vovuh's blog

By vovuh, history, 18 months ago,

1328A - Divisibility Problem

Idea: MikeMirzayanov

Tutorial
Solution

1328B - K-th Beautiful String

Idea: MikeMirzayanov

Tutorial
Solution

1328C - Ternary XOR

Idea: vovuh

Tutorial
Solution

1328D - Carousel

Idea: MikeMirzayanov

Tutorial
Solution

1328E - Tree Queries

Idea: MikeMirzayanov and vovuh

Tutorial
Solution

1328F - Make k Equal

Idea: MikeMirzayanov

Tutorial
Solution

• +127

 » 18 months ago, # |   0 Thanks for the editorial
 » 18 months ago, # | ← Rev. 2 →   0 Could someone please elaborate the tutorial for question B? Specifically, from where did we get the equations k<=n-i-1 and n-i-1?
•  » » 18 months ago, # ^ |   +7 Starting at the end, for every index, let's imagine some string such that it looks like aaaaaabb.Now imagine the left-most b moving to the left. For each move to the left, the right-most b has one more possible location. For example, aaaaaabb has 1 position for the second b, aaaaabab has 2, and so on.This remaining k value can also help us calculate the position of the right-most b because of the fact that it's remainder once it is less than n-i-1 hints at where it should go.
•  » » » 18 months ago, # ^ |   +2 Got your point. Thank you! :)
•  » » » » 18 months ago, # ^ |   0 Can you please explain it again. I still did not get why are we using 'k'. 'k' is just the position of string that we need to print right ?. Why are we checking if k<=n-i-1 to find the left occurrence of 'b'?
•  » » » » » 18 months ago, # ^ | ← Rev. 2 →   0 With every K , position of B is changing and notice how...look at first 10 permutation provided in the problem itself. It's not arbitrary there is a pattern. Now value of K will tell us how the final change might look like if you get the pattern.
•  » » » 18 months ago, # ^ |   0 I still didnt got your point. How we are forming these equations k<=n-i-1 and n-i-1?
•  » » » » 18 months ago, # ^ |   +2 Consider a string s of length n. First, number the positions of the letters from 1(rightmost letter) to n(leftmost letter).Lexicographically, the smallest string will have a 'b' in positions 2 and 1. It is easy to observe that there will be i - 1 strings having leftmost letter in the ith position. So, all you need to do is k -= (i - 1) as long as k > 0 and i goes from 2 to n. Suppose, j is the smallest value of i for which you weren't able to perform this operation. That means the leftmost b will be at position j. Since, this is j from the right, it will be equivalent to n - j + 1 from the left. This will be s[n - j].Now, if you would have k = 1 then rightmost b will be at position of 1. In general, for some remaining value of k, the rightmost b would be at position k. So, that would mean n - k + 1 from the left. That would mean s[n - k].For eg: n = 7 and k = 5. You can do k -= 1 and k -= 2 to get k = 2. j will be 4. That means s[7 - 4] = 'b'. Also, if k = 2 then s[7 - 2] = 'b'. Thus, s = "aaababa" for k = 5 and n = 7.I hope this helps. It will be easier to understand it if you work it out a bit more on paper with examples of your own.
•  » » » » » 18 months ago, # ^ |   0 Ok. Nice explanation!
•  » » » » » 10 months ago, # ^ |   -8 This is amazing explanation. Thank you!
•  » » » 12 months ago, # ^ |   0 Thank You for the explanation !
•  » » 18 months ago, # ^ |   0 If you notice carefully here k <= n-i-1 is really k <= (n-1)-i. And this (n-1)-i indicates the available position for right most 'b'.Now lets try to find a pattern from the given example in problem statement.when the left most 'b''s position is 3 we have 1 available space for rightmost 'b'when the left most 'b''s position is 2 we have 2 available space for rightmost 'b'when the left most 'b''s position is 1 we have 3 available space for rightmost 'b'when the left most 'b''s position is 0 we have 4 available space for rightmost 'b' K STRING POS. 1 aaabb 1 2 aabab 2 3 aabba 2 4 abaab 3 5 ababa 3 6 abbaa 3 7 baaab 4 8 baaba 4 9 babaa 4 10 bbaaa 4 Notice if k > available position for 2nd 'b' that means we have to increase the available position and to increase it we have to shift our 1st 'b' 1 position left. And if we shift our 1st 'b' 1 position left that means we have already considered all the position for 2nd 'b' for the available position. So we simply just subtract available position from k
•  » » » 18 months ago, # ^ |   0 very very nice explanation..thank you
•  » » » 18 months ago, # ^ |   0 thank you
 » 18 months ago, # |   0 nice contest :)
 » 18 months ago, # |   0 I am unable to understand third part of D's code . Any hint .. it seemed very easy ,but still i got wa at tc 2.
•  » » 18 months ago, # ^ | ← Rev. 3 →   +2 you can always get an answer if you color every odd vertex with 1 and every even with 2 but when n is odd then first and last will have same color.so if you have a consecutive pair with same value then you can color both of them with same value which will change parity of all subsequent numbers.EX: 1,2,2,3,4 can be colored as 1,2,2,1,2 instead of 1,2,1,2,1. and if no such pair exist then you have to use 3rd color.
•  » » 18 months ago, # ^ |   0 u must have forgotten to check if first ans last are same.
 » 18 months ago, # | ← Rev. 2 →   0 I had a different approach for 1st problem but it did not accept And when I compile on my idea it worked good.please help me
•  » » 18 months ago, # ^ |   0 I checked your code, the idea is absolutely correct. However, you need to print a newline after every distinct answer otherwise answers for all test cases would be shown side by side. cout<
•  » » » 18 months ago, # ^ |   -28 \n is faster than endl
•  » » » » 18 months ago, # ^ |   +14 Yes, but that's a different topic. Since he wrote the previous terms in CPP manner, I suggested using endl. And for this problem, this is not even a big deal!
•  » » » 18 months ago, # ^ |   0 even apply this I think his code gives TLE because of constraint 10^9.
•  » » » 18 months ago, # ^ |   0 will definitely timeout
•  » » 18 months ago, # ^ |   +10 Your solution will not work for large cases. Let me expand the solution given, i.e how this equation came up b - a%b — let's derive it.We know we can write a as a = b * k + a%b where k is the quotient. k can also be written as k = (a - a%b)/b. We are interested to find the next k for which remainder is zero, so we need k+1. So, a + c = (k+1)b + 0 where c is the count/number of steps (+1) to add to a to make a%b zero. Substituting k from above to find c:a+c = b * ((a- a%b)/b + 1)) = (a - a%b) + ba+c = a - a%b + bc = b - a%b. There it is!
 » 18 months ago, # |   0 The constraints for problem B were weak many o(k) solutions passed. Example — https://codeforces.com/contest/1328/submission/74509641. I tried to hack it but can't
•  » » 18 months ago, # ^ |   0 They'd probably FST.
•  » » » 18 months ago, # ^ | ← Rev. 2 →   0 Let's hope.
•  » » » » 18 months ago, # ^ |   0 It passed o_0
•  » » » » » 18 months ago, # ^ |   0 This is actually unfair to those who actually solved it :(
 » 18 months ago, # |   +3 I did B in O(1) per test case
•  » » 18 months ago, # ^ |   0 sqrtl() is actually O(lg n)
•  » » » 18 months ago, # ^ |   0 sorry. Overlooked that. Thanks bhavsar
•  » » 18 months ago, # ^ |   +12 Printing the string is O(n)
•  » » » 18 months ago, # ^ |   0 Oh Yes. I was so much excited to solve t[he main part in O(1). I completely forgot to consider that. Thanks neat
•  » » » » 18 months ago, # ^ |   0 Yes, I also used quadratic equation to solve the main part. Cheers!
•  » » » » 18 months ago, # ^ |   0 nah, I also think that I did it in O(1) till reading these comments :D
