### gyh20's blog

By gyh20, history, 16 months ago,

Sorry for the slow Editorial, I am new using Polygon.

Special thanks to TechNite for his help.

About one hour before the contest Retired_cherry found the same problem, then we changed it to $k=4$ and he found another same one again. We are running out of time so we didn't have time for a new one, so we changed $k$ to $5$. But we didn't know there is still a similar problem. Sorry again.

idea:gyh20 solution:gyh20 tutorial:TechNite

Jury solution:92671575

idea:feecIe6418 solution:feecIe6418 tutorial:feecIe6418

Jury solution:92671590

idea:gyh20 solution:gyh20 tutorial:feecIe6418

Jury solution:92671713

idea:isaf27 solution:gyh20 tutorial:TechNite

Jury solution:92671763

idea:gyh20 solution:gyh20 tutorial:gyh20

Jury solution:92671740

• +173

 » 16 months ago, # | ← Rev. 2 →   +14 Thanks for the fast editorial! It was an amazing contest (besides problem B)! UPD: Problem E editorial has a typo: tutotial -> tutorial.
•  » » 16 months ago, # ^ |   +9 Hi, in 1406B — Maximum Product, after sorting in descending order, i think it is feasible to check only 3 values that are 5 entries from top and none from bottom, 3 from top and 2 from bottom, 1 from top and 4 from bottom and then simply find their max :)
•  » » » 16 months ago, # ^ |   +2 Yes, I did that way, you can see my submissions. I just said that because problem B could be easily googled.
•  » » » » 16 months ago, # ^ |   +8 Oh okay, i thought you got stuck somewhere :)
•  » » » 15 months ago, # ^ |   0 How to prove that this will give the correct answer always? pls help
•  » » » » 15 months ago, # ^ |   0 Ok so we need to find a maximum product of five numbers, right? Now, sort the given array in descending order so that big positive numbers(if any) are in front and big negatives are in last(if any)and consider the following cases:- 1. If all numbers are negative, then the product of the first 5 would be the answer. eg. -1,-2,-3,-4,-5,-6 2. If all numbers are positive, then also the product of the first 5 will be the answer. eg. 5,4,3,2,1,0 Now see if there is a mixture of even and odd numbers, for the product to be maximum negative numbers should be multiplied even number of times, so I will simply consider negative numbers 0 times, 2 times, 4 times, and 5 times(if only all numbers are negative). You can check my solution also for effective implementation. https://codeforces.com/contest/1406/submission/92589865
•  » » » » » 15 months ago, # ^ |   0 thanks! and yeah, easy proof, was having bad day and couldn't prove this for corner cases that time!
 » 16 months ago, # |   +26 Couldn't solve E, but it was interesting(atleast to me).
 » 16 months ago, # |   +16 Auto comment: topic has been updated by gyh20 (previous revision, new revision, compare).
•  » » 16 months ago, # ^ |   +8 Link is still wrong now.
 » 16 months ago, # |   +8 The solution link is not working sir!!!.. Please check it once. Thank you
•  » » 16 months ago, # ^ |   +16 Fixed.
•  » » » 16 months ago, # ^ |   +15 I don't think it is fixed.It says that you are not allowed(even I am logged in)could you please check?
•  » » » 16 months ago, # ^ |   +1 still not working
•  » » » » 16 months ago, # ^ |   0 fixed
 » 16 months ago, # |   +16 Auto comment: topic has been updated by gyh20 (previous revision, new revision, compare).
 » 16 months ago, # |   +14 nice problems, liked the centroid concept!! .... hope to see more like this
 » 16 months ago, # | ← Rev. 2 →   +8 I used a different approach for C: 1. First I found the tree diameter. 2. Then I rooted the tree at one of the centres (if there are 2) and calculated the subtree sizes. 3. Then for each node of the tree diameter, I calculated the maximum subtree size that would be generated when I remove that node. 4. I found the minima over all these subtree sizes, the node/s which give this minima are the centroids. 5. If there is a unique centroid, I removed any edge and rejoined it. Else, I took a leaf, removed the edge between it and the parent, and joined it to one of the centroids. I don't know why my solution fails at test case 6. https://codeforces.com/contest/1406/submission/92644162 Thanks in advance.
•  » » 16 months ago, # ^ |   +32 The centroid(s) does not necessarily lie on diameter.
•  » » » 16 months ago, # ^ |   +1 Thanks, Can you please explain me C, I didn't understand the editorial.
•  » » » » 16 months ago, # ^ |   +1 If there are two centroids, they are adjacent and divide the tree into two components of size $\dfrac{n}{2}$ and $\dfrac{n}{2} - 1$.Pick a leaf from the subtree ( not containing the other centroid ) of any one of the centroids and link it with the other.
•  » » » » » 16 months ago, # ^ |   +3 why are there at most 2 centroids? Also, why do they have to be adjacent? An informal proof will be much appreciated! Thanks in advance.
•  » » » » » » 16 months ago, # ^ |   +4 If you agree with the fact that 2 centroids must be adjacent to each other, then if there exists another centroid it must be adjacent to the previous 2 centroids but that will lead to a cycle. Therefore, there are at the most 2 centroids.
