### Vladosiya's blog

By Vladosiya, history, 17 months ago, translation,

1800A - Is It a Cat?

Idea: Vladosiya, MikeMirzayanov

Tutorial
Solution

1800B - Count the Number of Pairs

Idea: myav

Tutorial
Solution

1800C1 - Powering the Hero (easy version)

Idea: Vladosiya

Tutorial
Solution

1800C2 - Powering the Hero (hard version)

Idea: Vladosiya

Tutorial
Solution

1800D - Remove Two Letters

Idea: MikeMirzayanov

Tutorial
Solution

1800E1 - Unforgivable Curse (easy version)

Idea: Aris, talant

Tutorial
Solution

1800E2 - Unforgivable Curse (hard version)

Idea: Aris, Vladosiya

Tutorial
Solution

1800F - Dasha and Nightmares

Idea: Gornak40

Tutorial
Solution

1800G - Symmetree

Idea: Vladosiya

Tutorial
Solution
• +46

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 » 17 months ago, # | ← Rev. 2 →   +12 P.S. I haven't found good articles about hashing root trees so I'll post one soon.UPD: Here it is
•  » » 17 months ago, # ^ |   +1 Isn't that blog enough for understanding hashing root trees? http://rng-58.blogspot.com/2017/02/hashing-and-probability-of-collision.html
•  » » » 17 months ago, # ^ | ← Rev. 2 →   +7 I just hope to make a bit better topic with showing approaches as main point.
•  » » 17 months ago, # ^ |   0 could u check my solution for e1 it gives a wrong answer when a char array is used but accepted when a string is used
 » 17 months ago, # |   -32 C1 was easier than A and B.
•  » » 17 months ago, # ^ |   +6 no it wasnt, A is easier
•  » » 9 months ago, # ^ |   0 why?? i spent apporximately 2 hour solving C1 by using dp 2-D.however, i can not complete it :)) maybe i thought too much
 » 17 months ago, # |   +22 StringForces
 » 17 months ago, # |   0 i dont understand BFS solution of task E can anyone explain please?
•  » » 17 months ago, # ^ |   +1 There is an observation that if you turn the string to a graph where the edges are like this: $(i,i+k)$ when $i+k \le n$ , $(i,i+k+1)$ when $i+k+1 \le n$. So you can swap characters in indices in the same component however you like. You can try a few examples if you want.
 » 17 months ago, # |   0 F. i guess it should be bj = bi XOR (1<<26 -1) XOR (1<
 » 17 months ago, # |   0 Except C1, C2 and F, all other problems are about strings
•  » » 17 months ago, # ^ |   0 F is also string
•  » » » 17 months ago, # ^ |   0 ah sorry, I meant except C1, C2 and G
 » 17 months ago, # |   0 So many FST's on A, such a tricky easy question !
•  » » 17 months ago, # ^ |   +3 Amazing, I have never seen so many FST on a problem A
•  » » » 17 months ago, # ^ |   0 So many FST's on F too, jiangly and SSRS got hacked, I don't feel too bad now after getting hacked, XD
•  » » » 17 months ago, # ^ |   +1 what is FST . sorry i m a newbie
•  » » » » 17 months ago, # ^ |   +2 Failed System Test
 » 17 months ago, # |   +9 What does count(all(cnt), 0) == 26 mean in solution of E2? What is the logic behind this? Can anyone tell me, please
•  » » 17 months ago, # ^ | ← Rev. 4 →   +8 He is simply checking if count of all letters that are not in the interval [n-k; k] is equal. Count(all(cnt), 0) returns number of zero elements in the array.So, with this line: cnt[s[i] - 'a']++; he is counting letters in string s, and with this: cnt[t[i] - 'a']--; he is deleting letters from cnt, and when all letters were processed there will be only zeros in cnt if counts of letters were equal
•  » » » 17 months ago, # ^ |   0 Thanks!)
 » 17 months ago, # | ← Rev. 3 →   +2 D&G two hashing problems!
 » 17 months ago, # | ← Rev. 2 →   +19 E can be solved using disjoint set union. Just check whether each connected component are the same.
•  » » 17 months ago, # ^ |   0 can u plz elaborate more on this?
•  » » » 17 months ago, # ^ |   0
•  » » » » 6 months ago, # ^ |   0 Can you please explain it, its really hard to understand the code.