•  » » 18 months ago, # ^ |   0 hey harsh_joeyit can you explain your solution of B Problem
•  » » » 18 months ago, # ^ |   0 Sure:) K STRINGS pos of 1st 'b'(from left) 1 aaaabb - 1 2 aaabab - 2 3 aaabba - 2 4 aabaab - 3 5 aababa - 3 6 aabbaa - 3 7 abaaab - 4 8 abaaba - 4 9 ababaa - 4 10 abbaaa - 4 ...... ... .. Observation - Position of first 'b' follows a sequence of 1, 2, 3, 4, 5...... Similarly, the position of second 'b' can be traced out if we find the position of first 'b'. for every block(i.e 1, 2, 3, 4.....) second 'b' starts from 0 and goes to 1 less than the position of first 'b'. for sequence 1, 2, 3, 4, 5...... so we use (n+1)*n/2 = k ---- for finding position of first 'b' 
•  » » » » 18 months ago, # ^ |   0 I am still having a problem understanding. What I understood for first 'b' is that it will be 1 time in 1st position, 2 times in second position and so on, but how do I find out that what will be the answer when I have only the info of 'k'.Though thanks for the observation.
•  » » » » » 18 months ago, # ^ |   0 Position of first b stays at - 1 - for 1 time , k = 1 2 - for 2 times, k = 2, 3 3 - for 3 times , k = 4, 5, 6 4 - for 4 times , k = 7, 8, 9, 10 ... .. this suggested me to use the formula for series 1+2+3+4...n = n(n+1)/2 to find which block does the k belong to 
•  » » » » » » 18 months ago, # ^ |   0 I understood why you used this formula, but Sorry, I can't understand how you used this formula when only 'k' and 'n' are given.
•  » » » » » » » 18 months ago, # ^ |   0 we have to run n from 1 to find the value.
•  » » » » » » » 17 months ago, # ^ | ← Rev. 2 →   0 @Kosei_Arima See the position of the first/left 'b' from the right is 2 for the first term, 3 for the next two terms etc.,i.e., for n = 5aaabbaabab aabbaabaab ababa abbaabaaab baaba babaa bbaaaIterate from i = 2 to n from the right side. Stop when i*(i+1)/2>k; then the value of i from the right gives the position of the left 'b'. Then do val = k-i(i-1)/2. The position corresponding to val from the right side gives the position of right 'b' .For example, in the above case where n=5, take k= 4; then i = 3, is the min number such that i*i+1/2>4. So the position of left 'b' is 4th from the right Now, 4-i*i-1/2 = 4-3= 1 So the position of the right 'b' is 1st from the right. Here the indexing from right is considered to start from 1.Hope this clears
•  » » » » » » » » 17 months ago, # ^ |   0 Your explanation is great. It very much clears everything. Thank you very much from the bottom of the heart.
•  » » » » » » » » 17 months ago, # ^ |   0 from where you find the expression k-i*i+1/2to find the position of right b;
•  » » » » 18 months ago, # ^ |   0 Thanks harsh_joeyit nice explanation, got ur method
 » 18 months ago, # |   +4 Beautiful approach for E!
 » 18 months ago, # |   +3 Is there a name for the technique used in E? Or where can I read more on it (formal proofs, etc.)?
•  » » 18 months ago, # ^ |   +8 LCA using euler tour
•  » » 18 months ago, # ^ |   0 Actually, you can solve some tasks about LCA such as https://codeforces.com/blog/entry/43917?locale=ru
•  » » » 8 months ago, # ^ | ← Rev. 2 →   0 I tried to solve E using LCA, but getting WA. Any Help would be appreciated. Link : https://codeforces.com/contest/1328/submission/105254170Approach :for each query : I find the deepest Node(D), I then find the LCA of every Node(Vi) with that Node(D), If for any Node(Vi) in the query, the absolute_diff_of_depth(LCA(D, Vi),Vi) > 1, ans is "NO" else "YES".
 » 18 months ago, # |   +2 I have a different approach for B. We know that after n*(n-1)/2 permutations, both b's will be together. So we can basically make an array of all such numbers (1,3,6,10,..),and for a given k, we can find an n s.t n*(n-1)/2<=k (by binary search). Let j be the pos of that n. We can clearly see that for this value the no. of a's after the 2 b's will be exactly j (i.e, we can determine that particular permutation). So, from that permutation, we can lexicographically check next permutations till we reach k and print that kth permutation. C++ soln:74461692
•  » » 18 months ago, # ^ |   0 why did u go upto 8*10000 not upto 100000
•  » » » 18 months ago, # ^ |   0 K is a t max 2*10^9. So n would be < 2*sqrt(2*10^9) which is approximately 89000. And as we had n*(n-1), i took 8*10^5 in the estimate. You could have put a higher value.
•  » » 18 months ago, # ^ |   0 I had the same idea with you bro! my code: https://codeforces.com/contest/1328/submission/74525103
•  » » 18 months ago, # ^ |   0 Even I did the same with very easy implementation. Solution
•  » » 18 months ago, # ^ |   0 Yes, I used binary search too.
 » 18 months ago, # |   +16 " Write some LCA algorithms and other hard stuff or write about 15 lines of code and solve the problem " — One of the best editorial (E problem)!
 » 18 months ago, # |   0 can anyone explain me problem F in simple words like how to solve the problem. i am not able to clearly understand the tutorial please help thanks in advance
 » 18 months ago, # | ← Rev. 4 →   0 I believe E can be solved by just maintain an array of vertex while do dfs. When you enter a vertex mark it and all it son in the maintained array, when you leave a vertex remove all it sons from the array. Now we can answer the queries in the order we visit the vertexes in the dfs. When we enter to a vertex v , we know exactly which vertexes appear on the path to it or in dist 1 and we can answer all the queries where u = v. My mistake, I changed the problem in my head and made U alway the last vertex of the query.
•  » » 18 months ago, # ^ |   0 when you leave a vertex remove just all it sons from the array but not the vertex you just left.
 » 18 months ago, # | ← Rev. 2 →   0 I don't understand the solution for E after it explained how to figure out if vertex u is parent of vertex v. Can somebody explain it please ? (I don't want the explanation for checking if one vertex is ancestor of another, I want explanation for AFTER it).
•  » » 18 months ago, # ^ | ← Rev. 4 →   +1 if Vertex A is ancestor of Vertex B, on the way of reaching B we will cross A In dfs(), we mark every vertex by its parent. (say u is parent of v, we mark v with u). On the path to reach the deepest vertex(say vertex x) in each query: 3a. we mark all vertices with parent vertex 1(because vertex 1 is always on the path , therefore we can reach all of the immdediate children of vertex 1, since they all have distance of '1 unit' from the path )3b. we mark all the ancestors of 'x'( cause we will cross them to reach 'x')3c. we mark all the immdediate children of these ancestors we found( because we still can reach them as maximum distance from the path to x is '1')
 » 18 months ago, # |   +3 My solution for Question D without using Graphs and DP.https://codeforces.com/contest/1328/submission/74499296
 » 18 months ago, # |   +9 The site is too f#cking slow today, it's taking minutes to logout/login, all links are loading really slow, is this normal or is this new??
•  » » 18 months ago, # ^ |   0 because of the quarantine that is in place in a lot of countries a lot of people have more time to practice CP so the site gets a lot more people on it.
 » 18 months ago, # | ← Rev. 5 →   +2 Anyone else used Quadratic equations to solve the main part of Question B in O(1) for each test case??My solution :- https://codeforces.com/contest/1328/submission/74439092
•  » » 18 months ago, # ^ |   0 Yeahh...me....that give the position of "b"'s in O(1)