•  » » » » » » » 16 months ago, # ^ |   +2 Ah! Very smart! But again, why do the centroids have to be adjacent?
•  » » » » » » » » 16 months ago, # ^ |   0 If not, choose any other vertex on the path from one of the centroids to the other, and the size of the largest connected component after cutting it will be smaller than the original ones.And that is a contradiction.
•  » » » » » » » » » 16 months ago, # ^ |   0 great thanks!
•  » » » » » » 16 months ago, # ^ |   +3 you can look up any centroid decomposition tutorials online. these are briefly explained over there.
•  » » » » » » » 16 months ago, # ^ |   0 Right! Thanks a lot!
•  » » » » » 16 months ago, # ^ | ← Rev. 3 →   0 I get that when two centroids are there we have this property — each centroid has a max of size n/2 component(necessarily the one including the other centroid).IN editorial it is written that x will have max component size of n/2 still after operation. But it can be of n/2-1 also right? n/2-1 would be when x has more than two edges, one to the other centroid and the other edges to the other nodes. I even worked through an example for this. Although max disconnected of x can at most be reduced to n/2-1 and really has only these two choices{n/2,n/2-1}.
•  » » » » » » 16 months ago, # ^ |   0 I agree with you
 » 16 months ago, # |   +9 A few typos in the first line of the solution explanation for problem D. It should be:Since sequence $b$ is non-decreasing and sequence $c$ is non-increasing, we need to find $max(c_1, b_n)$.
•  » » 16 months ago, # ^ | ← Rev. 3 →   +3 @ivanzuki can you explain me is there any need of suml in the tutorial part, please? and what is wrong with my solution can you help me to find out? solution link : https://codeforces.com/contest/1406/submission/93036847
•  » » » 16 months ago, # ^ |   0 Change int check() to long long check() and you get AC :)
 » 16 months ago, # |   +22 I had a different and much simpler approach to B.Firstly, sort the array from smallest to largest (no absolute value). After the sort, it can be proved that the answer is either a[n-1]*a[n-2]*a[n-3]*a[n-4]*a[n-5], a[n-1]*a[n-2]*a[n-3]*a[1]*a[0], or a[n-1]*a[3]*a[2]*a[1]*a[0]. Take the maximum of these three values as the answer. The logic is pretty self explanatory. My submission: 92587656
•  » » 16 months ago, # ^ |   0 I had a slightly more complicated to code but really easy to understand.I took the first 10 numbers and the last 10 numbers of the sorted array(handling the case when $1 \le n \le 20$). Then I bruteforced through all the combinations of 5 numbers from the 20 numbers. I then took the maximum of all of those products.Link to submission 92660785. Please tell me if you don't understand my code because of all of the macros.
•  » » 16 months ago, # ^ |   0 I also thought the same idea : 1. Think like this that the maximum answer should always be positive, so this is only possible when all numbers are positive or negative numbers exist even times.
•  » » 16 months ago, # ^ |   0 Hey , I also did same ( got AC too) , but my doubt is after sorting isn't the order of values is changed , so how the given i
•  » » » 16 months ago, # ^ |   0 before sorting the five values have some indices. After sorting, those same five values now got new indices. As we are taking the product, the values of indices don't matter. Any five different indices will satisfy that condition.
 » 16 months ago, # |   0 "maxinum component size of x" it should say "maximum component size of x" (for problem C)
 » 16 months ago, # | ← Rev. 6 →   +41 I have a slightly different solution for E using binary search that might be more intuitive. First, brute force all prime factors which are $\leq \sqrt{n}$, for $n = 1e5$, there are only $65$ such primes, and we can check each exponent for all in ~$200$ queries(even less if we binary search on the exponent). Then, for the higher primes only one such prime can appear, so we binary search on the rest of the primes with the following algorithm:Let $num$ be the amount of numbers left in the set.Consider some search space of primes $[l, r]$Ask B for all primes $[l, mid]$, subtract the result of the query from $num$Ask A 1If the result of A 1 and $num$ differ, the last prime is in the range $[l, mid]$, otherwise it is in the range $[mid+1, r]$This part takes at most $m + log(m), m = \pi(n)$ queries so it easily passes.
•  » » 16 months ago, # ^ |   0 After you ask B for all primes in [l,mid], how can you find the right factor if the result of A 1 and num differ?
•  » » » 16 months ago, # ^ |   0 Cause corresponding to that factor the query A *(that number) will give 1 as the output and we can be sure that it is that number as the largest prime number factor
•  » » » 16 months ago, # ^ |   0 Harshlyn94's method does work, but you can also continue to binary search that range, and so you don't need to write another check. Its not much of an implementation save, but it is a little easier.I found a submission that uses this idea: 92651206
 » 16 months ago, # | ← Rev. 2 →   +2 I thought that there should be atmost most 2 centroids in a tree is it correct?If it is how to prove it ? And if it's not then why, I was not able to find an example.
