### gepardo's blog

By gepardo, history, 3 days ago, translation,

• +79

 » 3 days ago, # |   +5 Thanks for the editorial ^^
 » 3 days ago, # |   0
•  » » 3 days ago, # ^ |   +11 Input5 1 ()()) 2 3 4  Expected Output1  Your Output3  Comment$S[3 \dots 4] = ()$ Clearly, there's just 1 way to permute it.
•  » » » 3 days ago, # ^ |   0 Thank it was mistaken with which pointer(mo_left/mo_right), should we move first. Moved Mo_right first and then Mo_left for query, finally AC. Thanks
 » 3 days ago, # |   0 My O(max(n,q) * sqrt(n) * log(max(q,sqrt(n))) solution to E1This solution works even when the segment asked in query is not a RBS. Note that log factor can be removed on using pointers and using prefix sum(for heavy chains).
 » 3 days ago, # |   +12 Can someone explain the Binary Spiders solutions.
•  » » 3 days ago, # ^ |   0 +. more detail if it's possible. cause I'm dealing with tries first time. and i watching about tries there. i got the gist of the idea. but here the process of finding the maximum subset for i in the trie is little bit complicated. thanks in advance
•  » » » 3 days ago, # ^ |   +12 let's give it a trie. this is my solution which is a little different than the one in the editorial. First, let's focus on the MSB of K and try to group the number based on bits starting from the MSB position to the end. for example, if we have K=00010101 so we're interested in the prefix part till the first 1 bit, split k like that. 0001 | 0101 and we split numbers at the same position to height bits and lowest bits 0110 | 1000 0110 | 1001 1010 | 1100 1011 | 1010 0001 | 1001 0111 | 1000 group numbers by the first part.observation-1 look at the left part (highest bits) if this part is equal then XOR will be zero and it's bad as it will make our XOR lower than K, so the max we can get from this group is only one element. observation-2 if the left part only differs at the lowest bit then we can get one element from each group, we just need to know that the XOR between these two elements is >=K and we can do this using trie (later I will explain how). observation-3 if the bits are different then it's guaranteed that it will produce a number greater than K. so getting to the Trie part, the trie is only needed in observation two, if we found two numbers have the same prefix and differ only in the first bit. so to check that we're checking two groups A, B and trying to find if we have any two pairs that have XOR >=k. put all the numbers in group A in a trie, our binary trie will have left as zero bit, right as 1 bit. now iterate over numbers in group B, for each number try to find the MAX XOR you can get, basically you iterate over the bits in this number from left to write and try to go to the opposite if possible. if the current bit is 1 then try to go to 0 in the trie (this will maximize the XOR value). hope this helped out. my submission for ref: https://codeforces.cc/contest/1625/submission/142621030
•  » » » » 3 days ago, # ^ | ← Rev. 2 →   0 thank you very much
 » 3 days ago, # |   +11 Do anyone have the implementation of problem D, using the approach explained in the editorial?
•  » » 2 days ago, # ^ |   0 I followed the editorial's approach 142894864
 » 3 days ago, # |   0 Sorry for copying my previous comment from the round announcement page. Hello, can anyone point out the error in my code: 142652368 for problem B, it seems I'm calculating less than it's expected to be but I can't find an error. I'm using the same approach of finding the pair of equal elements whose distance is minimum. Any short testcase that yields wrong output for my code would be appreciated. Thank you.
•  » » 3 days ago, # ^ |   0 Input1 6 8 19 8 15 8 8  Expected Output5  Your Output4  CommentThe segments $\ 8, \ 19, \ \ 8, \ 15, \ 8$ $19, \ \ 8, \ 15, \ \ 8, \ 8$are harmonious because the last $8$ matches.
•  » » » 3 days ago, # ^ |   0 Thanks a lot, I got that my approach to the third repetition of a number was incorrect. I modified my code (142746835) and it works against your testcase but it still fails on the exact same testcase as it did before.
 » 3 days ago, # |   +13 Please upload the codes also.
 » 3 days ago, # |   0 For C, I think "min(dp[n][j]) over all 0 <= j <= k" is not the final answer, because that is forcing the n-th sign to be taken (not removed). To get the final answer, we have to enumerate the last taken sign, add the remaining road section's time to each one, and finally take the minimum value among all of them.
•  » » 3 days ago, # ^ |   0 We are indexing from $0$ so element at at index $n$ is destination city (or simply it's $l$).
•  » » » 3 days ago, # ^ |   0 Oh, thank you. That makes sense.
 » 3 days ago, # | ← Rev. 2 →   +11 I think problem D memory limit should've been higher if bit-trie solutions were intended to be accepted. Fitting $O(n \log A)$ memory into 256 MiB is quite a challenge, especially for Java and Python