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 Me. But as someone suggested above, sqrt is O(logN). Even I got excited and overlooked this tiny detail . No O(1) but AC nevertheless , so who cares !!
•  » » 18 months ago, # ^ |   0 How can quadratic equation be used to solve 1328B - K-th Beautiful String ? can you explain the logic ?
•  » » » 18 months ago, # ^ |   0 Let q and w be positions of "b" in the final string from right side.Then refer to this image given below.
 » 18 months ago, # |   +3 Can someone suggest some good resources to learn how to make changes in DFS methods amd efficient graph algorithms...like many a times I see dfs1 and dfs2 kimd of functions.... Expecting genuine help!!
 » 18 months ago, # |   +17 Dear MikeMirzayanov and vovuh, nice problems. But Someone stole my rating after this round. I was dark blue color, so i should be unrated for this Div 3 !
 » 18 months ago, # |   +10 Thank MikeMirzayanov and vovuh for interesting problems and fast editorial!
 » 18 months ago, # |   0 I had used StringBuilders for the problem C. Here is my submission https://codeforces.com/contest/1328/submission/74463191 Can any java coder tell me why did it give a TLE?
•  » » 18 months ago, # ^ |   0 Use PrintWriter for fast output. System.out.println() is too slow for large output. My submission: 74442693
 » 18 months ago, # |   +3 this is my solution 74533033 to problem C in yesterday's contest. i dont understand what's wrong with it. it seems to work fine. can anyone help?
•  » » 18 months ago, # ^ |   0 I have fixed your solution: 74544016. You need only to resize your strings such as x, a and b.
•  » » » 18 months ago, # ^ |   0 thanks:)
•  » » » 18 months ago, # ^ |   0 vaaven I generally take whole string as input at once and not by each character. So I have a doubt that while taking character wise input(cin>>s[i]), why resizing the string to n, was not needed for smaller inputs(n~5) as in test cases 1 and 2. Runtime error came in test case 3 for n=250.
•  » » » » 18 months ago, # ^ |   0 Even I have the same doubt
•  » » 18 months ago, # ^ |   0 you should take input "string x" outside the for loop I guess, and then access each letter of x by x[i]. It will work.
 » 18 months ago, # | ← Rev. 2 →   0 Can anyone explain the concept behind Problem 'F' in Editorials or any simpler version that you know, I looks Easy But hard to connect logics?
 » 18 months ago, # | ← Rev. 2 →   0 For problem F, I sorted the array and used the difference between consecutive elements and the number of moves required to make all the numbers before a particular number to make all of them equal to that number. I did this for both prefix and suffix part and only counted the required number of moves. 74541622
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 how do you sure that you have " at least k equal elements in the array." in each step?
•  » » » 18 months ago, # ^ |   0 for that part i take the sum of steps from both left side(represented by 0) and right side (represented by 1) on a particular element(representing that all n-1 elements are converted to that element )and subtract it by n-k (meaning that these elements are extra element that were converted). for initialization part i took first k elements that means that first k-1 elements are converted to the the kth element and subtracted the answer with the number of occurrences of the kth element. i also did the same thing for the kth element from the last.
•  » » » » 16 months ago, # ^ |   0 Why you subtracted the number of occurrences of kth element?
•  » » » » » 16 months ago, # ^ | ← Rev. 6 →   0 Suppose k = 5,Ar = 2 4 5 6 6 6 7 8m[6] = 3dp[0] = 0 cost for all element to become 2dp[1] = (4-2 * 1) + 0 = 2 cost for all element to become 4dp[2] = (5-4 * 2) + 2 = 4 cost for all element to become 5dp[3] = (6-5 * 3) + 4 = 7 cost for all element to become 6dp[4] = (6-6 * 4) + 7 = 7 cost for all element to become 6Now m[6] = 3-2 = 1our answer is dp[4] — m[6] = 7 — (1) = 6How answer is 6 ?Keep in mind that we already have three 6's so we need not to convert 5 -> 6Final answer = 6 , 6, 5, 6, 6, 6, 7, 8 Total cost = 6-2, 6-4, 5, 6, 6, 6, 7, 8We only need to convert those which are required so if our Kth element is x and we are suxh more x already present ? we can use that too.Similarlyk = 4for ar = 1,1,1,2,3,33,3,2,2,3,3answer = 5We can do this for both sides and select the minimum one
•  » » » » » » 16 months ago, # ^ |   0 Right. That's what i had in mind for this solution
 » 18 months ago, # |   0 Please explain what exactly is happening in the B problem?how are we knowing the position of b by decreasing by n — i — 1?
 » 18 months ago, # |   0 Can someone please help me figure out why am I getting timeout even if my complexity is O(N + sum(K)) for problem E? Here's a link to my solution. https://pastebin.com/y9Fx5day. Thanks in advance!!
•  » » 18 months ago, # ^ |   0 Maxi can be equal n in worst case. So your solution is O(n^2)
•  » » » 18 months ago, # ^ |   0 Ohhh... Thanks a lot!!
 » 18 months ago, # |   0 can somebody please explain how the scoring works in codeforces? start off by explaining what's does it mean by +,+1,+2.
•  » » 18 months ago, # ^ |   0 Number of attempts
 » 18 months ago, # | ← Rev. 2 →   0 Anyone who has done D using DP ? Please share your solution
•  » » 18 months ago, # ^ |   0 I know that my code is not the most beautiful code :D, I would like if the tutorial had a DP solution for this problem.my code
•  » » 18 months ago, # ^ |   0 Here is my dp solution for proble D https://codeforces.com/contest/1328/submission/74586478
•  » » » 18 months ago, # ^ |   0 Hi Thanks for the perfect solution i was looking for. Thanks alot :)
 » 18 months ago, # | ← Rev. 10 →   0 This was my solution for B.Solution- 74469154i used the concept of "Sum of first n terms" So if the string size is n then the the total lexicographical strings produced will be sum of first n-1 terms.for ex- n=5 so total number of lexicographical strings produced will be sum of first 4 numbers 1+2+3+4=10 Then I made a vector pair consisting of lower range and the upper range of k. 1 contains (1,1) leftmost b will in the position n-2 2 contains (2,3) leftmost b will in the position n-3 3 contains (4,6) leftmost b will in the position n-4 4 contains (7,10) leftmost b will in the position n-5 (The concept is For every movement of leftmost b , Following will be the movement of rightmost b till the its position one more than the position of the leftmost b. For ex-string-abaab the rightmost b will have positions 3 and 4. ) Rightmost b's position will depend on the distance between k and its lower range for ex n=5,k=8 5 will lie in the range(4,6) leftmost b will in the position 1 (1 based indexing) rightmost b will in the position n- (difference(n-lower range)) ie 5-(5-4)=4 so the string is 'baaba' Complexity — O(n)
•  » » 18 months ago, # ^ |   0 Me too My simple Implementation for problem Blet make vector sum: O(size = 63246) = O(1) | for (ll x = 1, i = 1; x <= 2e9 + 1e5; x += ++i) | sum.push_back(x); Now for each query: O(n) | input >> n >> k | | int p1 = lower_bound(sum.begin(), sum.end(), k) - sum.begin() + 1; /// O(log size) | int p2 = p1 - (sum[p1 - 1] - k) - 1; | | for (int i = 1; i <= n; ++i) /// O(n) | putchar('a' + (n - i == p1 || n - i == p2)); My Submission
•  » » » 18 months ago, # ^ |   0 Nice code!! SPyofgame
 » 18 months ago, # |   0 Hello and thanks for contest. I was EXPERT before the contest DIV3 but my rate change -20, WHY? Thanks again.
•  » » 18 months ago, # ^ |   +2 Similarly I've seen an EXPERT coder get his rating increase for about 20 in the last div.3 round.Maybe it is a bug?
