wait actually since when did Radson pass me in rating :>

Good luck , I wish the same

i didnt sell

Sleeping early instead of CP makes a better life

You can't beat our orz CarViz!

You 2 graph profile pictures, is the same person.
Isn't it?
One is seeking downvotes and the another one came from future.

time traveler detected?

W flag in your pfp

As an enchanted golden apple, you are not a contest

You can (

cry's just exaggeratin' (or should i say 'joking'?), but hopefully nobody really ends up in his basement, which, i shall say, really D-I-S-G-U-S-T-I-N-G (especially the part of 'no toilets'). Second, i can't possibly travel there, can i?

ابعتهالو

So that higher rated people can’t create a new account or lose rating to participate in Div. 3 rounds and absolutely destroy everyone in the standings. It doesn’t affect you rating changes after the contest, that’s only the leaderboard feature, so you should really not be worried about that.

Same, but Carrot doesn't agree :(

I'm crying over these problems

this was my second contest, got destroyed by B TUT

I'm crying over these problems

just curious, how are you checking (live or post-contest) performance stat?

Thanks bro :)

My D failed system tests :(

Still ranked up

OH HECK I FORGOT TO CONSIDER DFN WHILE I AM DOING FENWICK TREE PUSHES.

OOOOOOOOOOOOOHHHHHHHHHHHHHHHHHHH HHHHHHHHHHHHEEEEEEEEEEEEEEEEEEECCCCCCCCCCCCCCCCCCCKKKKKKKKKK!!!!!!!!!!!!!!!!!!!!!!!!!!!

Please use spoilers

<spoiler> your code </spoiler>

yes it was

just had to check count of ones on even and odd places are same or not for C1

for C2 just add the (differnce between ith index of even index 1s and odd index 1s in a and b)/2

I think is analyzing the invariants of the operations, for example the invariant in C1 and C2 is that doing the operations in a substring of size 3 is the same that swapping the first and third character in that substring, so the character moves 2 positions to the left or to the right

I think that it suffices to start with all nodes in their own set, then at each step merge the cheapest node/set into the lowest set that is entirely above it. But I didn't finish coding it so idk for sure lol

Didn't solve during contest because misread problem, but my idea is roughly like this:

If $$$k$$$ is less than the number of leaves in the tree, then it is impossible because no two leaves can be in the same set. When $$$k$$$ is equal to the number of leaves in the tree, we try to solve the problem for this $$$k$$$ (Initially, each leave occupies a set. We go up the tree. For each subtree $$$x$$$, we try to merge $$$x$$$ into the set $$$S$$$ in its subtree that has the lowest $$$\max_{i\in S} a_i$$$, so $$$\max_{i\in S} a_i$$$ for all sets get maximized generally. We find this set $$$S$$$ for each subtree $$$x$$$ using DSU on tree or some DS I think.) Finally, for $$$k$$$ larger than the number of leaves in the tree: we maintain a heap of the values of the vertices that are not the maximum of the set they belong to. These nodes are unchosen, and as $$$k$$$ increases, we can choose and add them into the answer. So we just maintain a heap of these values, and each iteration just add the max to the answer.

The problem comes down to finding the best way to create the sets for smallest k such that it is possible. Then to get the anwer for k+1 you can put the biggest element that is not the biggest one in its set in a new set.

Now we have to find the best sets for smallest solvable k.

A key observation is that each leaf will always be in a different set meaning the smallest k is the count of leaves.

Now we have to decide how to decide the set for non-leaves. Say we are are deciding the set for vertex u and its subtree has some sets that could be chosen. You will always choose the set with the smallest maximum value since thats the biggest potential increase — greedy choice.

You want to know all the sets and their maximum values in u's subtree. This can be done by using small to large merging and saving the pairs of values in a set.

My solution is something along these lines.

for me E was much harder, in fact i guessed that the first element must be the min element that is more than zero. For C1 i think the most important part to notice was that by applying operations you never, ever, mess up the element in the middle and this means that elements in even pos will never have to deal with elements in odd pos.

i guess its better for you to solve problems rather than being sad for performing bad at contest(im not saying that dont participate in contests, im saying that be focused on solving problemset instead of contest for this level(at least this is what our teacher says))

wair for a few minutes until system testing finishes.

start from elements whose value in B is zero, if there are none is it possible to construct A ?

Now that we know this, i ask another question, if multiple elements in B are equal, will they have equal value in A ?