### gepardo's blog

By gepardo, history, 4 weeks ago, translation,

1420A - Cubes Sorting

Hint
Solution
Code in C++ (Wind_Eagle)

1420B - Rock and Lever

Hint
Solution
Code in C++ (Wind_Eagle)

1420C1 - Pokémon Army (easy version)

Hint
Solution
Code in C++ (gepardo)

1420C2 - Pokémon Army (hard version)

Hint
Solution
Code in Java (gepardo)
Code in C++ (Wind_Eagle)

1420D - Rescue Nibel!

Hint
Solution
Code in Java (gepardo)

1420E - Battle Lemmings

Hint 1
Hint 2
Hint 3
Solution
Code in C++ (gepardo)

• +314

 » 4 weeks ago, # |   +8 Auto comment: topic has been translated by gepardo (original revision, translated revision, compare)
•  » » 4 weeks ago, # ^ |   -48 Video Solution B — https://youtu.be/4U9WZVi4gdY Video Solution C(easy)- https://youtu.be/IWpnF3oUdR8
 » 4 weeks ago, # | ← Rev. 2 →   +172 Actually, many people say that now there are practically no data structure or DP tasks, only ad-hoc and constructive. Well, this contest disproves this. There were no constructive tasks, but there were Segment Tree in C2 (there is a solution using it) and DP in E.
•  » » 4 weeks ago, # ^ |   +245 One contest doesn't disprove a trend.
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   0 I am actually better at ad-hoc and constructive currently and loved the trend :sobs:, But I do like these rounds where I learn a lot.
•  » » 4 weeks ago, # ^ |   +3 Please could you explain/hint about the Segment tree solution of C2. How will we merge answer of left and right for getting answer of current node ? Thanks in advance!
•  » » » 4 weeks ago, # ^ |   +19 You can maintain four quantities in the segment tree node i.e maximum strength given that 1. first element has $plus$ sign before it and last one also has $plus$ sign 2. first element has $plus$ sign and last has $minus$ 3. first element has $minus$ sign and last has $plus$ sign 4. first element has $minus$ sign and last has $minus$ sign You can easily implement merge and update operations. Here is my implementation https://codeforces.com/contest/1420/submission/93716090. There may be easier solutions but this one was more obvious and intutive
•  » » » » 4 weeks ago, # ^ |   +1 Got it bro. Thanks a lot fenwick123.
•  » » » » 4 weeks ago, # ^ |   +18 Tch tch, "fenwick123" had to use segment tree. xD
•  » » » » 10 days ago, # ^ | ← Rev. 2 →   0 Hey, I'm trying to implement your approach, but I'm getting TLELink to my code: https://codeforces.com/contest/1420/submission/95420456
•  » » 4 weeks ago, # ^ |   0 Hi Sir, I wanted to know is there any site that focusses more on data structures and less on ad-hoc?
•  » » » 4 weeks ago, # ^ |   +5 I think atcoder contests are better in this matter.
•  » » » 9 days ago, # ^ |   0 Leetcode problems and its contests.
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   0 I implemented segtree and received tle in C2. someone willing to help? https://codeforces.com/contest/1420/submission/93731337
•  » » » 4 weeks ago, # ^ |   0 Massive output, use \n instead of endl.
•  » » 4 weeks ago, # ^ |   0 Also, if you exepected segtree, the constraints could've been relaxed (:
•  » » 3 weeks ago, # ^ |   0 @Wind_Eagle can you give the solution of problem c2 using segment Tree.
 » 4 weeks ago, # |   +62 For E, if we accept solution in O(n^5), I think it may be better to choose a smaller n. For me, I was too afraid that an O(n^5) solution will TL and just tried to think of a better solution.
•  » » 4 weeks ago, # ^ |   +26 We initially thought of making $n \le 100$ in this problem, because my solution in $O(n^5)$ passes with such constraints. Anyway, the DP solution is pretty fast and $n^5$ can be easily divided by some constant factor, so the constraints may seem too big for $O(n^5)$.
•  » » » 4 weeks ago, # ^ |   +34 I can understand your choosing n<=80. Maybe I just need to accept the fact that computers are much faster nowadays.
•  » » » 4 weeks ago, # ^ |   0 BTW, my solution at https://paste.storage.boleyn.su/cfr672d2e.cpp should have a better time complexity. I have not analyzed its TC yet so this is just my guess.
•  » » » » 4 weeks ago, # ^ |   0 Can you explain your approach? Is it true that you use a DP similar to the one described in the editorial, but you optimized the code by skipping invalid DP states?
•  » » » » » 4 weeks ago, # ^ | ← Rev. 4 →   0 For the last dimension, it is non-decreasing, so we can express it using a vector> and reduce some useless states.My guess is that there may be a bound better than O(n^2) for it. It would be nice to prove or disprove my guess.UPD: I tested it with larger constraints. It turns out that my guess is wrong. The size of vector> is still O(n^2).
 » 4 weeks ago, # | ← Rev. 2 →   0 I felt like D had hard time limits. But, later realized after contest with precomputation of inverse of each factorial it could have been made faster.
•  » » 4 weeks ago, # ^ |   0 Can you(someone) explain how to do that? Cause I got a TLE on pretest-11.
•  » » » 4 weeks ago, # ^ |   0 precompute fact 1 to n and mod inverse of those instead of computing them every time.
•  » » » » 4 weeks ago, # ^ | ← Rev. 2 →   -6 Actually, I did that but still got a TLE(UPDATED: NEVER MIND I PASSED IT)
•  » » » » 4 weeks ago, # ^ |   0 you can calculate the mod inverse at run time also, pre-computing the factorials is enough
•  » » » 4 weeks ago, # ^ |   +2 Precompute factorials from 0 to N (here $3*10^{5})$ initially in an array, say $fact$.Then calculate inverse factorial of N using formula $inv_f(N) = fact(N)^{MOD-2} \% MOD$ which takes $O(\log_2 (MOD))$ time. Using this value, now fill the remaing values from N-1 to 0 using recurrence : $inv_f(i) = (i+1)*inv_f(i+1)$. Thus we can precompute inverse factorials in $O(N)$ complexity. Solution for reference : 93729779
•  » » » » 4 weeks ago, # ^ |   0 Hey, I precomputed the factorials and inverse factorials similar to your solution but I am getting TLE.
•  » » » » » 4 weeks ago, # ^ | ← Rev. 2 →   0 Use binary exponentiation in your powmod function to calculate $a^b$ in log(b) complexity, you are calculating it in O(b) time. You can read it here : cp-algo