•  » » 18 months ago, # ^ |   +4 probably because you had registered yourself for this contest before rating change of last round (when you were<1600)
 » 18 months ago, # |   -9 It says time limit exceeded but is the logic correct?void solve(){ int n, k; cin >> n >> k; int l = n * (n - 1) / 2; string s(n, 'a'); int b1 = n - 2, b2 = n - 1; for (int i = 1; i < k; i++) { if ((b2 - b1) == 1) { b1--; b2 = n - 1; } else { b2--; } } s[b1] = 'b'; s[b2] = 'b'; cout << s << endl; }
•  » » 18 months ago, # ^ |   +4 Yes, logic is correct but it works very slow because you have O(k) where k can be equel 1e9.
 » 18 months ago, # |   0 Can anyone tell me what's the reason of getting TLE for the following solution of problem E? https://codeforces.com/contest/1328/submission/74550098
•  » » 18 months ago, # ^ |   0 Max depth can be n. So you will make n iterations to check only 1 node. Your solution work in O(nm)
•  » » » 18 months ago, # ^ |   0 got it. thank you.
•  » » 18 months ago, # ^ |   0 Because you iterate over the whole longest path every query, so the complexity may be O(mn) instead of O($\sum\limits_{k = 1}^mk_i$)The testcase may be a linear chain tree with every query contains a far node from the root
•  » » » 18 months ago, # ^ |   0 got it. thank you!
 » 18 months ago, # |   0 Problem : E My submission : 74555362 Verdict: WA on TC 56 Can anyone tell what is the mistake? I cant figure it out. Used the same method as told in Editorial. My code is very short. Thanks a lot :)
•  » » 18 months ago, # ^ |   0 I fixed your solution:74558555I just add this: reverse(temp.begin(), temp.end()); and edit this: if(temp[i].F>st or temp[i].S
 » 18 months ago, # |   0 Anyone who has done D using DP ? Please share your solution
 » 18 months ago, # | ← Rev. 2 →   0 C) Why if x[i]=2 or x[i]=0, then a[i]=b[i]=1 or a[i]=b[i]=0 relatively? Who can explain it? Please.
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 because max(1, 1) < max(2, 0) when x[i]=2 and max(0, 0) < max(1, 2) when x[i]=0
 » 18 months ago, # |   0 Can someone please explain this part Of problem EIt lies on this path if the root is the parent of u (it is always true) and u is the parent of fv. This approach can be used for each vertical path (such a path from x to y that lca(x,y) is either x or y).And this code of problem Eint u = v[0]; for (auto it : v) if (d[u] < d[it]) u = it; for (auto &it : v) { if (it == u) continue; if (p[it] != -1) it = p[it]; } bool ok = true; for (auto it : v) ok &= isAnc(it, u);
•  » » 10 months ago, # ^ |   0 See Mashup F: https://www.youtube.com/watch?v=mw2J6lvZZJ4&t=9712s. The explanations are really nice.
 » 18 months ago, # | ← Rev. 2 →   0 Help needed in E: My approach-: Check if there is path from root containing all the parents. Sort all the parents of given nodes according to their height. Then find lca of every adjacent pair, suppose x and y are adjacent , then if lca(x,y)==x then path exists else not. I am getting runtime error in test case 100. Can anyone help me out in this? My submission 74567267
 » 18 months ago, # |   0 ****_Problem B_ Can someone tell me the logic behind k
 » 18 months ago, # | ← Rev. 2 →   0 I used rint(sqrt(k * 2)) to solve B. What do you think? scanf("%lld %lld", &n, &k); firstb = rint(sqrt(k * 2)); secondb = k - (((firstb * (firstb - 1))) / 2) - 1; for(int i = 0; i < n - firstb - 1; i++){ printf("a"); } printf("b"); for(int i = 0; i < firstb - secondb - 1; i++){ printf("a"); } printf("b"); for(int i = 0; i < secondb; i++){ printf("a"); } puts("");
 » 18 months ago, # |   0 somebody plz explain B...
•  » » 18 months ago, # ^ | ← Rev. 2 →   +3 I also didn't know it before, got the solution by generating few cases while contest.Generate 6/7 cases and you will find the pattern. See this case: n = 5; k = 5; "a B a b a" Take minimum integer x so that k <= (x)(x+1)/2; For the first B , pos1 = (n — x)th index from left side. Keep = (x)(x+1)/2 — k; For the second b , pos2 = (pos1 + keep + 1)th index from the left side. Without pos1 and pos2 all positions will be a.
 » 18 months ago, # | ← Rev. 2 →   +1 Editorial solution for E is nice, but there is a solution that requires much less thinking. Handle the queries offline, and then you simply need to do a dfs from the root and add all the neighbors you pass on the way down to your current set (to add to your current set, just mark them as visited in all the queries that contain this vertex. if a query has everything marked as visited, then it is good) and remove the neighbors on your way back up. This way you can just check all the paths in O(n+m) time (fairly high constant but still runs < 1s).74575633
•  » » 18 months ago, # ^ |   0 can you plz elaborate how you use add and remove in your code
•  » » » 18 months ago, # ^ | ← Rev. 3 →   +6 He has made a vector of vectors qs where each entry(qs[i]) is a vector which stores queries in which i-th vertex has been mentioned. Then in add function, for a vertex v, all the queries are traversed in which v-th vertex was mentioned and another vector qc is being increased for those queries (++qc[query];). When value of qc for a particular query reaches number of vertices in that query(denoted by qn[query]), it means that all the vertices in that query are neighbours of some path from root to some vertex. This count of vertices which are neighbours, was kept by qc[query].Now remove fn is easy to understand as it is basically removing the neighbours count.
•  » » » 18 months ago, # ^ | ← Rev. 2 →   +1 Yes, amangupta's description is correct. Just to clarify a bit more on the specific variables:qn[i] — the total number of vertices in query i (qn = query number)qc[i] — the number of vertices in query i that have been currently marked (qc = query current)qg[i] — 1 if query i is achievable, 0 otherwise (qg = query good)qs[i] — a list of queries that vertex i appears in (qs = queries)
 » 18 months ago, # | ← Rev. 5 →   0 My simple Implementation for problem Blet make vector sum: O(size = 63246) = O(1) | for (ll x = 1, i = 1; x <= 2e9 + 1e5; x += ++i) | sum.push_back(x); Now for each query: O(n) | input >> n >> k | | int p1 = lower_bound(sum.begin(), sum.end(), k) - sum.begin() + 1; /// O(log size) | int p2 = p1 - (sum[p1 - 1] - k) - 1; | | for (int i = 1; i <= n; ++i) /// O(n) | putchar('a' + (n - i == p1 || n - i == p2)); My Submission
 » 18 months ago, # |   +3 Thanks MikeMirzayanov and vovuh for awesome editorial specially problem E.
 » 18 months ago, # | ← Rev. 2 →   +1 vovuh MikeMirzayanov how can we prove this fact for problem F.** The second observation is that we first need to take elements from one end (only less or only greater) and only then from the other (if needed).**why it will give correct answer and why we can't expand from both side. say if we need x, and we take y from begining first then we will take x-y from end. similarily for the same index, if we take a from beginning , we again take x — a from right.
 » 18 months ago, # |   0 in q3 , in the else statement how did the statement a[i] = b[i] = '0' + (x[i] — '0') / 2; work?please explain
•  » » 18 months ago, # ^ |   0 When x[i] = '2', then x[i]-'0' will give 2 and this divided by 2 will give 1 and this added with '0' will give '1'. So a[i]='1' and b[i]='1', as required. When x[i] = '0', then x[i]-'0' will give 0 and this divided by 2 will give 0 and this added with '0' will give '0'. So a[i]='0' and b[i]='0', again as required.
 » 18 months ago, # |   0 I feel like C was too easy :/
 » 18 months ago, # |   0 I am not getting approach of problem F. Can anyone explain it?Thanks in advance:)
»
18 months ago, # |
-6

