### pikmike's blog

By pikmike, history, 5 months ago, translation,

1380A - Three Indices

Idea: BledDest

Tutorial
Solution (Ne0n25)

1380B - Universal Solution

Tutorial

1380C - Create The Teams

Idea: Roms

Tutorial
Solution (Roms)

1380D - Berserk And Fireball

Idea: Roms

Tutorial
Solution (Roms)

1380E - Merging Towers

Idea: Roms and BledDest

Tutorial
Solution 1 (BledDest)
Solution 2 (pikmike)

Idea: Roms

Tutorial
Solution 1 (pikmike)
Solution 2 (Roms)

1380G - Circular Dungeon

Idea: BledDest

Tutorial
Solution (pikmike)

• +135

 » 5 months ago, # |   +14 Fastest editorial for an educational round I think.
 » 5 months ago, # |   +15 Btw, in E merging without “smallest to biggest” optimization somehow passes the tests :)
•  » » 5 months ago, # ^ |   +1 Nothing surprising. There is even such a heuristic: randomly connecting sets. It works for O(nlog(n)).
•  » » » 5 months ago, # ^ |   +5 Oh, that makes sense, thanks
•  » » » 4 months ago, # ^ |   +16 No it's not $O(n \log n)$. Let's say we have $n$ one element sets and in $i$-th round we merge $i+1$-st set to su of first $i$. Then on $i$-th turn we have $\frac 12$ probability to do one insert, and $\frac 12$ to do $i$ inserts, which means on avearge $O(i)$ inserts on $i$-th turn, which adds to aveeage of $O(n^2)$ inserts.
•  » » » » 4 months ago, # ^ |   0 That's one way to look at the probabilistic formulation. If you have broader probability space, for example, if operations are also chosen randomly, then the probability of big set being in the operation can be small, and I think it won't be O(n^2). I'm not sure how to show it, but my solution passed in the beginning, until they added a counterexample as yours, so if tests were generated using random at first, I think it shows that average complexity is not so bad.
 » 5 months ago, # |   +135
•  » » 5 months ago, # ^ |   +1 First time solving a div 2D in a contest and this happens :(
•  » » 5 months ago, # ^ |   +6 Haha!
 » 5 months ago, # | ← Rev. 2 →   +22 In the solution of problem E, visualising the towers and its merged states as tree is quite impressive. Thanks for the LCA solution. To me, Small-to-large merge sounds like a greedy approach. It would be helpful if you can elaborate the alternate solution.
•  » » 5 months ago, # ^ | ← Rev. 2 →   +13 If you are looking for dsu solution.... If you always merge smaller set to larger set then it is o(n log(n))... Because a single element will not change it set more than log(n) time..... Let say we have a element 'a' which is in set 1 of size x we merge set 1 to set 2 of size greter than x suppose y.... Than we have atleast 2*x element which get in same set as y>x..whenever we are merging we are atleast doubling the size of. set
 » 5 months ago, # |   +7 Statement of problem B was difficult to understand until watching the image at last as didn't know about the game.
•  » » 5 months ago, # ^ |   +10 Seriously! You don't know stone paper scissor! LOL!
•  » » 5 months ago, # ^ |   +16 r u patrick living under a rock?
 » 5 months ago, # |   0 Can anyone explain to me why greedy works in C? It was kind of intuitive to me, I coded it up and it passed but I still can't figure out that why's it optimal?
•  » » 5 months ago, # ^ |   +8 It's always optimal to take more skilled programmers over less skilled ones, and there is no advantage to expanding the team size once you've reached the minimum requirement of $x$ (since you risk adding a low-skill programmer that decreases the overall team skill below $x$, and you take programmers who could have otherwise formed more teams and improved the answer).
•  » » » 5 months ago, # ^ |   0 https://codeforces.com/contest/1380/submission/86713545Can you please explain why i am getting memory limit exceeded on this solution of problem A?
•  » » » » 5 months ago, # ^ |   0 Sorry to bother, I found the problem.
•  » » » » 5 months ago, # ^ |   0 I am trying some different approach,can anyone please tell me what is wrong in this approach. https://codeforces.com/contest/1380/submission/86678593
•  » » » » » 5 months ago, # ^ |   0 you are sorting it in non-decreasing order, sort it in non-increasing order
•  » » 4 weeks ago, # ^ |   0 We can solve with greedy in O(n) also. If we start from last index and keep the number of unused programmers and whenever for a particular index if no of programmers needed == no of unused we increment count and make unused 0 again.
 » 5 months ago, # |   0 I don't understand problem-B.Anyone help me to understand..And I don't know about this game.
•  » » 5 months ago, # ^ |   0 Seriously bro? There is an image explaining it at the end of the problem. If you still don't understand : wiki page
•  » » » 5 months ago, # ^ |   0 Got it, now..
