Good luck to you and me. Good luck to everyone

wu

I've made it! :D

Good luck for becoming specialist

I wish the same. Good luck dude.

What is so educational about them? They don't seem so different from normal div. 2 rounds.

Tough, but thank you

Wow that's a wonderful delta isn't it

Me too. (I only need 7 to reach green) Good luck to you, me, and everyone!

That's a really big jump though considering you are 1446

I don't have much experience, but I can say that only practice helps. It can be useful to review the problems and solutions of past competitions (but at the level you feel you can understand at the moment). My mistake was that I analyzed the solutions of the more difficult problems, although I could not solve the easier ones on my own. So I recommend solving gradually, it is better to solve an easier problem yourself than to understand or remember the solution of a harder one

Also at your level I recommend to pay more attention to past div 3 and div 4 contests, usually problems there easier than div 2

I have faith in you :)

Yeah, that was a good contest for me, I managed to do ABCD. Hope it won't get hacked

You are lucky to get it unrated.

You updated your comment 3 times.

problem B seems much harder before updates

I agree that it is pretty classical but this is an Educational round.

If problem D has 0 creativity , why you donot solve it

because of people like you, who copy paste ABCD as soon as they get them from telegram.

Y are a cheatforces

Use global variables, they are faster. Preallocate memory, rather than allocating runtime memory for better performance.

There are disadvantages also, you need to make sure, you are not allocating more than required, or you will get MLE.

so true lmao

more people solved C than B, despite B having a way easier algorithm

Guess what, aryanc403 bhai has tried C-problem 4 times. He is quite experienced and knows most of the things in CP. If it takes him 4 tries then the problem must be little challenging. and yet 12k people solved it.

Definitely C is leaked. xD .

C was easy compared with other Div 2s.

c was literally valid parenthesis from leetcode. It is one of the foundational problems for stacks. I think 12k submission is not justified. But it should be higher than usual div2 c. I solved it using rust as I am learning it.

agree, such cheater ruined codeforces

no, only for people with rating less than 2100

no bro, only till 2099

Because it is very easy and classical, however I won't deny the presence of cheaters.

its standard problem, cheaters are there tho

IT is a pretty straight forward Tree DP Question. And not everybody learns mostly off codeforces, I just do codeforces for the contests, but rarely for Practice Purposes (except if they appear on USACO Guide)

But yeah, 5k on D is still quite a lot.

Tbh as a newbie i found D easier than other div2 rounds but sadly i was not able to write code

Yes

What is the exact tc?? I also faced the same :/

In your code, val can grow exponentially. Keep a check, if it is greater that 1e9 return false.

If you passed the number Y to dfs, which must be subtracted from all the numbers in the subset, then Y may have become greater than 2^64

Me too, no f... idea what's wrong in that case 19...

Wait this causes an overflow? :0 Can you explain why?

It was not that hard. You just have to find the free cell that serves as the only bridge to 3 free cells. You can look at my post contest submission to figure it out. I misread the problem statement and I thought there could be more than 1 sets. I solved it in that way in contest.

It's ok it took me more than a year for expert

You are specialist so you are definitely not retarded. I looked at your profile and 70% of your solved problems are below 1400, so maybe practice harder problems.

Note that tokens behave like fibonacci sequence, you have $$$f_1$$$ $$$f_2$$$ and can create some $$$f_i$$$ for $$$i>=3$$$ using $$$f_{i-1}$$$ and $$$f_{i-2}$$$ or to change $$$f_i$$$ for $$$f_{i-1}$$$ and $$$f_{i-2}$$$. You can also freely change $$$f_1$$$ to $$$f_2$$$ and vice versa. So $$$f_i$$$ means token on position $$$i$$$.

First convert all tokens to $$$f_1$$$ and $$$f_2$$$, let the number of them be $$$s$$$. Note that, because you can change $$$f_1$$$ to $$$f_2$$$ and vice versa, all that matter is their sum ($$$s$$$). You want to use minimum number of tokens to represent $$$s$$$. One token means some $$$f_i$$$. If you have $$$1000$$$ $$$f_{10}$$$ tokens maximum $$$s$$$ you can get is $$$55000$$$.

Now you do greedy, for every sum $$$s <= 55000$$$, you go through $$$f_{30}$$$ to $$$f_1$$$ and try to fit sum $$$s$$$ in minimum number of tokens (there is proof for fibonacci numbers that greedy is optimal).

Now for some sum $$$s$$$ you have $$$min[s]$$$ which represent minimum number of tokens to represent it. We want such $$$s$$$ that $$$min[s] = m$$$.

Lastly we need to calculate in how many ways can we get sum $$$s$$$ using $$$n$$$ tokens on first $$$x$$$ $$$f$$$s. Define $$$dp[i][sm][p]$$$ as in first $$$i$$$ $$$f$$$s you placed $$$p$$$ tokens and their sum is $$$sm$$$, this dp has at most $$$10*55000*1000$$$ states, transitions are in $$$O(1)$$$.

Now you just sum $$$dp[x][sm][n]$$$ such that $$$min[sm] = m$$$

273592555

really I thought A was really easy I just thought that chose the substring with longest repeating chars and just insert the different char than the present one in between of that substring

I also WA in test1 2 times and chuckles. I know today im gonna screw up by not having enough time again

I was thinking why the time limit was 2 seconds. I literally did not make any sense.

do it in different ways when $$$x \le \sqrt n$$$ and when $$$x \gt \sqrt n$$$. The former can be pre-calculated, while the latter can be solved by binary searching on prefix-sum

You can precalculate an array of indices for each $$$k$$$, where the indices show where the last update of level happened. I did it with merge sort tree in $$$O(n \cdot log^3_2{n} + q \cdot log_2{n})$$$.

i have an idea that we can group the queries by k
then for each can simulate the game by using segment tree in $$$O((n / k) log(n))$$$
so we can do this in $$$O(n * log^2(n))$$$
someone please correct me if i'm wrong.

Notice that the level you can reach by the $$$i$$$-th monster is non-increasing as $$$k$$$ increases.

Let $$$need_i$$$ represent the minimum $$$k$$$ for which our level will be $$$\leq a_i$$$ when we reach the $$$i$$$-th monster.

We can binary search on the value of $$$need_i$$$. To check if $$$need_i \leq k$$$, we want to know if the number of monsters we would have fought till $$$i - 1$$$ is $$$\lt k \cdot a_i$$$, or formally $$$\text{cnt}(need_j \leq k$$$ where $$$j \lt i) \lt k \cdot a_i$$$.

To check this $$$cnt$$$ quickly iterate in increasing order of $$$i$$$ and store the values of $$$need_i$$$ in a fenwick tree, then $$$\text{cnt}(need_j \leq k$$$ where $$$j \lt i)$$$ is just $$$sum(k)$$$.

Code — 273568637

Check my comment.

for what reason? like the problems were easier than div 2?

Questions A to D are easier than usual,but E and F are still challenging