 » 3 days ago, # |   0 I was trying O(n^2) solution for Problem C by having additional DP state for propagating last selected speed limit. I am not looking for solution, but if someone can take a look and tell if this would work or not, that would help. Thanks.142748551
•  » » 3 days ago, # ^ | ← Rev. 2 →   0 Hey, I am also doing this only. But unable to figure out the mistake. Did you get to know why this will not work. Or anybody else can help us out! Please!! Thank you so much!My Code -> 142787581 (Similar to @i_will_be_expert's)PLEASE PLEASE, help us out!
•  » » » 3 days ago, # ^ | ← Rev. 2 →   0 This might help.
 » 3 days ago, # |   0 I find E1 really hard for me.
 » 3 days ago, # |   +24 On D To solve the problem, we need the following well-known fact. Suppose we have a set of numbers and we need to find minimal possible xor over all the pairs. It is true that, if we sort the numbers in non-descending order, the answer will be equal to minimal xor over neighboring numbers. Is it actually well known? Does anyone have a proof for why that is the case? Afaik the most common way to solve this problem is by making a bit trie.
•  » » 3 days ago, # ^ |   0 In bitwise xor, same bits give us 0, while different bits give us 1. If we try to find minimal xor pair, neighboring numbers have more of the same bits starting from the MSB, which makes the xor value smaller.
•  » » » 40 hours ago, # ^ |   0 Not always true.Consider 6 7 8, (7^8) is greater than (6^8).But still the overall minimum comes out as (6^7) which are still the neighboring elements. Its so confusing lol.
•  » » 3 days ago, # ^ | ← Rev. 2 →   +10 Let a,b and c be three numbers such that a
•  » » » 40 hours ago, # ^ |   0 Thanks for explanation.
 » 3 days ago, # |   +16 I think in E2,it is more natural for me to think up the $O((n+q)\log n)$ solution than the $O((n+q)\sqrt n)$ solution.
•  » » 3 days ago, # ^ | ← Rev. 3 →   +11 Another thing I've noticed, converting RBS to tree is pretty much the inverse operation of DFS/finding the Eulerian path of a tree. So instead of building the tree explicitly, I found it much more efficient to simply compute the statistics for each node/open bracket in place.142753393 link to my submission
 » 3 days ago, # |   +15 E2 can still be solved if not only leaves are erased, but each pair must form a matching.If we build the bracket tree, the erase operation can be transformed to erase a node and link all its sons to its father. The queries can be transformed to sum up $k*(k+1)/2$ for each node in the interval. We can use unionset to deal with father changes and a Fenwichtree to deal with subtree sum except the root. Degree of the root of each interval can be transformed to occurrences of minumum prefix sum in the interval. It can be maintained by segment tree.Here is the code.
 » 3 days ago, # |   0 In question B, the answer seem to be n - min(v - u).
•  » » 3 days ago, # ^ |   0 Yes, you are right. I will fix it soon.
 » 3 days ago, # |   0 Can someone tell me why doesn't greedy algorithm work for problem C in which we remove the sign which causes a maximum decrease in time. If someone can provide a short test case which I can visualize that would be helpful. Here is my code 142511895
 » 3 days ago, # |   0 In problem C I ran a nested loop. Outer one from 0 to k-1 and inner from 1 to n-1(because we can't remove the first sign). I then checked that by removing which sign I will save maximum time and then adjusted the array accordingly for removing another sign. But this approach was not working. Can someone give me some test cases where it will not work?142507173
 » 2 days ago, # |   +6 Some hints for people stuck at debugging problems B and C.Iizy Problem: B Input1 7 4 1 4 1 3 4 4  Expected Output6  Your Output5  CommentA possible sequence is $4 \ 1 \ 4 \ 1 \ 3 \ 4$ $1 \ 4 \ 1 \ 3 \ 4 \ 4$Obviously, the last 4 matches, hence it's a valid one.i_will_be_expert nitigya Problem: C Input5 5 2 0 1 2 3 4 3 2 4 6 9  Expected Output17  Your Output18  CommentThe optimal move is to remove the the signs $3$ and $4$, resulting in a total cost of $(1-0)*3 + (2 - 1)*2 + (5 - 2)*4 = 17$umanggupta1975 Problem: C Input4 5 2 0 1 2 4 1 3 3 5  Expected Output9  Your Output11  CommentThe optimal move is to remove signs at $1$ and $2$, resulting in a total cost of $(4 - 0)*1 + (5 - 4)*5 = 9$HaidRam Problem: C Input5 6 2 0 1 2 3 4 2 5 3 5 6  Expected Output19  Your Output22  CommentThe optimal move is to remove the signs $3$ and $4$ resulting in a total cost of $(1 - 0)*2 + (2 - 1)*5 + (6 - 2)*3 = 19$
•  » » 2 days ago, # ^ |   0 Thanks a lot, I got AC!
•  » » 2 days ago, # ^ |   0 Thanks a lot.
•  » » 40 hours ago, # ^ |   0 Thank u very much.
 » 6 hours ago, # | ← Rev. 2 →   0 .