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

Excited for this round. Best of luck to all contestants :)

Green testing crying in the corner

Can I give this contest and become lgm?

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

CR7 > MESSI

good luck y'all

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！！！

I will win Codeforces Round #909 Div3!

If you love your GOD like my comment

If everything okay, i will get green after this contest. I've been waiting a lot for this moment. My canonical event... is coming :'0

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

yes sir

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?

Solved A,C,D,E but can't understand B. B>>>C,D,E

I easily solved A,C and D

but B took me most of the time and still couldn't solve it , how to solve ?

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

The contest is very good. Hope I solved more

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

I had done more problems on this contest, than anyother, one of them being D, which boosted my confidence and my hunger for working more. Thank you for such an amazing round!!!

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

dont feel like div3, only 1 task was for div3 or div4

D will appear in my nightmares

other than that, the contest was great

Spent 1.5 hour for B,C. For D, you just need to understand the condition and play roulette. E is rofl.

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 $$$a_i > a_j$$$, 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!

what's intuition behind E?

Really great round! In my opinion, the condition of problem D was too confusing, but the problem is still good

What will i receive when i hack a problem successfully?

Problem D is fun to understand the statement and enjoyable to solve, I really like it!

Thank you guys for the contest!

Problem B was difficult to understand

can anyone tell me why people having fewer solve count and people having same solve count and high penalty have better rank than me. Thanks!

Which data structure do I need to know to solve G?

El_Gemmy

Agmed tester <3

Thank you so much for contest :)

what all concepts are needed to solve G?

It was a nice round. Hats off to you guys (editorial panel). Really enjoyed as a newbie coder.

It was really an awesome contest.! Thanks all writer's and tester's.

Don't know why problem B was one kinda boooooommm..

Found a cheater Salman_Johir. This 233189111 is his submission. He copied this code from this video's code. Look at the both code. He just has changed the variables' name.

very week pretest in problem C

Thank you for the problems!

It's nice there is a short and witty solution to each one (except G).

For problem C 1899C - Yarik and Array , just modify the Kadane's Algorithm to incorporate the extra condition that parity of consecutive elements should be different.

Rest of the algorithm remains the same.

void testcase() {
  int n;
  cin >> n;

  vector<int> a(n);
  for (int i = 0; i < n; i++)
    cin >> a[i];

  i64 max_so_far = INT_MIN, max_ending_here = 0;

  for (int i = 0; i < n; i++) {
    if (i == 0 || (abs(a[i - 1]) % 2 == abs(a[i]) % 2))
      max_ending_here = a[i];
    else
      max_ending_here = max_ending_here + a[i];

    if (max_so_far < max_ending_here)
      max_so_far = max_ending_here;

    if (max_ending_here < 0)
      max_ending_here = 0;
  }

  cout << max_so_far << "\n";
}

For Problem D 1899D - Yarik and Musical Notes, we need to count the number of pairs (i,j) such that

Taking logarithm to base 2 on both sides

We only need to take integral part of $$$\log_2{a_i}$$$ to avoid double/precision errors.

The only edge case to take care of here is power of 2 ($$$2^k$$$) and 1 less than power of 2 ($$$2^k-1$$$) both yield the same value yet they do not satisfy the original equation, so we need to separate powers of 2 from the other non-powers.

Also, club all the $$$1$$$s into $$$2$$$s. Now all the $$$a_i$$$, $$$a_j$$$ which hold the above equation can be stored in a map and number of such pairs will be : $$$ \displaystyle C_{2}^{N} $$$

void testcase() {
  int N;
  cin >> N;

  set<int> powerOf2;
  int mult = 1;
  for (int i = 0; i < 31; i++) {
    powerOf2.insert(mult);
    mult *= 2;
  }

  map<int, int> powers;
  map<int, int> others;
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;

    if (x == 1) {
      powers[2]++;
      continue;
    }

    if (powerOf2.find(x) != powerOf2.end())
      powers[x]++;
    else {
      int y = x - (int)log2(x);
      others[y]++;
    }
  }

  i64 ans = 0;
  for (auto x : others) {
    if (x.second < 2)
      continue;
    ans += 1LL * x.second * (x.second - 1) / 2;
  }

  for (auto x : powers) {
    if (x.second < 2)
      continue;
    ans += 1LL * x.second * (x.second - 1) / 2;
  }

  cout << ans << "\n";
}

Could anyone please explain how to solve B?

i got an O(n) solution that was hacked for problem D (TLE). The official solution is O(nlogn). Anyone throw some light on this? https://codeforces.com/contest/1899/submission/233176732

When I check the total number of participants who officially solved the problem A , it shows there are about 10k such participants , when I see the official participants who solved problem A , it shows there are about 17k of them .. so ideally , there should be about 17k people who have appeared for the contest , not near 10k...asking it because it would affect my rating..

can someone pls explain ?