•  » » 16 months ago, # ^ |   +15 Assume that node x is one of the centroids of the tree.Make x the root of the treeHere, we denote sz(a) as size of the subtree of aAccording to definition, sz(child of x) ≤ (n / 2). (1)We have another observation: The size of the subtree of a node < size of the subtree of its parent (2)If there are one centroid => We are doneElse, let y be another centroid of the tree.When we delete y, the subtree that doesn't contain the subtree of y is connected, so its size must ≤ (n / 2) according to definition=> sz(y) ≥ (n / 2) (3)From (1), (2) and (3), sz(y) = (n / 2) and y is a direct child of xIf there is another node z which is another centroid of the tree (now we have 3 centroids), then sz(y) + sz(z) + 1 (node x itself) = n + 1, and since y and z are directed childs of x, their subtrees don't share any node.=> The graph has ≥ (n + 1) node (contradiction)=> There are at most 2 centroids in a tree (Q.E.D)
•  » » » 16 months ago, # ^ |   0 "**When we delete y, the subtree that doesn't contain the subtree of y is connected, so its size must ≤ (n / 2) according to definition**"I don't get this, what is the subtree that doesn't contain the subtree of y ?
•  » » » » 16 months ago, # ^ |   +3 The subtree that doesn't contain the subtree of y is the given tree without the subtree of node y.
•  » » » » » 16 months ago, # ^ |   0 thanks
•  » » » 14 months ago, # ^ |   0 When we delete y, the subtree that doesn't contain the subtree of y is connected, so its size must ≤ (n / 2) according to definition=> sz(y) ≥ (n / 2) (3)Can you elaborate on this more I'm not getting it?
 » 16 months ago, # |   -42 SpeedForces GoogleForces CopyPasteForcesB and C seemed rather standard. I recognized B as an E problem from a recent ABC round. However, I didn't bother to go looking for it and then copy and paste a solution. For D, I googled "how to find centroid of a tree" and "how many centroids are in a tree". The latter gave me SpoilerA tree may have one centroid or may have two centroids. If it has two centroids, they are always connected (otherwise, the tree can't have n vertices). You can find these vertices by checking the size of each subtree, doing DFS.Which basically trivialized the problem.Nevertheless, A, D, and E seem quite nice. I found this round rather interesting (especially problems D and E).
•  » » 16 months ago, # ^ |   +26 We are sorry for making problem B.But for C, the point of the problem is not to find the centroids, but what to do next. So in my opinion, I don't think C is very standard.
 » 16 months ago, # |   0 Can anyone tell me why my code is giving runtime error on test 1?https://codeforces.com/contest/1406/submission/92659872
•  » » 16 months ago, # ^ |   +4 it's because your centroid finding function is not working, centroid vector is empty and consequently indexing empty vector causes runtime error
•  » » » 16 months ago, # ^ |   0 I know that some values are inserting to the vector, but sometimes this vector is empty. I struggle with fixing this bug. Do you know what exactly is not working in this function?
•  » » » » 16 months ago, # ^ | ← Rev. 2 →   +4 There might be other problems too but one of the obvious thing is if(n - graf_size[u] > n / 2) this check should be outside loop for(auto v : graf[u]) so that you check condition only when you have calculated actual subtree size for node u.Edit : I made some more changes to Centroid function. now it seems to work. correct submission : solution
•  » » » » » 16 months ago, # ^ |   0 What a stupid mistake with bracket. Thank you very much for your help.
 » 16 months ago, # |   +6 In E I believe you should add $\sqrt{m}$ to your estimate. You do $\sqrt{m} \cdot \sqrt{m} + 1 \cdot \sqrt{m}$ and in at most 2 buckets you'll find a prime that divides $x$, so you need to add $2 \cdot \sqrt{m}$ for that. And later there is that $\log_2(n)$ and one for answer, so counting this way we get $m + 3\sqrt{m} + \log(n)$.Also I don't understand why the first 3 paragraphs of its tutorial are there, in my opinion they are quite confusing.
•  » » 16 months ago, # ^ |   +5 I don't understand why we can't find the smallest prime number by using the given method. Shouldn't it be largest prime number which is not possible to find.
•  » » » 16 months ago, # ^ |   +9 We can find all prime divisors this way, editorial is just confusing.
 » 16 months ago, # |   +1 Can someone explain this line from the editorial of question E. After finishing asking a group, ask A 1 and check if the return value same as it supposed to be without x.
•  » » 16 months ago, # ^ |   0 Cause corresponding to A 1 we will know how many numbers are there in the set now and then if there is a change then it it's not same then it means one of the prime is the factor of x . You can check this video solution : https://www.youtube.com/watch?v=49GhzA5eWyE
 » 16 months ago, # |   0 plz explain solution of problem c..
 » 16 months ago, # | ← Rev. 2 →   +6 First, if all numbers are less than 0, then you should print the product of the five biggest numbers of them. Otherwise, the maximum product must be non-negative. Its false. How about -1 1 2 3 4?
•  » » 16 months ago, # ^ |   0 It will only make sense if n > 5.