•  » » » » » » 4 weeks ago, # ^ |   +1 Thanks, didn't know about binary exponentiation.
•  » » » » » » 4 weeks ago, # ^ |   0 Can you tell is problem D is some standard type based because getting that intution was tough at first tym as mentioned in solution
•  » » » » 4 weeks ago, # ^ |   0 Note that precalculateing the inverse factorial does not give a better runtime. We never need to calculate more than $3*10^5$ inverses. So precalculating them just uses more memory, but not less time.
•  » » » » » 4 weeks ago, # ^ |   +3 But to calculate $3*10^5$ inverse factorials you'd have to calculate them in $O(log_2(Mod))$ individually resulting in $O(N*log(Mod)$ time. But here I am precalculating them in $O(N)$, so I am unable to understand your logic behind why this won't give better runtime? Can you please elaborate.
•  » » » » » » 4 weeks ago, # ^ |   0 You are right. I was irritated by the long runtime of your solution. But that is caused by the use of the map instead of sorting an array once.
•  » » » » 4 weeks ago, # ^ | ← Rev. 2 →   +3 I encountered this for the first time, tried to dig around but can't figure out why this works- invf[i] = invf[i+1]*(i+1)%mod; Can u help or tell me where to look.
•  » » » » » 4 weeks ago, # ^ | ← Rev. 3 →   +3 n! = n*(n-1)! (n!)^(mod-2)%mod = ((n * (n-1)!)^(mod-2))%mod = (n ^(mod -2) * (n-1)!^(mod-2))%mod (invf(n)*n)%mod = (n!^(mod-2) * n)%mod = (n^(mod-1) * (n-1)!^(mod-2))%mod = 1 * (n-1)!^(mod-2)%mod = invf(n-1)%mod Does this makes sense?
•  » » » » » » 4 weeks ago, # ^ |   +4 $(n+1)!^{−1} = (n!*(n+1))^{-1}$ $(n+1)!^{−1} = n!^{−1}*(n+1)^{−1}$ $(n+1)!^{−1}*(n+1) = n!^{−1}$
•  » » » » » 4 weeks ago, # ^ |   0 Watch from 16:00 onwards Lecture on Combinatorics By Kevin Charles Atienza
•  » » » » 4 weeks ago, # ^ |   0 Hi there. Please, tell me about inverse factorial of N. What is it?
•  » » » » » 4 days ago, # ^ |   0 the inverse of any number n is just 1/n
•  » » 4 weeks ago, # ^ |   0 I precomputed the factorials and the inverses, thus my solution was linear, though I think $O(n\log mod)$ should fit in time.
 » 4 weeks ago, # |   +53 Great problemset, like it . waiting for your next contest,thanks
 » 4 weeks ago, # |   0 For some problem editorial is in english and for others it's in russian . Pls Fix.
•  » » 4 weeks ago, # ^ |   0 The English versions will be loaded in several minutes.
•  » » » 4 weeks ago, # ^ |   0 Can you tell if the elements of array in C1 are not distinct what will happen??
•  » » » » 4 weeks ago, # ^ |   0 I don't understand your question. The elements $a_i$ are distinct by problem statement. Or do you want to know if the intended solution works if equal elements are allowed?
•  » » » » » 4 weeks ago, # ^ |   0 Yess if we go by approach for maxima and minima approach for C1 problem then will it work??
•  » » » » » » 4 weeks ago, # ^ |   0 It is unclear how we will treat an element if its neighbor is equal to it. Or even worse, when both neighbors are equal to our element. Will it be a local maximum or local minimum? Also it's important that local minima and local maxima must alternate (this property is used actively in the proof).Anyway, it's an interesting question on how to fix the solution when equal elements are allowed.If you solve C1 using dynamic programming, you should not be affected by equal elements.
•  » » » » » » » 4 weeks ago, # ^ |   0 Yess C1 by dp will work.. And is there any specific reason for giving ai values from 1 to n??
•  » » » » » » » 4 weeks ago, # ^ |   +56 One may assume that out of equal neighboring elements, the left one is smaller. This is equivalent to adding $i \cdot \varepsilon$ to $a_i$, which makes all elements distinct while keeping the answer the same if $\varepsilon \rightarrow 0$.
•  » » » » » » » 4 weeks ago, # ^ | ← Rev. 2 →   +20 A trivial fix is to break ties by index, so that if $a_i = a_{i+1}$, we can simply pretend $a_i < a_{i+1}$. This is safe because the value of the optimal solution is continuous in the values of $a_i$ and we can imagine taking $a_i = \lim_{\varepsilon \to 0^+} a_i + i \cdot \varepsilon$ simultaneously at all indices. Alternatively, the solutions based on the $\sum_{i = 1}^n \max (0, a_i - a_{i-1})$ formula or on segment trees are completely unchanged, since they do not make use of the distinctness constraint in any way.EDIT: I was beaten to the punch! So it goes.
 » 4 weeks ago, # |   0 what is "Разбор"???? is it rated??
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   +5 It is a Russian word for "parsing".
•  » » » 4 weeks ago, # ^ |   +21 It's better to translate "разбор" as "analysis" :)
•  » » » » 4 weeks ago, # ^ |   0 thank you!!1!
 » 4 weeks ago, # |   0 Round was great. Kudos to author!!
 » 4 weeks ago, # |   0 My editorial is in Russian :(
•  » » 4 weeks ago, # ^ | ← Rev. 3 →   +1 Everything is OK now, except problem E.UPD: Problem E is also OK now.
•  » » » 4 weeks ago, # ^ |   0 Does "except problem E" mean this?
•  » » » » 4 weeks ago, # ^ |   +4 Yes.
 » 4 weeks ago, # |   +7 I could not solve a single problem in this contest, should I die? Why am I even alive.
•  » » 4 weeks ago, # ^ |   +14 Don't feel demotivated you have given just few contest.just keep practicing and keep motivated :)
•  » » 4 weeks ago, # ^ |   0 Same bro, not able to solve a single question. But after see the first questions editorial, i think that i thought too much for that question. Can anyone suggest how to approach a question when your attempts are wrong ( means when your pretests doesn't pass ).
•  » » » 4 weeks ago, # ^ |   0 @shivam_0105, first decide which one of the following is going wrongyour logic your approach or is it some specific test case? is your datatype overflowing somewhere? (I did not get a warning of overflow in contest this time that is why I could not submit my correct B solution during contest)if you are confident that 2 out of 3 are 100% correct then try to think what might be wrong with the 3rd one... if still you are not able to get it, just move on, read the next questiontry solving that unsolved question one more time during upsolving, still no? now its your choice to look at the editorial or solve more... (I would look at the editorial because sometimes you just need some specific maths formula/observation/trick that you don't know. But not this time though, all you had to know how bubble sort works to solve A and for B you had to know how the most significant bit is larger than sum of all bits before than it for ex. how 2^4 is greater than 2^3 + 2^2 + 2^1 + 2^0, if you knew this, question B was cake walk...)
