### kimden's blog

By kimden, history, 8 months ago, ,

Hi!

XIX Open Cup Grand Prix of Belarus takes place on Sunday, February 17, 2019 at 11:00 MSK.

The contest writers are GandalfTheGrey, gepardo, greencis, kimden, and progmatic2.

Links to the contest: division 1, division 2. Please do not solve the contest if you solved it in Petrozavodsk.

Let's discuss problems after the contest. Good luck!

UPD. Editorial

• +81

 » 8 months ago, # |   +72 On the contest page it says: "The virtual contest is in progress. You are not allowed to participate".
 » 8 months ago, # |   0 Is the contest not loading for anyone else as well?
 » 8 months ago, # | ← Rev. 2 →   +16 1) Was the intended solution for J just to precalculate somehow count of all primes and primes of form 4k + 3 up to 1011? If yes, how to do it fast enough?2) How to solve B? Is there any way to do it without calculation of polynom (x - c1)... (x - cn), where ci = di - dmin?
•  » » 8 months ago, # ^ | ← Rev. 2 →   +24 B: the cumulative distribution function of the answer is when , you need to calculate it multiplying polynomials with divide&conquer. Then you find in linear time.
•  » » » 8 months ago, # ^ | ← Rev. 2 →   0 Whoops, yeah, I'm definitely lacking calculus skills. Don't look into the first revision, please)
•  » » 8 months ago, # ^ | ← Rev. 2 →   +8 There were two intended approaches for problem J. One of them is to precalculate count of jimp numbers on blocks of length  ≈ 107 and count the rest using the segmented sieve of Eratosthenes. The second approach is quite complicated. It involves dynamic programming which does sort of "sieve of Eratosthenes" to calculate how many numbers are either prime or not multiple of any of the first k primes. More details may appear later; we would also appreciate if the teams shared ideas of their accepted solutions without precalc.Basically, you just need to run the segmented (or block) sieve of Eratosthenes. You can read more about it, for example, here. It works about 20 minutes for the authors to do all precalculations. Note that it's not necessary to calculate primes of form 4k + 1 and 4k + 3 separately, but just all jimp numbers together.
 » 8 months ago, # |   +18 What's the expected complexity of C? My O(NlogNlogMAXVAL) gets TLE at test 55. Is there an O(NlogN) solution?
•  » » 8 months ago, # ^ |   +13 Did you write binary search + segment tree with arithmetic progression update for TLE solution?
•  » » » 8 months ago, # ^ |   +8 No.I used static convex hull and binary search.
•  » » » » 8 months ago, # ^ |   -23 what is static convex hull , from where u study these things . i have never heard of it
•  » » » » » 8 months ago, # ^ |   -22 You are too noob to know this. Play more MOBA or BR, then you will eventually know.
•  » » » » » » 8 months ago, # ^ |   0 what are moba or br ?
•  » » » 8 months ago, # ^ |   +8 Well, is it possible to write a segment tree that supports: Initializing values (i.e. build function) Range update with arithmetic progression (a[l] += d, a[l+1] += 2d, ...) Range min/max query My code could do the second and third type of queries but I don't know why my code didn't work. The idea was keeping the first element of AP, the difference of AP, and min/max of range.
•  » » » » 8 months ago, # ^ |   +10 Can you please explain (or provide some links) how to do both 2nd and 3rd type of queries? I tried to find anything about that but failed. The main question is how to recalculate min/max of range after adding new progression?
•  » » » » » 8 months ago, # ^ |   0 After updating with AP, I updated the min/max element with the first and last element of range but that is totally wrong. However, in the task, I only needed to update range [1, n] in which the aforementioned method works because adding new AP to some AP derives new AP. Though initializing elements ruined everything since they are not AP every time.So, the question remains open, I also couldn't find any information from the internet.
•  » » 8 months ago, # ^ |   0 Some implementation issues, I presume? The authors' solutions have complexity , too, and we do not know faster solutions. If someone does, please tell us here.
•  » » » 8 months ago, # ^ |   +13 Can you please tell the author's approach?