 » 17 months ago, # |   +10 Thank you for the contest, loved all the questions. Kudos to authors, coordinators. I may finally become expert after being a specialist for 6 months now, :')
 » 17 months ago, # |   0 Good tutorial. Thanks for explaining the solutions properly.
 » 17 months ago, # |   0 G is too obsivious for people who know hash of trees.
 » 17 months ago, # |   +6 I thought my solution of G could fail on system test by hash-collision, but unexpectedly my solution of A got FST. There should be something like "eow" in the pretest of A.
 » 17 months ago, # |   0 Looks like this would have been my expert round had I not made a stupid error and got a penalty for overflow on problem C. Still, I won't cry about +100 delta :)
 » 17 months ago, # |   +2 Can not understand why O(26⋅n⋅log n) for F got TL. Can anyone suggest what I am doing wrong? https://mirror.codeforces.com/contest/1800/submission/195673951
•  » » 17 months ago, # ^ |   0 all you need is to lower your constant. change bitset to long long or using unordered_map instead of map. it finally runs 2000ms
•  » » 17 months ago, # ^ | ← Rev. 2 →   0 same my solution of O(26.n.log n) is also gave TLE on test 26 195813129 Can somebody please explain why is this happening? (also unordered map is also not working)
•  » » » 17 months ago, # ^ | ← Rev. 2 →   0 In your case I think that problem is that it is O(26 * 26 * n * log n) instead of O(26 * n * log n). Upd, actually it is O(26 * 26 * n + 26 * sum_of_len + 26 n * log n), but probably it also should be fine.
•  » » 17 months ago, # ^ |   +1 You are right it is because of the constant. If I use find from std::map instead of operator[] it pass tests. https://mirror.codeforces.com/contest/1800/submission/195806091 But anyway too close to TL.
•  » » » 17 months ago, # ^ |   0 yes, I also tried find instead of [] and it passed, can you please tell why is this happening shouldn't both take same O(log n) time?
•  » » 17 months ago, # ^ |   -10 Vladosiya Could you consider changing the time limit for F? The map based solution should also pass tests. It is div3 you should not care about constant too much.
 » 17 months ago, # |   +1 did not understand the "abudance" example in E1. Can anyone explain it to me?
 » 17 months ago, # |   0 D is a nice problem. I worried about the cases where 2 removals are not overlapping. Did it worry you too? Consider: R1+R2+S+R3+R4 where S is a string, and removals of R1+R2 and R3+R4 gives the same results: S+R3+R4 = R1+R2+S then S[even index]=R1 and S[odd index]=R2. If S has even length, R3=R1 and R4=R2; if S has odd length, R3=R2 and R4=R1. Thankfully, this shows checking overlapping removals will cover the non-overlapping ones too.
 » 17 months ago, # | ← Rev. 2 →   0 A,B,C1,C2,D were leaked on YT. This isn't fair. people think and try to solve and may get negative delta and others just copy and paste and get positive delta. I think Codeforses must prevent the new accounts and that solved less than N of problems in problem set to participate in Div 4 or 3like that one : 195674461 which is for Mohamed_712 and got +44 and this answer is 100% simmilar that was leaked !
 » 17 months ago, # | ← Rev. 2 →   0 [deleted] Oh, my method is the same with editorial.
 » 17 months ago, # | ← Rev. 2 →   0 Could someone explain why we need to do "1&" and "-1&" in calc() in the solution of problem F? I thought that they wouldn't take any effect since 1 & 1 = 1 and 1 & 0 = 0, but removing them indeed resulted in WA. Here's the WA submission: 195819749
•  » » 17 months ago, # ^ |   0 Oh I think I know why. The "-1&" is unnecessary but the "1&" is not. Sorry for the stupid question
•  » » » 17 months ago, # ^ | ← Rev. 2 →   +1 in the -1& case, i guess it just trying to find the mask where only the position of letter c is unset. So in my view, -1& is completely not required.
