### Endagorion's blog

By Endagorion, history, 4 years ago,

Thank you for waiting! I hope you've enjoyed the problems. Let me know what you think in the comments!

UPD: I almost forgot but here are some notes, as well as challenges for all problems. For a lot of them I don't have a solution and would be happy to hear your ideas.

Challenge (awful)
Challenge (?)
Notes
Challenge (probably doable)
Challenge (?)
Challenge (???)
Notes
Challenge (probably doable)
Challenge (?)
Notes
Challenge (running out of ideas)
• +335

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 » 4 years ago, # |   -7 In 1368A;C+= a=5, b=4,n=100 why the output is 7??
•  » » 4 years ago, # ^ | ← Rev. 2 →   +25 0 a=5 b=4 1 a=5 b=9 2 a=14 b=9 3 a=14 b=23 4 a=37 b=23 5 a=37 b=60 6 a=97 b=60 7 a=97 b=157 Hope that helped. 
•  » » 4 years ago, # ^ |   -22
•  » » » 4 years ago, # ^ |   +10 Please give feedback on my video.
•  » » » » 4 years ago, # ^ |   0 you are doing great job sir
•  » » » 4 years ago, # ^ | ← Rev. 2 →   +15 The explanation is awesome. I don't know why so much negative review on it .
 » 4 years ago, # |   0 Can someone please explain the problem B solution ? I am not able to get from editorial.
•  » » 4 years ago, # ^ |   +2 For simplicity, consider 'abc' subsequences. An optimal string would look like aa...abb...bcc...c. The number of 'abc' subsequences in this string is = count('a') * count('b') * count('c').We need to find these counts such that their product >= k while also minimizing the string length. This can be achieved if the counts are as equal as possible (this is just greedy, you can try out some examples). So we can initialize the counts to 0 and increment the counts by 1 in cyclic order until product becomes >= k.For example, if k = 18, a possible solution could be 3, 3, 2 and the string would be 'aaabbbcc'.
•  » » 4 years ago, # ^ |   0
•  » » 4 years ago, # ^ |   0 Try This video Solution: — https://youtu.be/xpFsAU4wTLg
 » 4 years ago, # | ← Rev. 2 →   +3 Great contest! Though, I had a doubt regarding a different version of B, which I initially misunderstood the question for. What will we do when we will want exactly k subsequences? (Assuming that an O(N^(1/2)) solution is good enough) Can we simply keep on multiplying the count of a letter by the smallest prime factor of the remaining k(dividing k at each step) and then move on to the next letter(of the cycle) and keep on doing so until k=1?
•  » » 4 years ago, # ^ |   +8 I misunderstood the question during my first try. I thought that we have to find exactly K subsequences. So I calculate all the prime factors of the given number, multiplying the count of a character by the prime factor in a circular manner. I think this will be the optimal answer. You can check my wrong submission if you want to have a look.
•  » » » 4 years ago, # ^ |   0 Yes, but my question’s if your method is right for the question you miasassumed at first!
•  » » » » 4 years ago, # ^ |   +3 I think so but I am not sure.
•  » » » 4 years ago, # ^ |   +7 Circular manner won't work. If the count of all the prime factors (multiset) is greater than 10 then we must keep multiplying the smallest two numbers and replace them with the product, untill the size of our multiset is exactly 10. And in case if the size was less than 10, then add the required number of 1s in it.
•  » » » » 4 years ago, # ^ |   0 It makes much sense now thanks
•  » » » » 4 years ago, # ^ |   0 Are you sure about this or we have to apply some sort of dp to calculate the minimum sum of number, multiplication of those lead to k?
•  » » » » » 4 years ago, # ^ |   0 I think so, this should work. Take an example:k = 2*2*3*5*7 String = abcSo, we want 3 integers, such that their product is k and their sum is minimum possible.Multiplying 2 and 2 gives us k = 3*4*5*7 Multiplying 3 and 4 gives us k = 5*7*12Length of the string = (5+7+12) = 24. This is the best answer we can achieve.