 » 16 months ago, # | ← Rev. 2 →   +4 Problem D. Three Sequences Youtube Video Explanation in Hindi. Solution Explanation Youtube LinkIf SomeOne is Upsolving Problem D and needed some help in solution explanation in hindi. please watch above video and give feedback to improve quality. Thanks
 » 16 months ago, # |   0 can anyone explain solution of 3rd problem? I can't understand why dfs is used here?Can it be solved without dfs like if we can find out 2 centroids by checking the largest component which is smallest by removing each vertex one by one and than if there are 2 or more centroids we can remove one edge from one centroid and connect to other centroid??
•  » » 16 months ago, # ^ |   0 How do you find a centroid without doing dfs?
•  » » » 16 months ago, # ^ |   0 I think by removing each vertex one by one and than checking for largest component than does not contain removed vertex and out of these components the centroids will be the ones that has minimum size component (obtained after removing those vertex). Please correct me if i am wrong.I can't understand how dfs is applied here to find centroid.It will be of great help if u could provide any source to solve this type of problems
•  » » » » 16 months ago, # ^ |   0 It is well explained in this blog https://codeforces.com/blog/entry/57593I did not know this algo before the contest, and implemented an alternative. Just find foreach vertex the maximum size of the subtree of its adjent vertex. Then find among all vertex the one or both with minimum value. 92620542
 » 16 months ago, # |   0 I think the problem C and E is interesting,but problem A is not easy enough to put it on A.It can be B.
 » 16 months ago, # |   0 Can anyone tell me why my code is wrong?(Problem E)https://codeforces.ml/contest/1406/submission/92640804
 » 16 months ago, # |   0 About centroids. Editorial says "If not, choose any other vertex on the path from x to y and the size of the largest connected component after cutting it will be smaller than x and y". Looks like this is impossible and centroids are always connected. If I am correct, please remove this confusing comment
•  » » 16 months ago, # ^ |   0 yeah ... I also think so.
•  » » 16 months ago, # ^ |   0 The point of that statement is proving, by contradiction, that "centroids are always connected".
 » 16 months ago, # |   0 I cannot see the Jury's solution for E. It shows "You are not allowed to view the requested page".
 » 16 months ago, # |   +6 I think there are some typos in the editorial of problem D: sequence $b$ is non-decreasing and $c$ is non-increasing (as in the problem description). and then, it should say $\max(c_1, b_n)$.
•  » » 16 months ago, # ^ |   +5 Will fix that soon
 » 16 months ago, # |   +8 About problem E:I didn't notice that $1\le a\le n$!What a pity!I almost made it to the top 20!
•  » » 16 months ago, # ^ |   +5 emmm, you are sure that you can solve E if you notice $1\le a\le n$ that time?
•  » » » 16 months ago, # ^ | ← Rev. 3 →   +4 I passed the problem after I saw $1\le a\le n$. :(
•  » » 16 months ago, # ^ |   0 Same here. I noticed the constraint after the contest. The solution I submitted during the contest ACed when I added the condition.
 » 16 months ago, # |   0 Auto comment: topic has been updated by feecIe6418 (previous revision, new revision, compare).
 » 16 months ago, # |   0 I can't view pC's solution(submission), does it only happen to my account or is it private ><
 » 16 months ago, # |   0 Though I've lost so many ratings,I enjoy the contset very much.THX for such a wonderful contest!!!
 » 16 months ago, # |   +9 Does anyone have a neat proof that definition of the centroid given in C matches the standard definition of centroid (disconnecting it separates the tree into connected components all <= n/2 in size)?
•  » » 16 months ago, # ^ | ← Rev. 2 →   +13 Let us root the tree at centroid (curren't problem's definition of centroid), and sz(x) be the size of subtree of vertex x, now sz(x) is the smallest possible largest connected component upon deleting a vertex s, now assume sz(x) > (n/2), now the largest connected component upon deleting x would be max(n-sz(x), max(sz(y1))) where y1 are x's children, sz(y1) n/2), so either way the sz(x) would not be the smallest possible largest connected component and hence s is not the centroid, hence a contradiction. We can extend this to say that both the centroids should by adjacent, with the equation sz(x) + sz(y) = n (where y is another centroid)
•  » » » 16 months ago, # ^ |   +5 Excellent. Although hard to wrap my mind around "We can extend this to say both centroids must be adjacent". Could you explain a bit more? Thanks!
•  » » » » 16 months ago, # ^ |   +5 Tree is still rooted at s, there is some other centroid out there y, let it's children be yi, now the largest component upon deleting y is max(sz(yi),n-sz(y)), now sz(yi) < sz(x) so if this also centroid, then n-sz(y) = sz(x), as both produce the smallest possible largest component, hence sz(x) + sz(y) = n, but sz(x)<=n/2, and sz(y)<=sz(x) as upon deleting the root, the largest component we get itself is sz(x), the only scenario where the above eqns can be satisfied is sz(x) = sz(y) = n/2 and y is child of s, but if y!=x then sum of subtree's of child itself becomes n, so the only way out is y equals x. So there can be atmost 2 centroids, both adjacent.
