Another Mac Book for tourist? He has like 10 of those.

For me ,the blog looks really attractive and fascinating

Another Deltix round. Just really love these rounds. Quality problems, quality prizes.

Who else was surprised by looking at that waving hand gif.....

I am really looking forward to these prizes which I can never win.

100 T-shirts for the top 101 to 1000. There is 1/9 chance if I get in there :D

150 minutes or 135?

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Loved your previous contest. It's great to see you back.

100 in (100,1000], probability=1/9=0.1111.... :)

Hope problems statement will be less confusing then the last time. ʕ•ᴥ•ʔっ

Div1 + Div2 is really good opportunity for anybody to develop programming skills

It's time to upsolve Deltix Round, Spring 2021 (open for everyone, rated, Div. 1 + Div. 2)

i hope i dont win nullptr again.

did anyone got those randomly distributed t-shirts ?

Visit the site guys, you won't regret. Also, good luck to everyone!

*Wind_Eagle's round announcement* Am I a joke for you?

Tourist is going to open an apple store after this contest.you know why XD

Nice rewards but i don't think that I can reach top 1000th or less . :))

The animation that links to the main site is so cool

EDIT — Commented on this forum by mistake, wanted to comment on today's competition's forum

I might be wrong but I thought initially they offered 150 minutes and now I see 135 minutes now.

Probability of Me getting T shirt 100/1000=1/10 Probability of Me coming under 1000 rank 0.0000000000000000000000001

Tourist's Macbook has a update feature, he doesn't have to buy the lastest version.

Is it possible to get MacBook Air from Tourist?

To tell the truth, when opening codeforces, this moving arm is really scary.

the problem set of the last delix round was hard to me D:

I really want to reach CM in the last contest in my summer hoilday TAT

I love rotavirus

I'm curious about scoring distribution. Any predictions?

Please update the scoring distribution fast

As a tester, I recommend you to solve all problems!

As a tester, I confirm that problem statements are to the point, no long stories.

as a participant I couldn't even solve 2nd problem :"(

great contest !!

Being a noob, I don't know how people solve interactive problems like D... :(

Problem D is effectively an easier version of 1451E1 - Bitwise Queries (Easy Version) (use sum = or + and to get other elements as well since we don't have xor here, it works since we have 2n queries anyway).

Thanks for the interesting problems! Really enjoyed problem E

Could only solve A, -91 incoming KEKW

Big dislike for Bs "funny" overflow twist.

Feedback on the problemset.

• A: Cute cakewalk problem.
• B: Ok standard easy problem (though a bit hard for its position). I don't see why you decided to consider parities instead of giving an array of zeroes and ones.
• C: Amazing problem. The statement is clean and asks a very natural question. The difficulty is appropriate for its position. For some reason, I liked this one a lot (most likely, because the solution is just so clean and does not require any advanced technique). I was wondering if it can be solved in pseudo-linear time.
• D: Nice problem. For me, this required some serious thinking, but the standings suggest that it is not as hard as I perceive. Maybe I perceived this one as hard because I did not realize for a long time that if I know $$$x, x\text{ and }y, x\text{ or }y$$$ then I know also $$$y$$$.
• E: Nice problem (but a bit standard). I was pretty annoyed by the fact that the word "even" was not highlighted in the statement and the first pretest which catches the bug of allowing also sequences of odd length is the 17th. Moreover, I don't see how asking for answers on intervals makes the problem more interesting than asking just for the answer on the whole array.
• F: Very nice problem. Knowing the chain-structure of the strongly-connected components of a tournament, this one felt easy for its position.
• G: I just read the statement, which is amazing.
• H: Not read.

Overall, there was not a single problem I disliked, and more than one that I really liked. The only complaint I have is that the difficulty-gap between F and G was huge, which made the contest a speed-contest for those with a rating similar to mine.

As a final remark, the images were awesome (it's the first time ever I appreciate images in statements). Whose the author of the images?