•  » » » » » » 4 years ago, # ^ |   0 In k = 2 * 2 * 3 * 5 * 7 we can multiply 5 * 2 = 10 and 3 * 2 = 6 so our final result is 6 * 7 * 10 = 23 is the minimum answer
•  » » » » » » » 4 years ago, # ^ |   0 Oh then my approach is wrong . We'll have to find another way to minimise the sum
•  » » » » » » » » 4 years ago, # ^ |   0 I think the only way is now using dynamic programming. I will try to find the solution using dp.
•  » » 4 years ago, # ^ | ← Rev. 2 →   +7 O(N^(1/2)) solution does not exist let alone good enough. If I give you a prime number K > 1e9 then you would have no choice but to print exactly K of any letter in "codeforces". The only prime factor of K is itself. Other than that I think Akshay has a correct solution.
•  » » » 4 years ago, # ^ |   0 Right, thanks!
•  » » » 4 years ago, # ^ | ← Rev. 2 →   +3 It is wrong. You have to take a min heap and multiply the two smallest element until the size of heap is greater than 10.
 » 4 years ago, # |   +4 Btw, there is an extra challenge for problem B. (I wrote about it as my puzzle.)As mentioned in the editorial, the approach may not work well for subsequence other than "ABCDE" and "CODEFORCES". Could you find such string?
•  » » 4 years ago, # ^ |   0 Take str as “well” and k=3. “welll” should suffice here, but according the aforementioned algorithm, we have to use “wellll” which isn’t optimal
•  » » » 4 years ago, # ^ |   0 Yep, that will do (tho the original algorithm will have "wweell" instead).Any idea if two adjacent letter must be different? :)
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +3 Take the string as "abcbc" and k=5. Now, according to the algorithm we have "aabbccbc". But, due to the repeated sequence of "bc", we can take the string as "abcbcbc"(which has just two characters extra). This will give us exactly 5 subsequences of "abcbc".
•  » » » » » 4 years ago, # ^ |   0 In your example,There are 5 subsequences instead of 6
•  » » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 You're right. I'll correct it at once. Taking k=5, the counter example still holds!
•  » » » » 4 years ago, # ^ |   +10 On the string "codeforces" the algorithm will work but what should be done if we had a same question with any string in general?
 » 4 years ago, # |   +10 I made a video explaining problem D because I really liked the problem — https://www.youtube.com/watch?v=oHgcHjk2fIMThere are many people with these videos, so let me know if it's worth doing more, thanks!(Well, I liked problem E too, but I didn't solve it during the contest!)
 » 4 years ago, # |   0 The 'smaller value of k' leads one to think, the question requires the minimum value of k in Problem C. Otherwise a good problem set!
 » 4 years ago, # |   0 Are the rating changes adjusted for the fact that this was open for both div. 1 and div. 2? Personally it felt a bit unfair as difference between d and e was huge and i saw many expert coders who got under 2000 but still had negative rating change. Doesn't it discourage div. 2 coders not to participate in combined rounds?
•  » » 4 years ago, # ^ |   0 Cause they are very close to Div1((1880-1899) what I found in rate changes). They are in top 2000, but for them, it is not good enough.
 » 4 years ago, # |   +14 "We divide them into three sets $V_0$, $V_1$, $V_2$."How?
•  » » 4 years ago, # ^ | ← Rev. 2 →   +3 Look at tourist's solution : 84233462, if path[u] is 0, u is in V0, if path[u] is 1, u is in V1, if path[u] is 2(later -1, for generalizing max operation), it is in V2.
•  » » » 4 years ago, # ^ |   0 Among other things, he sorts the the vertex by label. Why? What is the meaning (in words) of path[]? More formally, what have the labels of the vertex to do with the solution? Or more common, what is the idea of this solution?I completly do not get it.
•  » » » » 4 years ago, # ^ |   +6 He sorted the adjacency list g, but did not use it(he used gr instead). So, My guess would be that it was contained in his previous solution, and hence has no use. Also, note that he uses reversed graph to like relax every node on the basis of incoming paths(and not outgoing paths).