•  » » 4 weeks ago, # ^ |   0 change ur dp then u will solve questions
•  » » 4 weeks ago, # ^ |   +1 just take your meds : one tourist stream every day ! starting with : https://www.youtube.com/watch?v=97tieEKfvBs
 » 4 weeks ago, # |   0 I don't understand the solution for problem D. This is my understanding so far :-For example: if there is a starting point on 42 and our count of segments excluding this is 10 and the value of k is 3. then I found out the solution is to add ncr(10,k-1).Now the count is 11 and if we again reach to another starting position on 45 so again we will add ncr(11,k-1).Now count is 12.So if I understood it correctly, then we will choose one out of k to be the current one and choose rest k-1 from the previous segments right?
•  » » 4 weeks ago, # ^ |   +1 If we fixed a starting point, we may pick any $k$ segments that contain it. But there must be at least one segment that start in this point (note that more than one one segment may start in the same point).
•  » » » 4 weeks ago, # ^ |   0 Actually, I was talking something like this solution.I tried this concept after the contest.
•  » » » » 4 weeks ago, # ^ |   +4 I understood your solution. It's mostly the same as in the tutorial, but it considers the segments that start in the same point in a shorter and nicer way.
 » 4 weeks ago, # |   +15 How to solve C2 using segment tree?
•  » » 4 weeks ago, # ^ |   +27 I used an ad-hoc segment tree with 4 values for each range: The max value you can get by using an even number of elements in that range The max value for an odd number of elements The min value for an even number of elements The min value for an odd number of elements The value (max/even) of a node is the max from these values: left child's (max/even) + right child's (max/even) left child's (max/odd) – right child's (min/odd) The rest of the values are obtained similarly. Updating the tree is the same as a regular segment tree.Here is my code.
 » 4 weeks ago, # |   +3 I am still a bit surprised to see the no. of submissions on the Problem D, Is it some sort of a standard algo or question? Because some were able to this in a matter of just few minutes, if it is then please do tell me
•  » » 4 weeks ago, # ^ |   0 I don't know weather it's standard or not but many interval related questions have similar approach.
•  » » » 4 weeks ago, # ^ |   0 What is the approach?
•  » » » » 4 weeks ago, # ^ | ← Rev. 2 →   +8 My approach for D. Rescue Nibel : I will tell you the method i used to solve, i don't know weather its similar to tutorial's but mine one is easy.I will create a array and put inside L and R + 1 for all intervals. So array will have size = 2 * N. Then sort the array. We keep a variable cnt which will tell the number of lamps which are currently ON. Initially cnt = 0. Now we will iterate over all element of array (in chronological order). Let the current element be x. Now x will either be starting time or (ending time + 1) of some interval. If it is starting time and cnt >= k - 1 then we add NCR(cnt, k - 1) to the answer. We take k - 1 because the kth element will be the lamp whose starting time is x and remaining k - 1 we take from our cnt(lamps which are already in ON state). Then we increment our cnt by 1. If x is (ending time + 1) of some lamp then we just decrement our cnt. This is my code --->  vector > v; int L[n], R[n]; int ans = 0, cnt = 0; for (int i = 0; i < n; i++) { cin >> L[i] >> R[i]; v.push_back({L[i], 1}); // starting time of lamp v.push_back({R[i] + 1, -1}); //ending time + 1 of lamp } sort(v.begin(), v.end()); for (int i = 0; i < 2 * n; i++) { int S = v[i].second; if (S == 1) { if (cnt >= k - 1) { ans += NCR(cnt, k - 1, MOD); ans = ans % MOD; } cnt++; } else { cnt--; } } cout << ans; 
•  » » » » » 4 weeks ago, # ^ |   0 Can you please share the code for NCR(cnt, k, MOD)I did follow the same approach but end up having a run time error which was certainly "Arithmetic Exception" while computing this CombinationIt fails over 11th test case....could you please share some insightsMy Submission
•  » » » » » » 4 weeks ago, # ^ |   0
•  » » » » » » » 4 weeks ago, # ^ |   0 Thank you for that bro, is your code working vese?
•  » » » » » » » 4 weeks ago, # ^ |   0 Nice Code man
•  » » » » » 4 weeks ago, # ^ |   0 For anyone looking for his submission.
•  » » » 4 weeks ago, # ^ |   0 Is D problem has to be solved in that way how to get that intution someone should have solved 1 question at least of that type I think
•  » » 4 weeks ago, # ^ |   0 I agree. People did it quickly because I think the intuition was quick for someone who has solved similar stuff or questions based on such counting. I too thought it resembled something but couldn't put it into code :(
•  » » 4 weeks ago, # ^ |   0 I think you should try ** Min Stations for trains ** problem on GFG.
•  » » 4 weeks ago, # ^ |   0 https://www.geeksforgeeks.org/number-of-ways-to-choose-k-intersecting-line-segments-on-x-axis/ I found it on geeks for geeks after the contest.
•  » » » 4 weeks ago, # ^ |   0 Have you converted it yet or not?
 » 4 weeks ago, # |   0 I had a Java solution TLE on D which I fixed by just using a hashmap to cache the results of modular inverse :/
 » 4 weeks ago, # |   0 Solution for Problem C2Suddenly it turns out that a single request will change the state of no more than six elementsDidn't get intuition for this statement! Someone Please elaborate
•  » » 4 weeks ago, # ^ |   +24 Only the neighbour elements affect local maximum and local minimum. So, if you swap $l$ and $r$, you will need to recalculate only $l$, $r$ and their neighbours (in the worst case, it will be $l-1$, $l$, $l+1$, $r-1$, $r$ and $r+1$, thus six elements).
