, Bad E, it is too common so that people who know kruskal reconstruction tree can pass it very fast but it is very diffcult to people who don't know this algorithm like me :(. If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive. If you use $$$O(1)$$$ query lca methods (such as euler tour + sparse table) and use another sparse table to query lca of a range, you can achieve $$$O(1)$$$ time per query. If you are international, message me and I will get a shipping quote. To construct a persistent DSU data structure we need to keep snapshots for updates for every change for nodes parent. 2) D. Red-Green Towers DP, Codeforces Round #273 (Div. Then we can continue in the same way using a range max query structure. Initially, you place block $$$1$$$ at $$$(0, 0)$$$. I realized brute force was an option only after being fruitless in finding a clever number theory manipulation. I am not able to find why the code is giving tle for one case?? You're right, that's my mistake. The first line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$n \le k \le 10^9$$$) the number of towers and the restriction on slices, respectively. Instantly share code, notes, and snippets. The second line contains $$$n$$$ space separated integers $$$h_1, h_2, \dots, h_n$$$ ($$$1 \le h_i \le 2 \cdot 10^5$$$) the initial heights of towers. There are many ways to do this. Could anyone help me explain why that the number of distinct $$$\lfloor\frac{a}{q}\rfloor$$$ is at most $$$O(\sqrt{a})$$$ ? Think about how many times a particular range will have to change sets, i.e., be merged. Webhttp://codeforces.com/problemset/problem/626/C. Why does this pass? You can sort by queries by midpoint of binary search interval, and add edges that are needed between queries. This set of problems is very good, harvest a lot. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. Recalculate ai / pi. Encuentre este nmero. If you cannot form any tower of color r, the r-th integer should be 0. standard output. Am I correct? For each test case, output $$$n$$$ integers. We're essentially only doing $$$\frac{a_n}{v}$$$ operations for every $$$v$$$ (This is also written in editorial). tower Number (and weirdness) of comments is proportional to amount of skill issues I have on this problem :( I think I also didn't care about typos at all lol, I wrote this solution at first but got TLE, After that I tried to use Heavy-Light Decomposition to get LCA of two nodes and it was 4 times faster =_=. For 2in, if the (i1)-th block is placed at position (x,y), then the i-th block can be placed at one of positions (x+1,y), (x1,y), (x,y+1) (but not at position (x,y1)), as long no previous block was placed at that position. The color of the i-th block is ci, an integer between 1 and n. You will place the blocks down in sequence on an infinite coordinate grid in the following way. When it becomes impossible for a segment to contain any elements of the array we can stop early. Thanks for noticing it. The second line of each test case contains n integers c1,c2,,cn (1cin). Why did my code get WA https://codeforces.com/contest/1706/submission/164908500. The students dont want to use too many blocks, but they also want to be unique, so no two students towers may contain the same number of blocks. 2) B - Making Towers, tql, https://edu.csdn.net/skill/algorithm?utm_source=AI_act_algorithm, https://blog.csdn.net/weixin_62765166/article/details/125869853, Codeforces Round #807 (Div. All he can do is to perform the following operation several times: take the top cube from some tower and put it on top of some other tower of his set. How can we have continuous minimums? All problems in this contest are very nice! Sorry, I meant max node R[i] where everything [i, R[i]] is connected. 2). So this means we can build a segment tree over all the nodes. Each time you do this operation, the total number of towers will decrease by 1 (nonsense). I'm getting the wrong answer, however I might have a bug somewhere. As such, after finding this pair for a segment (by using segment tree), one could then easily get their LCA and that would be the answer for the query, It's also funny because now reading back this comment I realised one could make O(1) per query, because the "leftmost/rightmost" pair is denoted by a minimum/maximum value on a segement, as such using RMQ for getting the pair and RMQ for LCA you could have O(1) per query. The way using dsu to find the