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Jury solution: 54047380

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Jury solution: 54047665
Hard B, trivial C and D, cool E. And I completely understand that it's easy to misjudge the difficulty of B.

I don't like limits in F and I want setters to take some things into consideration in the future. Here, the jury solution (attached in the editorial) takes 40% of TL what is quite tight, especially considering the fact that it does all modulo efficiently — avoiding the

`%`

operator if possible, doing`if(x >= MOD) x -= MOD`

instead. You should do that in brute force solutions and make sure they still don't pass, but the author solution shouldn't be written very efficiently. Remember that participants won't necessarily implement everything in the same way. Do you really want to penalize them for writing a code that is twice slower? And let's remember about a bit slower Java, though maybe it wasn't an issue here — because modulo was the heaviest operation. TL 8-10s and ML 512MB look much better to me (the author solution takes 118MB and it would barely pass after replacing ints with long longs).Tight limits can be forced by a dangerously fast brute force and maybe everything was chosen as good as possible in this problem, I don't know. Even in that case, I think that my post still can be of use for future setters.

Remember this: limits should be chosen in such a way that naive solutions don't pass. If this condition is satisfied, it's great to make TL to be much much more than twice the running time of your solution.

And limits in C were too tight as well. Reading $$$500\,000$$$ numbers and possibly doing something with a set or a segment tree? That takes some time, especially in slower languages. It's good if easy problems are solvable with Python. Just make TL to be 2s or 3s. Why not?

I don't agree that B was difficult. If you implement brute force solution which just runs through all possible strings of length n and play around with it for a while you can clearly see the pattern.

That is not you expect from B in six problem contest.

I did B as a specialist. So I would just agree with oxytocyna

In problem C, segment tree/set takes $$$\mathcal{O}(n\log n)$$$ time, and author's solution is $$$\mathcal{O}(n)$$$, so such a tight TL could be the result of one of following considerations: 1) the author didn't know about $$$\mathcal{O}(n\log n)$$$, so he didn't see any difference between $$$500\,000$$$ and $$$200\,000$$$ and chose arbitrarily, or he tried to avoid participants sending unoptimal unpredicted solutions; 2) the author knew about $$$\mathcal{O}(n\log n)$$$ solutions and didn't want them to pass. Actually, in this case he didn't perform the clipping perfectly as the segment tree approach turned out to be productive for some participants, but my segtree failed the system testing (though changing the vectors to arrays and switching compiler from С++17 to C++14 let me gain an ok in upsolving in 998ms).

I don't know about your claim — I implemented C in $$$O(nlogn)$$$ using priority queue, and I got AC in 327 ms.

Code is here

I implemented $$$O(nlogn)$$$ using set and passed (in 265 ms).

C can be solved in O(n)

I'm still beginner, but I think that any Div1 participant must be able to optimize his code to the fullest

PS: In fact I love down-votes

Amazing Problemset <3 Fast Editorials <3

can anyone explain me div 1 C with more detail how to solve this with set/segment tree.

When will the editorials of rest of the problems be translated to english?

BTW, is it only me who is thinking that after AtCoder in japanese and Codeforces in Russian, Codechef will start publishing editorials in Hindi and will refuse to translate them?

that was a good one

Please translate

UPDATE: Answered hereIn Div2E/Div1C.

Does anybody know why this 54049263 gives MLE? When I change stack to set I get AC in this 54047364

My Approach: Construct a directed graph where edge (u,v) means a[u] < a[v] and then do topological sort on that graph.

The way I construct it is if nxt[i] = j that means there is an edge(i, j), also there is an edge between all elements [i+1, j-1] and i since all of them must be less than i. I only keep track of the least previous element.

Here's an example where ix, jx means nxt[ix] = jx:

i1 i2 i3 ... j3 j2 j1. Edges are (i1, j1), (i2, j2), (i3, j3), (i2, i1), (i3, i2). I don't add (i3, i1) since it's redundant.

In a case like i1 .. i2 .. j1 .. j2 In the MLE submission I exit immediately and say that it's impossible. In the AC submission with set I let the topological sort handle it.

isaf27 why no english editorial ?

I'm sorry, before the round I've written only Russian editorial. Now I should translate it, but I have a very hard time at university. Before the end of the week, I will finish. You will see the update in "Recent actions" section.

First of all, thank you for having prepared a CF round. It helps the community and it requires a lot of work. ... But, why couldn't you write the editorial in English from the beginning? In that case you wouldn't have to spend extra time translating.

Yes, I wanted to do this. But I started working on test cases shortly before the round and was busy doing test cases. Next time I will definitely begin from doing statements and editorials because it is usually hardest work for me :)

Thanks

Where could I find the solutions in English pls

English solutions please!!

