I think the you won't be rated in this situation.

I think after rating is updated, you can deregister and register back

I guess you are lucky (?) :D

After each contest, there is a time of 12 hrs where if we find someone's solution incorrect, then we can give input the case at which the submission fails, if hacked you are rewarded with some points sometimes.

1)After this contest (the same for following div.3 rounds and educational rounds), there's a period of 12-hour hacking phase.

That means, you can hack anyone you want(while in normal rounds you can only hack your "roommate" during the contest).

2)If you find something wrong in other's solutions(but passed the pretests), you can give an input which will make the solution fail (wa,tl,...)

In div.2 you get 100 points from each successful hack and lost 50 points from each unsuccessful one

In div.3 there's no prize or penalty for hacks.

50iq is not bad actually idk

But comparison of other websites code forces server is good

Did you ever participate in codechef cook off or luch time...they can't handle even 5000 participants properly...whereas codeforces handles nearly 20K participants...comparitively codeforces servers >>>>> any other cp server

The girl is the middle would rather be saying that "Give the lengthy problem A and engage participants."

Did you ever participate in atcoder contests It servers cannot hold even 5000 participants. Codeforces servers when it is +5000 submissions at time it will get some time but when it is less it works good!

When div3 was first proposed then it was said that a beginner should waste time in hacking instead of solving problem. So contest time hacking is not available in div3 rounds. And there is 12 hours hacking phase for improving hacking skill. But i think 12 hour is too long. It can be reduced.

rated for div 3(below 1600 rating)

lol

Because of my invaluable suggestions :)

You should be more careful with some corner cases

That's more important and valuable than strong pretests.

good luck and high rating bro

00:01 nice network

You just risked there by some basic observation, I don't think that on minute 2 you were able to prove your solution. So don't be happy by that.

The color spectra doesn't quite match Codeforces band XD

I don't think so

For you, yes. For expert or more, no.

If successful:

You will lose a problem solved

If unsuccessful:

Nothing changes.

So it's obviously meaningless to hack yourself unless you've found some mistakes in your code after the contest.

architb_12 why the hell are u giving this contest ,if u have such a mentality,and posting this nonsense doesn't show that u are cool

How can one read the problem, figure out the solution and write down the code in 1 minute.....

I'm sorry, what about you? I don't see any of your submissions were judged for more than 1.5 minutes.

Observation:

The contribution of a number is limited by number of occurrences of prefix and number of occurrences of suffix, In other words, if you have a prefix of 7 occurrences and suffix of 2 occurrences, Then you're only contributing to the final answer with just 4(2 from the left side which has 7 occurrences, and 2 from the suffix), So contribution = min(prefix,suffix) * 2.

Knowing the above, We can just try for each number x, A possible prefix that has 1 occurrence.. 2 occurrences .. 3 and so on, you can fill the middle with an element that occurs the most, you have only 200 distinct element, just go over them all.

store position of every element in a vector v[203].for each number between 1 to 200 run two pointer from start and end of its position vector and between them find maximum time occuring integer frequency using prefix sum 2-d array.

Hack via status page,not standing page

Oh man the alphabet is only size 200, Mo's was definitely overkill.

Yeah, I used Mo's too, don't know how others got that right

Here's my approach:

1. Loop through all $$$k$$$ possible "symbols". This will be the symbol forming the prefix and suffix (aaa...aaa)
2. Put two pointers on the first ($$$i$$$) and last ($$$j$$$) occurrences of the symbol.
3. Find what's the best score for the middle part (...bb...) between $$$i+1$$$ and $$$j-1$$$ using a precomputed table of counts for each prefix and each symbol.
4. Move pointers towards each other to the next two occurrences of the symbol from #2. At every time, you know exactly how many of these symbols there are to the left (= to the right) of the middle part, so the score for this partition is middle part score + 2 * symbol count.
5. Repeat until they meet, keeping track of the maximum score.

$$$O(n\cdot k^2)$$$ in total, but the operation #3 is rare in practice, so it passes the pretests ¯\_(ツ)_/¯. Got MLE on the first attempt though, lol. Don't define int long long where it's not necessary.

What is Mo. I thought of doing binary search but failed to do so.

Could you please explain how you applied Mo's?

You can enumerate a and the number of a, and then enumerate what's b, it seems 200*200*n, but you can find the really time complexity is 200*n

Author's solution is to replace all 2 with 1.

a[1][1] = a[2][1], a[2][4] = a[1][4], a[3][7] = a[2][7], a[4][2] = a[3][2], a[5][5] = a[4][5], a[6][8] = a[5][8], a[7][3] = a[8][3], a[8][6] = a[7][6], a[9][9] = a[8][9] This was my idea. Later, got that replacing all 2's with 1's was good enough.

This is author's solution :)