Next time just try to say "As an" and look how many contribution you get

Thankyou for the contest :)

To be honest triple__a worked very hard for this contest and I am more on a supportive role, so he should receive more contribution than me xd

As a participant...

your comment has your name.... triple_a ("As An Author")

my first comment

As an Um_nik, ***** div1A.

Wait Nezzar are you RB the villager? (pls don't ghost me)

Nice catch! The one in contest tab will be changed to 135 minutes.

n minutes (0 <= n <= +inf)

Sure thing

We can actually make the constraints better. Since we will anyways get the scoring distribution after this certain point of time the range of $$$n$$$ can be better written as $$$0 \le n \le f$$$ where $$$f$$$ is a finite number which I don't have the finiteness to calculate.(but it is finite).

• as a tester

Let's hope that MikeMirzayanov does not have any doctor's appointment scheduled for Thursday.

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I am sure all of you work very hard and prepare such contests. Thank you for all the efforts you put!

I can understand you. Different coordinators have different review standards. Some coordinators are very strict, and they have a very low pass rate for the questions they review. This does not mean that your topic is not interesting enough, maybe it is just that the level of interestingness does not meet the standards of this coordinator. Different coordinators have different definitions of interesting, so your topic does not seem interesting enough for this coordinator, but another coordinator (possibly) will find your topic interesting. Then, even if your topic is not selected, I think your contribution is enough to make us thank you. Looking forward to solving the problem of your proposition, although my level is not high, I will try my best.(One more thing to say, if your dream is to be an excellent writer, don't be disappointed in yourself. I believe that one day, you can impress the coordinator with your interesting topic.)

Thankyou for the contest :)

In every Div1 + Div2 round, Candidate Masters are considered Div1 participants.

No problem, just extend the round by 30 mins

Since when $$$17.35 \text{ (UTC)} = 22.35 \text{ (UTC+8)}$$$ :( ? Did you mean $$$17.35 \textbf{ (MSK)}$$$?

VOT ETO DAAAAAAAA

long queues started even before the beginning of the contest

Thanks for replacing "!" with "."

they said that it took half-year preparing it so they had put a lot of effort to make it. so I think why problem F gives 4000 points and I think div2 will have a subproblem.

u r right

Good luck

im the first upvote of your first comment :dab:

Yes, the queue is very large. I hope it will be fixed soon.

The latest submission is considered.

suppose n = 3 , then suppose array a will be of form $$$a_1<a_2<a_3$$$ and their negative values also . Then array d in increasing order will be (note that every number in d will be repeated twice so we are only considering n numbers) say $$$d_1<d_2<d_3$$$ then $$$2*(a_1+a_2+a_3) = d_1$$$ , $$$4*a_2 + 2*a_3 = d_2$$$ , $$$6*a_3 = d_3$$$ .You can show this by using formula in question. Now you can find $$$a_3,a_2,a_1$$$ one after the other using the equations .

Similar generalization for any n.

Exactly, there was no need of them to be distinct.

totally agree, id rather have repetitive algorithmic problems than these find the pattern after 100 examples problems.

Ya I don't know who likes these problems if most of them diskikes this type of problems then these puzzles are made for whom a big question mark on the community?

It is not atcoder no!

I think div2D/1A was a great problem. It was refreshing to see Bezout's Theorem, and it was a cute observation.

Div 1B on the other hand was just standard lazy seg-tree so that wasn't so exciting but it was cool still.

How to solve Div1C?

I don't have any ideas about D, but I think the observation for C was not hard to stumble upon. I would say C was more intuitive and easier for me than B.

Basically, take some small arrays, then try writing the sum of absolute difference for them, the way it's defined — but don't calculate the value, instead just observe the terms themselves in the sum of absolute difference.

For C i solved by using the definition of how array d is constructed (by getting n equations).So it wasn't puzzle for sure.

In D i observed (don't know if correct) that we require at most 2 numbers to create k .

In B , i had idea that after a point all numbers can be created by lucky numbers but was not able prove it and didn't solved it.

So i can't say anything regarding D and B.

Let's hope tomorrows Educational round will be a good one.

I disagree with you . They were really beautiful problem with some really good observation . I couldn't solve D but you can see my solution to C which is I think really clean and nice .

These day if you want to get good in codeforces contest then it appears practicing puzzles is more beneficial than studying algorithms to

agreed. Solving C was not at all fun. it was just some boring observations.

Well, here is how you could navigate a labyrinth even blind without any weird observations or luck: click

for problem B, you can use lucky number multiple times. means for d= 7, ai = 52, 7+7+7+7+7+17 is allowed answer

In problem B, we need to observe that for every number greater than 10*d, it is always possible to get the sum, for number < 10*d we can brute force and check.

pp maps :(

The result of formula should not repeat and should not be negative.

You can view the operation as taking a number in the array and adding the difference (with your choice of sign) between any two other numbers of the array to it. Note that you can repeat the operation as many times as you want.

Assume we have set $$$(a_{1}, \dots a_{n})$$$ and $$$g=gcd(a_{2}-a_{1}, \dots , a_{n}-a_{n-1})$$$. Then the closure of operation $$$2x-y$$$ is set $$$(\dots a_{1}-g, a_{1}, a_{1}+g, a_{1}+2g, \dots)$$$.

I solved it using segment tree, don't know if there's an easier solution

Yes, it was segtree with: "set all values in a range to $$$x$$$ " and "find sum of a range".

Yes, you consider the queries in reverse order, and then note that the starting string is uniquely determined by the conditions (if at a point you have exactly half the elements in a range to be 0 and the other half to be 1, then this is invalid, and the answer is NO, otherwise you can just set the whole range to the majority element of the range). If the array so found is the same as the given initial array, then the answer is YES, else it's NO.

both segment tree and chtholly tree is ok

Assuming we did the same thing, i.e. creating the sorted positive half of array a, I was getting it when I was checking if the array is non-decreasing, it was resolved when I checked the array to be strictly increasing.

Very close, but no. The observation that works is that if you take the difference of all elements with the first element, and call the resulting array $$$b_1, \dots, b_{n - 1}$$$, then performing the operation corresponds to finding the sum/difference of a few of these elements, and you can find an operation where it corresponds to finding the sum/difference of any two elements of $$$b_i$$$. Hence, the gcd of all $$$b_i$$$'s is achievable, and this is necessary and sufficient.

deleted

We are really sorry that D1C has been appeared before. None of the team know the existence of this problem and we come up the whole problem ourselves (the idea comes from the lore actually)