 » 5 months ago, # |   -9 Why are submissions low on problems? Were they hard, imo first 3 were pretty easy?
•  » » 5 months ago, # ^ |   +17 You didn't even gave the round......Better first participate in the live round and then give your wishful verdict on whether the problems were easy or tough and why there are more or less submissions on a particular problemset
•  » » » 5 months ago, # ^ |   -12 Do you even realize that this is an alt/troll handle
•  » » 5 months ago, # ^ |   +21 Codeforces was down for the first half-hour of the round; by the time it came back up, the round was declared unrated and many competitors left.
 » 5 months ago, # |   0 problem G. 'Let the values be x and y. If they differ by at least 2 (x≤y−2), then the smaller result can always be achieved by moving a regular chest from the larger one to the smaller one.' I could not understand the proof below. I indeed turn a coefficient y into a smaller one (x + 1), but how can I assign every regular chest to the old value, with the changes of this two interval lengths?
•  » » 5 months ago, # ^ |   +5 Sorry for the misread of the statement. Every chest's position can be changed.
 » 5 months ago, # | ← Rev. 2 →   0 https://codeforces.com/contest/1380/submission/86713545Can anyone explain why i am getting memory limit exceeded error? It is the solution to problem A and is not much different from what explained in editorialUPD: I have found the problem. Sorry to bother.
 » 5 months ago, # |   +1 Can someone explain why the second solution for problem A works?
•  » » 5 months ago, # ^ |   0 @spectre_gd I had the same doubt but I have figured it out while I was trying to contradict second solution approach. Go ahead post a test case(if you can find one) where this approach fails.
•  » » 5 months ago, # ^ |   0 imagine array as a mountain and we have two situations when traversing 2nd to 2nd last element: 1. we find element lower on left side and element lower on right side (we got answer) 2. we dont get element lower on left side (means our current element is lowest and array is descending upto current element so it cant be the peak), do the same thing on right side
•  » » 4 months ago, # ^ |   0 Second solution is based on negation of the problem. We are exploring those cases where answer does not exists and it will be when array is either non-decreasing or non-increasing. So, to break this condition, there must be local maxima i.e. an index j where both (j-1) & (j+1) elements are less. And this index is our answer!
 » 5 months ago, # |   0 Can someone explain in problem B why picking the greedy ci for the most frequent will somehow work for the rest of string? How does that make sense? Can't there be a value in the string where ci fails?
•  » » 5 months ago, # ^ |   0 Your choice string is played agains the given string on every position. This means, all positions of the two Strings are paired once. Or, in other words, it does not matter in which order you put the symbols into the choice string.Since the order does not matter, every single position in your string conributes independend of any other position. Finally, for a single position it is obvious that the best choice is based on the frequency of the symbols.
•  » » 5 months ago, # ^ | ← Rev. 2 →   0 you have to understand that we are comparing each of element of s with c1 BUT we are doing the same thing with c2, c3, c4.... after just rotating the sfor eg s = RSPRR R is most frequent so we choose P for c1 for most wins Now for c2 we are doing the same comparisons and P will give best result overall again (cause R is most frequest) so its basically the same
 » 5 months ago, # |   0 My Video Solution of B and C where i have tried to explain my thought process and why the solution worked .
 » 5 months ago, # | ← Rev. 2 →   0 n = 5, x = 10, 7 11 2 9 5how can we choose 2 teams from this? shouldn't it be just 1 team cause min = 2 multiplied by 5 members = 10 so only one team?
•  » » 5 months ago, # ^ | ← Rev. 2 →   0 choose 1 member for one team....this team is formed only 11...bcz 11*1=11 > x... then choose 9 & 7 ..then you get min(9,7)*2=14>x... another way to choose is (11,9) & (7,5)
 » 5 months ago, # | ← Rev. 3 →   -6 hi, problem D: how to select a segment? like for k=3,x=1,y=100,n=9 a=1,2,2,3,3,3,3,3,4 b=1,3,4 
•  » » 5 months ago, # ^ |   0 It is given that elements are distinct
•  » » » 5 months ago, # ^ |   +1 If the elements are distinct just suppose, then how we can solve it?
