### Ari's blog

By Ari, history, 3 years ago,

Note: I can't figure out how to place the tutorials inside spoilers. If someone is familiar with how CF spoilers work and can help I would really appreciate it. For now, be warned that the tutorials are visible by default (but everything else isn't)

UPD: I figured out how to use spoilers! Also added implementations for all problems.

Thanks for participating in our contest!

1509A - Average Height

Author: Kuroni
First solve: Nutella3001 at 00:01:02

Hint
Tutorial

Implementation

1509B - TMT Document

Author: Ari
First solve: blue at 00:04:51

Hint
Tutorial

Implementation

1509C - The Sports Festival

Author: Ari
First solve: Dukkha at 00:05:32

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Tutorial

Implementation

1508A - Binary Literature

Author: Ari
First solve (Div. 2): traxex2 at 00:22:48
First solve (Div. 1): tourist at 00:05:04

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Hint 3
Tutorial

Implementation

1508B - Almost Sorted

Author: Both of us!
First solve (Div. 2): shenmadongdong.qaq at 00:29:52
First solve (Div. 1): tourist at 00:08:19

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Tutorial

Implementation

1508C - Complete the MST

Author: Kuroni
First solve (Div. 2): deepspacewaifu at 01:42:36
First solve (Div. 1): maroonrk at 00:33:17

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Tutorial

Implementation

1508D - Swap Pass

Author: Ari
First solve: ksun48 at 00:57:13

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Tutorial

Implementation

1508E - Tree Calendar

Author: Kuroni
First solve: ecnerwala at 01:02:26

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Tutorial

Implementation

1508F - Optimal Encoding

Author: Kuroni
First solve: ecnerwala at 01:52:02 (The only contestant who solved this problem!)

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Tutorial

Implementation

Finally, here's some funny moments that happened while we were working on the round :)

