As a problemsetter, I can certify that I did not test in any shape or form

would be funny if tourist regains no 1 in a Benq round

As a tester, I can confirm that this is a great contest and has great problems, much like another amazing contest by the name of CerealCodes :D

As a problemsetter, I hope you all enjoy the problems (especially mine).

As a tester, I have enjoyed the round and I hope you also enjoy it.

As a tester, I can confirm that this contest is truly one of the contests of all time

it has been scientifically proven that if you play genshin impact you will gain rating on this round!!!

Time to enjoy -167 delta in a div.1 round!

"The Div. 2 round will be rated for participants with rating below 1900 , while the Div. 1 round will be rated for participants with ratings which are at least 1900"

Does this mean that I, as a purple, am forced to participate in Div1 round?

Scared for my first div1 round. Hopefully I can actually solve 2 problems and avoid instant demotion to blue.

As a tester, I hope u enjoy the round as much as I did

I'm afraid to participate cz of your handle.

Hoping to become purple. ( Please make this my most memorable contest ever! )

Problems are high quality. Definitely try to participate!

As a tester I can confirm that this round is just as good as Oshi no Ko chapter 123.

As a problemsetter, I can confirm at least one of the problems will feature Ninomae Ina'nis from hololive-EN Myth!

As a tester, I can confirm the problems will be problems.

i hope to be blue this time

Very rare to see this many colors in the author list

Hoping for a good div2 round.

as a problemsetter who didnt problemset, i can assure you that this will be a great contest :D

OMG!! Benq is the writer. Sounds Good!

As the president of CerealCodes and a contest organizer, I can assure you that the problems are of high quality (just like CerealCodes problems).

As a participant i recommend you to participate.

Why didn't PurpleCrayon test when Purple Cry is an author?

As a tester, I can confirm that this round will not only give you lots of rating but ginormous muscles too :D

Is div.2 also 2.5 hours?

Am I the only one who noticed this announcement was made 7 weeks ago?

Hi everyone,

We would also like to mention this round was mostly made possible by problemsetters from the CerealCodes initiative!

As a tester, I can confirm that problemsetting might have happened.

Benq!This round would be very interesting

hope I can be a candidate master.

as a cyan participant, i hope to be blue after this contest

Excited to see tourist in round prepared by Benq

Do you mean that you will make us cry in this contest :(

will be fun to see if tourist takes back rank 1 in a Benq round

The panda looks like Firefox)

As a participant I hope I'll get over this damn pain this round T_T

Hope for color change!

Are you saying Div1 has easier problems?

Hello Benq. This is a fellow United States of America Computing Olympaid competitor, and wishes for you to not write any problems in the USACO 2023 December Gold contest. If this is possible, I will certainly be elated. Please consider others emotions before problem setting for the USACO 2023 December Gold contest.

Hoping to become a green , also good luck for everyone

I LOVE YOU CRY :) MY BEST FRIEND, THANK YOU FOR PREPARING A CONTEST <3, good luck to all!!!!.

you fool me, i fool you！ Good luck

Red panda is so cute!

It's even a little insulting — authors have all possible colours except green...

Anyway, why there is such a variety — is it some university project or so on?

i hope to be blue this time

Feeling excited to give this contest.

good luck for all

I have a question, why the problems of Div.2 have higher score than the problems of Div.1?

QaQ

Whoever came up with div2 problems B and C I hope that for the rest of their life they would only get pairs of integers as their birthday presents.

Imagine 1434ing us

to my cyan color, i will miss you

C>>>>>>>A,B

Disgusting problem b and c

Nightmarish for me..atleast

very strange and hardest contest I have never seen ever...

Don't know what this round's authors were thinking, but I feel like there is a huge gap between A and B. As a div 2 participant I feel like this round would have been a nice round if B would have been C and C would have been D.

ty guys for the another speedforces round

div2 c>>b and b>>a. Doesn't feel like this is a well balanced contest :(

does the problem ratings depend only on the ratings of the people who solved it or it also depend on the position of problem ??

like consider same problem in div2C and div4H, then comparitivley div2C will have higher solves than div4H

so, if only the ratings of the problem solvers is considered then div4H will have higher rating than div2C.

As a div1 contestant, i gave up A after reading it :(

Yes, We are cry ing :(

More then a half div1 contestants gave up after reading Div2C. Sounds like something that shouldn't happen... Hardest div2C... Spent an hour, but still have as few ideas as after i read it for the first time...

Just want a honest opinion , Really 7000 plus people were able to solve B , got me into thinking , for a hard time

My approach for getting strong idea for C:

I failed to impress Ntarsis moreover i failed to impress myself.

Div1C

How to solve div1C?

Wow I like Constructive and MathForces so much!!!

IMHO Problem B to me was like you either know it or you don't. Like, I can't get a piece of paper and start working on it and finally get an answer, but 7000+ people solving it really got me questioning my existence.