•  » » » » 4 years ago, # ^ |   0 The idea part: After relaxing every vertex u, path[u] can only contain one of {-1, 0, 1}(Already denoting three sets {V2, V0, V1} respectively). Notice that he is using reversed graph(for incoming edges). He is relaxing in topological order(from 0 to n-1) in gr(and not g). For a vertex 'v' if all the vertices 'u', which have a an outgoing edge to 'v'(i.e 'u' ---> 'v'), are deleted(i.e they are in V2 or their path[u] is equal to -1). In this case, maximum incoming path length to 'v' is 0, hence it should be in V0. For a vertex 'v' if it has an edge incoming from a vertex in V0 and no incoming edge from vertices V1, it means that the maximum length path to 'v' is 1, and hence should be in V1. For a vertex 'v' if it has an incoming edge from any vertex from V1, then max path length upto 'v' is 2, and hence should be in V2(deleted).
•  » » 4 years ago, # ^ | ← Rev. 2 →   +3 You can traverse them in topological order and assign each vertex distance mod 3. let say dist[i] represents distance of ith vertex mod 3, for all adjacent vertices j, dist[j] = max(dist[j],(dist[i] + 1)%3)V2 is the set of vertices with dist[i] = 2,My submission
•  » » 4 years ago, # ^ |   0
 » 4 years ago, # |   -10 what is contribution in profiles?
 » 4 years ago, # |   +1 Better editorial to problem Dhttps://codeforces.com/blog/entry/78987?#comment-644649
 » 4 years ago, # |   +6 problem D was beautiful.And the part which i liked about the contest is you have to come up with a solution after that coding part is really easy and short.
 » 4 years ago, # |   0 In problem 5 what does "in topological order" means and in which set a vertex with no incoming edge belong to ?
 » 4 years ago, # | ← Rev. 2 →   0 Can Someone please explain E in a simpler fashion?
•  » » 4 years ago, # ^ |   +33 In E, we can make a directed graph with given nodes and edges. Here I will call a closed node as deactivated and an open node as activated.So, a node 'x' will be activated if none of its activated parent nodes are having any activated parent , otherwise it will be deactivated. (Greedily)Now suppose there are y activated nodes whose parents are also activated. No. of distinct parents they have must be greater than (y/2) (As each node has atmost 2 children). Now no. of distinct child they can have is 2*y. So for y + (y/2) = (3*y/2) activated nodes we can have at max 2*y (All childs) deactivated nodes.So for total (3*y/2) + 2*y = (7*y/2) nodes we can have at max 2*y deactivated nodes. Hence For n nodes we can have at max 2*(2/7)*n = (4*n/7) deactivated nodes
•  » » » 4 years ago, # ^ |   0 Thanks a lot for the explanation!! It's much clearer now!
•  » » » 4 years ago, # ^ |   0 cool ! I love your explanation!
•  » » » 4 years ago, # ^ |   0 Thanks
 » 4 years ago, # |   -27 Do check out our video editorials-A: C+=B: Codeforces SubsequencesC:Even PictureD:AND,OR and Square Sum
 » 4 years ago, # |   0 Can someone help me understand the meaning of the line "At that point, all numbers should be submasks of each other." (question D). An implementation of the solution will also be appreciated. Thanks.