•  » » » 4 weeks ago, # ^ |   -8 Amazing explanation...
•  » » » 4 weeks ago, # ^ |   0 Awesome explanation mate
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   +1 Whether an element is a local maximum (or minimum) or not solely depends on its relation to the neighbors, as the required condition for a local maximum is a_i > a_{i-1} and a_i > a_{i+1} (reverse signs for a minimum). Now when swapping elements at given positions l and r, this will affect the status of the elements at positions l-1, l, l+1, and r-1, r, r+1 (note that these triplets may partly overlap).
•  » » 4 weeks ago, # ^ |   +3
 » 4 weeks ago, # |   0 can't we solve problem b using tries
•  » » 4 weeks ago, # ^ |   +43 Why cut a apple using sword?
•  » » 4 weeks ago, # ^ |   0 Always try to think of simplest logic. Don't think too hard. Obviously there might be a solution using different logic also.
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   0
•  » » » 4 weeks ago, # ^ |   0 Can you please explain your logic?
•  » » » 4 weeks ago, # ^ |   0 thanks
 » 4 weeks ago, # |   +1 I am looking details description of solution D or a video tutorial
•  » » 4 weeks ago, # ^ |   0 Isn't the solution of problem D detailed enough?
•  » » » 4 weeks ago, # ^ |   0 may be..but i can't understand this..
•  » » » » 4 weeks ago, # ^ |   +1 You can try looking at secondthread's youtube channel , I guess it may help.Youtube channel name is "SecondThread".
•  » » » » 4 weeks ago, # ^ |   0 Also tourist explained it on his youtube channel.
•  » » 4 weeks ago, # ^ |   0 /* * For every endPoint, add the number of ways to choose k segments * Such that the segment(call it p) with that endpoint is one of the k segments * Which basically leaves you to choose the k-1 segments that pass through * that endpoint, other than p. */ This is the implementation. Hope it helps.
 » 4 weeks ago, # |   +30 I've succesfully uphacked myself. Turns out I forgot to check if all ai are different and still passed systests :D
•  » » 4 weeks ago, # ^ |   0 Never knew that we can hack our own answers!
 » 4 weeks ago, # |   0 Can anyone please explain why my code for Problem D got Accepted when I used fastI/O but was getting TLE without using fastI/O even after that fact that the question had just one testcase per testfile?
•  » » 4 weeks ago, # ^ |   +2 fast I/O is so powerful that it can even reduce your acceptance time to half. Even I was amazed when I first observed it, few days back.
•  » » 4 weeks ago, # ^ |   -49 It is because of the large input file, you have to read 6 * 10^5 numbers
•  » » » 4 weeks ago, # ^ |   0 Thanx a lot buddy!
•  » » » » 4 weeks ago, # ^ |   -47 Your welcome my friend.
•  » » 4 weeks ago, # ^ |   0 I'm a bit new...please mind my silly questionBut what is fast I/O anywaysI did get Runtime error while computing nCr
 » 4 weeks ago, # | ← Rev. 2 →   -63 C1 was already on internet. Make this contest unrated https://tkramesh.wordpress.com/2009/10/08/maximum-alternating-sum-subsequence/
•  » » 4 weeks ago, # ^ | ← Rev. 3 →   +31 C1 is a standard dp problem, you can't argue on that basis to make the contest unrated. It is difficult to make new problems and very easy to make new accounts.
•  » » » 4 weeks ago, # ^ |   +4 I agree making a new problem is a difficult task but at least we shouldn't use the same language as has been used on other sources. Just googling "maximum alternating sum subsequence" gives you the solution (I googled after seeing this).
 » 4 weeks ago, # |   +3 Can C1 be solved w/o dynamic programming i.e greedy?
•  » » 4 weeks ago, # ^ |   0 See solution for C2.
•  » » » 4 weeks ago, # ^ | ← Rev. 7 →   0 can u help me find where am i going wrong in my C1 submission. 93713126 UPD : I didn't use long long .
•  » » 4 weeks ago, # ^ |   +2 yes you just need to select each local maxima and minima, and remove the last minima.
•  » » » 4 weeks ago, # ^ | ← Rev. 2 →   0 I did the similar thing, first calculated the local maxima then if difference of subsequent minima and maxima is greater then zero then add it to answer and got AC, But still I can't get the intuition of why it worked. Here is my solution. Can I get help with this.Update-Understood after reading the editorial again :).
•  » » 4 weeks ago, # ^ |   0 Yes,it can be solved using greedy
•  » » 4 weeks ago, # ^ |   0 My C1 solution without DP 93696961
 » 4 weeks ago, # |   +11 Is there a typo in the summation of $p(A)$ for E? Isn't it $\displaystyle \sum_{1 \leq i,j \leq k} f_i \cdot f_j = \left( \sum_{i=1}^k f_i \right)^2 \neq \left( \sum_{i=1}^k f_i^2 \right)^2$
•  » » 4 weeks ago, # ^ |   0 Yes, this is a typo. Thanks for finding it out.
 » 4 weeks ago, # |   +26 Segment tree solution for C2We assign 4 values to each node (corresponding to a segment of the array) in the segment tree, which are the maximum value obtained if: 1/ The sequence begins with a (+) sign (addition) and ends with a (+) sign. 2/ The sequence begins with a (+) sign and ends with a (-) sign (substraction). 3/ The sequence begins with a (-) sign and ends with a (+) sign. 4/ The sequence begins with a (-) sign and ends with a (-) sign. For each node, these values can be deduced from its two children. My submission for more details: 93687329
 » 4 weeks ago, # |   0 i think the time limit for D is really bad, i made this submission in the contest and i got a TLE after it ,i don't know why my code gets TLE is the problem with factor 2? https://codeforces.com/contest/1420/submission/93716363
•  » » 4 weeks ago, # ^ |   0 Same happend with me bro see this, can you help why its happening in my code?
 » 4 weeks ago, # |   +27 I really like this kind of editorial where hints are provided rather than directly displaying the solution approach completely. I wish more than more authors adopt this strategy. Also, the problem statements were quite nice. Thank you for the good round and helpful editorials.
•  » » 4 weeks ago, # ^ |   +8 You are welcome :)
 » 4 weeks ago, # | ← Rev. 2 →   0 Can A be solved with ordered_set. What is the problem here 93823197
•  » » 4 weeks ago, # ^ |   0 One thing you can do is check if array is strictly decreasing, then answer will be NO in all other cases answer will be YES