First time solving d1ABC!

great problems d1C&D!

mathforces

I think this contest is not for div 2 coders

hardest Div2C I've ever seen

Don't make contests just to Show-off.

Is there anyone who can give some hint on 1C? I have absolutely no idea.

wow. It was really ( i mean really ) interesting contest ( for me, as div2 user). Amazing second & third problems, like, for me, they was hard, but really interesting. Second was on idea mostly ( like not many fibonacchi numbers ) & third was on idea & realization ( as for me i mean )

cry (The author of this blog) really made me cry during the contest.

S/He literally gave the spoiler of the full contest in the handle.

Div.2(x) Div.1.2(√)

Problem B can be easily solved thanks to this post.

So we can iterate over all values of $$$f_1$$$ from $$$0$$$ to $$$n$$$ and check if there exists integer $$$f_2$$$ such that $$$f_2 >= f_1$$$ since we need a non decreasing sequence.

We don't even need to check for $$$k > 30 $$$, as $$$f_{30} > 2 * 10^{5}$$$ . So the answer for all such values of $$$k$$$ will be 0.

Time complexity : $$$O(n)$$$

Div2C was so difficult but once you solve it ,it seems easy.

Please somebody tell me how to do div2 C and how you came to this conclusion :(

For Div2 B, I got that we want number of integers in the interval $$$[nx_{k-1},nx_k]$$$ where

Here $f_i$ is $$$i$$$-th fibonacci number with $$$f_1 = 0, f_2 = 1$$$. It can be shown that

with $x_1 = 0$.But when I tried to evaluate this, I was getting TLE on pretest 5; is the way to go?

While I'm waiting for solution, here is what I wrote in $$$B$$$ and just now noticed, that this is incorrect.

Corner cases: all are $$$0$$$, all are $$$n$$$, there are $$$0$$$ and $$$n$$$.

For simplicity, there is $$$0$$$. It not, do $$$a_i = n - a_i$$$.

Some number are positive, others are negative. Sort all values. Some suffix is positive and has all edges. Let size of suffix be $$$cnt$$$, sum on suffix be $$$s2$$$ and sum on prefix $$$s1$$$. If $$$s2 - s1 = cnt^2$$$, this is correct split position, answer is $$$YES$$$. Now to build it. Look at suffix, let it be $$$a_1, \dots, a_k$$$. Subtract from all of them $$$cnt$$$. Let $$$idx = 1$$$. Take $$$a$$$ values by blocks of same values. For each block put $$$-idx$$$ to front of result vector $$$a_i$$$ times, then put $$$idx+1$$$ to back of result vector size of block times, add $$$idx += 2$$$. At the end append left $$$-n$$$ to front of result vector.

Further steps:

1. Write stress

2. See, as it immedeately asserts, investigate test

3. Cry, write this

div 2 c Is a good problem to think about for 3 days after contest

Problem B and C were good, amazing contest!!!.

This was hard. Managed to lose only 21 (according to carrot), i'll get purple in the next context.

Also my brain farted a little on B when i tried using diophantine equations or double for to find the coefficients. Anyways the problems are nice, B is a very cool property of fibonacci-like sequences

2*Div 1

I got some inspiration from this task to solve D1A/D2C

GoodBye my chance to become master :( I really have hard time

Hardforces for div2

Nice problems!

Solution for B (if anyone needs):

-> Form the Fibonacci Series up to 200005, call it arr

-> For a Fibonacci Series of length k and last term n, it looks like this:

a, b, b + a, 2b + a, 3b + 2a, 5b + 3a, ....., arr[k — 1] b + arr[k — 2] a

-> The equation is arr[k — 1] * b + arr[k — 2] * a = n

-> Fix the values of a from 0 to n and find if there exists an integral solution b for the equation

-> if exists, ans++

-> One edge case: n is only up to 200000, so k can't be > 200000, if so answer is 0

:

When every problem in the contest seems solvable but you're still not able to solve all :(

I overkilled problem B by doing fast general fibonacci calculation to determine the last element of the sequence in log(k) time by doing fast squaring of matrices for the fibonacci sequence.

Then I bruteforced over all of the first element, binary searching for the second element using my fast_fibo function to see which one is closest to n. If the fast_fibo(i, j, k) == n, then i increase the count

Time complexity: O(n log(n) log(k)) ??? LOL

I was like: "Theres no way this can be a problem B..."

EDIT: Just failed system tests LOL

I came up with a pretty easy and intuitive solution for div2 B after about an hour.

bool check(int n, int difference, int k) {
    if(n < 0) return false;
    if(k == 0) return true;
    return check(difference, n - difference, k - 1);
}
 
void solve() {
    long long n, k; cin >> n >> k;
    long long ans = 0;
    for(int i = 0; i <= n; i++) {
        if(check(n, i, k)) ans++;
    }
    cout << ans << endl;
}