•  » » » » » 16 months ago, # ^ |   +5 Thanks!
 » 16 months ago, # |   0 Can someone find me a video editorial for problem C. Thanks in advance!
•  » » 16 months ago, # ^ |   0 Watch this video by SecondThread
 » 16 months ago, # |   0 It's easy to see that the size of y's subtree must be exactly n/2. After cutting and linking, the maxinum component size of y becomes n/2+1 while the maxinum component size of x is still n/2.I think they exchanged x and y in tutorial of problem C if x and y are two centroids and x is y's father, then initially their sub tree size would be n/2 and n/2.this also tells that if n is odd then only 1 centroid exists. Now if we take leaf of y and add it to x, x sub tree size would be n/2 + 1 and y's would be n/2 — 1, but in editorial they have exhanged x and y i think !!
 » 16 months ago, # |   0 Can you anyone help to identify a failing test case for the below code for problem B ?https://ideone.com/D5Wq6zApproach : Seperate integers less than 0 (l0) and greater than equal 0 (g0). Sort both and reverse g0. Try all possible combinations in (l0,g0) that add upto 5 numbers i.e, (0,5),(1,4),etc. wherever the size allows. Take max of their products.
 » 16 months ago, # |   0 can anyone help ,where my logic get wrong for B. Maximum Product https://codeforces.com/contest/1406/submission/92678443
 » 16 months ago, # |   +2 In Maximum Product(B) it is written that i
•  » » 16 months ago, # ^ | ← Rev. 2 →   +4 If i>j,then you can see j as i,and i as j.
•  » » » 16 months ago, # ^ |   0 It doesn't makes any sense to me
•  » » » » 16 months ago, # ^ | ← Rev. 2 →   +3 i
•  » » 16 months ago, # ^ |   0 Does taking product require order of numbers. That's why. just ignore that relation(i
•  » » » 16 months ago, # ^ |   0 now i get it thanks !...
 » 16 months ago, # |   0 Can someone tell what is wrong in my solution for problem B. It is giving me WA on test 2 for the 143rd input.Thanks in advance.https://codeforces.com/contest/1406/submission/92649674
•  » » 16 months ago, # ^ |   0 consider the test case : -1 -2 -3 -4 -5 -6 your code gives -720 but the answer is -120.
 » 16 months ago, # | ← Rev. 2 →   0 There is another solution for problem B with dynamic programming. That approach seems not to be the easiest one, but I would like to share it just in case it may be interesting for someone.Submission: 92600395.
•  » » 16 months ago, # ^ |   0 Hey bro please can you write a recursive dp or memoized form of B because i don't know bottom up Thanks in advance.
 » 16 months ago, # |   0 will DP work for problem B?
•  » » 16 months ago, # ^ |   0 Yes, dp solution is possible. But, div 2 — B is not made for this. I saw i
•  » » » 16 months ago, # ^ |   0 can you tell me what's wrong with this DP solution, submission link
•  » » » » 16 months ago, # ^ |   0 I guess your initializations are wrong. For pos dp, each element should be initialized to number less than min negative product possible (-1e15 since each element could be up to -1e3). A similar analogy for neg dp.
 » 16 months ago, # |   0 Nice round! Thanks~
 » 16 months ago, # |   0 Plag Checker is dead ![Proof] 92607281 92620817
 » 16 months ago, # |   0 Generalisation of problem B has already been asked quite recently on atcoderHere's the link: atcoder ABC 173 E
 » 16 months ago, # |   0 In Problem B, I am getting wrong answer on test 2, more specifically on 143rd input — answer is not matching . Can anyone please point me out where I am going wrong? Link to submission — https://codeforces.com/contest/1406/submission/92680690
•  » » 16 months ago, # ^ |   0 I am also getting wrong answer on the same input. Still not able to figure it out.
•  » » » 16 months ago, # ^ |   0 As kennethyonghw has told, We are making the same mistake. When your product in all cases turns out to be negative, then we need to minimize it, on contrary following line maximizes it: prod=pos[i-1]*neg[4-i];
•  » » 16 months ago, # ^ |   +3 Consider the test case: -6 -5 -4 -3 -2 -1. Your code gives -720 but the answer should be -120.
•  » » » 16 months ago, # ^ |   0 Thanks kennethyonghw for pointing out that bug, I made a few changes and it got accepted. Thanks a lot.
 » 16 months ago, # | ← Rev. 2 →   0 In task C , is it necessary for x to be y's father to cut a leaf from y and attach it to x?I am not able to understand this .
 » 16 months ago, # |   +1 My solution for C :If there are two centroids, I pick one of the centroids and disconnect one of its non-centroid subtree and connect that subtree to the other centroid. Will this always work?My submission : 92632074
•  » » 15 months ago, # ^ |   0 can anybody tell this?
•  » » 14 months ago, # ^ |   0 I also have the same doubt. Can we remove any edge or just the one connecting the leaf?
 » 16 months ago, # |   0 Isn't B's editorial wrong?Consider input as : 1 1 1 1 -1All numbers are not less than 0 yet maximum product is negative.
 » 16 months ago, # |   0 Does anyone knows any good tutorial/blog for finding the centroid of a tree?