Memes
• +413

| Write comment?
 » 3 years ago, # | ← Rev. 2 →   +9 Longest editorial I ever had the chance to read. Anyways thanks, it's awesome <3
•  » » 3 years ago, # ^ | ← Rev. 2 →   +23 Sorry, as we stated the spoilers were not working to our favor :(
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +7 It's pretty well documented editorial, thank you once again!!
 » 3 years ago, # |   0 For Div 2 D, can somebody find a case where it fails? submission, I think on one case it doesn't print any string because my answer gets over n*3.
•  » » 3 years ago, # ^ |   0 If you find any mistake or corner case please reply me, my solution fails in the same testcase with the same error (In the test 2 case 43)Thanks!
•  » » » 3 years ago, # ^ |   0 I found one: 1 3 001111 110110 000000
•  » » » » 3 years ago, # ^ |   0 https://mirror.codeforces.com/contest/1509/submission/113292585 i did according to the tutorial but my submission gives WA on test 9. Where am I going wrong? please someone help
•  » » » » » 3 years ago, # ^ |   +3 Your mistake, is a common mistake, since a I had the same mistake use a variable ans and print it just once, your mistake is in the fisrt if, you forgget the endl before the return.skipping this you code have a good implementation good look bro!PSD: I submmit your code for be sure that im right and sorry for my bad english
 » 3 years ago, # |   -7 gg
 » 3 years ago, # |   +11 i think problem E div1 can be done with HLD & treap by reversing the operations :thinking:if any one solved the problem with the similar solution please share it : )
 » 3 years ago, # |   0 This contest was amazing and specially it's editorial after all Thanks for preparing such contests and we're waiting for more contests from you! :D
 » 3 years ago, # |   +4 I don't understand why my B solution is wrong, pls help 113225551
•  » » 3 years ago, # ^ |   +3 you go through forward and taking the count of Ts and Ms do the same thing for backward it can't be less Ts from Ms from froward and from backward
 » 3 years ago, # |   +59 My solution to F runs in $O(n^2)$, and is quite simple. Here's my simple condition for an edge being in the graph:For any vertex $v$, there are 4 candidates for edges involving this vertex: consider the intervals containing $v$ which extend the furthest left and the furthest right; let those bounds be $l_v$ and $r_v$. Then, $v$ can have edges to its predecessor in $[l_v, v]$, its successor in $[l_v, v]$, its predecessor in $[v, r_v]$, and its successor in $[v, r_v]$. Furthermore, an edge exists iff it is a valid candidate for both endpoints.Then, we can just maintain these candidates and the counts in $O(n^2)$: starting from the initial state, $l_v$ decrements and $r_v$ increments at most $O(n^2)$ times in total. The code is very short, and runs in less than 1 second.
•  » » 3 years ago, # ^ | ← Rev. 2 →   +26 I'm actually surprised about your solution; I had an $O(nq)$ solution but I didn't know that $O(n^2)$ exists (probably the reason is because I abruptly changed the constraint from $n = q = 50000$ to the current one like 1 day before the round).Anyway, this solution is very clean and also interesting, thanks.
 » 3 years ago, # |   +1 Why this solution for Div2 C gets TLE despite that fact that it's $O(n^2)$ ? It is because it should be written iteratively ?
•  » » 3 years ago, # ^ |   +38 Array bound
•  » » » 3 years ago, # ^ |   +2 Kuroni Yup, AC. Thanks a lot!
 » 3 years ago, # |   +8 Is there any greedy solution for Div2C? It was tagged greedy in the problemset, hence wondering :)
•  » » 3 years ago, # ^ |   +16 Maybe the fact that the minimum and maximum should be at the end is a greedy intuition?
 » 3 years ago, # |   +1 I dont think my div1A/div2D is supposed to pass. Can someone hack ?
•  » » 3 years ago, # ^ | ← Rev. 2 →   +9 Done, random generator found a test that fails.Here is also a smaller test case (n=6), there is no example with smaller n. Maybe it helps: 000001011110 111100100000 111111111111 
•  » » » 3 years ago, # ^ |   +2 How you are generating these cases? (some online website or you have your own generator)
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 For the original test I just generated a bunch of random inputs and tried to see if the solution works (using my own program, basically just a while loop that generates a random input (50% chance for 1 and 50% for 0 at each position) and then runs a simplified version of his code to see if it finds an answer).For finding the n=6 case, I went over all the possible inputs for n=1...6 and tried each one. There arent too many possible inputs for n=1,2,3,4,5 and for n=6 it found one quite quickly.