 » 17 months ago, # |   0 How to proof C? I am not convinced even the solution magically works.
•  » » 17 months ago, # ^ |   +6 As we know all the cards before we take action, we can just discard those that aren't good enough. That ensures the greedy solution using priority queue.
 » 17 months ago, # |   0 I solved G without using hashing, but instead I "sorted" the subtrees in a certain order. Did I do a fakesolve? or is it a valid method?
•  » » 17 months ago, # ^ |   0 I also solve G without hashing, it seems like we have similar idea, I scan nodes in depth ascending order to maintain some equivalent groups among them.
 » 17 months ago, # |   +2 Really good contest!
 » 17 months ago, # | ← Rev. 2 →   +1 https://mirror.codeforces.com/contest/1800/submission/195826604, isn't the tc of this code same as the tc suggested for problem F? Why did it TLE? Can anyone help?
•  » » 17 months ago, # ^ |   +4 When viewing your code, I saw two variables: ev and od. They eat a lot of time and memory, because this is a hashmap array, try to pull them apart a little and reformat the code. If it's not clear what I'm saying, then here's an idea: do not cycle(1...n) cycle (0...26) and vice versa cycle(0...26) cycle(1...n) then you can remove huge arrays and put the use of hashmap into the cycle itself, made the code faster. Sorry for grammatical errors, I do not know English.
 » 17 months ago, # |   +5 Literally StringFORCES!
 » 17 months ago, # |   0 Someone pls explain this conclusion in the last paragraph of tutorial of problem F:To count the number of pairs that include our word, we need to count the number of words with the characteristic $b_j=b_i\oplus(2^{26} − 1)$.
•  » » 17 months ago, # ^ |   0 Letter with odd count will be represented as digit 1, and that with even count will be digit 0. To find pair (i, j) which meets the condition, we should check if they miss a certain letter and all the remaining 25 letters have odd count. Note that $odd = odd + even$ is similar to $1 = 1 \oplus 0$. So the pairs we are finding are (i,j) which have $b_i \oplus b_j = 2^{26}-1$ (the right side is 26 1's). (In the implementation, we actually count words meeting $b_j = b_i \oplus ((2^{26}-1) \oplus 2^{c})$ to handle the missing letter $c$.)
 » 17 months ago, # | ← Rev. 2 →   0 Great tutorial!!!!
 » 17 months ago, # |   0 What is the meaning of this line in solution F if (brr[i] >> c & 1 ^ 1)
•  » » 17 months ago, # ^ |   0 It aims to pick words that miss the certain letter c
 » 17 months ago, # |   0 import java.util.*; import java.io.*; public class RemoveTwoLetters { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out)); int t = Integer.parseInt(br.readLine().trim()); while (t-- > 0) { int n = Integer.parseInt(br.readLine().trim()); String s = br.readLine(); Set set = new HashSet<>(); for (int i = 1; i < n; i++) { StringBuilder sb = new StringBuilder(s.substring(0, i - 1)); sb.append(s.substring(i + 1)); set.add(sb.toString()); } pw.print(set.size()); pw.print('\n'); } pw.flush(); pw.close(); br.close(); } } I wrote this java solution to D, I got out of memory exception on test case 5, this could be optimized, reading insights about my solution would be fun. Please comment.
•  » » 17 months ago, # ^ | ← Rev. 2 →   0 this would have given mle in cpp as well
•  » » » 17 months ago, # ^ |   0 Why so?
•  » » » » 17 months ago, # ^ |   0 You are trying to store 200 000 strings of size 200 000 in the memory. You would need 4 * 10^10 bytes of memory to do that, which is about 40 GB. Not to mention that substring has linear time complexity, so your overall solution is quadratic and would TLE even with enough memory.