•  » » 16 months ago, # ^ |   +4 The answer you seek is one google search away
•  » » » 16 months ago, # ^ |   0 I am only finding about centroid decompostion(other than 1 CF blog). Is it the same ?
•  » » » » 16 months ago, # ^ |   0 This blog is quite nice. Very short and has interesting problems. Try to solve all. You will learn much quicker via solving problems than just reading theory.
 » 16 months ago, # |   0 Hi, in 1406B — Maximum Product, after sorting in descending order, i think it is feasible to check only 3 values that are 5 entries from top and none from bottom, 3 from top and 2 from bottom, 1 from top and 4 from bottom and then simply find their max :)
 » 16 months ago, # |   0 Am I the only one who tried solving D using range updates and range maximum queries? I feel stupid.
 » 16 months ago, # | ← Rev. 2 →   0 For problem D, how to prove that this arrangement will be optimal and gives us the least possible max(c1,bn) ? For example say d = a(i) — a(i-1), when d > 0, b(i) = b(i-1) + y(aribitary y), c(i) = c(i-1) + d — y(arbitary y), how to prove that in the optimal solution y = 0 ?
•  » » 16 months ago, # ^ |   0 For every i>1, you just want to minimize b(i) and maximize c(i), when y=0, this is the optimal.
•  » » » 16 months ago, # ^ |   +3 Thanks, but in the editorial, shouldn't it be a(r+1)-a(r) instead of a(r) — a(r-1) ? both a(r) and a(r-1) change by x
•  » » » » 16 months ago, # ^ |   +3 will fix that soon,thanks.
 » 16 months ago, # |   0 In a tree, the center of the diameter and the centroid are different points, right? Please give me an example, I am unable to come up with one... Thanks in advance
•  » » 16 months ago, # ^ |   +6 N = 7  Edges: 1 2 2 3 3 4 4 5 4 6 4 7  Center of diameter: 3 Centroid: 4 
 » 16 months ago, # |   0 Is it always that tree can have atmost two centroids? If so, then why not more than 2? Please help, just heard about centriod in this contest and am confused even after reading other blogs.. can someone explain centroid in simple terms. :(
 » 16 months ago, # |   0 please someone explain this line written in editorial of problem C, Proof: It's easy to see that the size of y's subtree must be exactly n/2why it's n/2 always??
 » 16 months ago, # |   0 problem b is interesting because it has many solutions, there is a very easy way to understand that is to break down 6 cases : - - - - - - - - - + - - - + + - - + + + - + + + + + + + + + 
 » 16 months ago, # |   0 Hello, For 1406B — Maximum Product : I believe we can answer the question by checking 3 values after sorting. The first is the product of the 5 largest numbers, the second the product of the 2 least and 3 largest numbers and finally the third will be the product of the 4 least and the 1 largest number.The solution will be the maximum of these 3 values. The solution tries to keep the answer +ve and that can be only done by selecting an even number of -ve numbers(least numbers). If there are no -ve numbers or all are -ve then the solution will simply contain the product of the largest 5 numbers[first value] which will be the max possible product.
 » 16 months ago, # | ← Rev. 2 →   0 For question 3can anyone give me an example tree with more than 2 centroids? my assert(n<=2) statement is failing.
 » 16 months ago, # |   0 Video editorial for Problem E : Deleting Numbers for anyone interested : https://www.youtube.com/watch?v=49GhzA5eWyE . Note : This video is a discussion and not take this as a tutorial .
 » 16 months ago, # |   0 Can someone please tell me where my solution is failing for problem C?Link to my submission : https://codeforces.com/contest/1406/submission/92731735
 » 16 months ago, # |   0 Can this solution be used for any k (which is 5) in the problem B? 92677568 , why not?
 » 16 months ago, # |   0 for problem C, how a tree can contain no more than 2 centroids?
 » 16 months ago, # | ← Rev. 2 →   +3 I solved [problem:1460E] in another way: "B i" for every prime in (n/2, n], then "A 1", check is your cnt equal to this: if not: "A i" == 1 for every prime in (n/2, n] -> ans *= i; else take (n/4, n/2] ...And in the end check primes from (1, n / ans)I mention that primes in (1, n/2] > (n/2, n], that's why you won't have more than enough queriesBut the author's solution with sqrt-decomposition is better and easier — Sorry for the waste of time ))) Wanted to let you know about other solution :)
 » 16 months ago, # |   0 I think the total number of operations for E is around $M + \sqrt{M} \cdot log(M) + log(M)$
 » 16 months ago, # | ← Rev. 2 →   0 Can problem D find legal arrays of $b$ and $c$? upd:Oh, yes, it can be done.
 » 16 months ago, # |   0 In Problem C in the case of 2 centroids, instead of finding a leaf in a centroid and attaching to the other centroid I found an edge from a centroid other than the one joining the two centroids and attached it to the other centroid. My solution passed system tests. Is it right or is there any test case for which my solution fails?
•  » » 16 months ago, # ^ |   +3 It's right.
 » 16 months ago, # |   0 For Problem D, i am not able to see why x should be (a1+K)/2.Can someone please explain?
»
16 months ago, # |
0

