C) Why if x[i]=2 or x[i]=0, then a[i]=b[i]=1 or a[i]=b[i]=0 relatively? Who can explain it? Please.

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    3 years ago, # ^ |
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    because max(1, 1) < max(2, 0) when x[i]=2 and max(0, 0) < max(1, 2) when x[i]=0

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3 years ago, # |
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Can someone please explain this part Of problem E

It lies on this path if the root is the parent of u (it is always true) and u is the parent of fv. This approach can be used for each vertical path (such a path from x to y that lca(x,y) is either x or y).

And this code of problem E

int u = v[0]; for (auto it : v) if (d[u] < d[it]) u = it; for (auto &it : v) { if (it == u) continue; if (p[it] != -1) it = p[it]; } bool ok = true; for (auto it : v) ok &= isAnc(it, u);

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Help needed in E: My approach-: Check if there is path from root containing all the parents. Sort all the parents of given nodes according to their height. Then find lca of every adjacent pair, suppose x and y are adjacent , then if lca(x,y)==x then path exists else not. I am getting runtime error in test case 100. Can anyone help me out in this?
My submission 74567267

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3 years ago, # |
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somebody plz explain B...

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    3 years ago, # ^ |
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    I also didn't know it before, got the solution by generating few cases while contest.Generate 6/7 cases and you will find the pattern.

    • See this case: n = 5; k = 5; "a B a b a"
    • Take minimum integer x so that k <= (x)(x+1)/2;
    • For the first B , pos1 = (n — x)th index from left side.
    • Keep = (x)(x+1)/2 — k;

    • For the second b , pos2 = (pos1 + keep + 1)th index from the left side.
    • Without pos1 and pos2 all positions will be a.
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Editorial solution for E is nice, but there is a solution that requires much less thinking. Handle the queries offline, and then you simply need to do a dfs from the root and add all the neighbors you pass on the way down to your current set (to add to your current set, just mark them as visited in all the queries that contain this vertex. if a query has everything marked as visited, then it is good) and remove the neighbors on your way back up. This way you can just check all the paths in O(n+m) time (fairly high constant but still runs < 1s).

74575633

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    3 years ago, # ^ |
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    can you plz elaborate how you use add and remove in your code

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      3 years ago, # ^ |
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      He has made a vector of vectors qs where each entry(qs[i]) is a vector which stores queries in which i-th vertex has been mentioned.

      Then in add function, for a vertex v, all the queries are traversed in which v-th vertex was mentioned and another vector qc is being increased for those queries (++qc[query];). When value of qc for a particular query reaches number of vertices in that query(denoted by qn[query]), it means that all the vertices in that query are neighbours of some path from root to some vertex. This count of vertices which are neighbours, was kept by qc[query].

      Now remove fn is easy to understand as it is basically removing the neighbours count.

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      3 years ago, # ^ |
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      Yes, amangupta's description is correct. Just to clarify a bit more on the specific variables:

      qn[i] — the total number of vertices in query i (qn = query number)

      qc[i] — the number of vertices in query i that have been currently marked (qc = query current)

      qg[i] — 1 if query i is achievable, 0 otherwise (qg = query good)

      qs[i] — a list of queries that vertex i appears in (qs = queries)

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3 years ago, # |
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My simple Implementation for problem B

My Submission

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3 years ago, # |
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Thanks MikeMirzayanov and vovuh for awesome editorial specially problem E.

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3 years ago, # |
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vovuh MikeMirzayanov how can we prove this fact for problem F.

** The second observation is that we first need to take elements from one end (only less or only greater) and only then from the other (if needed).**

why it will give correct answer and why we can't expand from both side. say if we need x, and we take y from begining first then we will take x-y from end. similarily for the same index, if we take a from beginning , we again take x — a from right.

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I feel like C was too easy :/

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CAN ANYONE EXPLAIN THE ERROR I HAVE REPLACED V[] WITH THEIR PARENT // TO OVERCOME THE DISTANCE FACTOR OF 1 I HAVE SORTED THE V[] ACCORDING TO THE LEVELS THEN CHECKED EVERY CONSECUTIVE WHETHER THEY LIE IN SAME PATH OR NOT

include<bits/stdc++.h>

using namespace std;