 » 4 weeks ago, # |   +3 thanks for giving hints before solutions, would expect the same for future contests
 » 4 weeks ago, # |   0 I heard that D was copied from gfg. I was surprised at number of submissions of D
•  » » 4 weeks ago, # ^ |   -11 I too first thought that only C1 was standard and hence had a lot of subs. After seeing a few solutions for D, I realized it was standard as well. Two pretty standard problems in a single contest is definitely a bit sad.
 » 4 weeks ago, # | ← Rev. 3 →   +24 An observation free solution for C1 and C2:First let's try to solve this problem using divide and conquer. On every divide stage, we need the following information for each halve: $1 =$ Maximum sum subsequence starting with a positive element and ending with a positive element $2 =$ Maximum sum subsequence starting with a positive element and ending with a negative element $3 =$ Maximum sum subsequence starting with a negative element and ending with a positive element $4 =$ Maximum sum subsequence starting with a negative element and ending with a negative element For the state where $R - L = 0$, $1 = A_l$ $2 = -10^{16}$ $3 = -10^{16}$ $4 = -A_l$ In the conquer stage, we need to merge these properties. We can easily do this as follows: $1 =$ max(1L, 1R, 1L + 3R, 2L + 1R) $2 =$ max(2L, 2R, 1L + 4R, 2L + 2R) $3 =$ max(3L, 3R, 3L + 3R, 4L + 1R) $4 =$ max(4L, 4R, 4L + 2R, 3L + 4R) Our final answer will be $max(1, 2)$ over the entire array. This divide and conquer can solve C1. The question is, how do we keep updating this?Well, in this 'divide and conquer', the conquer part is only $O(1)$. Therefore, this is a decomposable search problem which can be handled fast by segment trees. Therefore to AC C2 we can create a segment tree where each node in the segment tree stores the 4 properties mentioned above and the merge function performs the max operations as described above. After this, it's standard.Time Complexity: $O(N + Q \cdot LogN)$You can view my submission for clarity:https://codeforces.com/contest/1420/submission/93674507Note: There was a much easier solution using some problem specific observations, but it's always better to have a solution that works on a wider range of cases. I hope this helped :)
•  » » 4 weeks ago, # ^ |   0 To make it cleaner change 2 and 3 to "Maximum sum subsequence of even size starting with a positive element" and "Maximum sum subsequence of even size starting with a negative element", this removes isolated 1L 1R 2L 2R from the max, because the base case turns into 0 (empty subsequence has even size) instead of -10^16.
•  » » » 4 weeks ago, # ^ |   0 Yeah GG
 » 4 weeks ago, # |   -9 Solution Video :Problem A : https://www.youtube.com/watch?v=hFTC6ED6opgProblem B : https://www.youtube.com/watch?v=Zp0be0LmLs8Problem C : https://www.youtube.com/watch?v=KD4DN5KN74U
 » 4 weeks ago, # |   +7 I liked the fact that hints are provided before the full solution. Thank you so much!
 » 4 weeks ago, # |   +3 What is event method mentioned in D's solution?
•  » » 4 weeks ago, # ^ |   +3 We intend to keep track of the number of lamps which are switched ON at a particular instance of time,let's call that variable x. Now consider turning ON and turning OFF of each lamps as seperate events.It's intuitive that the numbers of lamps switched ON (variable x) only changes at the time of an event(switching ON or switching OFF of some lamp). Hence we sort all the events in non-decreasing order of their occurence and process them one by one. (Note that for some time T if there are multiple events occuring ,we place the switching ON events before the switching OFF events because the question mentions that lamps are switched ON from li to ri inclusive).As we process the events 1by1, if we encounter a switching ON event, x is incremented, otherwise x is decremented.(quite intuitive). Whenever there is a switching on event and x>=k-1 , there are X choose K-1 ways of selecting k-1 lamps ,which cobmined with the current lamp being switched on gives us a new solution of k lamps.(My Submission)If you are new to this technique, https://www.geeksforgeeks.org/minimum-number-platforms-required-railwaybus-station/ is a good introductory problem.
•  » » » 4 weeks ago, # ^ |   +1 Thanks for your explanation. I had this thought first but hesitated in doing it.
•  » » » 4 weeks ago, # ^ | ← Rev. 2 →   +1 Thank you for the explanation. For anyone who did not get it, an alternate explanation could be that we just have to chronologically sort all that happens and iterate over it while maintaining a counter of the number of bulbs still switched on. Just keep in mind to switch off a bulb after switching on all the bulbs at that instant.
 » 4 weeks ago, # |   0 nice editorial
 » 4 weeks ago, # |   +25 I found an AC submission to problem B in my room:https://codeforces.com/contest/1420/submission/93659129It has O (n^2) complexity but runs surprisingly fast due to the use of intel vectorization.Is this possible on AMD processors?
•  » » 4 weeks ago, # ^ |   +5 AMD processors have the same instruction set as Intel and support vectorization, so I think it's possible.
•  » » » 4 weeks ago, # ^ |   0 Does this mean that we can just pass simple n2 loops like these with this vectorization technique?
•  » » » » 4 weeks ago, # ^ |   0 Sometimes it helps to fit into time limits, but these cases are not so often. Vectorization improves speed only in constant time, so if I make $200'000$ instead of $100'000$, then the $O(n)$ solution will be two times slower, but vectorized $O(n^2)$ will become four times slower, because it's still $O(n^2)$. Additionally, not every solution is good to vectorize.
•  » » 4 weeks ago, # ^ |   +5 What is intel vectorization and why this O(n^2) solution passing??
•  » » » 4 weeks ago, # ^ |   0 In short, it allow the core perform multiple instructions simutanenously.
•  » » 4 weeks ago, # ^ |   0 OMG!! I tried my luck with the exact same approach but got tle.....Can you plz explain what is Intel vectorization?