•  » » 4 years ago, # ^ |   +5 https://youtu.be/54lWTHW3DBE?sub_confirmation=1 Check it out
 » 4 years ago, # |   -8 Your code here... #include using namespace std; int ADD(int a,int b,int c,int n) { if(a>n||b>n) return c; if(a>b) return (a,a+b,c+1,n); if(b>a) return (a+b,b,c+1,n); } int main() { int T,a,b,n,c; cin>>T; while(T--) { c=0; cin>>a>>b>>n; cout<
•  » » 4 years ago, # ^ |   +3 You forgot to make the recursive call in the 2nd and 3rd if statements,you have simple returned (a,a+b,c+1,n) and not made the function call. :)
•  » » » 4 years ago, # ^ |   +6 thank you very much
•  » » » » 4 years ago, # ^ |   0 Happy to help :)
•  » » » » » 4 years ago, # ^ |   0 but still it is giving error
•  » » » » » » 4 years ago, # ^ |   0 Your code is not working for the case where a=b, I have modified your code you can check it out here
•  » » » » » » 4 years ago, # ^ |   0 in main function call ur function with c=0
•  » » » » » » » 4 years ago, # ^ |   0 i m calling the function with c=0;
•  » » » » » » » » 4 years ago, # ^ |   0 ohh sorry ,my bad,didn't saw c=0
•  » » 4 years ago, # ^ | ← Rev. 3 →   0 Just need a equal in the 3rd if and i got accepted with your code...Your Edited accepted code
 » 4 years ago, # |   0 Great editorial for problem F, Thanks!
 » 4 years ago, # |   -6 I need help in understanding, why my E approach does not work. I kinda proved it should work under constraints (If it does what I want it to do).Algorithm as follows : First you check for all nodes without incoming edges ($V_0$). Then you make a set of nodes that these nodes touch ($V_1$). Then make ($V_2$) out of those accordingly. ($V_0$) Will never be spoiled, because no edges enter them. But ($V_1$) might point to another ($V_1$), making it ($V_2$) maybe.So I keep track of votes for being ($V_2$). votes = how many edges from ($V_1$) enter this node. Nodes with nonzero votes are the ones we remove. Now go from top to bottom, and if we meet some node that used to be ($V_1$) but got voted by someone else, I put it in ($V_2$) and undo its votes, since it does not count as good any more. Since I do this from top to bottom, I only take necessary nodes in greedy way.The answer should give different combinations of binary trees that are less than 3 in depth, and all of them have number of bad nodes $\leq$ $\frac{4}{7}$, and each node can be associated with one such binary tree, so the answer must be also less than constraint.After the cycle, we remove $V_0$ $V_1$ and $V_2$, also changing incoming nodes from $V_2$ since they touch the rest of the graph and continue till we have $V_0$ emptySumbission : 84272118
 » 4 years ago, # |   -77 How is making a pattern a CP question?
•  » » 4 years ago, # ^ |   0 You have to find the pattern first
 » 4 years ago, # |   0 I had a doubt in Problem E. I tried solving it like this: Form the graph and remove only those vertices which have both incoming and outgoing edges. As I remove a vertex, I also update the graph accordingly. I tried to analyze the number of vertices that I will be required to remove in the worst case. For that, we can consider the whole graph as a binary tree and what my approach is doing is that it is removing all the vertices at even levels (considering 1 based indexing of levels). In this case, the maximum vertices that will be removed will be n/2 which is less than 4n/7. However, I am getting the wrong answer on Test Case 7. Can somebody give me a counter-example or tell me if I am making a mistake somewhere. Thank you. Here is my submission link: https://codeforces.com/contest/1368/submission/84306393
•  » » 4 years ago, # ^ |   +3 But the whole graph is not a binary tree.The verdict of your submission is clear. You deleted 6 nodes in Case 137 of Test 7, while there are 10 nodes, so you can delete no more than 5.
•  » » » 4 years ago, # ^ |   0 Thanks for the reply. Yeah, that is the issue, I am not able to understand how come the answer is exceeding n/2. I think I will have to work out some test cases to check that. I have not been able to find any test case that fails until now. If you have a test case that you can suggest, it would really help.
•  » » 4 years ago, # ^ | ← Rev. 3 →   +11 1 — 5 1 — 6 2 — 7 2 — 8 3 — 10 3 — 9 5 — 11 6 — 11 7 — 11 8 — 11 9 — 11 10 — 11
•  » » 4 years ago, # ^ |   +3 i also tried the same thing ....but later understood that in this case no. of deleted nodes is2*(n-1)/3 .. so it fails for n=10.