PROBLEM B

# method not working

def partition(arr, low, high): i = (low-1)
pivot = abs(arr[high])
for j in range(low, high):

if abs(arr[j]) <= pivot:

i = i+1
arr[i], arr[j] = arr[j], arr[i]

arr[i+1], arr[high] = arr[high], arr[i+1]
return (i+1)

def quickSort(arr, low, high): if len(arr) == 1: return arr if low < high:

pi = partition(arr, low, high)

quickSort(arr, low, pi-1)
quickSort(arr, pi+1, high)

if name == "__main__": for i in range(int(input())): l2 = [] n = int(input()) l1 = list(map(int,input().split())) quickSort(l1,0,n-1) #print(l1) sm = 1 for i in range(len(l1)-5,len(l1)): l2.append(l1[i]) sm *= l1[i]

if sm >= 0:
print(sm)
else:
smm = sm
for i in range(0,5):
for j in range(n-6,-1,-1):
sm = 1
for x in range(0,i):
sm *= l2[x]
sm *= l1[j]
for z in range(i+1,5):
sm *= l2[z]
if sm >= 0:
smm  = max(sm,smm)

print(smm)

https://codeforces.com/contest/1406/submission/92847774 test 2 fails ,It will be great if someone help me understand what have i done wrong here or what am i missing.Thank you in advance.

