As a tester, omg round by coordinator.

Why multiple contests on same date ?

Why are ratings 1600-2099 so lucky they can participate in two Div.2 competitions.

The problems are very good. Hope you will love solving them. (:3 first time testing)

I'll never forget exciting rounds coordinated by you. Hope the round will be fine.

Looking at the score distribution this seems like a speed run. Gonna register with the motto Don't participate if you can't dominate.

But why two div2 rounds on the same day?

Deleted

pshat orz

Only five questions and the score distributin is also high why so?

you forgot NOTE THE UNUSUAL STARTING TIME in bold to make people aware

Why is the registration for Codeforces Round 905 (Div. 2) saying "Rating should be between 1,600 and 2,099 in order to register for the contest"?

I think it's better if you write in the post that only >1600 can register and get rated

Seems like Codeforces Round 905 (Div. 2) gonna be blue-purple war!

I really like this timing.

Why is it said here that the competition will be for rated participants with a rating of no more than 2100, but at registration this is a rated competition for participants with a rating from 1600 to 2099

Это какое-то новое правило, что див2 можно писать только с 1600?

I guess at CFR904 and 905, there will be chaos because of the back-to-back rounds and tricky format of CFR905. So I suggest some unusual rules for them:

• The rating change of CFR904 won't be updated until CFR905 ends so that avoiding double-register (and getting double-rated mistakenly) for CFR905
• Everyone is rated by the division in which they registered even if the participant gets division promotion from CFR904.
  • Another choice is to update the rating rapidly and remove double-register before the round starts.
• Rated participants of Div1 or Div2 of CFR905 should be kicked out from Div3 of CFR905, Otherwise some penalties will be avoidable by submitting their solution for Div3 first.
• Open hacking is disabled for CFR905(Div3). Only div3 participants get open hacking is unreasonable unless the problemsets are disjoint.
  • Another choice is to introduce open hacking for all 3 rounds. Anyway, I don't like the parallel rounds with different rules without suitable reason.

When the registration for division 905 will be open? Usually I get an email with invitation.

Why there was said in the announcement that "this round will be rated for all the participants with rating strictly less than 2100.", but there was said in the register page different "You are registering "out of competition", reason: rating should be between 1,600 and 2,099 in order to register for the contest"? I'm getting confused, can anyone help me to figure out it?

The rating changes for a round are sometimes cancelled and then recalculated, which takes about 1 day.

Round 905 starts just 2 hours after round 904 ends. Consider the following situation.

You participate in round 904, your rating changes, and then you participate in the corresponding division of round 905. After a few hours, rating changes for round 904 are cancelled and recalculated, causing your division to change and meaning you haven't participated in your division of round 905.

The chance of this happening is very small, and it's interesting to know what would happen in this case.

Why there are so 2 contests on the same day?

I registered in both 904 (in div 2) and 905 (in div 3). In case, for 904 my rating increased to be >=1600 and rating update happens before start of contest 905 but after registration end time. So in this scenario, will my registration in 905 as div 3 still be considered valid even if my rating now becomes >= 1600. Wont it create any confusion?

(13-21) — 9 days 0 contest. (22) — 1 day 4 contest!

Why there are 4 contests in one day, and for the next week there's no contests?

would love to have more rounds like this that are not late in the night(in my country) :)

Hoping to become CM in this contest!Only 9 ratings!

It's time to rage(100%)

Please note the non-standard start time for the round

This is my first contest. Can i join it?

Yes, I love you too;

Dangerous round 3rd & 4th questions will be difficult as the points are very high and 1st question is of 500 points which is more than some earlier rounds. ( Means div 2 B === this round A)

More suitable for Chinese baby constitution

So intersting

Hats off to US contestants who will join at 00:05 am and 04:00 am today.

Nice start to a Sunday afternoon (╥﹏╥)

How to solve problem C?

How to solve problem D?

pE too hard, can someone help me :(( thanks

I found $$$D$$$ so much easier for $$$2500$$$ points. If $$$a_k$$$ divides $$$a_i$$$ and $$$a_j$$$, this means it must divide $$$gcd(a_i,a_j)$$$. So, iterate over $$$gcd(a_i,a_j)$$$, Let $$$gcd(a_i,a_j) = m$$$, find $$$num[m]$$$, denoting number of pairs having $$$gcd = m$$$, can be found easily using recurrence $$$num[m] = cnt[m]\cdot(cnt[m]-1)/2 - \displaystyle \sum_{r=2}^{\lfloor{n/m}\rfloor} num[r \cdot m]$$$, where $$$cnt[m]$$$ is the number of indices $$$j$$$ such that $$$a_j$$$ is a multiple of $$$m$$$ in $$$O(n \cdot logn)$$$.

Now, after fixing $$$m$$$, we just need to check if some index $$$k$$$ exists such that $$$m$$$ is a multiple of $$$a_k$$$ or not. This can also be checked easily.

WA on test #3,Would someone like to give me the hack data?

pE too hard :(( can someone help me, thanks

Problem B: https://ideone.com/GV1ZXe why wrong answer on pretest (4) ?

How to solve B? Spent lot of time but couldn't think of correct approach. Any hint?

pE too hard :((( can someone help me, thanks

C was very similar to this problem: link

is there a non brute force solution for A if k>10 ?

I am a beginner but problems A and B were very interesting. Enjoyed solving it

Couldnt debug C in time I just need 5 more minute :((

Made a screencast for this round: link. Very unedited and just made on a whim.

Got A-D and was almost done implementing E. Probably would've got E if I didn't spend so much time explaining stuff. Lmk any suggestions.

I a beginner find it really tough, I cant even solve C, and on B cant get my head around failing to pass the contest 4, but really really enjoy it, challenging myself to make better

Thanks guys for the contest!

Now let's get ready to continue with Codeforces Round #905!

Can anyone please show me why my code goes wrong

Very trash Problem E with 0 thought but +inf boring implementation.

I feel like such a loser for trying to debug 229180556 for an hour!!!

I was using 32 bit int instead of 64 bit long long to count the num of operations.

I've made the same mistake couple of times in just this year. I think I should give up on cp.

trash contest. The passage is poorly written and difficult to understand, with many omissions.

Ratings updated preliminary, it will be recalculated after removing the cheaters.

Is it rated？

so i recently got a notification that ive been caught for plagiarism, I am absolutely shattered after seeing it. I am assure you that i wouldnt use cheap tricks to increase my ratings, to prove that i had a common source published before the contest

int findClosestRecursive(vector arr, int left, int right, int target) { // base case: when there is only one element in the array if (left == right) { return arr[left]; }

// calculate the middle index
int mid = (left + right) / 2;

// recursively search the left half of the array
int leftClosest = findClosestRecursive(arr, left, mid, target);

// recursively search the right half of the array
int rightClosest = findClosestRecursive(arr, mid + 1, right, target);

// compare the absolute differences of the closest elements in the left and right halves
if (abs(leftClosest - target) <= abs(rightClosest - target)) {
    return leftClosest;
}
else {
    return rightClosest;
}

} this part was taken from a popular coding website geek for geeks i never knew this would be such a problem

It was a nice round! Congrats to all participants!