 » 17 months ago, # |   0 What does the -1 & do inside the F solution? For me it looks like no-op because -1 has 1111...111 binary representation and bitwise-and with such number shouldn't change the second numberIt's just my understanding so that someone can point out what is wrong with it — I don't know much about bitwise operations
•  » » 17 months ago, # ^ |   0 It does nothing and removing it won't have any problem. Here's a such AC code: 195838110
•  » » » 15 months ago, # ^ |   0 Can you explain why do we need to keep distinct maps based on parity of lengths? As per my understanding, if we can find two strings such (b_i ^ b_j) = 25, won't the odd length of the s_i.s_j concatenation hold already? In that case, do we really need to count masks of odd length strings and even length strings separately?
 » 17 months ago, # |   0 Can someone please tell me why my submission 195888152 with unodered_map doesn't work while the solution with map 195888396 works.
•  » » 17 months ago, # ^ |   0 Actually, it has nothing to do with hash collisions. When you iterate over your map you add new elements by using operator[]. This leads to visiting some characters multiple times in the case of unordered maps. By iterating over a copy of the map your code gets accepted: 196522048
 » 17 months ago, # |   0 F is easy but I didn't solve it ontime, otherwise I would be Expert now :V
 » 17 months ago, # |   0 1800F - 31 - Dasha and Nightmares I tried solving it using bitsets in C++ but it ended up giving WA on TC3.Can anyone help me with my approach. 195952648
 » 17 months ago, # |   0 Is the argument given in the tutorial for D sufficient? i.e. what about potential over counting (the argument only considers consecutive positions)? Or is this obvious?
 » 17 months ago, # |   0 In problem 1800E2 — Unforgivable Curse (hard version) it is enough to check if index of mismatching character in string s and string t is within the limit k <= index <= n — (k + 1). accepted Code
•  » » 5 months ago, # ^ |   0 can you prove it?
 » 17 months ago, # | ← Rev. 2 →   0 Can someone explain why am I getting Wrong Answer in Problem F by using unordered_map and Accepted using map.UNORDERED MAP Solution
•  » » 17 months ago, # ^ |   0
•  » » 17 months ago, # ^ | ← Rev. 3 →   0 Actually, it has nothing to do with hash collisions. When you iterate over your map you add new elements by using operator[]. This leads to visiting some masks multiple times in the case of unordered maps. By iterating over a copy of the map your code gets accepted: 196526230
•  » » » 17 months ago, # ^ |   0 Thanks. But I'm still wondering that order of elements in unordered map changes while traversing as I was not updating the map, just traversing it.
•  » » » » 17 months ago, # ^ |   0 If you call operator[] on an element that didn't exist before in the map, it gets created. As the map is unordered, you don't know where it gets inserted in the traversing order.
 » 17 months ago, # |   0 Here is my solution for problem F. Can anybody tell me where am I getting wrong? My submission link
 » 16 months ago, # |   0 So many problems based on strings
 » 14 months ago, # |   0 In Promblem D, Lets X(i,i+1) = string obtained after chars at index i,i+1 removed Then Why X(i,i+1) and Y(j,j+1) are guarenteed to be distinct ? Why cant they be same ?
•  » » 8 months ago, # ^ |   0 Have the same doubt
 » 6 months ago, # |   0 There is a much easier solution to E. You don't need to use anything except one map container. Refer my submission for the same:- 244026758
 » 5 months ago, # | ← Rev. 2 →   0 I dont understand for problem E when i >= k or i + k < n it doesn't guarantee the possibility of moving the letter in any direction by 1 because if i = k we aren't guaranteed to be able to move i to the right if i also equals n.Can someone prove it to me? Thank you in advance