•  » » » » 5 months ago, # ^ |   0 I suppose it can be done in many ways. Store the index of each element occurring in a. Let's call it apos. Also make a copy of array a. Let's call it astrip (I know the naming is bad). Now while iterating with b, mark all positions in strip where this number occurred. astrip[apos[b[i]]] = some_identifier. Because all elements are unique, this'll work. Now just iterate through array astrip and figure out the segments. Along with this you also need to make sure apos[b[i]] > apos[b[i - 1]] because you want b to be a subsequence of a.86693530
•  » » » 5 months ago, # ^ |   0 Elements are pairwise distinct. Doesn't that mean something like 1,2,1,2,1,2 could be a possible test case?
•  » » » » 5 months ago, # ^ |   +3 My bad :(
•  » » » 5 months ago, # ^ |   0 my bad.... thanks all for responding.....
 » 5 months ago, # | ← Rev. 2 →   0 in question E merge towers: for the given test case [[5,1],[2],[7,4,3],[6]] are the towers before any query on merging tower 1 and 3 according to me it should be done like: from tower [5,1] ,1 goes to tower 3-> [5],[7,4,3,1] -> 1st operation from tower [7,4,3,1] , 1,3 and 4 goes to tower 1 -> [5,4,3,1],[7] -> 2nd operation from tower [5,4,3,1], 5,4,1 and 3 goes to 3rd tower -> [],[7,5,4,3,1] -> 3rd operation it can be done in three operation but it shows 5 as a result . Where am i wrong?
•  » » 5 months ago, # ^ |   0 Put some clothes on and format the question ;)The result must be one tower [7,6,5,4,3,2,1]. Somehow you are loosing the 2 and 6. The rule for merging is fairly simple: Take the top segment from the tower where the 1 is, and put the whole segment onto the tower where it fits. The moved segments in the example would be: [1] [1,2] [1,2,3,4] [1,2,3,4,5] [1,2,3,4,5,6] The last move creates the complete tower.
 » 5 months ago, # |   +8 We can also solve F by matrix-DP .. First we calculate the matrix for every dp transition .. then we just need the product of these matrices .. (To simplify append 0 at the begining and 9 at the end ) and build segTree in [0,n] and then remains only two point updates at index(x) and index(x-1) .. (Make sure to have dp[n+1]=1 (even if its nine ))
 » 5 months ago, # |   -8 Can someone explain C more elaborately?
 » 5 months ago, # | ← Rev. 3 →   +3 .
 » 5 months ago, # |   0 In D problem, if the elements are not distinct then how we can solve it??according to me.i think complexity will be >=qudratic
 » 5 months ago, # |   0 There's another solution O(n) for A. As per the given condition, we can see that the element with value n will definitely satisfy this condition as middle element and the ends of the array as first and last element. (Because n will be bigger than anything and that's what we want here). If n is at the ends of the array, then n can't be the answer. We ignore that element, update ends of the array and do the same thing again. The highest element in the array is now n - 1 and if it's somewhere in the middle, that should be the answer. We'll keep trying the same for all elements and we'll know if the answer exists at all. 86668426
•  » » 5 months ago, # ^ |   0 Can also be done with prefix and suffix max in 0(n)
 » 5 months ago, # |   0 can anyone please explain what is this line doing in problem D solution res += (len - k) * y + x; . this line will be executed when we cannot perform berserk because one of the elements in segment is greater than both l and r and cost of berserk is lesser than fireball to delete everything. i thought this should have been res+=len/k*x so that we delete everything remaining using fireball. Thank you!!
•  » » 5 months ago, # ^ |   +4 The tutorial does the operations in irritating order. However, if the biggest monster is bigger than l and r we need to use Fire at least once. And independend of anything else, we need to use Berserk on at least len%k elements. The order of these two operations or the others operations does not matter.
•  » » » 5 months ago, # ^ |   0 thank you so much !! understood now
 » 5 months ago, # |   0 In problem C, I calculated minimum number of programmers required starting from programmer at position 'i'. This is will basically give me n (or less) segments. I then run an algorithm same as activity selection problem on this array of segements, to find the maximum number of teams. https://codeforces.com/contest/1380/submission/86686449 This comes out as wrong on test case 8. My answer is higher than the correct answer. I am not able to figure out why is this method wrong. Some help?
 » 5 months ago, # |   +10 In problem G why can we ignore the initial order of chests and sort? I was thinking on a case like 4 40 400 400 600 and k=1. Where can I put the mimic so that the answer gives 330 (as it does with the editorial solution)?
•  » » 4 months ago, # ^ |   0 You misread the problem: "Print n integers — the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.""the chests are placed into the rooms in some order and exactly k of the chests are mimics." so you can choose any order.