•  » » » 4 weeks ago, # ^ |   0 You need a new generation of intel cpu e.g. 9th or 10th gen. Older cpus do not support this feature.
•  » » » » 4 weeks ago, # ^ |   0 So it means who has high processor laptop dont need to worry about time complexity??
•  » » » » » 4 weeks ago, # ^ |   +10 No, having a better algorithm is always better.
•  » » » » » » 4 weeks ago, # ^ |   0 Indeed
•  » » » » 4 weeks ago, # ^ |   0 Vectorization is available for long time. According to Wiki, AVX2 is supported in Intel CPUs since 2013, and SSE is used for even longer period. Why do you think that only 9th or 10th generations are supported?
•  » » » » » 4 weeks ago, # ^ |   0 I guess so because this code runs on my computer in 6 seconds (intel xeon 6th gen).
•  » » » » » » 4 weeks ago, # ^ | ← Rev. 2 →   0 1) What compiler are you using?2) Did you see the generated assembly?If your CPU doesn't support AVX2 instructions and the compiled binary has them, the program will simply crash.
•  » » » » » » » 4 weeks ago, # ^ |   0 I am using GCC 8.2 Windows, C++14. The compiler bypass all directives and generate the output exe as the same as normal
•  » » » » » » » » 4 weeks ago, # ^ |   +5 The compiler bypass all directives and generate the output exe as the same as normal Then your compiler didn't produce such instructions. It doesn't mean that CPU is not capable of executing AVX2.If you try to compile https://codeforces.com/contest/1420/submission/93659129 with -O2, everything should be optimized and AVX2 will be used.
•  » » » » 4 weeks ago, # ^ |   0 So does this mean that someone with 9th/10th gen processor can get AC easily by doing 10^10 operations in like 1 second?
•  » » » » » 4 weeks ago, # ^ |   0 It's not true for every program. Vectorization improves the speed much only in the specific parts, like short loops which iterate over the array. You cannot just write #pragma GCC target("avx,avx2") and make any program two times faster. But there are some problems on Codeforces where solutions in $O(n^2)$ with vectorization passed with $n \le 100'000$.
 » 4 weeks ago, # |   0 Clean solution to C1 using maxima minima approach CODE#include #define int long long using namespace std; void solve(){ int n,q ;cin >> n >> q ; vectora(n) ; for(int &x:a) cin >> x ; a.insert(a.begin(),(int)-1e9) ; a.push_back((int)-1e9) ; int ans = 0 ; for(int i=1;ia[i-1]&&a[i]>a[i+1]) ans+=a[i] ; if(a[i]> t ; while(t--) solve() ; } 
•  » » 4 weeks ago, # ^ |   0
•  » » » 4 weeks ago, # ^ |   0 Nicely written keep making these type of blogs for upcoming contests also
 » 4 weeks ago, # |   +35 For C2, the answer can be written as $\sum_{i=0}^{n}{max(0, a_i-a_{i+1})}$, where, $a_0 = a_{n+1} = 0$. Viewed this way, updates become very easy.
 » 4 weeks ago, # |   +11 Actually , problem E has a solution in only $O(N^4)$ with simple convex hull trick . You can check my solution here .
•  » » 4 weeks ago, # ^ |   +8 Fantastic!
 » 4 weeks ago, # |   0 Can anyone explain to me the problem b's code? why j from 29 to 0?
•  » » 4 weeks ago, # ^ |   0 It is because 1<= Ai <=10^9. So Ai can have maximum length of 29 in binary form.
•  » » » 3 weeks ago, # ^ |   0 ok thx!
 » 4 weeks ago, # | ← Rev. 3 →   +14 For 1420C2 - Армия покемонов (сложная версия), instead of storing the local maxima/minima, we actually only need to store the current value of each element.Why?Because the best answer we can get can also be represented in $\sum\max(0, a_i-a_{i+1})$If we add an extra $0$ to the end of the array.So for each query, we need to check at most 4 pairs: $(l-1,l)$, $(l,l+1)$, $(r-1,r)$, $(r,r+1)$. Note that there might be duplicates, so we need to use a set to deduplicate.If the original pair is counted in the current sum, then we subtract it from the sum.Then we do the swap and check the pairs again. If the new pair should be counted, we just add it to the sum.I think this solution is clearer than the official one.AC submission: 93743846
•  » » 4 weeks ago, # ^ |   0 i am getting wrong answer for test case 44 in problem D. please help me out. i have done the same thing as mentioned in editorial most probably. submission : 93742154Please help lucifer1004
•  » » 4 weeks ago, # ^ |   0 This solution is really nice :) Do you have an AC submission?
•  » » » 4 weeks ago, # ^ |   +8 I have added my submission.
•  » » 4 weeks ago, # ^ |   0 I think it should be $\sum max(0, a_{i+1} - a_{i})$, $i = 0$ to $n$, $a_0 = 0$ and $a_{n+1} = 0$
•  » » » 4 weeks ago, # ^ |   0 The two forms are equivalent.Note that I only add one $0$ at the end, while you add $0$ at both front and end.
 » 4 weeks ago, # |   0 i am getting wrong answer for test case 44 in problem D. please help me out. i have done the same thing as mentioned in editorial most probably. submission : 93742154
 » 4 weeks ago, # |   0 I really like this editorial providing hints not only solutions.
 » 4 weeks ago, # |   0 Can someone help me to find a corner case for my solution to Problem A.My logic is similar to the editorials.I tried creating a generator for the problem but I am still not able to find the test case on which the solution fails.Here the link to the solution 93669466 .
 » 4 weeks ago, # |   +13 Actually DP solution with $O(n + q\sqrt{n})$ is also conceivable for C2 although because of hard constraints it gets a TLE. My approach, drawing inspiration from Mo's algorithm, was basically to split the pokemons into $\sqrt{n}$ segments of size $\sqrt{n}$ each. For each segment we can calculate following 4 things in $O(\sqrt{n})$ time (using the approach in C1). 1. Max +- sequence with even number of terms. 2. Max -+ sequence with even number of terms. 3. Max +- sequence with odd number of terms 4. Max -+ sequence with odd number of terms.Finally we can run the dp on these $\sqrt{n}$ segments to get the answer in $O(\sqrt{n})$ time. Now, for every query just swap the pokemons and then recompute the (atmost) 2 segments in which the swapped pokemon lie and finally recompute the DP. For each query this can be done in $O(\sqrt{n})$ time.My solution, TLE ofcourse - 93717594
 » 4 weeks ago, # |   +8 Can someone Please explain how to do c2 with segment trees?
