minum susu gaming

minum susu gaming

No, of course.

Definitely no

agree

very excited to wake up at 5 AM for this wonderful contest!

Yea.. Two times in a row now

I know about it already. But this contest is based on COMPFEST 14, the only possible way to avoid the conflict is to reschedule it after ABC (clashing with ABC would have more complaints), but by then the contestants of the onsite contest would have gone home and would have access to internet which is obviously not ideal.

Therefore, it is impossible to reschedule due to these constraints.

I think your answer should be in contest problems.

i think it means rama_pang drank milk (cmiiw)

Hoping your wish comes true.

Good luck!

...

why?

Everytime this happens my adrenal glands are confused. That can't be healthy for them... :(

Same, m2.codeforces.com is working fine

I had multiple different brain-fade solutions in E, and each of them gave WA 6, so likely it's just a normal case catching many of those WAs. The correct solution is to take the max height you can, or allow your children to take the max height they can, and so on.

Example: if you have a tree

9
1 2 3 3 5 6 1 8

the answer is $$$7$$$, because you can assign

$$$a = [8, 5, 4, 9, 3, 2, 1, 7, 6]$$$ and remove them in the order $$$[7, 6, 5, 4, 3, 2, 9, 8, 1]$$$.

Since you can only remove leaf node, consider the vertices in topo-sorted order. You have two option: 1. continue the sequence from some child. 2. End the sequence here.

You can solve it with some kind of tree DP. At each vertex, you decide if you want to go further up the branch and if you want to stop at the vertex and merge the longest branches of the child nodes.

The explanation seems a bit abstract, so I upload 178381518 together.

Solution for E:

Let $$$dp[i]$$$ be the answer to subtree i.

Consider in two cases：

Case 1:node i is not in the answer.In this case,we can easily get:$$$dp[i]=\Sigma dp[j]$$$,$$$j$$$ is son of $$$i$$$.

Case 2:node i is in the answer.Through observation, the answer must be a chain.

In general,$$$dp[i]=max(\Sigma dp[j],depth[i]-max(depth[k])+1)$$$,$$$j$$$ is son of $$$i$$$,$$$k$$$ is descendant of $$$i$$$.

for each node calculate (a, b) where a is LIS endint at this node and b is LIS not ending at this node (considering this node's subtree)

transition for a is max of childrens' a +1

transition for b is sum of all childrens'(max of a or b)

Consider painting each node with black or white.

• If you paint node $$$x$$$ with black, then its score will be the maximum point among its children plus $$$1$$$.

• If you paint node $$$x$$$ with white, then its score will be the sum of all its children.

The answer will be max(black(1), white(1)).

Use this test case:

1
6
2 3 12 25 27 28

Expected output: 36

Video Solution for Problem C

Yeap, dp[i] = max(depth[i], sum_of_all(dp[children])

yes, transitions are pretty simple

dp(i, 1) = max(dp(u, 1) + 1), over all child u if i

dp(i, 0) += max(dp(u, 1), dp(u, 0) for all child u of i

i, 1 -> if we include i as next next element of longest subsequence

i, 0 -> if we exclude i and only take its children in longest common subsequence

base case is as following dp[leaves][0] = 0

dp[leaves][1] = 1

It's optimal to always place the block so that it's height is bigger or equal than width .

Now placing the blocks in asc/desc order will be enough.

1. Rotate the box such that x axis is aligned with minimum length side (Reason: Its perimeter cannot be minimised by overlap, so try to minimize the length)
2. Sort the (length,breadth) of all pairs
3. Calculate perimeter by excluding the overlap

178390392 Could someone provide test for this?

Rotate blocks in such a way that width of block be minimum. Then place them in ascending order of their hight and calculate perimeter.

Figuring out minimum number of cells without a card needed to carry a card from cell (1,1) to cell (n,m)

No. Consider frequency array of original one is [3,1,1,3]. M=3, D=4. But we cant make sizes of sets to be [4,3,1].

Wrong.

Just don't code in python)

Probably this counts as "magic with acceleration", but after getting TLE a couple of times with Python because of slow input (in earlier contests), I found that adding

import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

as the first two lines of the code helps a lot. You just add these two lines, no other changes required for inputting integers. With this I passed B in Python in 233 ms. I now have these lines standard in my Python template.

Note that when inputting a string, you now need to do input().decode() instead of just input().

Not dp

Suppose you have an array a of 3 elements 1,3,5. How you should rearrange these elements so that abs(a[0] — a[1]) + abs(a[1] — a[2]) becomes maximize?

consider array: 1 2 10 12

if you put 1 in bag 2 alone, the best answer is |1-2|+|1-12|. you see you can always subtract the number you want (in this case is 1) from the biggest number in the array (in this case it's 12) but the problem is nuber 2 because you can't get rid of it if you have chosen 1.

but if you put 1 and 2 in bag 2, now you have the value |2-10|+|2-12| which is much better. so the answer is max(|a[i+1]-a[i]|+|a[n]-a[i]|) for all i.

and you need to consider the case when the pivot is the biggest so max(|a[i]-a[i-1]|+|a[i]-a[1]|

I hope you got it now.

I found COMPFEST problems to be nice every time , there was a compfest round (unrated) few months ago, that had great problems too

Nostalgic! (◦'⌣'◦)/

I also thought of this same scale when solving problem D lol.

The statement says $$$3\le n,m$$$, so this kind of case doesn't exist.

By the way, when I read problem E, I went to reading the sample explanation (because I accidentally confused myself thinking that we remax instead of remin), and then I just coded the solution without proving it (well, I kinda convinced myself that it is intuitively obvious, but point stands)

what is the speedup from N^3?

I'm lazy to provide an actual code but one possible solution is using threads. You can set an increased default stack size when you make a new thread.

w2 as 45 and w1 8 and w2 19

1

3 3 6

1 2 3 4 5 6

this case must give "YA".

I think this is the worst case in the problem.