CAN ANYONE EXPLAIN THE ERROR I HAVE REPLACED V[] WITH THEIR PARENT // TO OVERCOME THE DISTANCE FACTOR OF 1 I HAVE SORTED THE V[] ACCORDING TO THE LEVELS THEN CHECKED EVERY CONSECUTIVE WHETHER THEY LIE IN SAME PATH OR NOT

include<bits/stdc++.h>

using namespace std;

define mp make_pair

ll in[200007],out[200007]; vector adj[200007]; ll level[200007]; ll par[200007]; ll k=0; ll k2; ll timer=0; int flag; pair<ll,ll> p[200007]; ll v[200007]; void dfs(ll u,ll lv) { level[u]=lv; ++timer; in[u]=timer; for(int i=0;i<adj[u].size();i++) { par[adj[u][i]]=u; dfs(adj[u][i],lv+1); } ++timer; out[u]=timer; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); ll n,m; cin>>n>>m; level[0]=0; in[0]=INT_MIN; out[0]=INT_MAX; for(int i=0;i<n-1;i++) { ll x,y; cin>>x>>y; adj[x].pb(y); } dfs(1,1); par[1]=0; for(int idx=0;idx<m;idx++) {

    cin>>k;

flag=0;
for(int i=0;i<k;i++)
cin>>v[i];
for(int i=0;i<k;i++)
{
v[i]=par[v[i]];
// cout<<v[i]<<" ";

}
//  cout<<endl;
for(int i=0;i<k;i++)
p[i]=mp(level[v[i]],v[i]);
sort(p,p+k);
// cout<<flag<<"\n";

//  cout<<in[p[0].se]<<" "<<out[p[0].se]<<" "<<p[0].se<<endl;

int nop=0;
for(int i=1;i<k;i++)
{
if(in[p[i].se]>=in[p[i-1].se]&&out[p[i].se]<=out[p[i-1].se])
nop++;
else
{
flag=1;
break;

}
}

if(flag==0)
cout<<"YES\n";
else
cout<<"NO\n";