define ll int

define pb push_back

define fi first

define se second

define mp make_pair

ll in[200007],out[200007]; vector adj[200007]; ll level[200007]; ll par[200007]; ll k=0; ll k2; ll timer=0; int flag; pair<ll,ll> p[200007]; ll v[200007]; void dfs(ll u,ll lv) { level[u]=lv; ++timer; in[u]=timer; for(int i=0;i<adj[u].size();i++) { par[adj[u][i]]=u; dfs(adj[u][i],lv+1); } ++timer; out[u]=timer; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); ll n,m; cin>>n>>m; level[0]=0; in[0]=INT_MIN; out[0]=INT_MAX; for(int i=0;i<n-1;i++) { ll x,y; cin>>x>>y; adj[x].pb(y); } dfs(1,1); par[1]=0; for(int idx=0;idx<m;idx++) {

    cin>>k;


    flag=0;
    for(int i=0;i<k;i++)
       cin>>v[i];
    for(int i=0;i<k;i++)
    {
       v[i]=par[v[i]];
    // cout<<v[i]<<" ";

    }
//  cout<<endl;
    for(int i=0;i<k;i++)
       p[i]=mp(level[v[i]],v[i]);
    sort(p,p+k);
    // cout<<flag<<"\n";

//  cout<<in[p[0].se]<<" "<<out[p[0].se]<<" "<<p[0].se<<endl;

    int nop=0;
    for(int i=1;i<k;i++)
    {
       if(in[p[i].se]>=in[p[i-1].se]&&out[p[i].se]<=out[p[i-1].se])
         nop++;
       else
       {
         flag=1;
         break;

       }
    }

    if(flag==0)
       cout<<"YES\n";
    else
       cout<<"NO\n";

}

}

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    3 years ago, # ^ |
    Rev. 3   Vote: I like it +4 Vote: I do not like it

    Sharing your code via one of the text storage sites like pastebin or ubuntu paste might significantly increase your odds of getting a help

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

    Or simply include the submission id like 74594978

    This gives you TLE, wich means your solution runs out of time, hence is to slow. I think for every query you recreate the whole path from root to deepest node. This makes it basically $$$O(n^2)$$$ wich is to much for very deep trees.

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3 years ago, # |
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hello i submitted a solution for kth beautiful string problem.i got runtime error for that during the contest.but today i submitted that same file from my pc without any edit and it got accepted.where can i complain regarding this? this is my first contest on codeforces.please help.

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

    I think it is not the same code, you changed the types of your integers. Most likely the runtime error was caused by overflow.

    Aside from that, would be useful for such question to include the both submission ids.

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

      i didnot change anything bro...i just submitted the file that was there in my pc.without even opening it in the editor.and it got accepted this time

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

Hi @vovuh, I have a question regarding 1328F and sorry if disturbing you.

I think I may have triggered some wired behavior of Java 11, with which Arrays.sort(int[]) becomes extremely slow. From my attempted submission https://codeforces.com/contest/1328/submission/74614186, you will see that it's TLE with Arrays.sort(a) at the top part of the solve(n,k,a). If I switched from Arrays.sort(a) to use an ArrayList of tmpA to sort and assign back to a, which becomes submission https://codeforces.com/contest/1328/submission/74613801, it's accepted.

I tried to reproduce this on my local machine by randomly generating 200000 numbers between 100000 and 200000 and failed, it's super fast with Arrays.sort(int[]). This may due my usage of openjdk 11, which differs from that used by codeforces, or due the data itself.

Could I have the 3rd testcase so I can verify it on my local machine?

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

    Quick sort is known to have $$$O(n^2)$$$ in worst case.

    see some link

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

      Yeah I know that basic info.

      But:

      1. from what I see, the 3rd testcase is not already sorted or even mostly sorted.
      2. as we all know the worst cases of traditional quicksort, sure does the designers of java standard library. In fact, there are lots of technics to deal with the worst case of quicksort in Arrays.sort() or Collection.sort(). And from Java 7, the dualpivotquicksort for primitives is introduced to further improve the performance under traditional worst cases, as the implementation notes said: "The sorting algorithm is a Dual-Pivot Quicksort by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm offers O(n log(n)) performance on many data sets that cause other quicksorts to degrade to quadratic performance, and is typically faster than traditional (one-pivot) Quicksort implementations."

      But you made your point, and I already suspect that this specific testcase may be one of the rare bad case for this implementation. That's why I need this case to experiment on my local machine. And if it is I may report this to jdk team as a bug.

      Anyway, thanks for the reply.

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

        Hi, MikeMirzayanov and vovuh, I did further experiment on this issue.

        As in submission https://codeforces.com/contest/1328/submission/74698017, I deliberately force the 3rd test case to use ArrayList.sort(...) and let rest of the test cases use Arrays.sort(int[]), everything goes well and it's accepted. This makes me more confident that the 3rd test case happens to be one of the rare bad case for DualPivotQuicksort in Java 11's standard library, which actually has been there since Java 7.