 » 3 years ago, # | ← Rev. 3 →   +2 Since so many people were asking for problem C (div2), I decided to make a video tutorial for it.http://youtube.com/watch?v=7CPgT3gYPocHope you like it :)
 » 3 years ago, # |   +1 i failed to recognise these types of dp problems during short contests.so i know i need to practise more of these kinds of problem .it will be a great help if you people suggest me some similar kind of problems .
•  » » 3 years ago, # ^ |   +1 Search DP Tagged problems on CF and Leetcode. Some other problems:Longest Palindromic Substring (link)Longest Palindromic Subsequence (link)Matrix chain multiplication (similar problem to div2C)Burst the balloons (link)
 » 3 years ago, # |   +13 Auto comment: topic has been updated by Ari (previous revision, new revision, compare).
 » 3 years ago, # |   +1 I cannot understand div2C, help me!!. we have to minimize discrepancies so why are we adding the element having greatest discrepancies. like An-A1???
•  » » 3 years ago, # ^ |   +2 I will try to explain my approach. First thing to notice is that discrepancy for i = n-1 ( 0- indexing ) will be fixed (max element — min element). Now we put either the max element or the min element at the (n-1)th position that is the only possible way to change the discrepancy for (n-2)th index. Now choosing greedily won't work because we won't know whether our current best move would be a better decision in future also. This is very obvious that we will require to check all possible combinations hence dp is used to optimize the same. So let's talk about,1. dp States : sort the given array, let's say 'L' = any index to left and 'R' denotes any index to right( greater than 'L')- this would mean we have the option to choose between Lth element or Rth element for some index 'i'.2. Transition function : this is simple as you can choose between L or R so dp[L][R] = min(go(L+1, R) , go(L, R-1))3. Base case : L should be less than R Codeint go(int l, int r){  if( l > r) return 0; if(dp[l][r] != -1) return dp[l][r]; dp[l][r] = 0; dp[l][r] = (a[r] - a[l]); dp[l][r] += min(go(l+1,r), go(l,r-1)); return dp[l][r]; } void solve(){ int n; cin>>n;  for(int i =0;i < n;i++) cin>>a[i]; for(int i =0 ;i < 2001; i++){ for(int j = 0;j <2001;j++){ dp[i][j] = -1; } } sort(a,a+n); cout<
 » 3 years ago, # |   0 Problem C : What exactly do those transitions represent ? and if anyone has some intuitive or any other way to deduce the state of that dp ?
•  » » 3 years ago, # ^ |   +2 I explained it in my youtube video But I'll try to give a brief description of my solution (slightly different from editorial)Basically as mentioned in the editorial we start from the back because we have complete knowledge of d[n]d[n]= maximum element in a — minimum element in a (Coz it is for the whole array)Now to minimize d[n-1] we have to remove maximum element or minimum element coz otherwise d[n-1]=d[n]So after sorting at every stage we have two options either we remove from right side or left side. This gives us a feel for a dp approach because the same state of array can be reached by taking different paths(overlapping subproblems)So we define dp[i][j] as we have removed elements from left (minimum) i times and elements from the right (maximum) j times. We can obviously deduce the maximum and minimum in the current array as jth maximum and ith minimum in original arrayThis gives us the transition cost -> jth maximum element — ith minimum element As we can reach state (i,j) from (i-1,j) and (i,j-1)So basically dp[i][j]=min(dp[i-1][j],dp[i][j-1]) + cost (Very much like grid DP ) The answer is minimum sum we can get after removing exactly n-1 elements
•  » » » 3 years ago, # ^ |   +3 Beautiful it is.