}


}

•  » » 18 months ago, # ^ | ← Rev. 3 →   +4 Sharing your code via one of the text storage sites like pastebin or ubuntu paste might significantly increase your odds of getting a help
•  » » 18 months ago, # ^ | ← Rev. 2 →   +1 Or simply include the submission id like 74594978This gives you TLE, wich means your solution runs out of time, hence is to slow. I think for every query you recreate the whole path from root to deepest node. This makes it basically $O(n^2)$ wich is to much for very deep trees.
 » 18 months ago, # |   0 For Problem F i didnt understood what is needl=min(need,prefcntprv) i:e prefcntprv
 » 18 months ago, # |   0 hello i submitted a solution for kth beautiful string problem.i got runtime error for that during the contest.but today i submitted that same file from my pc without any edit and it got accepted.where can i complain regarding this? this is my first contest on codeforces.please help.
•  » » 18 months ago, # ^ |   +1 I think it is not the same code, you changed the types of your integers. Most likely the runtime error was caused by overflow.Aside from that, would be useful for such question to include the both submission ids.
•  » » » 18 months ago, # ^ |   0 i didnot change anything bro...i just submitted the file that was there in my pc.without even opening it in the editor.and it got accepted this time
•  » » » » 18 months ago, # ^ |   0 Your submission history does not look like that.
•  » » » » » 18 months ago, # ^ |   0 ok let it be
•  » » » » » » 18 months ago, # ^ |   0 You're welcome.
 » 18 months ago, # | ← Rev. 2 →   0 Hi @vovuh, I have a question regarding 1328F and sorry if disturbing you.I think I may have triggered some wired behavior of Java 11, with which Arrays.sort(int[]) becomes extremely slow. From my attempted submission https://codeforces.com/contest/1328/submission/74614186, you will see that it's TLE with Arrays.sort(a) at the top part of the solve(n,k,a). If I switched from Arrays.sort(a) to use an ArrayList of tmpA to sort and assign back to a, which becomes submission https://codeforces.com/contest/1328/submission/74613801, it's accepted.I tried to reproduce this on my local machine by randomly generating 200000 numbers between 100000 and 200000 and failed, it's super fast with Arrays.sort(int[]). This may due my usage of openjdk 11, which differs from that used by codeforces, or due the data itself.Could I have the 3rd testcase so I can verify it on my local machine?
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 Quick sort is known to have $O(n^2)$ in worst case.see some link
•  » » » » 18 months ago, # ^ |   0 Hi, MikeMirzayanov and vovuh, I did further experiment on this issue. As in submission https://codeforces.com/contest/1328/submission/74698017, I deliberately force the 3rd test case to use ArrayList.sort(...) and let rest of the test cases use Arrays.sort(int[]), everything goes well and it's accepted. This makes me more confident that the 3rd test case happens to be one of the rare bad case for DualPivotQuicksort in Java 11's standard library, which actually has been there since Java 7.It would be great if I can have the 3rd test case and confirm it on my local machine. Anyway I can have it?
•  » » » » » 18 months ago, # ^ |   0 Link to the generator.
•  » » » » » » 18 months ago, # ^ |   0 So you deliberately made such a case, haha :PGood to know.
•  » » » » » » 13 months ago, # ^ |   0 Dude this is some next level stuff.
 » 18 months ago, # | ← Rev. 2 →   0 Could anyone tell me what's wrong with string c_str in GNU C++?My two submissions of problem C:https://codeforces.com/contest/1328/submission/74623565https://codeforces.com/contest/1328/submission/74623526They're exactly the same code with only difference is the lang(one is GNU C++17, the other is Clang++17 Diagnostics). I get WA with GNUC++, but passed with Clang++. Why did this happen?And I get AC using cout to print string with GNU C++:https://codeforces.com/contest/1328/submission/74623915That's weird... at least to me...
 » 18 months ago, # |   +1 74633006 TLE on tc 58 can someone help me
•  » » 18 months ago, # ^ |   +1  while(note!=-1) { mpp[note]=1; note=par[note]; } This loop could reach O(n) so your solution has a worst complexity of O(n*m)
•  » » » 18 months ago, # ^ |   0 Thanks
 » 18 months ago, # |   0 where are the ratings of this contest?? My rating was incresed by 57 but it is not showing now.
•  » » 18 months ago, # ^ |   0 Don't worry it will be back soon
 » 18 months ago, # |   +1 i need help in problem E. I'm getting wrong answer in test 43 and i don't know the reason why here is my code: https://codeforces.com/contest/1328/submission/74652500 my answer is 253243 but expected answer is 253245
•  » » 18 months ago, # ^ |   0 if you know the problem can you tell me.
•  » » » 18 months ago, # ^ | ← Rev. 2 →   +2 finally i found it after few hours! if you are having same problem as me you have to check this test my previvious solution was giving me answer 10 but answer is 11 test: 5 4 1 3 5 7 9 good luck!
•  » » » » 18 months ago, # ^ |   0 hey, thanks! it helped me a lot <3
•  » » » » » 18 months ago, # ^ |   0 no problem good luck!
•  » » 18 months ago, # ^ |   0 the same thing is happening to me :'v
 » 18 months ago, # |   0 Hello, this is my first message on Codeforces, so apologies if it violates any community guideline. I am getting wrong answer on test case 2 for Problem D. My submission id is 74659654. I have looked at it a lot but I am unable to find the problem. Can someone please help me. Thanks in advance. Happy coding !
•  » » 18 months ago, # ^ |   0 drag you submission page to the buttom, there's checker comment to show which case you didn't pass.
•  » » » 18 months ago, # ^ |   0 Thank you !
 » 18 months ago, # | ← Rev. 2 →   0 In problem E i am getting runtime error on test 100. I am not able to find the mistake in my code. Can someone help me out with error. link to my code:-74663766
 » 18 months ago, # |   +1 In question E why we are choosing depeest node among the tree please explain
•  » » 18 months ago, # ^ |   0 You can Simply refer my approach if you want Solution linkSolution
 » 18 months ago, # |   0 Could someone help me with C? I keep getting a runtime error and I don't know what's causing it. Here's my submission:https://codeforces.com/contest/1328/submission/74690659
 » 18 months ago, # |   0 Can anyone explain how we are forming the equations k<=n-i-1 and n-i-1 in B(kth beautiful string)??
 » 18 months ago, # |   0 I know my approach should result in TLE, but I am getting wrong answer in Problem E. Could someone please help. 74696152
 » 18 months ago, # |   0 My Fair simple approach for E Prerequisites:-LCA of Two nodes in log(N) time in a tree. Step1:- Running dfs traversal on tree and noting down level of each node in the tree and also the parent of each node in the tree and also obtaining the sequence of dfs traversal with which you can get LCA of two nodes tutorial link:LCA Finding tutorialStep2:-Now for each query we have to answer in YES or NO .So what I did is for each node in a particular query find parent of that node and if node is 1 skip it simply. Store two things for a node particularly its parent number and its level . Sort the obtained vector accorsing to level of parents of all nodes in decreasing order.Step 3:-Start from top and find lca of top 2 nodes and then with this obtained lca find lca of nodes successively coming and check if obtained lca value is similar to coming value and if it is then move forward till end and else break and print NO otherwise print "YES".Solution link Implementation
•  » » 18 months ago, # ^ |   0 Could you explain why did you skip the node if its parent is 1.
•  » » » 18 months ago, # ^ |   0 Because when parent is 1 after that LCA of every node is gonna 1 as 1 is root and also as we started in bottom up fashion
•  » » » » 18 months ago, # ^ |   0 Could you please check what is wrong with this piece of code. I'm creating a dummy root, I expected the code to get TLE, but it's giving wrong answer. Thanks in advance. 74696152
•  » » » » » 18 months ago, # ^ |   0 Hello __Apocalypse__.@Line 48: g[x].push_back(y);You forgot add this g[y].push_back(x);.I added it here: 75222286 and now it gets a TLE instead of a WA.Hope this helps. :)
•  » » » » » » 18 months ago, # ^ |   0 My submission Can u tell why this giving tle. I have sorted the vertices which are input in m queries according to the height of the tree given by 'dp[i]',which is the only thing different from the solution ,so my complexity is mlog(m),but still tle
•  » » » » » » » 18 months ago, # ^ |   0 Hi Chodermal1.@Line 6: void dfs(vector> ma,ll dp[],ll par[],ll src,ll p){ Here, you're passing ma by value although it is not being modified there. Hence, it is taking more time to copy the data.I changed it to: void dfs(vector>& ma,ll dp[],ll par[],ll src,ll p){ Notice the & used to pass ma by reference to prevent copying of the data and it gets an AC here.Hope this helps. :)
•  » » » » » » » » 18 months ago, # ^ |   0 Thanks bro new thing for me :)
 » 18 months ago, # |   0 I dont know whats wrong with the code,It passed all the testcases and verified the logic.It seems to give an error as unintialised value at line 56 due to which i am getting an wa.Please help!!!!here is my soln: https://codeforces.com/contest/1328/submission/74730546
•  » » 18 months ago, # ^ |   0 Initailise your array before using it..
 » 18 months ago, # |   0 Can anyone explain why this answer for test case in D(Carousel) is wrong? Input4 5 1 2 1 2 2 6 1 2 2 1 2 2 5 1 2 1 2 3 3 10 10 10 my output 2 1 2 1 2 1 2 1 2 1 2 1 2 3 1 2 1 2 3 1 1 1 1 Jury's output 2 2 1 2 1 1 2 1 2 1 2 1 2 3 1 2 1 2 3 1 1 1 1
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 Input: 1 2 1 2 2 Your output: 1 2 1 2 1 ^ ^ | | These are the same but not in the input. Assume that after the $n$-th figure the figure $1$ goes.
•  » » » 18 months ago, # ^ |   0 Thanks
 » 18 months ago, # | ← Rev. 3 →   +1 In problem F (1328F - Make k Equal — Make k Equal), Can someone please explain why the answer to the following custom test case is 29 and not 21?12 81 3 3 6 6 6 11 11 11 11 11 11Answer as per given Editorial Code = 29. Possible answer = ( 1*(6-1) + 2*(6-3) + 2*(11-6) = 21 ). {by making all 8 elements equal to '11'}
•  » » 18 months ago, # ^ |   +1 After making all the elements 6, you can't directly increase two elements to 11, because once you increase it by 1, it becomes 7 and is no longer the smallest element in the array....
•  » » » 18 months ago, # ^ |   0 My bad! Thanks a lot for pointing out!
 » 18 months ago, # |   0 Why in the solution of ternary xor we are adding '0'+(x[i]-'0')/2 is it to convert it in to string so that the number is in the string format
 » 18 months ago, # |   0 I submitted the same source code to problem C twice, the first time it failed, and the second time it is accepted. Has anyone got similar problem before?
 » 18 months ago, # | ← Rev. 2 →   0 can anyone help me find what i am doing wrong in this code of problem E75035885
 » 18 months ago, # |   0 I have solved "Problem E" using the Lowest Common Ancestor Concept. Straight forward lowest common ancestor logic.
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 My submission Can u tell why this giving tle. I have sorted the vertices which are input in m queries according to the height of the tree given by 'dp[i]',which is the only thing different from the solution ,so my complexity is mlog(m),but still tle
 » 18 months ago, # | ← Rev. 2 →   0 In problem D. why if n is even coloring always can be done like [1,2,1,2,..2] For n = 6 , 1 2 2 1 2 3 should give 3 , 1 2 2 1 2 3 instead of 2 , 1 2 1 2 1 2
•  » » 18 months ago, # ^ |   0 u can have different colours for same number. So when n is even just alternatively put 1 and 2 so each adjacent position will have different colours regardless of the fact that they are same or different
»
18 months ago, # |
0