maximum weight in problem E was new to me. place block $$$1$$$ at position $$$(0, 0)$$$; place block $$$2$$$ to the right of block $$$1$$$, at position $$$(1, 0)$$$; place block $$$3$$$ above block $$$2$$$, at position $$$(1, 1)$$$; place block $$$4$$$ to the left of block $$$3$$$, at position $$$(0, 1)$$$; place block $$$5$$$ to the left of block $$$4$$$, at position $$$(-1, 1)$$$; place block $$$6$$$ above block $$$5$$$, at position $$$(-1, 2)$$$; place block $$$7$$$ to the right of block $$$6$$$, at position $$$(0, 2)$$$. I saw the first sample code for D2 and I wanted to show how to iterate the distinct values of $$$\lfloor{\frac{a}{b}}\rfloor$$$. My submission is a bit messy : 164790795. 1. Title link: http://codeforces.com/problemset/problem/478/D The meaning: There is a square of red and green, and the quantity is R and G. There are 0 <= r, g <= 2e5, R + g> = 1 We want to use C. Block Towers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Students in a class are making towers of blocks. In the first case, the student using two-block pieces can make a tower of height4, and the students using three-block pieces can make towers of height3,6, and9blocks. Building a tower for codewars Next update the answer if the differcence of the maximum and minimum value is smaller than before until a $$$p_i > k$$$ or maximum value of $$$\lfloor {\frac{a_i}{p_i}} \rfloor$$$ is $$$0$$$. Now some changes are required to make the landscape more beautiful, that is, the towers are arranged in a non-decreasing sequence from left to right. Initially we have {(vv), 0} for each v in the graph. floor(a/b) < q then a/b < q ?.can u explain a bit ? Now we also know that all the a[i]/p[i] values will be greater than or equal to v1 in the interation you are checking. Hay n personas apilando torres con tablas de madera de 2 metros de altura, ym personas apilando torres con tablas de madera de 3 metros de altura. time complexity will be O( log(maximum element of a[i]) * n * 108) 108 is maximum possible numbers we can make by a[i]/k. For example, if node 1 is connected to 5, and we're checking that [1, 3] are connected, this will give a wrong answer. A tower of color r is a tower such that all blocks in it have the color r. For each color r from 1 to n, solve the following problem independently: Find the maximum size of a tower of color r that you can form by placing down the blocks according to the rules. I'm wondering about the solution to D2. It makes the implementation slightly easier as you don't have to track whether v was achieved. Let $$$\lfloor{\frac{a}{b}}\rfloor = q$$$. Each student makes a (non CodeForces Title & Luo Valley Title Some people may ask me why I will explain this * 2500 D1C write questions. Working with inexperienced problemsetters over a 10 hour timezone difference is already difficult enough. In the first line print two space-separated non-negative integers s and m (mk). Got a message explaining the idea, instead of doing a DFS of the search tree instead traverse level by level, traversing each level from left to right. Tower with number i consists of ai cubes stacked one on top of the other. to Get Started with Codeforces If so how would you handle the DSU updates? Codeforces Round #884 (Div. #bigfoodchallenge #hugeburger #manvsfoodT-Shirts are only $18! If there are multiple correct sequences at which the minimum possible instability is achieved, you are allowed to print any of them. Codeforces is one of the best platforms for competitive coding and is usually known for its short challenges/contests where programmers from every corner of the world participate. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). WebProblem - 1065C - Codeforces C. Make It Equal time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There is a toy building consisting of n n towers. Don't understand solution of D1,can someone explain it to me in detail. Little Petya has n towers consisting of cubes of the same size. where $t_r$ is the number of consecutive "indices" differing by an odd number. Each student makes a (non-zero) tower by stacking pie ces lengthwise on top of each other. But lets just say if v is the minimum value that gives us the best possible result, so this means that we can say that there would be a v' > v that can be created by some ai = floor(ai / pi) that gives us the same result. Por ejemplo, ingrese una matriz de ioctl SiControlador de dispositivoPar de dispositivosCanal de E / SFunciones de gestin. In particular, want to share a sol for E that involves kruskal tree (reachability tree/line tree). I enjoyed the round! The problem statement has recently been changed. For example, if we take a_i=2v+1 then p_i=2. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. Note the way we construct the ranges: there are no intersecting ranges: the only possible case is one range contains the other. You have a sequence of $$$n$$$ colored blocks. It is built with the help of rare magic by levitating the current top of tower and adding rocks at its bottom. 1.) We will only merge smaller size sets into the larger size sets. The author says that this is only true for samples with the same "u", which does make sense, but at that point, aren't you just iterating through all values in the array regardless, since you need to check for the minimum of every value at every "u"? We had $$$\lfloor \frac{a}{b'} \rfloor < q$$$, which is equivalent to $$$\lfloor \frac{a}{b'} \rfloor \le q-1$$$ so that gives $$$\frac{a}{b'} = \lfloor \frac{a}{b'} \rfloor + r \le q-1 + r < q-1+1 = q$$$ since $$$r < 1$$$. Why would I use memset in Problem B? I'll leave it here if anyone is interested. Then the problem boils down to, find the smallest window containing all n elements in them. I liked the problems a lot, though B seems a bit hard for a Div2 B problem.. To answer all queries offline we can sort them so that first come the ones with smaller l and then iterate the queries preserving a set of not-yet-closed ranges with their second values. input. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. The $$$i$$$-th tower consists of $$$h_i$$$ cubes, so it has height $$$h_i$$$. In the second case, the students can make towers of heights2,4, and8with two-block pieces and towers of heights3and6with three-block pieces, for a maximum height of8blocks. Instalacin de TypScript de VIE-cli npm i -g @vue/[emailprotected] typescript Vue -V v bien conocidoujson Mejor quejson El mdulo es eficiente, pero la instalacin es demasiado problemtica. https://edu.csdn.net/skill/algorithm?utm_source=AI_act_algorithm, 1.1:1 2.VIP, Codeforces Round #809 (Div. First, we merge $$$[4, 4]$$$. Thanks bro for the explanation I will work it out the rest on my own! 2), [GYM] Al-Baath Collegiate Programming Contest 2023, [GYM] HIAST Collegiate Programming Contest 2023. In fact, this was quite easy to write. :) Shirts seen here: https://www.youtube.com/watch?v=Hy-8WVIBMUc To Order:1. Can someone explain solution 1 of problem D2 in details please? We no longer need to compute $$$R(i)$$$ for checking if $$$(i, i + 1)$$$ are connected so we can implement rollback easily. So by the Pigeonhole Principle, for the entire domain $$$x \le \sqrt{N}$$$, there exists at most $$$\sqrt{N}$$$ distinct "landing spots" comprised of a coordinate in $$$y$$$. Ultimately, I don't really see anyway to make this work without supporting $$$R(i)$$$ query in true $$$O(\log n)$$$ (or better) time. and how that will help us for a fixed maximum rather than a fixed minimum. Also, we need to do small into large merges instead of path compression to get the required complexity. In tutorial for D2, Solution 2 (AlperenT) "Continuing this logic, for all integers u=1,2,,k, we should check the elements ai satisfying (u1)v+1aiuv, and set all these pi=u. Define $$$f(u, v)$$$ as the maximum value of any weight on the path from $$$u$$$ to $$$v$$$ on the MST. I came up with the two-pointer method during the contest and got an MLE on test 7. Please correct me if I understood this question wrong. Output For each test case, output n integers. answer: Take the number of layers of the red ball, when D. Red-Green Towers There arerred andggreen blocks for construction of thered-greentower. My DP returns a pair of integers the first one is the maximum number of buildings that can be cooled and the second value returns the minimum cost we need to cool those buildings. Codeforces Codeforces (Max(a[i]/p[i]) v1). the answer for query l=1, r=4 is the edge i we used to unite the above sets). A segment consists of numbers $$$[(u - 1)(v + 1), u(v + 1) - 1]$$$. This way, we only ever add edges and don't need to roll back. Example inputCopy 6 7 1 2 3 1 2 3 1 6 4 2 2 2 4 4 1 1 5 5 4 5 3 5 6 3 3 3 1 3 3 8 1 2 3 4 4 3 2 1 outputCopy 3 2 2 0 0 0 0 0 3 0 2 0 0 1 0 0 1 1 1 1 0 4 0 0 0 2 2 2 2 0 0 0 0 Note In the first test case, one of the possible ways