- A: Author: Giove, Prepared by: Giove
- B: Author: antontrygubO_o, Prepared by: antontrygubO_o.
- C: Author: Petr, Prepared by: MZuenni
- D: Author: Petr, Prepared by: Petr
- E: Author: MZuenni and dario2994, Prepared by: MZuenni
- F: Author: dario2994, Prepared by: bicsi
- G: Author: Giove, Prepared by: bicsi
- H: Author: Petr, Prepared by: Petr
- I: Author: Giove, Prepared by: Giove
- J: Author: bicsi, Prepared by: Martin Kacer.
- K: Author: antontrygubO_o, Prepared by: antontrygubO_o.

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Auto comment: topic has been updated by Petr (previous revision, new revision, compare).OMG!Endless editorial!

Thank you!

Is there a dp solution for problem B

I am also interested.

Helpful. +1

Did someone tried to solve B using binary search and maximum bipartite matching? I tried but got TLE (253131171).

I did :) I would not say I am proud of that but I thought about doing a binary search + checking Hall's theorem statement on the graph. It can be shown that it is enough to check Hall's condition only for subsegments of appetizers in the sorted order. However, $$$O(n^2 \log C)$$$ would probably be too slow, so I just initially set the answer to infinity and then looking at all subsegments decrease this value as long as Hall's condition does not hold. This leads to an $$$O(n^2)$$$ solution. Huge overkill though...

My approach was also using Hall's theorem; I thought about how to optimize $$$O(n^2$$$ $$$log$$$ $$$C)$$$ to $$$O(n^2)$$$ for a bit before realizing it's easier to optimize it to $$$O(n$$$ $$$log$$$ $$$C)$$$.

For each element, let's say $$$x_i$$$ elements of $$$b$$$ are $$$\geq a_i+ans$$$ and $$$y_i$$$ elements of $$$b$$$ are $$$\leq a_i-ans$$$. Then, you want for all $$$i \leq j$$$, $$$x_i+y_j \geq j-i+1$$$, or $$$x_i+i \geq j+1-y_j$$$, which you can check by finding looking at the prefix minimum of $$$x_i+i$$$ and making sure that it's $$$\geq j+1-y_j$$$.

Thank you for replying. It would be really helpful if you can share your code.

I checked for hall in $$$O(n²\log{C})$$$ and it passed :). Maybe methods with a higher constant TLE.

For all elements in $$$a$$$ you calculate $$$(l_i,r_i)$$$ such that $$$a_i$$$ can match with everyone from $$$b_1$$$ to $$$b_{l_i}$$$ and from $$$b_{r_i}$$$ to $$$b_n$$$. Now for each $$$(l,r)$$$ you want to calculate how many $$$(l_i,r_i)$$$ there are such that $$$l_i \leq l \leq r \leq r_i$$$, which is quite easy to do with a 2d prefix sum. I did have to do that prefix sum with a global array instead of a vector of vectors to get AC though.

Poblem B, please, help me to figure out what's wrong with the code below:Thanks

Is there any solution for problem F that use the alternate method they propose here? $$$O(K\sqrt(K)$$$ wouldn't enter with the k=10^6 right?

you can use bitsets to make it $$$O(\frac{k\sqrt{k}}{w})$$$

Can someone explain the third paragraph of the proof in problem K("Now, note that if we swap free elements in A and C...")?I know the proof's idea is to swap the smallest $$$n_{b}$$$ numbers from A and C to B.I also agree with the solution about the truth that free elements in B and C can be swapped without break the conditions.But why the third paragraph swap free elements in A and C?I can't understand the necessity.I'm also doubt with the sentence at the end:"for the same reason.".Thank you.

Auto comment: topic has been updated by dario2994 (previous revision, new revision, compare).i am studying this solution

can anyone explain this how we are calculating value of H in this solution like in this loop..

for(int i=x;i<=N;i+=x) xx+=cnt[i];

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Can someone please explain the solution of problem B .

can anybody tell why this solution of Problem I(eye) is getting wrong answer on test 27 while this 269511518 getting accepted!!

what i believe is that for a connected component, you need to visit it completely and not prematurely return the answer, otherwise some of the nodes will remain unvisited and you may start a dfs for the same component again later with half of the component already visited giving wrong results..

Hello i know its a bit late but can someone help me with C. I am trying with multisource BFS. first we take the leaves then move them and so on. But there is probably a greedy way of doing it.

Imagine u is a leaf node after some operations with val[u]>val[v]. Now ofc it is possible that after more operations, val[v] >= va[u] is possible. I am not able to figure this condition in the BFS.

I thought of 2 things. First if val[v] < val[u] and the neighbours of v is less than 1 then it wont be possible. Another thing i can think of is if the val[v] has immediate connection to leaves all having value greater than itself.

Not exactly sure how to do :(