Orz

all testers and authors orzzz

stop this cringe orz thing please

What does orz mean?

beaten_by_ai orz , remember you from IICPC regionals

How to contribute as tester. is there any blog in codeforce to explain it?

My expectations are like: - Solve 2 problems - Close the browser

demoralizer orz

It's the third post on the front page

But I was blocked by human-machine verification when I was trying to submit B,and it took me about 20 minutes.Otherwise my rank could rise to about 500,and I won't go back to Expert today.QAQ

its part of the journey... GLHF

good internet + speedforces

same... it was a WA forces for me

Please do not discuss anything about the round before it finishes. Yes, it is 10 minutes before, but you still cannot discuss it.

We are aware of the situation on F and are looking into it.

the amount of newbies in top 50 is crazy

I agree there is more implementation than usual, but I hope not too much. We cannot do much about stopping the cheaters other than punishing them after they cheat.

I'm glad you liked them. But please do not make such comments before the contest is over.

what was C? was it like :

median <= k possible iff number of elements <= k are >= 2.

get the max and min index of these elements which are <= k. let them be l and r respectively.

check if any 0...i or i ... n (where i is between l and r) has a median <= k. if yes then ans is yes thats because we can always chose that (the range that i mentioned) subarray and take the other element as l or r and since 2 out of 3 elements are <= k the median is always <= k.

I didn't look into status page to check submissions: I saw disbalanced stats on F in overall stats and opened it as a next one. Based on scoring I thought they should be relatively similar. Even keeping in mind that someone is doing gpt-submissions, it still made sense for me to read it first: if the task is solvable by llm, maybe the statement is clearer than in E, and this is important for problems with close scoring.

I havent solved F but it was a close call. So I believe this statement I don't think banning everyone who solved F but didn’t solve E would be inaccurate. is a bit too much: some people's strategy definitely was affected by standings.

However I agree that the situation is a ridiculous one.

so because median is at ceil(m/2) and the limit is actually <= K ...

to quickly say for a range if median is <=k .. you can just count how many elements are <= k .. .if that >= len/2 this subarray has a median <= k

so imagine we call subarray with median <= k to be v and > k to be w

we need to find the pattern vvw or vwv or wvv

vvw and wvv are same... just reverse the array

for vwv I calculated smalles v in prefix and suffix

to make it easy to count elements <= k in a subarray ... I replaced all elements <= k with 1 and >k with 0

now v subarray has sum >= 0 .. and I solve everything I mention aboe using prefix sum

https://mirror.codeforces.com/contest/2103/submission/316591716

Why am I getting WA on test 2

my approach https://mirror.codeforces.com/blog/entry/141912?#comment-1268096

Think it from this perspective : for a segment to have median <=k , the count of elements <=k should be greater than the rest of the elements of this segment...

Try this:

1 
6 
1 2 1 2 1 -1

I think this is wrong: (o && x!=v[i][o-1]+1)

After 1 finished, you check if 2 is connected. You should use linked list.

After looking at it for a bit, it seems our solutions use the same basic idea. However, it seems you're missing some casework.

Firstly, in your for, you look for an index $$$i$$$ such that $$$cur_{i - 1} + 1 \lt cur_i$$$ between your indices as the reorientation point for the sum (right, cause we want to form a structure akin to the following: "\/" or "/"). But, what if the values between $$$cur_{i - 1}$$$ and $$$cur_{i}$$$ are all less that the current value that we're checking? In that case, you reorient yourself at the wrong point (as one of the values in that iteration could become either a local minima or a local maxima). Therefore, you need to find a mechanism of selecting that index $$$i$$$ such that the values between $$$cur_{i - 1}$$$ and $$$cur_{i}$$$ hold with the previously stated condition.

Secondly, I think you aren't handling the case when we shouldn't reorient ourselves at any point properly. Anyways, I think if you manage to rework that previous case (the solution to that part isn't as bad as it sounds) you will be able to find all additional janky cases.

try to break the string into components of 0's and 1's and count them rest is just casework