to form a tower of color 1 and size 3 is: place block 1 at position (0,0); place block 2 to the right of block 1, at position (1,0); place block 3 above block 2, at position (1,1); place block 4 to the left of block 3, at position (0,1); place block 5 to the left of block 4, at position (1,1); place block 6 above block 5, at position (1,2); place block 7 to the right of block 6, at position (0,2). Codeforces The size of the tower is $$$s$$$, the number of blocks in it. Find the minimum height necessary for the tallest of the students' towers. Hence sqrt(n) updates value is decreasing by one in every operation. Can anyone explain the problem no. Assume that your answer is 5 now to make answer 5 we can have minimum and maximum as (1,6) or (2,7) or (3,8) and so on. Usualmente usamos ApplicationContext como contenedores de primavera. That is to say, if we can get floor value of x by dividing ith element by some value, then I did this bSet[x].set(i, 1). time limit per test. WebNow, after the first occurrence of this particular color, we inspect the problem statement and see that the most optimal greedy solution would be to build a tower using a zig-zag formation. We find the minimum element in the segment with the help of the $$$next$$$ array in $$$O(1)$$$ so this is also $$$O(a_n log(a_n))$$$. from problem B. E.g. Easier explanation for Problem B: extract the indices of each color. time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output. Try to minimize $$$k$$$ for each right end of a range. Video Solution for Problem C. Will upload Problem D in the morning :), I think there is a slight mistake in the alternative solution for D: "Continuing this logic, for all integers u=1,2,,k, we should check the elements a_i satisfying (u1)v+1a_iuv, and set all these p_i=u. The problem statement has recently been changed. So lets say for each $$$i$$$, you gave the $$$i$$$'th edge wegiht $$$i$$$. 2), [GYM] The 2023 Damascus University Collegiate Programming Contest, Codeforces Round 888 (Div. Aqu, tambin estamos hablando del cicl Recientemente, nunca he escrito COCOJS para empacar Android Encuentro un pozo intermitente. lets say we take min-max as (1,6). Then there is a state transition equation: dp(i)=min(dp(j)+i-j+1), sum(i)-sum(j)>=last(j). There are a few things I don't understand. Then for every $$$i$$$ and $$$i+1$$$ nodes. If there is no division that gives $$$v$$$, there will be another lower bound $$$v' > v$$$ that will give you a better answer, and previously trying to update the answer from $$$v$$$ will not harm your result. A tower is formed by s blocks such that they are placed at positions (x,y),(x,y+1),,(x,y+s1) for some position (x,y) and integer s. The size of the tower is s, the number of blocks in it. To get 10 under the maximum, we must divide by 2, leaving us with 5, 6 --> 6 is not the maximum. But the good part of the solution is that , suppose you are in an interation for a v that is not attainable, you are assured a v higher than that (lets call it v1) is attainable. You can either use lower_bound or if you want to do it in $$$O(n)$$$ you can notice that for every $$$a_{i - 1} < j \leq a_i$$$, $$$next_j$$$ should be $$$a_i$$$. (at least [1, a[0]]?) If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. Time complexity : $$$O(N Log^{2}N)$$$ for merging sets using DSU. For example, in the first sample test case resulting array {4,5,6,4,5} has maximum value at i=3. Trying to solve an old Topcoder TCO06 problem, Invitation to Codeforces Round 887 (Div. Good food, great staff! For the minimim X such that A//X <= B it is not working. Second, AlpertenT says that the minimum value for our fixed maximum will come from the smallest Ai. BIG burger, plus fries, coleslaw and pickles! This could work out fine however, we need to support the $$$R(i)$$$ which uses path compression so we can't do this either. The function doesn't itterate all the edge between u and v. I would like to interject to add that tjere is yet another way to achieve $$$O(log n)$$$ per query, that is by not using RMQ to do LCA, but by finding the "leftmost" and "rightmost" element in the range you are LCA-ing. We could instead add the edges $$$[l, m)$$$ to $$$g$$$, do the check then roll back these changes when traversing further down. Codeforces Codeforces Clean Code Champions: Nominate Your Favorites! Hey, can anyone please help me clear doubt in the editorial of Problem D1? But due to a wide range of problems with the round, the preparation process took several months. I haven't seen the $$$t\le 100$$$ constraint in D2, so I only count the $$$a_i$$$/x or ($$$a_i$$$/x)+1 place, let the complexity be strictly $$$O(\sum \sqrt a)$$$. Once we have $$$f(i)$$$ we just need RMQ so that works out nicely. Here is how i understand it (I'll be so glad if anybody fix it if I get anything wrong). The order of merging must be from smaller weight to large weight. But feedback from testers suggested that their current placement would be better. Thanks. 