 » 4 years ago, # |   0 Thumps up for high quality descriptive editorial. By the way, in number c "The sample picture was a red herring". I stepped into that trap.
 » 4 years ago, # |   +174 That feel when you AC C with an ugly solution...
•  » » 4 years ago, # ^ |   +4
•  » » » 4 years ago, # ^ |   +12 You are not alone.
 » 4 years ago, # |   0 can any one explain the question part in D??? it says:  If before the operation ai=x, aj=y, then after the operation ai=x AND y, aj=x OR y, where AND and OR are bitwise AND and OR respectively 
•  » » 4 years ago, # ^ |   0 void operation(int& x, int& y) { int aux=x&y; y=x|y; x=aux; } ... operation(a[i],a[j]); ... 
 » 4 years ago, # |   +8 Beautiful solution for E.Anyone know any algorithms for minimizing k, or interesting bounds on k for different constraints on the graph? Does this problem have an existing name in the literature?
 » 4 years ago, # | ← Rev. 2 →   0 "The same solution can be simply rephrased as: go from left to right, and remove current vertex if it is at the end of a two-edge path"I am writing an explanation vis-a-vis the correlation of the sets $V_0, V_1, V_2$ to the above algorithm,Let, $V_2$ represent the set of all the removed vertices.Now, the vertices that only have incoming edges from $V_2$ are the "roots" of a sub-graph that has been disconnected from the initial graph, let they be represented by the set $V_0$, the vertices in $V_0$ no longer have any incoming edges, therefore, they can't possibly be the endpoints of a path of length 2 or more, so according to the algorithm they don't have to be removed.Similarly, all the "neighbours" of the vertices in $V_0$ need not be removed (they are the endpoints of paths of length 1), these "neighbours" form the set $V_1$.P.S after reading this you can refer to the editorial for the upper-bound on the number of elements in set $V_2$, it's exactly $4n/7$.
•  » » 4 years ago, # ^ |   0 Now, the vertices that only have incoming edges from V2 are the "roots" of a sub-graph that has been disconnected from the initial graph, let they be represented by the set V0, the vertices in V0 no longer have any incoming edges, therefore, they can't possibly be the endpoints of a path of length 2 or more, so according to the algorithm they don't have to be removed.How can you guarantee that the vertices in V0 do not have tracks to each other?
 » 4 years ago, # |   0 An Interesting Problem:In problem B Consider exactly k instead of at least k. I think it would be problem of finding number of factors <= 10 such that their sum is minimum. Prime value of k would be nightmare.
 » 4 years ago, # | ← Rev. 2 →   0 i'm a little confused about E, is it saying to just remove the node if it looks like this?0 — 0 — 0then remove middle to0 — X — 0cuz i think a lot of solutions failed that when trying to greedy like that
 » 4 years ago, # |   +3 Hello. I need help in problem E. My submissions: this close more than 4/7 n vertices and this may leave tracks with a length of 2 or more . I found counter test where this submissions don't work. Is this possible in this task? Why my solutions work? This test 20 191 22 42 51 33 63 77 88 99 119 128 1010 1310 1415 1717 817 2015 1616 1916 18
•  » » 4 years ago, # ^ |   +6 For all edges x->y the statement says that x
 » 4 years ago, # | ← Rev. 3 →   0 A simpler solution for E problem:Let's invert all edges in the graph, then for each vertex in graph no more than two incoming edges. Also for each edge u -> v, u > v. We will iterate over the vertexes in ascending order and for each vertex we will check whether there is a path of length 2 from it through not deleted vertexes, if there is then we will delete this vertex.Note that for each deleted vertex there is an edge to not deleted vertex, from which also exists edge to not deleted vertex. Since after we inverted the graph, each vertex contains no more than two incoming edges. Then each not deleted vertex contains a maximum two incoming edges from the deleted vertices, but at the same time for such vertex with incoming edges from deleted vertices there is an edge to not deleted vertex, which means that the number of deleted vertices <= 1/2(for 2 deleted vertices which have an edge to same not deleted vertex, exists 2 not deleted vertices)
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Here is the counterexample to your upper_bound = 1/2n. 1 7 6 1 2 1 5 2 3 2 4 5 6 5 7 Your algorithm removes 4 vertices = 4*n/7.
