Yes I feel the same. The problem set is great and recommended to several participants

Only one?

only one?:(

This is one of the most funny comments I have ever seen

That is true

That would be a spoiler for the onsite contestants (SWERC has no score distribution).

I wanted to comment that typical ICPC contest should have 10-15 problems so actually judges had to remove something from the set (and this is usually the case). While trying to verify that, I found https://icpc.global/regionals/rules that states:

At least six problems will be posted.

Which means technically a Codeforces div2 round with A-F problems might be held as a regional? That's funny.

it's div 1-2 for this reason

Problems A-F in div1+div2 is about the same difficulty with div 2 (sometimes even easier, especially D) ,so you'll probably get a +rating if you solve A-D within 1.5h or A-E within 3h.

GG

Yes!

Those who jc (use somebody else's account to do something bad) others will have such a bad luck that he or she would even not be able to solve A.

Yeah, a really kind star time to Chinese.

The onsite contest starts 90 minutes before the CF round.

If I reach 2300 FIDE Elo at chess within October 2025, I will start playing Honkai: Star Rail :))

They will be in difficulty order.

I trust you. Don't let me down

You can check previous Div. 1 + Div. 2 rounds (for example, Codeforces Global Round 30 (Div. 1 + Div. 2)) for reference.

give it a try!

Good luck you guys

fml

Ouch, just 2 points off

As your chemistry teacher, I will provide you with a computer.

We will make it public when the CF round starts.

I got stuck on it sadly, only started with an hour left in the comp but I breezed A but B felt like going round in circles.

This user is disabled

Just saw this. We will cancel solution discussion. Thanks for pinging me.

I think Shayan is starting problem discussion 5 minutes after codeforces contest ends.

Me too buddy. Guess we'll just have to accept our negative delta

there is a chrome extension named 'carrot'

that is a huge delta!

Congratulations to becoming Candidate Master!

yes,i tried it for two hour

certainly

I am happy that pretest = system test ... as I came up with some construction and it was correct .. but got cooked by D

just because you couldn't solve it? funny

why?? it was hard no doubt but why rubbish?

turly

Negative because I implemented a stupid binary search on segment tree.

Btw E<C even so.

It's not that hard to see that this problem is easier to do with a frequency map.

From observations you can get that when, for some $$$i$$$, $$$f_i \gt k$$$, then you have to move $$$f_i-1$$$ drones from energy $$$i$$$ to $$$i+1$$$, so basically $$$f_{i+1}:=f_i-1,f_i:=1^{(1)}$$$ ($$$:=$$$ means assignment).

Let's define a chain reaction when $$$f_i \gt k$$$, and we advance the drones of $$$f_i$$$ to the right (thus incrementing $$$i$$$) until $$$f_i≤k$$$, leaving a trail of $$$1$$$s behind. At the end, if we started from $$$s$$$ and ended at $$$i$$$, the total number of operations that chain reaction did was $$$i-s$$$. Let's save all trails of ones in an array $$$jump$$$, in the format $$$jump_s=j$$$.

Any chain reaction passing on an existing trail of $$$1$$$s (which starts on position $$$j$$$) will, after $$$jump_j-j$$$ operations, continue from $$$jump_j$$$ without changing the trail (ofc it will leave it's own trail of $$$1$$$s, but $$$1=1$$$, so the elements of $$$f_{j:jump_j}$$$ won't change).

So, our answer to the problem will be the longest chain reaction. Of course, every chain reaction has a constant rate, so no chain reaction can "catch up" to another ongoing, which means it's safe to simulate chain reactions from the right, to the left.

This means we can have a variable $$$s\in{2n,\dots,1}$$$, and we can check if $$$f_s \gt k$$$. If this is true, we start simulating the chain reaction from $$$f_s$$$, with $$$(1)$$$. If we get to $$$i$$$ where $$$jump_i$$$ is set, we can skip a lot of indexes, which saves runtime and reduces energy consumption, thus slowing down global warming. At the end of the chain reaction, if the length of this chain reaction is the largest ever, we save it ($$$m:=max{m,i-s}$$$).

Implementation: codeforces.com/contest/2157/submission/350326830

Same. Here on UTC+9 most of the contests starts at 11:30pm.

check mine!

You can do some metagame and realize that the solution is "optimal", in the sense that you cannot do better (if the array is fixed), then the solution must be correct.