•  » » 4 weeks ago, # ^ |   +3 https://codeforces.com/blog/entry/82978?#comment-701497Read the comments above before posting yours.
 » 4 weeks ago, # |   0 Why is my code getting runtime error on test 5 of problem c1. I have applied DP using 2d array where dp[i][j] tells answer when i elements are considered with j elements included in subsequence. https://codeforces.com/contest/1420/submission/93757697I will be obliged if someone helps.
•  » » 4 weeks ago, # ^ |   +3 It can happen that $n = 3\cdot10^5$ in this task. And $3\cdot10^5\times3\cdot10^5$ is too large to fit into memory on stack, so the program crashes.
•  » » » 4 weeks ago, # ^ |   -10 Can you also look at my code and tell me why i am getting WA on Problem A? I tried generating different test cases and testing but i am not able to find any test case where my solution fails.Link to my submission. 93762133.
 » 4 weeks ago, # | ← Rev. 2 →   +5 When will ratings update?Whats the reason behind the delay?
 » 4 weeks ago, # | ← Rev. 2 →   +4 For C2, we can use set to avoid categorical discussions. setnow{l-1,l,l+1,r-1,r,r+1}; for(auto &x:now) erase(x); 
 » 4 weeks ago, # |   +10 Can anyone tell me about the "event metho" which author is referring in solution of D? "It should be noted that p(x) and s(x) can be easily supported using the event method. Then, the total runtime will be O(nlogn)."
•  » » 4 weeks ago, # ^ |   0 The idea is as follows. You have segments $[l_i; r_i]$ and want to calculate (for example) the number of segments that contain specified points $x_i$. To do this, you create an array of events: a segment is started (at time $l_i$) a point is encountered (at time $x_i$) a segment is finished (at time $r_i$) Then, sort such events by time and process them in sorted order. Now it's easy to keep track of number of open segments in the specified points $x_i$.
 » 4 weeks ago, # |   0 I still don't understand the logic behind A Cube SortingIt is not difficult to see that the answer «NO» in this task is possible when and only when all ai are different and sorted in descending order.How can one derive such logic PS — newbie here, want to learn :D
•  » » 4 weeks ago, # ^ |   +3 If you know something about bubble sorting, you can think of the worst time complexity of bubble sorting ($\frac{n-1 \cdot n}{2}$) and its reasons
•  » » » 4 weeks ago, # ^ |   0 Yet to learn , thanks for info
 » 4 weeks ago, # | ← Rev. 2 →   0 In 1420B- Rock and lever can you explain what this part of the code is doing? for (int j=29; j>=0; j--) { int64_t cnt=0; for (int i=0; i=(1<
•  » » 4 weeks ago, # ^ | ← Rev. 2 →   0 `if (a[i]>=(1<
 » 4 weeks ago, # |   -13 Is there anyone else who is reading the name of this editorial's author as jeopardy?? xD
•  » » 4 weeks ago, # ^ |   -10 Why so many downvotes??? I didn't said that he is really in a jeopardy ¯_(ツ)_/¯
 » 4 weeks ago, # |   -9 Hello everyone !! I hope you are having a good day!i tried to make editorial for this round :- https://www.youtube.com/watch?v=uc6mN7BZVbI&list=PLrT256hfFv5WH3U33DuKUzL9fYnVsTGAjproblem solved. -> A , B , C1 , C2 , Dlanguage of communication -> Hindiprogramming language -> C++
•  » » 3 weeks ago, # ^ |   0 I have tried to solve A by counting inversions. Can someone please tell why it is giving wrong answer. submission
 » 4 weeks ago, # | ← Rev. 2 →   0 in problem c2 let y be the total number of local maximum and minimum we have now, can we change the parity of the y with an update(swapping two element)?
 » 4 weeks ago, # |   0 Can someone explain the dp which we have done in ques E. Thanks
 » 4 weeks ago, # |   0 Can anyone help me with the problem D. I got WA in test 11. Here is my code:93779028
•  » » 4 weeks ago, # ^ |   0 If I am not wrong you have compressed the values but you cannot compress the values in this question. You have to keep them as it is.
•  » » » 4 weeks ago, # ^ |   0 Can you explain why I cannot compress the values? I think the values are not important in this problem. For example, two segments [1,4] and [2,6] after being compressed will become [1,3] and [2,4], which will not affect the result.
 » 4 weeks ago, # |   0 I liked A question it was a good implementation of bubble sort if you know it's worst case.
 » 4 weeks ago, # |   +1 Here are video solutions for all problems, including a description of the weird segtree sol for C2
 » 4 weeks ago, # |   0 Very nice Editorial for problem E. Thanks :). Has anyone solved it using flows?
 » 4 weeks ago, # |   +1 problem D is a nice educational problem thanks :)
 » 4 weeks ago, # |   0 Why is my code for B failing on 2nd pretest? I have stored the count of occurrences of each MSBs in a map and then have iterated over it to calculate the final count. Link to submission: https://codeforces.com/contest/1420/submission/93796185
•  » » 4 weeks ago, # ^ |   0 Because when you are adding m(m-1)/2s, m can be up to 10^6, so if you set m as an int, it can overflow.
 » 4 weeks ago, # |   0 I today constucted a easy way to do the C2 problem or C1 problem. The ans is essentially the half of sum of absolute value of adjacent indices. (To work it, we have to add zero at start and end of the array). Now the update is really easy, we have to just remove the old absolute values and add the new ones. This can also handle updates like changing the strength of a indice. See the Ac code for implementation details- AC code
 » 4 weeks ago, # |   0 Can someone spot why I am getting WA on test 11 in Problem D. Here is my submission. 93837255
 » 4 weeks ago, # |   0 Lol, Missed the intended solution on problem A Did it using BIT .
 » 4 weeks ago, # |   +10 I like this step by step style of the editorial. Thanks.
 » 4 weeks ago, # |   +10 In the tutorial of problem E, should $z_i$ be $\sum^i_{j=1}{f_j}$? And could you please explain why it takes $|z_i-(s+h)|$ steps from $dp_i$ to $dp_{i+1}$?