 » 4 years ago, # | ← Rev. 2 →   +1 In problem E: sample test case 1: if we close both 3 & 4 ,how will we reach 4 from 2 since 4 is closed so there should be no incoming edge on 4? Kindly explain sample test cases of Problem E ?
 » 4 years ago, # | ← Rev. 2 →   0 In E question : Can someone please explain how paths and tracks are different from each other and if suppose this is our graph:1 21 32 4 does it mean that 1, 2 ,3, 4 are on the same line?
 » 4 years ago, # |   +3 https://codeforces.com/contest/1368/submission/84323789 can anyone tell me why is TLE coming in testcase 2
•  » » 4 years ago, # ^ |   +1 I had the same issue, you are initializing an array of size = 1e5, for every testcase, which is not necessary. Try using atmost linear space.
•  » » » 4 years ago, # ^ |   0 ok thx will try that
 » 4 years ago, # | ← Rev. 2 →   -8 Can anyone explain me the solution of Problem D elaborately. what's happening there ? I am unable to visualize it .
•  » » 4 years ago, # ^ |   0 I can try For D we have to play with bits, That is when you and a ,b and simultaneously or a,b Total bits of a and b are preserved in the ored number that is a|b Thus when we have an array of numbers Out first priority is to set all the bits in some number because consecutive sum of squares of (2^n-1)^2+(2^n)^2. << (2^(n+1)-1)^2 By this observation I tried to accumulate available bits in only some of the numbers Not distribute it in all the numbersSo we will accumulate the bits in only some numbers . rest all numbers can be 0 . Lets say we have 4 bits of 0th place , 4 bits at 1st place and 2 bits at 3rd place then in total we will try making 4 different numbers which are 11 11 3 3. And then finally we will find the square of these values. Carefully observe that by oring the bits and its frequencies are preserved. and by and the bits which are 1 in both the numbers stay preserved
•  » » » 4 years ago, # ^ |   +6 I appreciate your efforts but still i can't understand it ..
•  » » » » 4 years ago, # ^ |   0 Are you able to understand each of the following observations? Suppose w = x & y, z = x | y (where & is bitwise and, | is bitwise or). Then x + y = w + z w^2 + z^2 >= x^2 + y^2 The sum a_1 + a_2 + ... a_n is unchanged by any operation If so, try to think about maximizing the sum of squares while maintaining the invariant from #3
 » 4 years ago, # |   +3 Why in E removing middle vertex in 2-edge path will not work?
•  » » 4 years ago, # ^ |   +6 All of the middle vertices could have tracks to the same end vertices.Say you have this input 1 10 12 1 4 1 5 2 6 2 7 3 8 3 9 4 10 5 10 6 10 7 10 8 10 9 10 Removing the middle vertices would remove 4, 5, 6, 7, 8, and 9. This would be 6/10 vertices removed, and 6/10 > 4/7. Worst case on that strategy approaches 2/3 with more nodes (need 1 first node for 2 middle nodes, only need 1 final node total)
 » 4 years ago, # | ← Rev. 2 →   0 In problem 5, Ski Accidents, What would the answer for this test case: 3 3 1 2 1 3 3 2 According to me, (4*3)/7 = 1, and by deleting any one node would satisfy conditions. That is, the answer should be 1 1, 1 2, or 1 3. But in most of the accepted solutions, it is showing 0 as the output answer. Am I right? Even my accepted solution is also printing 0 as the answer.
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Sorry, my bad, x < y in every test case but test case I have given has x > y
 » 4 years ago, # |   0 I spent two hours trying to solve B, for exactly k subsequences, before i realized i miss read the problem. Anybody has an idea on how to solve this variation of the problem ?