•  » » 16 months ago, # ^ |   0 Sorry for how weird the post looks , this is my firs post
 » 16 months ago, # |   +3 problem B's tutotial: while the maxinum component size of x is still $\frac{n}{2}$.I think it should be: while the maxinum component size of x is $\frac{n}{2}$ or $\frac{n}{2}-1$.
•  » » 16 months ago, # ^ |   0 Since x is parent of y, won't maximum component size of x become n/2 -1 always? How will it be n/2?
•  » » » 16 months ago, # ^ | ← Rev. 4 →   0 If you link the leaf with x, that's n/2 — 1. but if you cut a leaf from the subtree of y that does not contain x and don’t link the leaf with x,the maximum component size of x will it be n/2
 » 16 months ago, # |   0 I am not sure I completely understand problem D. I have been trying for past 3 days but the queries part is not clear to me and the solutions I refer to also are not understandable Can someone please explain the queries part, how are we updating the elements in the range l..r or if not, then why not? I understood why we need only the two adjacent differences in every query but what about updating the elements? Please help me.I would appreciate it.
 » 16 months ago, # | ← Rev. 2 →   0 Can someone please explain how the updates are being made in editorial's solution? I looked at many codes but I am unable to understand anything from them? Please help me!!! I understood the rest of the solution.
 » 16 months ago, # |   0 can anyone explain the editorial of problem A. and why i got runtime error herehttps://codeforces.com/contest/1406/submission/93185539
•  » » 16 months ago, # ^ |   0 I posted an explanation: https://www.youtube.com/watch?v=bQ1jIuTNib0. Thank you!
 » 16 months ago, # | ← Rev. 2 →   0 ;
 » 16 months ago, # |   0 Is problem A wrong? Why not split the set (0 2 1 5 0 1) into (0 0 2 5) and (1 1) to have mex of 1 and 0 respectively?
•  » » 16 months ago, # ^ |   +12 Because in that case, the answer will be 0 + 1 = 1. Our goal is to maximize the sum of the Mex functions on the subsets. For a split (0, 1, 2) and (0, 1, 5) we will have 3 and 2 retrospectively and will get 5, which is greater than 1.I recorded a short video explaining this: https://www.youtube.com/watch?v=bQ1jIuTNib0Thank you!
 » 16 months ago, # | ← Rev. 2 →   0 Anyone can explain the algorithm to calculate the centroids of problem C.I don't get that. Thanks in advance.
 » 16 months ago, # | ← Rev. 2 →   +12 Hi guys, this is our solution for problem A. We would be happy to hear your feedback regarding the format. Thank you!
 » 15 months ago, # |   0 My Solution for BWA in test 143. I used vector pair (absolute value of array element, original array element) and sorted it in descending order. Took product of first 5(product of vector[i].second). If the product of first 5 is negative and total array size is more than 5, I tried to delete one element from first 5(priority of 5th element is highest and first element is lowest according to its absolute value) and replace it with one such number from n-5 numbers such that product is non-negative and store it in res vector. And output the max number from res vector as answer. Please help. Thank you.
 » 15 months ago, # |   0 For problem C, will this approach work? ApproachFirst I find for each vertex, in how many edges that vertex is involved. If there is only one vertex with the maximum value then I just remove any edge and add it back again. If there are 2 vertices (let's say "A" and "B") with the same maximum value, then I take one child (let's say "a") of one vertex (let's say "A"), remove that edge ("A — a") and make an edge between the other vertex, "B" and the removed vertex "a", i.e. an edge "B — a".
 » 13 months ago, # |   0 can someone help me with A problem
 » 7 weeks ago, # |   0 feecIe6418 For the problem 1406B - Maximum Product, you have asked to sort the given array based on the absolute values from big to small. And still how can the time complexity be O(N)?
 » 5 weeks ago, # |   0 in problem b solution it is written that if all the numbers are positive then only the answer will be negative but what about the case when there are 5 numbers among which one is negative and other four are positive ? can anyone help pls?