•  » » » » 3 years ago, # ^ |   +4 There is an explanation for the author's Dp State that I liked and suhail.loya described in a youtube comment. The author's solution says that in the optimal solution, suppose after reordering the elements it is a[p1],a[p2],a[p3],..a[pn] where p1,p2,..,pn is a permutation of 1 to n, every prefix of this reordered optimal sequence will be a subarray of the sorted sequence. This makes sense because it is always optimal to put the maximum or minimum in the current prefix in the last step. This is basically same as my solution in which I remove elements only from the "ends" of the sorted array so the result will always be a subarray. So the author's solution starts with a subarray of size 1 for which the answer is 0. Now to expand this subarray either you can add one element to the left, or add one element to the right so at every state we have two options just like my solution(where I remove one element from left or right). State Description: dp[l][r] is the minimum cost you can create such that the elements of the current array are a[l],a[l + 1],...a[r] in the sorted array a.Transitions : You can reach state (l,r) from (l+1,r) [ expanding to left] or (l,r-1) [expanding to the right]. Cost of expanding will be maximum-minimum in the subarray a[l]....a[r] which is a[r]-a[l] since it is sorted.So : dp[l][r] = a[r] - a[l] + min(dp[l + 1][r], dp[l][r - 1])Answer : = dp[1][n].
•  » » » » » 3 years ago, # ^ |   0 Glad you liked it :)
•  » » » » » 3 years ago, # ^ |   0 every prefix of this reordered optimal sequence will be a subarray of the sorted sequence. jalotra Can you please explain why it will always be true?
•  » » » » » » 3 years ago, # ^ |   0 Example: s[] = 29 30 67 68 82optimal[] = 67 68 82 30 29, optimal d = 121.See that for each prefix i from 0 to len(s) — 1, the optimal sequence is some contiguous subarray of the sorted sequence.Proof : Let's choose a single starting speed S : s[k] => d[1] = 0 here. Now I have to add one element at the back of it.2a) S : s[k] => Choose either s[k + 1] s[k - 1], because choosing some other element will make d[2] higher.3a) S : s[k] s[k + 1] => Choose either s[k - 1] or s[k + 2].3b) S : s[k] s[k — 1] => Choose either s[k + 1] or s[k - 2].suhail.loya Is this correct?
•  » » » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Example: s[] = 29 30 67 68 82 optimal[] = 67 68 82 30 29, optimal d = 121. jalotra consider prefix for i = 3 in optimal solution (67 68 82 30) .How is this a contiguous subarray of s?
•  » » » » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 The key observation is at every stage while expanding from the current sequence, you want to add an element which is either the minimum of the new sequence or maximum of the new sequence. So you want to either take one step left or one step right in the sorted sequence.What I meant by the statement "every prefix of this reordered optimal sequence will be a subarray of the sorted sequence." that every prefix will contain elements which are in a continguous subarray of the sorted sequenceSo prefix for i=3 (67 68 82 30) although is not a subarray of s[] in exact order but it contains only the elements in the contiguous subarray of s[] from index 1 to 4 (30 67 68 82)This result is more obvious in the solution I describe in this comment
•  » » » » » » » » » 3 years ago, # ^ |   0 suhail.loya Thanks for helping me out :).I was confused at the subarray part but now it is clear. Yeah, the exact order thing was creating the problem for me.
•  » » » 3 years ago, # ^ |   0 So basically we can imagine that we are adding elements to an array let's call it A we know that the last element of that array is either the maximum or the minimum (because otherwise it will make the cost of the function d a lot bigger hence d(n) = max(A) — min(A)) and we work backwards by asking which is the last element max(A) or min(A) and after that we ask the same question for the next biggest and smallest elements left in the original array and here we observe that it must be dp
•  » » » 3 years ago, # ^ |   0 In the last line i think you meant dp[i][j] = min(dp[i + 1][j], dp[i][j — 1]) + Cost. Nice explanation though.
•  » » » » 3 years ago, # ^ |   0 No, according to my dp states it should be min(dp[i-1][j],dp[i][j-1])My states are dp[i][j] -> min cost we can get after removing i elements from minimum side and j elements from maximum side. So to reach (i,j) you can remove one element from maximum side in (i,j-1) or one element from minimum side in (i-1,j)Code
 » 3 years ago, # |   +14 The second solution for Div1A/Div2D is beautiful.
•  » » 3 years ago, # ^ |   0 can anyone explain second solution for Div2D im still unable to understand the math over there :/