 » 4 years ago, # | ← Rev. 3 →   0 My Doubt is for Problem C. Can Anyone Explain the Meaning of this line in Problem Statement of C:"There are exactly n gray cells with all gray neighbours. The number of other gray cells can be arbitrary (but reasonable, so that they can all be listed)."What i Understood with this line is :if n=1 then a grid of four cells Can be the Answer(because it is satisfying all three conditions)I Know i am wrong somewhere,Can Anybody Point out the Catch i am losing!
•  » » 4 years ago, # ^ |   0 No, if $N = 1$, then the grid looks something like this:011111110where 1 represents gray cells. Here, only the middlemost cell $(2,2)$ [1-based indexing] has all gray neighbours. And notice all other gray cells have even neighbours. For extending to any arbitrary $N$, you can follow the pattern of editorial.
•  » » » 4 years ago, # ^ |   0 According to your example ,The (1,1)Gray cell [1-Based indexing] is also surrounded by all gray neighbours and the number of neighbours are even too.Right?
•  » » » » 4 years ago, # ^ |   0 No, notice that grid is actually infinite, so a gray cell needs to be surrounded by gray cells in all 4 directions (sharing an edge).
•  » » » » » 4 years ago, # ^ |   0 Oh! But According to this line in the problem statement,"assume that the sheet is provided with a Cartesian coordinate system such that one of the cells is chosen to be the origin (0,0), axes 0x and 0y are orthogonal and parallel to grid lines, and a unit step along any axis in any direction takes you to a neighbouring cell."I thought first cell would start from (0,0).This is a bit conflicting right? blitzitout
•  » » » » » » 4 years ago, # ^ |   0 Actually that thought never crossed my mind. The author mentioned that fact only when outputting the gray cells. But yes, you can clarify with the authors during the contest.
•  » » » » » » » 4 years ago, # ^ |   +3 Sure Brother, Thank You So Much for the genuine explaination blitzitout
•  » » » 4 years ago, # ^ |   +3 Thanks for making this grid for N=1, this observation made me change what was wrong in my approach, and then it got accepted.
 » 4 years ago, # |   0 For the solution to D, it's written that x^y+xANDy=x+y. But when you set x = 3 and y = 5 for example, the equation becomes 6+1=8 which is incorrect. Can someone help me understand this claim? Thanks.
•  » » 4 years ago, # ^ |   +3 The operations are OR and AND, not XOR and AND.
 » 4 years ago, # |   0 Though, I could clearly form an idea on C, I chose not to implement it. I hate those kind of problems.
 » 4 years ago, # | ← Rev. 3 →   +3 Hi all, 1368D — AND, OR and square sumWhy?2d(d+y−x), which is positive when d>0.Thank you very much.
 » 4 years ago, # |   +8 Auto comment: topic has been updated by Endagorion (previous revision, new revision, compare).
 » 4 years ago, # |   +3 For the solution to F, I think $x + k + \dfrac{x + k}{k - 1} \leq n$ transforms into $x \leq n - k - \dfrac{n}{k}$, not $x \leq n - k - \dfrac{n}{k} + 1$?
•  » » 4 years ago, # ^ |   +3 Endagorion Could you please answer this?
•  » » » 4 years ago, # ^ |   0 Yeah, you're right, I was off by one. The answer still stays the same since $x \leq n - k - \frac{n}{k}$ is applied immediately before $x$ can be increased, so it can be one larger in the very end.
 » 4 years ago, # |   0 Can anyone explain the side note for the editorial of problem D?Side note: an easier (?) way to spot the same thing is to remember that f(x)=x2 is convex, thus moving two points on the parabola away from each other by the same amount increases the sum of values.
 » 4 years ago, # |   0 Where is the "Contest materials" in the contest page? There must be something wrong.
 » 4 years ago, # |   0 One optimization for problem B could be initializing every occurrence of the letters in "codeforces" to n^(0.1). The resulting numbers (Nc*No*Nd*..)will always be greater than or equal to n^(0.1) you can realise it better if you know AM(arithmetic mean)>Geometric mean principle.