•  » » » 4 months ago, # ^ |   0 Oh. Now It makes sense. Ty
 » 5 months ago, # |   0 In B ,what if there is an extra constraint that "You will lose point if the bot's choice is superior",So doing the same thing which we actually did for the original problem,would it be still optimal ? For Ex :"RRSP" ,and output "PPPP", Win(1),Win(2),Win(3),Win(4)= (+1)+(+1)+(-1)+(0) = 1; Avg = 1 
•  » » 5 months ago, # ^ |   +3 It is still the case that every position is played against any other position. So the order of the symbols in the choosen string does not matter. Since the order does not matter it is still optimal to use one symbol for the whole string. As a pragmatic solution, just try all three possibilities and choose the best.
 » 5 months ago, # |   +4 Anyone please explain the update part in problem F .how we are doing updates ?
 » 5 months ago, # |   0 Just curious:Does there exist some kind of dp solution for D?
•  » » 5 months ago, # ^ |   0 Constraints are too high for a DP solution for problem C(assuming you mean by DP is to brute force and use dynamic programming to decrease number of total operations made).
•  » » 4 months ago, # ^ |   0 Disclaimer: It is a little messy because I tried to code as fast I could and submit it during contest.Here you go! 87810909 ExplanationThe idea is simple. We precompute minimum cost of deleting subarrays of size 1 to n such that either only single or zero element remains at last.When we are computing the answer we keep in mind that cost of deleting all elements but one with added cost of $y$ is lesser or the cost of deleting all elements is lesser i.e.$ans = ans + min(dp[size][1] + y, dp[size][0])$
 » 5 months ago, # |   0 problem C is enjoyable
 » 5 months ago, # |   0 86766761 Can someone help me identify what did I do wrongly here on problem C?
•  » » 5 months ago, # ^ |   0 I didn't understand your code completely...You can see this one might be helpful 86676269
•  » » » 5 months ago, # ^ |   0 I first consider all those with skill >= x because they can form independent group of size 1.I then have an array, say A, for the rest of those whose skillsets are strictly less than x, which means they need to team up with others to use the length of their team to compensante for x. I sorted this array A and loop it from the smallest value to the greatest.
•  » » » 5 months ago, # ^ | ← Rev. 2 →   0 (continue from above) Within each iteration in my array A, I calculate the corresponding length that I need such that I can form a team by taking the ceiling quotient num = x/a (where a represents the element of A and num represents the required team length)# Example: Array A = [1,1,2,3,4], given x = 5In my first loop, num = ceil(5/1) = 5 (which is reasonable, given that I would need a team of length 5 to fit the team formation requirement)# So, if in my iteration I find that my array A has the length greater than or equal to "num", then my group count will += 1 and I will delete the partition of my array from the original.If in my iteration I realize that given the element, it is impossible to form such a team from my original array A, then I remove the element from my original array A.Now that I think of this logic again, I feel like in certain cases it might undercount. But for now the failing case that I have is that the solution is 0 and my code gives 1, which is an overcount
•  » » » » 5 months ago, # ^ | ← Rev. 2 →   0 I have applied part of your logic here 86781167But without using the ceil part, which is not needed.And if you think of it carefully you will realize that there is no need for deleting elements from array A, since you are iterating throw elements...Note that ceil won't be useful if it is used with integers, and it can sometimes cause some troubles if not used properly, so avoid using it as you can.
•  » » » » » 5 months ago, # ^ |   0 Thank you, but I would be more enlightened if you could help me find what went wrong with my implementation or my thinking process
•  » » 4 months ago, # ^ |   0 If you start small, you are guilty of waste. M = 8, for example,,2,7,7,7,7,7,7 [1]; Instead of [7.7][7.7][7.7] [7.7] As for the answer greater than correct, if m=10,[4,4]; You only need 10/4=2, but (10/4)*2=8 does not equal 10
•  » » » 4 months ago, # ^ |   0 Thank you!
 » 5 months ago, # |   0 In C I miss read the question and was thinking that each team should have a product at least x. can anyone tell how to solve if this was the question ??
•  » » 5 months ago, # ^ |   0 In this case also, just sort the array and keep two variables. One containing the product of the present array and one containg the count of number of arrays. The code will look something like this: https://pastebin.com/zL2pxcfZ