•  » » » 3 years ago, # ^ |   +4 Let's say we have these strings, and we currently our pointer is pointed to bold ones in all three of them. (say p1, p2, p3) 0 10101 1 10011 1 00100 Since we have three pointers pointed to three different strings consists only of 0's and 1's, then atleast two of them must point to either 0 or 1.(In this case, p2 and p3 to 1) Now after adding the "1" to our answer, we increment pointer p2 and p3. 0 1 0101 1 1 0011 1 0 0100 Now p1 and p3 is pointed to "0". So in next step, we increment p1 and p3 by 1. 0 1 0101 1 1 0011 1 0 0100 We repeat this process until one of them is completely exhausted. 0101 01 1100 11 (exhausted) 1001 00 Our ans in this case comes to be 1010011 (len = 7) Now in simple terms, len + (one of the remaining in p1 (01) and p3 (00) ) is always less than or equal to 9 (3n). Complex termsLet's suppose our ans has k(7 here) characters, and thus the pointers have advanced by at least 2k(14 here), since at each step atleast two of them is advanced. We have exhausted the characters of s2, so we have advanced p2 by 2n, and the other two pointers have advanced 2k−2n (8 here), and thus one of them has advanced by at least k−n (4 here). Now just add the remaining characters of this string to t. There are at most 2n−(k−n)=3n−k of them, so in the end t has at most 3n characters.
•  » » » » 3 years ago, # ^ |   0 Thanks alot for your detailed explanation , now its clear
 » 3 years ago, # |   +9 Nice Editorial tho
 » 3 years ago, # |   0 Can anyone please help me where I am getting wrong?I have considered two cases the first and the last character should not be M 2 The number of T should be even and the number of M should be half of the number of TAm I missing anything?
•  » » 3 years ago, # ^ |   0 TTTMMT
 » 3 years ago, # |   0 Can anyone prove problem C can or cannot be solved with greedy? I use greedy and cannot pass example 3.
 » 3 years ago, # |   0 Can anyone explain E Solution? Not very clear from tutorial
 » 3 years ago, # |   0 In the Problem div2B, shouldn't this statement: "A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters" be corrected as: "A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters, without changing the order of the elements"? I wasted some time thinking if the order is important or not -_-
•  » » 3 years ago, # ^ | ← Rev. 2 →   +25 Why stop there?"... without changing the order of the elements, without inserting new characters, without turning the characters upside down,..." The list can go on.
 » 3 years ago, # |   0 editorial is awesome
 » 3 years ago, # |   0 I don't understand why my solution for Div1A is wrong. can somebody please help me check? 113203126. My method was just to try each pair of strings, and for every character of the strings, if they were different, my answer would add each of the characters once, and if they are the same, it would add just that one character. This generates three different solutions, and we take the shortest one. It should guarantee that the shortest string would have a length shorter than 3n. Please help, thanks
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 Sometimes the length of your ans string is less than 3*n. Test case : n=2 0101 1010 1011 I fixed that but still, it gave WA. Then I found this case out. It is possible to make n+1 mismatches(positions in which the bits differ) in between the pairs of the strings. Thus the final string would be of length 3*n+1. 4 00011111 11100111 01010000
•  » » 3 years ago, # ^ |   0 What about,000011110000 111100000000 000000001111BTW, Same mistake ):
•  » » » 3 years ago, # ^ |   0 oh wow. thank you guys. I can't believe I made this logic error
 » 3 years ago, # |   -21 Speedforces :))
 » 3 years ago, # |   0 Is there anyway to bound the maximum of minimum hamming distances of three distinct binary strings? I'd appreciate if someone can give out a proof or explain their intuition behind this. Thanks.
 » 3 years ago, # |   0 In TMT question I used counting. First incrementing count when T is found and when M is found then decrementing. Also another variable for counting m. If at anytime count becomes negative , solution is impossible. At the end I checked for if count of T == 2 * count of M. Can anyone help me why this solution was not working ?
•  » » 3 years ago, # ^ |   +1 Check your solution on: $TTTMMT$
 » 3 years ago, # | ← Rev. 3 →   0 Hi, can somebody tell me I got WA on test2(C++) in B? Code: 113309060. Thanks : The accepted solution in Python: 113304387
•  » » 3 years ago, # ^ |   0 AC solution c++ — 113310825 compare to find your mistake (this is your code)