        It would be great if I can have the 3rd test case and confirm it on my local machine. Anyway I can have it?

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

Could anyone tell me what's wrong with string c_str in GNU C++?

My two submissions of problem C:

https://codeforces.com/contest/1328/submission/74623565

https://codeforces.com/contest/1328/submission/74623526

They're exactly the same code with only difference is the lang(one is GNU C++17, the other is Clang++17 Diagnostics). I get WA with GNUC++, but passed with Clang++. Why did this happen?

And I get AC using cout to print string with GNU C++:

https://codeforces.com/contest/1328/submission/74623915

That's weird... at least to me...

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

74633006 TLE on tc 58 can someone help me

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it
     while(note!=-1)
            {
                mpp[note]=1;
                note=par[note];
            }
    

    This loop could reach O(n) so your solution has a worst complexity of O(n*m)

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3 years ago, # |
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where are the ratings of this contest?? My rating was incresed by 57 but it is not showing now.

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

i need help in problem E. I'm getting wrong answer in test 43 and i don't know the reason why here is my code: https://codeforces.com/contest/1328/submission/74652500 my answer is 253243 but expected answer is 253245

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

    if you know the problem can you tell me.

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

      finally i found it after few hours! if you are having same problem as me you have to check this test my previvious solution was giving me answer 10 but answer is 11 test: 5 4 1 3 5 7 9 good luck!

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

    the same thing is happening to me :'v

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3 years ago, # |
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Hello, this is my first message on Codeforces, so apologies if it violates any community guideline. I am getting wrong answer on test case 2 for Problem D. My submission id is 74659654. I have looked at it a lot but I am unable to find the problem. Can someone please help me. Thanks in advance. Happy coding !

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

    drag you submission page to the buttom, there's checker comment to show which case you didn't pass.

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

In problem E i am getting runtime error on test 100. I am not able to find the mistake in my code. Can someone help me out with error. link to my code:-74663766

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3 years ago, # |
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In question E why we are choosing depeest node among the tree please explain

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3 years ago, # |
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I know my approach should result in TLE, but I am getting wrong answer in Problem E. Could someone please help. 74696152

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3 years ago, # |
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My Fair simple approach for E Prerequisites:-LCA of Two nodes in log(N) time in a tree. Step1:- Running dfs traversal on tree and noting down level of each node in the tree and also the parent of each node in the tree and also obtaining the sequence of dfs traversal with which you can get LCA of two nodes tutorial link:LCA Finding tutorial

Step2:-Now for each query we have to answer in YES or NO .So what I did is for each node in a particular query find parent of that node and if node is 1 skip it simply. Store two things for a node particularly its parent number and its level . Sort the obtained vector accorsing to level of parents of all nodes in decreasing order.

Step 3:-Start from top and find lca of top 2 nodes and then with this obtained lca find lca of nodes successively coming and check if obtained lca value is similar to coming value and if it is then move forward till end and else break and print NO otherwise print "YES".

Solution link Implementation

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

    Could you explain why did you skip the node if its parent is 1.

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      3 years ago, # ^ |
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      Because when parent is 1 after that LCA of every node is gonna 1 as 1 is root and also as we started in bottom up fashion

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

        Could you please check what is wrong with this piece of code. I'm creating a dummy root, I expected the code to get TLE, but it's giving wrong answer. Thanks in advance. 74696152

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

          Hello __Apocalypse__.

          @Line 48: g[x].push_back(y);

          You forgot add this g[y].push_back(x);.

          I added it here: 75222286 and now it gets a TLE instead of a WA.

          Hope this helps. :)

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

            My submission Can u tell why this giving tle. I have sorted the vertices which are input in m queries according to the height of the tree given by 'dp[i]',which is the only thing different from the solution ,so my complexity is mlog(m),but still tle

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              3 years ago, # ^ |
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              Hi Chodermal1.

              @Line 6:

              void dfs(vector<vector<ll>>  ma,ll dp[],ll par[],ll src,ll p){
              

              Here, you're passing ma by value although it is not being modified there. Hence, it is taking more time to copy the data.

              I changed it to:

              void dfs(vector<vector<ll>>&  ma,ll dp[],ll par[],ll src,ll p){
              

              Notice the & used to pass ma by reference to prevent copying of the data and it gets an AC here.

