Video Solution to C and D

Live upsolving 5 mins after the round ends: https://youtu.be/Fcxm3iROflI

Good luck!

Video Editorial for Problems A, B, C

Looking forward to this round, man!

You know the deal, upsolving the problems 5 mins after the round ends: https://youtu.be/2jW2zTSoGCM

Good luck and have fun!

Looks like I really got lucky this round :D

GG bruh, I figured out during the livestream that I messed up one case :))

Oh that's right thanks!

Video Editorials to A + B(Div.1) / D + E(Div. 2)

Hey, guys!

Right after contest ends, we will discuss the problemset live on Algopedia: https://youtu.be/ZPQzYDf5A4Y

Good luck to all!

Video Editorials for A, B, C, D and E

Congrats to our indian problemsetters, I'm really excited about this round!

And talking about excitement, we will upsolve the problems 5 mins after the round:

https://youtu.be/yDdzbZhSYD0

See you on Algopedia!

Video Editorials for Problems C and D

As you got used to it, upsolving is on us 5 mins after the round ends:

https://youtu.be/KfiUrDLad4Q

On Algopedia. Bring your masks!

Video Tutorials for C, D, F, G, H

Good luck to all! Live solving the problems 15 mins after the contest finishes:

https://youtu.be/gFepyRGn1vs

Only on Algopedia!

Problems B + C1 + D + E

Problems B + C1 + D + E

Problems B + C1 + D + E

Live epic upsolving after the round:

https://youtu.be/SdwVz4qzhkY

We will be live upsolving the round 10 minutes after it finishes on Algopedia: https://youtu.be/E7fiCGAqNJM

Unfortunately the round got rescheduled, but this is a good opportunity to solve some other problems with you guys, so we will do that instead!

https://youtu.be/HneuObfyDtw

Do you have any preffenreces?

We will still be live at 19:00 with some other Codefores problems:

https://youtu.be/HneuObfyDtw

Q: Will it be some live stream on YouTube to up-solve the problems straight after contest?

A: Algopedia: https://youtu.be/HneuObfyDtw

Upsolving C and D live on YouTube: https://youtu.be/XeK6lYKd8W4

Good luck, guys!

We will be live upsolving the round when it ends:

https://youtu.be/0kr6YSroclg

Great round, guys!

We are going live now:

https://youtu.be/ppk9pXL9rR4

Good luck to you all and have lots of fun, coders!

As you got used to it, we will upsolve the most interesting problems of the round live at 8PM:

https://youtu.be/ppk9pXL9rR4

Thanks a lot bro, I'll keep them coming!

We will upsolve the Div.2 round live on Youtube after the contest is over: https://youtu.be/FgrB_bvZJ_A

Going live on YouTube explaining A, B, C: https://youtu.be/FDbd_sYP7hM

Here it is brothers: https://www.youtube.com/watch?v=SS5eONgsluE&feature=youtu.be

Thanks a lot, bro! See you on the next one

Auto comment: topic has been updated by RomeoFantastik (previous revision, new revision, compare).

Hey guys!

As promised, I will upsolve the problems in a livestream on YouTube. Hope to see as many as you there!

https://codeforces.com/blog/entry/75777

They see me codin, they hatin

I will provide you video tutorials for the round after it's done!

Hey guys, I made a video tutorial for problem F which, btw I really enjoyed solving! https://www.youtube.com/watch?v=oEPtZLUG4iQ

Auto comment: topic has been updated by RomeoFantastik (previous revision, new revision, compare).

That is pretty accurate in a lot of fields and in life in general. Everything is a marathon rather than a sprint. Could be getting good at competitive programming, landing a job or growing a Youtu... (okay, no more of that!) The point is that even if you achieve that short-term goal, on the long run it is going to kill you. I have a lot of friends who want to learn some coding interview questions quick and ask me for help, but I can't get them to realise how big of a mistake they make by having this mentality. Result: none of them has kept a job for more than 5 months... (that includes me too LOL)

I've written in the second paragraph I'm a little too tryhard..

There is definitely a good thing about raising awareness about the possible threat, but at the same keeping it balanced. While a lot of people want to keep the environment as safe as possible, a lot of people take advantage of this to hit the trends. Now you might say: yeah bro, but you just posted about Coronavirus on Codeforces, wtf!?

That's right, you got a point. But I'm confident this community can find that balance: be aware of what's happening while still be chill and bring down the servers of all coding platforms by compulsive problem solving :D

Keep coding guys!

Auto comment: topic has been updated by RomeoFantastik (previous revision, new revision, compare).