 » 3 years ago, # |   -6 Ari sorry if i'm wrong, but in my opinion 2D is much simpler than 2C. In D, the solution comes to mind immediately, and in 2C, vice versa.
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 You are not wrong, but many people will disagree with you because we have tested and testers rated 2D harder than 2C :) Heck, at first we even proposed 2D as 2B since we had the same opinion as yours (you can check our comments on the problem).
 » 3 years ago, # |   0 Please suggest some problems like div 2 C (matrix chain multiplication). Thanks in advance.
 » 3 years ago, # |   +1 Great set, even better editorial! Awaiting more rounds from you!
 » 3 years ago, # |   0 Ari, Please share with us what mistake were you doing in the first place that caused spoiler tags to overflow? Also, how did you rectify it?
 » 3 years ago, # | ← Rev. 3 →   0 In Problem C of div 2 could someone please explain and give a testcase why this approach fails ??? I did this and got WA on testcase 5 .Keeping the most frequent elements in the front and if frequency of two elements are same then keep the smaller among them in the front.
•  » » 3 years ago, # ^ |   0 your approach fails on test case 1 3 3 3 9 9. According to you answer should be 3 3 3 9 9 1, which gives a sum of d's = 20. But there exists another optimal solution 3 3 3 1 9 9 which gives a sum of d's = 18. So, check your solution on this test case.
 » 3 years ago, # |   +1 can somebody please tell me why i am getting tle in case 18 in 3 question if i initialize dp with 0 and ac when initiallizing with -1?please tell me when to initialize with -1 and when with 0
 » 3 years ago, # |   0 For this submission of div2-D https://mirror.codeforces.com/contest/1509/submission/113251552 can someone provide a small enough test case. I was able to make a test case using random generator but it was large and not useful to find mistake.And i tried to find one myself but failed to do so. So a test case or a logical prove to show that the solution is wrong would be very helpful :) .
•  » » 3 years ago, # ^ |   +4 4101111000011110100000000
•  » » » 3 years ago, # ^ | ← Rev. 2 →   0 thanks, also you found it so quickly, i was also trying to make a similar kind of case but failed :( Also string 1 is just reverse of string 2 XD
•  » » » » 3 years ago, # ^ |   +4 It's pretty easy to find it: Codeint main() { cin>>n; while (1) { string a,b,c; for (int i=0;i<2*n;++i) a.pb(rand()%2+'0'); for (int i=0;i<2*n;++i) b.pb(rand()%2+'0'); for (int i=0;i<2*n;++i) c.pb(rand()%2+'0'); if (check(a,b).empty() && check(a,c).empty() && check(b,a).empty() && check(b,c).empty() && check(c,a).empty() && check(c,b).empty()) { cout<
•  » » » » » 3 years ago, # ^ |   0 Oh ok , I actually tried generating some random cases but didn't got any small case so i thought it was not possible for small cases.Now i feel estupid XD.
 » 3 years ago, # |   +28 in Div2E, how to DFS on the unassigned edges ? In general, the number of unassigned edges could be up to 1e10 so trivial DFS will cause TLE
•  » » 3 years ago, # ^ |   +11 Refer to 920E - Connected Components?
•  » » » 3 years ago, # ^ |   0 Thank you !!
 » 3 years ago, # | ← Rev. 2 →   0 I solved Div2C using ternary search. Any idea for the proof. 113242501Update: hacked
 » 3 years ago, # |   +28 Let me highlight that d1D is really interesting to solve. Thanks for such a great task!(btw, I failed to spot that d1E was indeed an attempt to 1477D - Nezzar and Hidden Permutations xD)
•  » » 3 years ago, # ^ |   0 I remember when I came up with this idea to solve your problem I was soo excited, then I saw it was wrong x(
 » 3 years ago, # | ← Rev. 3 →   0 Can anyone give me a test case or tell me what's wrong with my solution in div.2 D My approach was to try every possible 2 strings and greedily make a resulting string such that it contains both strings and check if its size is less than 3*n.Submission link getting wrong answer on test 843 of test-set#6
 » 3 years ago, # |   0 can anyone tell why my submission getting wa in test case 3 113414122
•  » » 3 years ago, # ^ |   +8 5000000000000001111111110101010
 » 3 years ago, # |   0 How to solve Div2E/Div1B recursively? or how to find the length of the first block?