//TRee queries solution //BISMILLAH

include <bits/stdc++.h>

using namespace std;

define sz 2*100009

map < int, int> dd; map < int,int> ff; map <int , int > cost; vector adj[sz]; int vis[sz]; int T = 0; int par[sz]; void dfs( int source){ int i,j; int len; vis[source] = 1; dd[source] = T++; len = adj[source].size(); for(i = 0;i<len;i++){ if(!vis[adj[source][i]]){ par[adj[source][i]] = source; cost[adj[source][i]] = cost[source] + 1; dfs(adj[source][i]); } } ff[source] = T++; return; } int main(){ int len; int n,q,m,i,j; int u,v; int temp,maxx = -1; int y,w; int b,c; int flag = 0; int k; int res; cin >> n >> q; m = n-1; for(i = 1;i<=m;i++){ cin >> u >> v; adj[u].push_back(v); adj[v].push_back(u); } par[1] = -1; dfs(1); while(q--){ flag = 0; res = 0; cin >> k; int a[k]; for(i = 0;i<k;i++){ cin >> a[i]; a[i] = par[a[i]]; if(a[i] == -1) continue; else if(a[i] == 1) continue; else if(a[i] != 1 && flag == 0){ u = a[i]; flag = 1; } else if(a[i] != 1 && flag == 1){ if(cost[a[i]]>cost[u]){ v = u; u = a[i]; } else v = a[i]; if(ff[u]<dd[v] || ff[v]<dd[u]) res = 1; } } if(res) cout << "NO" << endl; else cout << "YES" << endl; } return 0; }

 » 18 months ago, # |   0 A doubt please on C part although i got accepted in python in test case 5 in C language i was getting Wrong output format Unexpected end of file — token expected....... my approach was correct........as with same logic python code was accepted........1328C][YOUR TEXT TO LINK HERE... - Ternary XOR(https://codeforces.com/contest/1328/submission/75352334)
 » 18 months ago, # |   0 Can anybody please explain the reason for doing this Let's take every non-root vertex (except fv) and replace it with its parent. Thanks
•  » » 12 months ago, # ^ |   0 There are two types of nodes in queries. Either a node can already exist in the simple path from fv to root or not. Let's see both cases -1. Exist in simple path from fv to root — Since it's parent will also exist in the path we are not making any harm by replacing it with parent.2. Doesn't exist in simple path from fv to root — Since at most distance of 1 is allowed, it's better to check whether parent exist in path or not.
»
18 months ago, # |
Rev. 2   0

can someone help me with the code when i was running it it was fine as hell but in codeforces it's showing a runtime error please help me with it.... This is the division 3 5th Question's Answer:

