As a tester I recommend you to participate in round, and don't forget to read all problems statements

As a tester, I recommend you to take part in this round!

as a tester, this round is awesome

As a tester, this round is very rofl.

as a not tester, I hope this round will be great

Does participating in at least five rounds count this round? Because I have been in exactly 4 contest before this :D

petition to give Gornak40 black handle (as he is, in fact, huge male gorilla) :spinning_gorilla_emoji:

Score Distribution?

I hope everyone enjoys this contest ^_^

Good thing that Gornak40 didn't create the tasks too, his problems are special

Green testing crying in the corner

In the contest page shows 2 hrs but in the blog page shows 2 hrs 15 minutes how is it possible?

CR7 > MESSI

Coming back to give contests after long time, feels good to be back!!

I hope this round will help me reach 1200+ rating. As a participant, I wish you good luck!

hope this round will help me reach cyan

I really enjoyed your first div 3 round (Codeforces Round 881 (Div. 3)). Very much looking forward to this one =)

Wish I could be a pupil tonight！！！

As a tester I recommend you to participate in round, and don't forget to read all problems statements, it can be usefully)))

m1, m2, m3 codeforces sites are not working.

Thx. Very nice round. Enjoyed problem G alot!

Didn't notice that only people competed in 5+ rounds are rated......

Already spent 1+ hour on the 7 problems

Sad QAQ

B,C >> D,E

What's the intended solution to G? The only two solutions I could think of seem completely overkill:

1. Merge Sort Tree on DFS order times of the inverse values of the permutation.

2. SQRT Decomposition on whether each node is a parent of at least one node in the buckets $$$[p_1, p_{\sqrt{N}}], [p_{\sqrt{N} + 1}, p_{2 \cdot \sqrt{N}}], \ldots, [p_{k \cdot \sqrt{N} + 1}, p_n]$$$. For the upto $$$2 \cdot \sqrt{n}$$$ remaining nodes that don't lie completely in a bucket check them each in $$$O(1)$$$ using dfs order ancestor checking.

Is there an easier approach?

E was easier than B wtf

Problem F,on example,2nd test case, in the third query already exist a path with distances 3 between 2 leafs , 5-4-2-3, but the answer wasn't -1 -1 -1.I'm receiving WA veredict for that?

Greedy-forces :)

oh f i just solve G 1 minute after contest time woooooooow

I finished the G at the last minute, and because I forgot to write "return 0" ,I got compilation error and I passed it only 1 minute after the contest.

In D it's hard to prove that 1 and 2 the only correct combination without same numbers. Bad problem in my opinion.

come onnnnnnn if the time was just one minute moreeeee ohhhhhhhhhhhhhhhhh

One of the easier contest I've had in a while

Problem Statements are too hard to understand.

low quality problems on Greedy_forces

In G, if It is guaranteed that the sum of n and the sum of q over all test cases do not exceed 1e5. shouldn't Binary Lifting work?

I got TLE on TC 10. My Submission

D will appear in my nightmares

other than that, the contest was great

Free hack : submission

What was the solution to D? I observed something that only 1 and 2 matters.. but couldn't get to the solution

O(n log(n)) solution of G using persistent segment tree 233188158

$$$O(nq)$$$ solution for G: 233198776

Limits $$$n, q \le 10^5$$$, TL=3sec should be non-existent in 2023 :)

D could be solved without knowing the fact that only the combination of 1 and 2 is valid.

The statement equivalents to: Count the number of pair (i, j) such that $$$a_i * 2^{a_j} = a_j * 2^{a_i}$$$. Suppose $$$a_i \gt a_j$$$, then we have: $$$a_i = 2^{a_i - a_j} * a_j$$$. Knowing this, we can loop for each $$$a_j$$$, how many $$$a_i$$$ could satisfy this equality, and add it to the answer. The case when $$$a_i = a_j$$$ should be handled separately. My solution: submission

First contest I could fullsolve, wow!