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 Let's suppose there are n elements. Then, if you place a 1 at the front, then there will be n-1 positions to fill; for 2, n-2; similarly for 3, n-3 and so on. Eg. if you put 3 as the first element, this is how the array will look like.3, 2, 1, ....n-3 elementsNow, if you have n elements, then there are 2 ^ n-1 almost sorted permutations. So, if you put 3 at front, then the minimum ranking of an almost sorted permutation that starts with 3 will be 2^n-1 + 2^n-2 (because permutations starting with 1 and 2 will precede 3). Eg. 1, ...n elements (total: 2^n-1) 2, 1, ...n-2 elements (total: 2^n-2) 3, 2, 1, ... n-3 elements (total: 2^n-3).Thus if the k is between 2^n-3 (inclusive) and 2^n-4 (exclusive), then the permutation definitely starts with 3, and now you need to find rest of the elements of the permutation in this way recursively. Hope this clears it up.Btw, did you understand how the second solution works?
•  » » » 3 years ago, # ^ |   0 Hey thanks for your reply! and yes I do understand how the second solution works.
 » 3 years ago, # |   0 For div1 F,in Tutorial:However, there will be cases when u→ur is actually not needed (I will call this phenomenon being hidden). What is the condition for u→ur to be hidden? That's when there's a path from u to ur with length ≥2! Suppose this path is in the format u→⋯→t→ur. We can prove that tu then that implies at>au but at
•  » » 3 years ago, # ^ |   0 You are not wrong, it's just that we have proven that if there is such t then it cannot lie between u and ur, so we only need to query in [l, u].
•  » » » 3 years ago, # ^ |   +18 I don't think you have answered my question correctly.The defination of $l$ should be different.In my opinion,$l=\min_{i,[u,ur]\in[L_i,R_i]} L_i$In the tutorial,$l=\min_{i,u\in[L_i,R_i]} L_i$
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Ah yeah I see that's true, typo on my part. I will update the editorial.Edit: similar typo on second part of editorial as well :(
 » 3 years ago, # |   0 Can't think of the problematic case in div1D, can anyone provide an explicit example?
 » 3 years ago, # | ← Rev. 2 →   0 It is really a ****ing enjoyment to read your code for F(div2)...orz %%%%
 » 3 years ago, # | ← Rev. 3 →   0 if the pivot point we chose is on the convex hull of the set of points, then one of the border segments may actually intersect some other segments! Hi, I have tried to solve problem D and pick the smallest y, smallest x point which should definitely be on the convex hull. However, it still passed the tests without encountering the case above.Are the tests simply weak or is that case really impossible to encounter? Actually, I also cannot think of how that case should happen there should be no segment that can intersect with a border segment. Is my understanding wrong?
•  » » 3 years ago, # ^ |   0 Right below that, listed as one of the workarounds: "Always choose the bottom-most and left-most point as a pivot. This makes this potentially troublesome segment predictable, and also allows us to slightly simplify the angular sorting code." That's exactly what you did.A segment can intersect with a border segment if, for example, you have a square ABCD, you choose A as pivot and your sort by angle puts B and D next to each other(for example, by sorting like C(0°),B(45°),D(315°)).
•  » » » 3 years ago, # ^ |   0 But in this case, between B and D still lies A, so BD is not a border segment (it's AB and AD the border segments). We can connect B, C and D by swapping B and C, and/or C and D only and then we are certain that all of them belong to 1 circle. Is there a case where we need to connect B and D directly?
•  » » » » 3 years ago, # ^ |   +3 A is the pivot, so BC, CD and BD are the border segments.
•  » » » » » 3 years ago, # ^ |   0 Ah understood, you exclude A. In my case, I consider A part of the polygon even after using it as a pivot so the troublesome border segment never appears. :) thanks.
 » 3 years ago, # |   0 I badly wanted to see a greedy solution for problem C. I spent so much time thinking greedily about it ;-;. Amazing problem anyway!
 » 2 years ago, # |   0 Almost Sorted is the coolest sorting problem I've ever seen, nice explanation
 » 7 months ago, # | ← Rev. 2 →   0 Can I know why this code submission is giving WA, though it is correct?For Testcase 2 (first subtestcase):6001010100100001100001011000010010000The output: 001010100100001011 seems to be correct as it contains both the first two strings.
 » 4 months ago, # | ← Rev. 4 →   +3 Can anyone pls give any counter example for this logic. This is the code.
 » 4 weeks ago, # |   0 div2 C was nice