•  » » » 5 months ago, # ^ | ← Rev. 4 →   0 I don't think it's that easy. here is the testcase 6 12 2 2 2 10 10 10 answer : 3 your code output: 2
•  » » » » 5 months ago, # ^ | ← Rev. 2 →   0 Ya, it's not easy. Sorry for the wrong explanation. Need a better approach for this.
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   0 I think the correct answer would be: 1. Because only the minimum element of the group and the size of the group matters. if I'm wrong I would be happy if u correct me.
•  » » » » » 4 months ago, # ^ |   0 I think i am not asking problem C. how about you read my comment again.
 » 5 months ago, # | ← Rev. 2 →   0 Can anybody help me with Problem D Berserk and Fireball I am facing some implementation issues? My submission — 86799781 Thanks
•  » » 5 months ago, # ^ |   -10 Do you really expect a meaningful answer to such a vague question?
•  » » » 5 months ago, # ^ |   0 Can you help me with my submission?
•  » » » » 5 months ago, # ^ |   0 You noticed the diagnostics at the end of testcase 7? Diagnostics ...runtime error: addition of unsigned offset to 0x13201820 overflowed to 0x13201818... 
•  » » » » » 5 months ago, # ^ |   0 Haa But I couldn't make out the error like what does it mean?
•  » » » » » » 5 months ago, # ^ |   0 I cannot spot the bug either. But I see some ways to greatly simplify your code.Do not use a set for index. The values you are inserting are distinct and sorted, just use vector.Additionally put a -1 as first element into that vector, and a n as last. So you can delete the copy of the main loops body.The first loop where you collect index is unnice. Still not sure if there is an off by one somehow. You better loop over a[i] and check every index if a[i]==b[j]. If yes insert i and increment j.
 » 5 months ago, # |   0 how's solution of 1380A-Three Indices correct w O(n), as we don't know that those three will be the consecutive no.s.
•  » » 5 months ago, # ^ |   0 If there exists a solution then it is guaranteed that three consecutive numbers are definitely a solution otherwise, the solution doesn't exist. e.g.-> Consider array --> [3,1,4,2,5] one solution is i=1, j=3, k=4 but instead, to make the algorithm O(n) we are more interested in i=2, j=3, k=4
 » 5 months ago, # | ← Rev. 2 →   0 upd:done
 » 5 months ago, # |   0 pikmike can you explain what 0,1,2,3 represent in the solution of problem F. Thank you.
•  » » 5 months ago, # ^ | ← Rev. 2 →   +6 0 = 00 — neither is taken1 = 01 — leftmost isn't taken, rightmost is taken2 = 10 — leftmost is taken, rightmost isn't taken3 = 11 — both taken
 » 5 months ago, # | ← Rev. 2 →   0 @pikmike Can you please explain me the problem -e test case when we merge the 3 and 1 tower how the ans coming out to be 4 instead of 3 . Ok I can tell you how i got 3.1st tower contain — 5 1 and 3rd tower contain — 7 4 31st move — you shift (1) from tower 1 to tower 3 now 3rd tower contain — 7 4 3 1 and 1st tower conatin — 52nd move — you shift (4 3 1) from tower 3 to tower 1 now tower 3 contain -7 and tower 1 conatin — 5 4 3 13rd move — now you shift (5 4 3 1) from tower 1 to tower 3 now tower 3 contain — 7 5 4 3 1 and tower 1 conatin nothing .So my answer coming out to be 3 instead of 4 . I know i am missing something could you please point out.Thankyou.
•  » » 5 months ago, # ^ |   0 Reread the problem statement carefully.We are not calculating the number of operations needed to merge the towers in the query. Instead, after each query, we want to calculate the number of operations required to merge all current towers into one.
 » 4 months ago, # |   0 why I'm getting Time Limit Error at test set 6 ? https://codeforces.com/contest/1380/submission/86946265 can anybody explain? thanks in advance
 » 4 months ago, # |   0 I think, there is a mistake in editorial of the problem 'C'. There should be 'Non-increasing' instead of 'non-decreasing'.
 » 4 months ago, # |   0 In problem E it is not at all clear how to achieve these answers.
 » 4 months ago, # | ← Rev. 2 →   +13 I have tried to make editorial for questions A-E . please have a look. Language :- Hindihttps://www.youtube.com/playlist?list=PLrT256hfFv5UUIBTJBUL7ZlZVHaWWEmGB
 » 4 months ago, # |   +1 Was F inspired by this meme?
 » 3 months ago, # |   0 I still don't understand the tutorial for F. Specifically, I don't understand how a segment tree could optimize this dp. What will the nodes store and what the merging looks like. I get the part about the dp but become totally confused when they start talking about segment tree.Can anyone explain it for me? Maybe some visual example would help. Thanks in advance.
 » 3 months ago, # |   0 4 6 1 2 5 3 This is the second testcase for problem A.. Why doesn't 4 < 6 > 1 give 1 2 3 as output ????