I really enjoyed taking part in the contest, which seemed to be really well-prepared. Thanks to the authors :)

Problem C test 13. Couldn't debug my code.

The only prerequisite for D is this equation -> a + b = a&b + a|b

Also, is there any way to solve E if we can choose an integer $$$x$$$ and increase / decrease the specified elements by $$$x$$$ instead of using 1 each time? I misread the statement and tried to solve this for quite some time.

My solution to G (which passed pretests) runs in 4.5secs and takes 1130MB of memory in the worst case. Oh, woe is me!

D is (almost) a copy of https://codeforces.me/contest/1451/problem/E1

C murdered my brain I felt I was so close :(

https://codeforces.me/problemset/problem/727/C (same as today's D because a+b = a&b + a|b)

found this by googling and solved.

In Problem E, I didn't find that k is always even, so I wasted my time for meaningless debugging. sad.

Is 4^n meant to pass in F?

C and D are just boring standard problems.

E is yet another prefix sum transformation problem.

F is basically a rip-off of this.

0 interesting ideas out of the first 6 problems.

At least there is no problem which requires constant optimization.

how to solve C

Felt really good when I submitted B and C, only to have that feeling snatched away by pretests 2 and 9 respectively :(

Great problems though, interested in knowing the corner cases.

problem D is same as this problem.

Why is my problem B code giving run time error? Can anybody please help me? https://pastebin.pl/view/95d9737e

For problem H, I think we need to fix the edges between the k special nodes (since there are at most 2^10 such ways) and then use min cost matroid intersection for the rest of the edges, thought about the solution, but couldn't implement it during the contest.

How to solve E, many people solved it ? I thought of this way, we can take, if a[i] < b[i], then consider there are b[i] — a[i] opening brackets are there and else a[i] — b[i] closing brackets. So, for each i, we want to know what is the closest j, where if l = i and r >= j, then it becomes impossible to solve. But it seemed very tedious and on top of that the query is also asked on segments.

Edit: Somehow the wrong solution also passed system test

Weak pretests for E

I found a counter-example for my approach that passed pretests but I'm sure will fail system test because I forgot to handle one impossible case. (When there exists a negative prefix sum of $$$b[i]-a[i]$$$ starting from $$$l$$$ in a query)

Can't believe that there are 20 pretests yet none that there are none that targeted this rather common mistake

4 1
0 2 0 1
1 0 2 0
1 4

Any hints for B?

Anyone wants to share his thought process for problem B.

When you finally get an AC on problem C with a linear solution, then you find that the problem can be solved in quadratic. QQ

I have a doubt. Why is the value of n given as 1000 for C, As this can be easily solved in O(N) time, is it to keep the answer within 64 bits ?

I hate weak pretests.

Look at my code for B and E the first time.
B:if (tot=n/2+1) cout<<chk(1)<<"\n";else {puts("-1");return;}
Noticed that there is only one = in the code,but it can passed the pretest!
It can be hacked by datas like 1 3 2 2 2 but it passed the pretests.

E:int Max=max(mx[l][k],mx[r-(1<<k)+1][k])-d[l-1];int Min=max(mx[l][k],mx[r-(1<<k)+1][k])-d[l-1]; Noticed that the Max and the Min are same!!!
it can be hacked by datas like 2 1 2 1 1 2 1 2 but it passed the pretests.

Luckily I noticed these mistakes and resubmitted them.

These mistakes will be only made by few people,but both of them are wrong obviously.
I wonder why the pretest are so weak that these two obviously wrong code can pass the pretests.

I can't imagine how mad will I be if I don't find these two mistakes and get FST!!!

I want to say,it's not funny to make weak pretest and see a lot of $$$\color{red}{\mathrm{-1}}$$$ on the standings.

update:It seems that my first code for E can pass not only the pretests but also the system tests!I don't know how the gen work.