              Hope this helps. :)

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3 years ago, # |
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I dont know whats wrong with the code,It passed all the testcases and verified the logic.It seems to give an error as unintialised value at line 56 due to which i am getting an wa.Please help!!!!

here is my soln: https://codeforces.com/contest/1328/submission/74730546

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3 years ago, # |
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Can anyone explain why this answer for test case in D(Carousel) is wrong? Input

4 5 1 2 1 2 2 6 1 2 2 1 2 2 5 1 2 1 2 3 3 10 10 10 my output 2 1 2 1 2 1 2 1 2 1 2 1 2 3 1 2 1 2 3 1 1 1 1 Jury's output 2 2 1 2 1 1 2 1 2 1 2 1 2 3 1 2 1 2 3 1 1 1 1

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    3 years ago, # ^ |
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    Input:

    1 2 1 2 2
    

    Your output:

    1 2 1 2 1
    ^       ^
    |       |
    

    These are the same but not in the input. Assume that after the $$$n$$$-th figure the figure $$$1$$$ goes.

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3 years ago, # |
Rev. 3   Vote: I like it +1 Vote: I do not like it

In problem F (1328F - Make k Equal — Make k Equal), Can someone please explain why the answer to the following custom test case is 29 and not 21?

12 8

1 3 3 6 6 6 11 11 11 11 11 11

Answer as per given Editorial Code = 29. Possible answer = ( 1*(6-1) + 2*(6-3) + 2*(11-6) = 21 ). {by making all 8 elements equal to '11'}

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

    After making all the elements 6, you can't directly increase two elements to 11, because once you increase it by 1, it becomes 7 and is no longer the smallest element in the array....

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3 years ago, # |
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I have solved "Problem E" using the Lowest Common Ancestor Concept. Straight forward lowest common ancestor logic.

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    3 years ago, # ^ |
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    My submission Can u tell why this giving tle. I have sorted the vertices which are input in m queries according to the height of the tree given by 'dp[i]',which is the only thing different from the solution ,so my complexity is mlog(m),but still tle

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3 years ago, # |
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In problem D. why if n is even coloring always can be done like [1,2,1,2,..2] For n = 6 , 1 2 2 1 2 3 should give 3 , 1 2 2 1 2 3 instead of 2 , 1 2 1 2 1 2

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    3 years ago, # ^ |
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    u can have different colours for same number. So when n is even just alternatively put 1 and 2 so each adjacent position will have different colours regardless of the fact that they are same or different

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3 years ago, # |
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Can anybody please explain the reason for doing this Let's take every non-root vertex (except fv) and replace it with its parent. Thanks

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    3 years ago, # ^ |
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    There are two types of nodes in queries. Either a node can already exist in the simple path from fv to root or not. Let's see both cases -
    1. Exist in simple path from fv to root — Since it's parent will also exist in the path we are not making any harm by replacing it with parent.
    2. Doesn't exist in simple path from fv to root — Since at most distance of 1 is allowed, it's better to check whether parent exist in path or not.

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3 years ago, # |
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Where could I find more information about the Technique used for E, such as proofs or different applications?

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3 years ago, # |
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can anybody tell what is wrong in my code for problem D https://codeforces.com/contest/1328/submission/75753056

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

test

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3 years ago, # |
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awesome contest...!!

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3 years ago, # |
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Now we have a beautiful structure giving us so much information about the tree

Mike got emotional here xD.

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3 years ago, # |
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In problem E, I created a set containing all the vertices from 1 to fv. Then I just check if parent[querypoints] are all in the set. I think this is sum(k_i)logn? Why would this give TLE while sum(k_i) and n are only 2e5

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    3 years ago, # ^ |
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    No, finding all values in set in not logn. In worst case which is when all node are connected in a line and leaf is fv mentioned in solution, it is O(n).

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5 months ago, # |
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Why topic tags of D is dp ?

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4 months ago, # |
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The editorial for F is way too complicated, the problem can be solved simply by observing that we either only increase the minimum and make it equal to (p[k] or p[k-1]), or we only decrease the maximum and we make it equal to (p[n — k + 1] or p[n — k + 1] + 1), or we increase minimum to (median or median -1) and decrease maximum to (median or median + 1).

(Median in the third case because in a sorted array, the median is the point at which sum of distances from all other points is minimised)

Implementation: https://codeforces.com/contest/1328/submission/193670280

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3 months ago, # |
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can someone please look at my code for E once, i even matched it with users it got ac and many were similar but still i got wa on 53. Thanks.

My Submission

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2 months ago, # |
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oversimpliedforces !!