Code Made By Rap_coder in python 3

75464011

 » 18 months ago, # |   0 Where could I find more information about the Technique used for E, such as proofs or different applications?
 » 18 months ago, # |   0 can anybody tell what is wrong in my code for problem D https://codeforces.com/contest/1328/submission/75753056
 » 18 months ago, # | ← Rev. 2 →   0 test
 » 18 months ago, # |   0 awesome contest...!!
»
18 months ago, # |
0

my solution for 1328C is written below but i get runtime error on test case 3. Could anyone please help me with it? Thanks in advance.

include <bits/stdc++.h>

using namespace std; int main() {
int t; cin>>t; while(t--) { int n,temp; cin>>temp; cin>>n; vector vec; long long int vec1=1; long long int vec2=1;

while(n>0)
{
vec.push_back(n%10);
n=n/10;
}
reverse(vec.begin(),vec.end());
int flag=0;
for(int i=1;i<temp;i++)
{   if(vec[i]==0)
{
vec1=vec1*10;
vec2=vec2*10;

}
else if(flag==0)
{
if(vec[i]==2)
{
vec1=vec1*10+1;
vec2=vec2*10+1;
}
else if(vec[i]==1)
{
vec1=vec1*10+1;
vec2=vec2*10;
flag=1;
}
}
else if(flag==1)
{
if(vec[i]==2)
{
vec1=vec1*10;
vec2=vec2*10+2;
}
else if(vec[i]==1)
{
vec1=vec1*10;
vec2=vec2*10+1;

}
}
}
cout<<vec1<<endl;
cout<<vec2<<endl;
}
return 0;

}

»
17 months ago, # |
0

This is regarding prob C of this contest. What is wrong with this piece of code?

define ll long long

using namespace std;

int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int t; cin>>t; int n; int x; string f; while(t--) { cin>>n; cin>>x; // cout<<n<<"\n"<<x<<"\n"; string f; f=to_string(x); int c=0; string a="1"; string b="1"; for(int i=1;i<n;i++) { if(f[i]=='1') c++;

if(f[i]=='1'&&c==1)
{
a.push_back('1');
b.push_back('0');
}
else if(f[i]=='1'&&c!=1)
{
a.push_back('0');
b.push_back('1');
}
else if(f[i]=='2'&&c==0)
{
a.push_back('1');
b.push_back('1');
}
else if(f[i]=='2'&&c!=0)
{
a.push_back('0');
b.push_back('2');
}
else if(f[i]=='0')
{
a.push_back('0');
b.push_back('0');
}
}
cout<<a<<"\n"<<b<<"\n";
}
return 0;

}

 » 17 months ago, # |   0 Can anyone help me why I am getting wrong answer for the question D. Carousel. Here is my code https://codeforces.com/contest/1328/submission/77173034
 » 17 months ago, # |   0 can anyone tell me which algorithm is used in B problem
 » 17 months ago, # |   0 In problem E,why are we not updating the deepest node to its parent while the rest of the nodes are updated to its parent?
 » 17 months ago, # |   0 Now we have a beautiful structure giving us so much information about the treeMike got emotional here xD.
 » 16 months ago, # |   0 found a solution for problem B that's a bit easier to understand and also runs in O(N) time: here instead of combinatorics, it uses quadratics. should be easier to understand the problem
•  » » 14 months ago, # ^ | ← Rev. 2 →   0 can you please explain your process??i can't understand your approach..
 » 15 months ago, # |   0 solution of E is hott!!
 » 15 months ago, # |   0 On E editorial."Let's take every non-root vertex (except fv) and replace it with its parent."I think that you mean: "Let's take the k vertices vi (except if vi is fv, and except if vi is the root) and replace it with its parent.".
 » 15 months ago, # |   0 My code appears to work up till test case 100 for problem E. I've spent a while looking at my code to debug it but I haven't been able to identify the issue. If anyone has any idea, I would appreciate suggestions. My Submission
»
13 months ago, # |
0

include <bits/stdc++.h>

using namespace std;

int main() { int t; cin>>t; while(t--) { int n,i,f=0; cin>>n; int a[n]; for(i=0;i<n;i++) cin>>a[i]; for(i=1;i<n;i++) { if(a[i]!=a[0]) { f=1; break; } } if(f==0) { cout<<'1'<<endl; for(i=0;i<n;i++) { cout<<'1'<<' '; } cout<<endl; } else if(n%2==0) { cout<<'2'<<endl; for(i=0;i<n;i++) { if(i%2==0) cout<<'1'<<' '; else cout<<'2'<<' '; } cout<<endl; } else { if(a[n-1]==a[n-2]) { cout<<'2'<<endl; for(i=0;i<n-1;i++) { if(i%2==0) cout<<'1'<<' '; else cout<<'2'<<' '; } cout<<'2'<<endl; } else if(a[n-1]==a[0]) { cout<<'2'<<endl; for(i=0;i<n;i++) { if(i%2==0) cout<<'1'<<' '; else cout<<'2'<<' '; } cout<<endl; } else { cout<<'3'<<endl; for(i=0;i<n-1;i++) { if(i%2==0) cout<<'1'<<' '; else cout<<'2'<<' '; } cout<<'3'<<endl; } } }

return 0;

}

 » 12 months ago, # |   0 In problem E, I created a set containing all the vertices from 1 to fv. Then I just check if parent[querypoints] are all in the set. I think this is sum(k_i)logn? Why would this give TLE while sum(k_i) and n are only 2e5
•  » » 12 months ago, # ^ |   0 No, finding all values in set in not logn. In worst case which is when all node are connected in a line and leaf is fv mentioned in solution, it is O(n).
 » 3 months ago, # |   0 I know it might be late, but I just encountered Problem B and honestly its an amazing problem. For those who want to try a more mathematical approach towards solving B unlike the editorial, we can calculate the inverse of triangular number (triangular number is n(n+1)/2) to directly calculate the position of the left most b.More specifically, position of the left most B can be calculated is: i = n - f(k) - 1 where f(x) is the inverse function of a triangular number, i is the index of the left most b.Through a similar line of thought, we can also calculate position of the right most position of b.The central intuition behind this mathematical formula was the patterns between the changing positions in the sample provided in the problem. More specifically, consider the index of the left most b as i.Then the values of first 10 i when n = 5 is: 1,2,2,3,3,3,4,4,4,4 From here it is easy to observe that the position in the series at which the value of i changes is a triangular number.