288 views Jul 19, 2022. If we ignore the floor operation for a while, n/p is integer only when p is a factor of n and there can be atmost sqrt(n) factors for a number n. Hence floor(n/p) has sqrt(n) values. It is guaranteed that the sum of n over all test cases does not exceed 2105. We start the same as we did in D1, by fixing the minimum, then bringing each element as close to the minimum as possible, but instead of checking each element individually, let's look at the various values of $$$p_i$$$ that we choose (We can easily handle the case where ideal $$$p_i$$$ >= $$$k$$$ separately).You want $$$a_i/p_i$$$ to be as close to the min as possible, so you should choose $$$p_i$$$ = $$$x$$$ for all elements of the array such that $$$x*min <= a_i < (x+1)*min$$$.Even in floor division $$$a/b >= c/b$$$ if $$$a > c$$$, so among all those values you divide by $$$x$$$, the max result would be given by the one that's just smaller than $$$(x+1)*min$$$.You can iterate over all possible $$$x$$$ from $$$1$$$ to $$$k$$$ and get the maximums in $$$O(log(n))$$$ this way, and if you break the loop when $$$x*min > 10^5$$$, it gives you an amortised complexity of I think $$$O(a_{max} log(a_{max}))$$$, Can anybody explain the logic of code of @BucketPotato problem C Qpwoeirut And The City, Solution of problem D1 using sets and lower_bound: 165982319. I can't think of a way to handle all the dsu simultaneously. Also it only works because in this union find we are not doing path compression. This can be solved by using a segment tree for the minimum. It's said that we iterate v as the minimum value, but what if it can't derived from the array p and the array a? Let's name slice as good one if its cost is lower or equal to $$$k$$$ ($$$k \ge n$$$). from where are you learn DP? Vasya wants to construct the minimal number of Submission: https://codeforces.com/contest/1706/submission/164765734. However I still don't get how you can solve the problem using just $$$R[i]$$$. 3)!!! The r-th of them should be the maximum size of an tower of color r you can form by following the given rules. Which means i screwed up my algo by iterating on k, instead i need to iterate on minimum value. The $$$r$$$-th of them should be the maximum size of an tower of color $$$r$$$ you can form by following the given rules. We do that and get a new range $$$[6, 11$$$. Este hoyo implica la alta versin de Android 6.0 o ms requiere permiso de aplicaci&oac Como nueva, la primera vez que present en CF, he experimentado 5 veces. The boy wants the instability of his set of towers to be as low as possible. More formally, we have $$$\frac{a}{b'} = \lfloor \frac{a}{b'} \rfloor + r$$$, where $$$0 \le r < 1$$$. Did I get that right? Students in a class are making towers of blocks. I mean complexity-wise. Next let's sort the ranges we have after all edges processed through the dsu. This will give us all the possible ranges of the form $$$[l, r, k]$$$, which means that it is possible to connect all vertices within $$$[l, r]$$$ using the first $$$k$$$ edges. Then $$$\frac{a}{b'} < q \Rightarrow b' > \frac{a}{q} \Rightarrow b' > \lfloor{\frac{a}{q}}\rfloor \Rightarrow b' \ge \lfloor{\frac{a}{\lfloor{\frac{a}{b}}\rfloor}}\rfloor + 1$$$, which is the formula you can see in the loop in the sample code. Finally I think it's hard to say (Max(a[i]/p[i]) v1) is greater than (Max(a[i]/p[i]) v), min(K, Floor(ai/V)) = min(K, Floor(ai/v1)), I think the proof is simple to work out.. You can take up some examples and see, Thx very much! WebCodeforces Round #809 (Div. Fast editorial.. Why D2 has an unusual memory limit? Now, after the first occurrence of this particular color, we inspect the problem statement and see that the most optimal greedy solution would be to build a tower using a zig-zag formation. In total it makes $$$O(a_n log(a_n))$$$ because this is the same as harmonic series * $$$a_n$$$.