Can anyone help, what is 'l' in the explanation of D above?

It is from the statements — any position of the string t occurrence.

Now editorials in English are available, sorry for waiting. If something isn't clear you can ask in comments.

in problem c (The Party And The Sweets) in test case 1 what is wrong in this ans: b1- 1 1

b2- 3 4

b3- 1 1

ans=11 ??

It was really the most popular question during the round :) Check the formal explanation of the problem, a minimal number in the second row is wrong.

can you pls elaborate

In the problem statement it has been stated that 'For all 1≤i≤n the minimal number of sweets, which i-th boy presented to some girl is equal to b[i] ', which means that b[2]

hasto present a girl with 2 sweets, therefore b[2] can either be 2 4 or 3 2 and not 3 4, and further on as 'for all 1≤j≤m the maximal number of sweets, which j-th girl received from some boy is equal to g[j] ' therefore if we choose 3-g[1] 2-g[2] for b[2] therefore some other boy (b[1], b[3]) has to give the other girl (g[2]) 4 sweets (which is the maximum she received from some boy)Thus the solution can be,

b[1]- 1 4 b[2]- 3 2 b[3]- 1 1

ans=12

Div2D:Editorial is not helpful at all.You explained the pattern to generate the answers.

But how to come up with that kind of pattern in first place.

Please share some ideas in what direction to think for problem like that to come up with patterns like that. Stating formula and why formula works is not helpful at all.

To solve this problem it's better to solve some cases, for example:

And after the last case, it's easy to guess the pattern. I think it is a good way to solve this problem.

About Div2-D, condition on the assumption of periodic

aand minimum unique lengthk, I got$$$ a >= (n - k + 1) / 2 $$$

doesn't it means the period not unique but have a lower bound? But I failed on program checking.

And when the parity of

nandknot same, even the lower bound ofais wrong. Why is that? Really doubtful.Hi, I guess, ( n%2 = k%2 ) ensures about parity being same. Unless you are talking about the bonus question provided at the end of solution.

Actually I have doubts to both problem. In the original problem I cannot justify the

uniqueofa, else on bonus I completely have no idea... Anyway, Thanks for your reply.Can someone please elaborate on the solution for 1159B (Expansion coefficient of the array)? Thanks:)

I binary searched the answer. Link to my answer

Thank you for your solution. I understand the binary search procedure, but could you elaborate on the math logic explained in the editorial that you used in your code?

Actually, even I didn't understood it. I rather used some diffrent way.

So, the maximum possible value of k is equal to the minimum of min(ai,aj)/|i−j| for all i<j. Let's note, that min(ai,aj)/|i−j|=min(ai/|i−j|,aj/|i−j|). So, we need to take a minimum of ai/|i−j| for all i≠j. If we will fix i the minimum value for all j is equal to ai/max(i−1,n−i) and it is reached at the maximum denominator value, because ai≥0.

Can someone please explain the last line of the paragraph? It seems to me that we need to look up the minimum value of input and then divide it by some index. I'm not getting the formula like, why we need to take max(i−1,n−i) as the denominator.

EDIT: It's for problem B. Expansion coefficient of the array.

In problem C, I think I solved it quite similar way as written above, but I keep getting wrong answer on testcase #7. Can somebody help me to find why? Code is written in Python.

code: n, m = map(int,input().split())

b = sorted(list(map(int,input().split())),reverse=True) g = sorted(list(map(int,input().split())),reverse=True)

if g[-1] < b[0]: print(-1) exit(0)

total = sum(b) *m

g_idx = 0 b_idx = 0 while g_idx < m: temp = 0 while g_idx < m and temp < m-1: if g[g_idx] == b[b_idx]: g_idx+=1 continue total += (g[g_idx] — b[b_idx]) g_idx+=1 temp+=1 b_idx +=1

print(total)

can anyone give me the binary search solution of problem C

Hello, I wanted to propose that there are problems that seek to minimize the coronavirus transmission curve. Perhaps problems where data are given on the populations by age of different parts of the world, even the least populated and their communication channels, run a simulation of the interactions over time and wonder what would be the best strategy. A first intuitive solution would be, for example, to divide the inhabitants by age or risk groups and those least prone to getting sick, to send them to the densest urban centers and the least prone to the least dense urban centers, once they have been infected and cured. healthier they can go back to less populated places, so there would be less spread. What do you think ?.

Can someone pls make me understand the approach to be used in Permutation Recovery Problem...pls..pls..pls

I think the editorial's proof is overly complicated for Problem E. We can just prove it by induction. (construction from the right side)