•  » » 8 months ago, # ^ |   +5 Our solution with dynamic CHT + parallel binary search also failed the time limit. After about 6 unsuccessful attempts of changing the number of iterations on the binary search, we decided to implement stack-like CHT + parallel binary search, which fit into time limit (even though it's the same complexity, binary searching on array vs set is faster).Is it really necessary to put such tight TL constraints, so that one O(Nlog(N)log(1 / eps)) solution fails and another O(Nlog(N)log(1 / eps)) solution passes? (especially as the intended complexity of the solution was this)
•  » » » 8 months ago, # ^ |   +21 My solution with 60 iteration runs in 2.4s, so TL is not very lenient. But I don't see any reason for setter to test and pass a solution which is much more slower (and harder) than reasonable implementation? I believe Dynamic CHT would fail even if TL = 10s.Btw, what I actually didn't liked was the ai ≤ 109, which makes n × n × ai to overflow int64. I was forced to use doubles because of it.
•  » » » 8 months ago, # ^ |   0 I didn't use dynamic CHT, rather building my convex hull was done in O(n). My complexity came to be O(NlogNlogC) due to a binary search and a seperate ternary search in CHT for the query.
 » 8 months ago, # | ← Rev. 2 →   +8 How to solve E?
•  » » 8 months ago, # ^ |   +31 E: The sum of distances is minimized on the centroid of the tree. So if an edge splits the tree into components of size k and n - k it adds min(k, n - k) to the cost. There are ways to split the vertices in 2 sets of size k and n - k, and kk - 2(n - k)n - k - 2k(n - k) trees with an edge connecting those 2 sets. For each such tree we add min(k, n - k) to the answer. The solution works in . It seems that the only reason for n ≤ 5000 is so I spend half the contest optimizing solution.
 » 8 months ago, # |   0 too many errors in samples :(
•  » » 8 months ago, # ^ | ← Rev. 2 →   0 Could you please tell which particular problems had errors in their samples? We are only aware of typos in Notes section in problem C, where 3 values in sample explanation are wrong. But it seems that it didn't cause any troubles for many teams in passing problem C. (UPD: Div2 problems had some typos, too.)Anyway, sorry for the inconvenience, we'll be more careful next time.
•  » » » 8 months ago, # ^ | ← Rev. 2 →   0 (Div2) For example O, P, i catch strange "Check failed" in G. It's not problem a big problem for a lot of people, that's just affected me enough, bcs my incorrect program gave the same result for incorrect sample :)
 » 8 months ago, # |   0 How to solve A?
•  » » 8 months ago, # ^ |   +16 First, ask about all the vertices. Then for each i ask about all the vertices except the i-th one. If the area decreases, then i-th vertex is the vertex of convex hull. Notice that vertices of convex hull are in the same order as the vertices of the polygon. Now we know that convex hull is p1, p2, ..., pk. To find the area of the polygon we need to subtract from area of convex hull areas of polygons of the form pi, pi + 1, ..., pi + 1 - 1, pi + 1. We can find those areas recursively with the same idea.
•  » » » 8 months ago, # ^ |   +8 Also, if the area of convex hull is 0, we should set the area of polygon to 0 and stop, otherwise we can ask about this polygon infinitely.
•  » » » 8 months ago, # ^ |   0 How do you show that this asks no more than (n(n - 1)) / 2 queries? We could show O(n2) by arguing about how many times a vertex can be a "hole" (ie. the vertex in the except clause) -- but couldn't get the constant.Also, did anyone else get Presentation Error? We checked number / validity of queries with asserts, and can't figure out why it's happening.
•  » » » » 8 months ago, # ^ |   +16 In each recursion call we ask one question about all vertices and one question about each vertex which is not on convex hull. So, for each vertex the number of questions about it is the depth of recursion when it becomes on convex hull. On the first recursion call we find at least 3 vertices on convex hull. Then on each depth of the recursion we find at least 1 vertex on convex hull. Also there are no more than n - 2 recursion calls, because on each recursion call we find at least one vertex on convex hull, but on the first one we find 3. So overall the number of questions is no more than 3 + 2 + 3 + 4 + ... + (n - 2) + (n - 2) = n(n - 1) / 2 + 1. Actually the number of queries is a bit less because when there are only 3 remaining vertices, we know that all of them are the convex hull.
 » 8 months ago, # |   +26 What is the solution to D?
 » 8 months ago, # |   +26 Auto comment: topic has been updated by kimden (previous revision, new revision, compare).
•  » » 8 months ago, # ^ |   +8 Will the problems be added to GYM?