•  » » 4 weeks ago, # ^ |   +8 In the tutorial of problem E, should $z_i$ be $\sum\limits^i_{j=1} f_j$? Yes, there is a typo in the editorial. And could you please explain why it takes $|z_i-(s+h)|$ steps from $dp_i$ to $dp_{i+1}$? Remember the reasoning above on how many steps it takes to transform the array $a$ into $b$. It's $\sum\limits_{i=1}^n \left|z_i - \sum\limits_{j=1}^n b_j\right|$, as shown above. In our DP, we know the array $a$ and build an optimal array $b$ step by step. After deciding that the value of the $i$-th element of the array $b$ will be $h$, we say that $\sum\limits_{j=1}^i b_j = s + h$. So, we add $\left|z_i - \sum\limits_{j=1}^i b_j\right| = \left|z_i - (s + h)\right|$ to the number of steps.Is it now clear for you?
•  » » » 4 weeks ago, # ^ |   +10 Yes, thanks. I didn't realize $dp_i$ starts from 0 so I mistook $h$ as $b_{i+1}$.
•  » » » » 4 weeks ago, # ^ |   +13 I think there is some inconsistency in the editorial text. $s + h$ should be of the same "size" as $z_i$, so there I would rather write $z_{i+1}$.
 » 4 weeks ago, # |   0 Can someone please tell me why I am not getting TLE on my solution to problem E?It seems like magic. I really don't understand how a not very well optimized code like this is working.Here is the solution link: 93898124.
•  » » 4 weeks ago, # ^ |   +10 Isn't it just $O(n^5)$? You have very straightforward loops, so it should be pretty fast. Switching to plain arrays might speed it up, but otherwise this looks like an intended solution.
•  » » » 4 weeks ago, # ^ |   0 Yes, but $80^5 > 32 \times 10^8$ and therefore, should not work, unless it is optimized heavily.
•  » » » » 4 weeks ago, # ^ |   +10 There is a small constant in front, rep(i,1,cnt+1) rep(j,i,n+1) is $n^2/2$, rep(p,0,j) further gives $n^3/3$. $10^8$ is quite okay, it is also very simple operations inside the loops.
 » 4 weeks ago, # |   0 Can anyone help me with the reason for this solution getting TLE? Problem c2 https://codeforces.com/contest/1420/submission/93901635
•  » » 3 weeks ago, # ^ |   +10 Use '\n' instead of endl and use fast I/O.
 » 4 weeks ago, # |   -7 The outputs of the editorial codes themselves do not match for the same input test cases. For example, Input 1 4 0 5 2 2 5Output (C1 Editorial Code) 8Output (C2 Editorial Code) 10I know it's a bit late, but the editorial remains as is. So, this should be corrected.
•  » » 3 weeks ago, # ^ |   0 Is the input correct? The numbers in the input must be distinct, according to the statement.
•  » » » 3 weeks ago, # ^ |   +10 Oh, I'm really sorry for that. Thanks for correcting me.
 » 3 weeks ago, # |   +18 Can anybody explain how this formula is deduced in solution of problem E?$p(A) = \sum_{1 \le i < j \le k} f_i\cdot f_j = \frac 12 \left( \sum_{1 \le i, j \le k} f_i\cdot f_j - \sum_{i=1}^k f_i^2 \right) = \frac 12 \left( \left(\sum_{i=1}^k f_i \right)^2 - \sum_{i=1}^k f_i^2 \right)$
•  » » 3 weeks ago, # ^ |   0 We can get this equation in an alternative way:$\displaystyle\left(\sum\limits_{1 \le i \le k} f_i\right)^2 = \sum\limits_{1 \le i \le k,\ 1 \le j \le k} f_i f_j = \sum\limits_{1 \le i \le k,\ 1 \le j \le k,\ i < j} f_i f_j + \sum\limits_{1 \le i \le k,\ 1 \le j \le k,\ i > j} f_i f_j + \sum\limits_{1 \le i \le k,\ 1 \le j \le k,\ i = j} f_i f_j = $$\displaystyle = 2\sum\limits_{1 \le i < j \le k} f_if_j + \sum\limits_{i=1}^k f_i^2 \implies 2\sum\limits_{1 \le i < j \le k} f_if_j = \left(\sum\limits_{1 \le i \le k} f_i\right)^2 - \sum\limits_{i=1}^k f_i^2.In simple words, we take \left(\sum\limits_{1 \le i \le k} f_k\right)^2, which is a sum of pairwise products. Split these pairwise products onto three parts: where i < j, where i > j and where i = j. The first two parts are equal to \sum\limits_{1 \le i < j \le k} f_if_j, and the last part is a sum of squares. From this we get that 2\sum\limits_{1 \le i < j \le k} f_if_j = \left(\sum\limits_{1 \le i \le k} f_k\right)^2 - \sum\limits_{i=1}^k f_i^2, therefore the original formula is also true. •  » » » 3 weeks ago, # ^ | ← Rev. 2 → +8 I did not get this part: "\left(\sum\limits_{1 \le i \le k} f_i\right)^2 = sum of pairwise products". Can you explain how? •  » » » » 3 weeks ago, # ^ | ← Rev. 2 → +8 Okay, do you agree that(f_1 + f_2 + f_3)\cdot(f_1 + f_2 + f_3) = f_1(f_1 + f_2 + f_3) + f_2(f_1 + f_2 + f_3) + f_3(f_1 + f_2 + f_3) =$$= f_1f_1 + f_1f_2 + f_1f_3 + f_2f_1 + f_2f_2 + f_2f_3 + f_3f_1 + f_3f_2 + f_3f_3$?(So, we got that $(f_1 + f_2 + f_3)^2$ is a sum of pairwise products of $f_1$, $f_2$ and $f_3$.You can do this for arbitrary number of elements in the same manner.
•  » » » » » 3 weeks ago, # ^ |   +10 Yeah got it. Thanks.
 » 13 days ago, # |   0 I have an alternative DP solution to problem E. Observe that the answer is at most C(# zero , 2), in stead of counting the good pairs, I count the bad pairs. (The number of bad pairs is the sum of C(len[i] , 2) for each consecutive 0s with length len[i])Firstly, I considered interval DP, however, we need also keep track with the number of orders, so we have a state at least DP[l <= 80][r <= 80][k <= 80 * 79 / 2], when merging interval, the time complexity is at least O(N^6), so this doesn't seem like a good approach. Then, I consider a left to right solution, let DP[i][j][k][l] = the minimal bad pairs for the first making length of i sequence such that, we have used j orders, and have a length k full of consecutive 0s suffix, and have used l ones. For transition, DP[i][j][k][l] -> DP[i + 1][j][k + 1][l] + k if we put 0 on the (i + 1) th position. DP[i][j][k][l] -> DP[i + 1][abs(pos[l + 1] — (i + 1))][0][l + 1]. where pos[i] is ith one's position in the initial sequence. Time complexity O(N^5 / 8) Runtime is some thing like O(80 * 80 * 80 * 79 / 2 * 40 * 40) which should be ok for 2 seconds.
 » 12 days ago, # |   0 thank you very much for again add the hints option in the editorial.Well Done, good Job
 » 10 days ago, # | ← Rev. 2 →   0 I'm using segment tree approach for C2, but I'm getting TLE. I think time complexity should be O(qlogn). Any help would be appreciated :)Link to my code: https://codeforces.com/contest/1420/submission/95420456
 » 10 days ago, # |   0 I need help in debugging Div2 D. I'm getting runtime error on testcase 11.Link to my code: https://codeforces.com/contest/1420/submission/95433102
 » 10 days ago, # |   0 can anybody please help me why i get tle in problem c1. i used a top down approach to do that.here it is