There is an algorithm which does this even in linear time, based on Quick Sort. Thinking about doing a video on that to

But, do you know how to solve it with Divide and Conquer approach?

That is right, it was "DZY loves squences".

Auto comment: topic has been updated by RomeoFantastik (previous revision, new revision, compare).

Of course one problem a day is the biggest joke in order to achieve great success! At first, I will post the one that offers the most value, the focus these days is so dispersed that posting more than one initially could not be effective.

We always have the min_valid_start for our current R, right?

We may when we propagate the answer to need this information for older R's.

Yes, min_valid_start are also ok, but I was talking about the solution for every node, this soluiton we process by lazy update, but when we have to propagate it, we will eventually modify it by some R1 — min_valid_start + 1 , but this min_valid_start is not the one we have for the current R, is the one for R1 and we no more have that value.

I also have a question. When you want to make lazy propagation, you have some R values which for some segments should be updated, but these R values are others that our current R, so the min_valid_start in those nodes are others than they were for that R values, and we can't update the node with our current values, we need the old ones. How do you solve this? Do we need a persistent tree or something?

Awesome! Thanks a lot!

At problem f I used suffix arrays with lcp precalculations,then just a stack and that's all. I still can't figure why I get wa on test 12.Is anybody who had the same problem? Can you help me fix this?

I really don't know how it is to become a red coder. But I think succes means only 2 things : 1. Hard work!!!! 2. A good mentality and to relax when it is needed. You can look at my rating how it decreased as hell; the point is that I really don't care and I know someday I will be as good as I want in contests if I keep working, but I'll have to wait until the right moment and NEVER GIVE UP!! It's not all about innate skill.

If you find the solution please let us know

In your problem it is clear that the brute force solution is dp[i][j] = the total minimal cost for the first i elements to split them in j sub-lists, and for that you fix the left side of your last list. O(n*n*k) There is an optimization that can transform you solution in O(n*k* logn ) , noticing than for your dimension j , once we iterate through all i s , the optimal left side for dp(i,j) is >= or <= then the left side for dp(i-1,j) . You can do a divide and conquer algorithm for all possible values of j. Last Edit : This solution doesn't work, you can brute force the left sides and the propery isn't satisfied :(

Some ideas, please? :)

Auto comment: topic has been updated by RomeoFantastik (previous revision, new revision, compare).

Auto comment: topic has been updated by RomeoFantastik (previous revision, new revision, compare).

Can anyone offer a brief explanation about problem E please? I really don't know what is the basis of a vector and why the answer is 2 ^ (b.size()) ...

Can anyone offer a brief explanation about problem E please? I really don't know what is the basis of a vector and why the answer is 2 ^ (b.size()) ...

How to solve problem H?

I forgot a return 1;

The windows returns it automatically,but on linux is a big problem. Sorry guys for the misunderstanding

at problem d i output 3

2 2 23 i have wrong answer on pretest 1 , which has n=27 why? LE : my bad, sorry

I am very sure that your second solution to the problem D is wrong. Me myself I thought of that solution. It isn't always esentially to take the minimum. Sometimes it's good to save some '?' for the future the other columns

Hello, BAM! Good to see you ;) In Romania, "Mo's algorithm" is the RangeMode "smen". You just solve the queryes offline sorting by some points which are marked from sqrt to sqrt(the most left one). In case of the same point, you sort by the right limit. In this way it is very easy to process the queryes using a Fenwick Tree. Read the solution to that problem, maybe it will shed some light, bro.

There is also an sqrt-deomposition solution that our good friend MirceaFF didn't mention. We can divide into sqrt segments. When an element comes, you see the bucket where he belongs and put it there, for this of course we will use sqrt Queues. And from sqrt to sqrt Updates we kind balance the buckets because some of them will become too big. When [l,r] comes, we take the maximmum from the whole buckets and the rests from the other possible 2 buckets. o(m * sqrt(n+m)). But of course FF's solution is good and quite intuitive too, you just solve the queryes and updates backwards.

Nice. Thanks!

Very nice. Thanks a lot! I have read one solution, but this is really awesome.

Thanks!

Thanks a lot! It's very helpfull. Do you also have some implementation of it?

Yes, it is rectangular and can be.

Tell me your solution if you know please.

I need o(n^3) :p

Can you get it in o(n^3) ?

Nope, i am the best in this bussines, have you just seen my haircut and my muscles? Way better than yours, but I still like your style of telling stories and your videos of 3 hours lentgh, maybe we will teach together these newbies how it should be done. VALORE!!!!!!!!!!!!!

hey, I am Romeofantastik, i got #valoare and Morena, I am a boss.

Around 8 things in particular?