Is there a separate category for Iranians, or will all participants be grouped together?

Sorry man, but I have to admit that I'm kinda glad you didn’t know that 🫂

For the 1st time in my life, I found the tester's comment funny!

he already did unfortunately last contest

I predict tourist will become tourist once again.

yes i think so

He lost to orzdevinwang again

Let me tell u sth about ArshiaDadras:

I officially started competitive programming and programming classes with Arshia Dadras four years ago. He is a very capable and ethical teacher. Thank you, Arshia, for being so great <3

We will start inviting participants from the top of the list while considering the country limit (a maximum of three participants per country) until we fill all 60 spots. Each invited participant will have a specific time frame to confirm their attendance. If someone declines or cannot attend, their spot will be given to the next eligible participant from the scoreboard who meets the criteria.

Why do you like Persian contests?

gpi

If you look at their website, the onsite part is sponsored by a university and from what I can tell an entire ministry, and these prizes are for the onsite part. This round is only a selection round before the onsite one.

But I guess we'll have someone from Iran in the top 10... P.S. Let’s bet lunch on it!

With pleasure—he’s more than welcome!

We’re planning to do that, but this is Rayan’s first year!

We need to evaluate the feedback and reception from participants before making a clearer decision.

Nah he will become reverse nutella again

Why not? At the very least, you’d get to enjoy a wonderful trip to Iran, stay in one of the best hotels here for free, and experience some amazing hospitality! (if you can make it to the top 60 😁)

Yes, as I said here! (if you'll be one of the next candidates to invite)

And I would like to thank the medalists who designed the questions and all the people involved and hardworking in this event!

I think it is a good thing that the famous programmers of Codeforces come to our country.

I hope to see the tourist here!

Good luck to us.

It's like a normal rated contest in CF. Also, participation is enough for registration and you don't need to register on the website or somewhere else.

иди ученика апни прежде чем захватывать кого-то

No, buddy!

c>d for sure bro

in all cases, a solution will exist if n and k are odd except for k=1 and k=(n-1)!

1
5 3

YES
1 2 3 4 5 
3 4 5 1 2 
5 3 1 4 2

Actually, C seemed difficult to me not because of solving the problem, but because of the implementation. What edge case did you get stuck on? I can share my solution if you want.

Can you share it with us common folk?

why is this getting downvoted

Happy birthday tho!

Well a gain of 0 is better than any loss I suppose. Happy birthday!

You're welcome!

The max size of the answer is 6. let $$$x=2\cdot3\cdot5\cdot7\cdot11\cdot13$$$, then $$${\frac{x}{2},\frac{x}{3},\frac{x}{5},\frac{x}{7},\frac{x}{11},\frac{x}{13}}$$$ is a valid set

Here is the rough sketch for H.

This is how we can characterize all possible linearly independent sets $$$a$$$.

First find a set of distinct primes $$$p_1,p_2,\ldots,p_k$$$ and let the gcd of all the $$$a$$$ be $$$g$$$. We will write $$$P = \prod p_i$$$ as the product of all the primes here.

Then we can construct a linearly independent set $$$a_1,a_2,\ldots,a_k$$$ where $$$a_i$$$ is a multiple of $$$\frac{gP}{p_i}$$$ but not $$$gP$$$.

For each linearly independent set, you can find a corresponding $$$(g,p_1,p_2,\ldots,p_k)$$$ that satisfy the above constraint.

Then after brute forcing everything, you realize the the number of possible $$$(g,p_1,p_2,\ldots,p_k)$$$ with $$$k \geq 3$$$, $$$g \mid P$$$ and $$$\frac{gP}{\min(p_i)} \leq 10^5$$$ is at most 500k or something. So you brute force everything. I had to use a bunch of constant opts to pass because my code is shit.

Checking answer is at least $$$2$$$ can be done in $$$O(n \log n)$$$.

Nevermind, they all passed the system tests.

Firstly, for simplicity, redefine the special conditions to be in terms of the number of rubies/sapphires in your satchel (not chest). Then add dummy states $$$(0, 0)$$$ and $$$(n, m)$$$ for convenience (we won't double value at these though).

Define dp states to be $$$dp[i]$$$, which is the total sum of value of your satchel across all ways you can reach the state defined by the $$$i$$$-th condition.

To solve this dp you will firstly need to define some ordering on the conditions so that when evaluating some dp state, all the states that it depends on have already been evaluated. Just sort by $$$x + y$$$.

Then, the transition from state $$$i$$$ to state $$$j$$$ will correspond to the ways you can go from state $$$i$$$ to state $$$j$$$ without visiting any other special condition state in between.

Computing the number of ways you can go from state $$$i$$$ to state $$$j$$$ (where $$$i$$$ occurs before $$$j$$$ in the topological ordering) without visiting any intermediate special condition state is easy and can be done using inclusion exclusion in $$$O(n^3)$$$. Call this $$$w[i][j]$$$.

Also define the total number of ways for us to go from state $$$i$$$ to $$$j$$$ to be $$$g[i][j]$$$. This can be computed in $$$O(1)$$$ using simple combinatorics.

Define the addition in value for going from state $$$i$$$ to state $$$j$$$ without any intermediate doubling to be some $$$h[i][j]$$$.

Then, the dp formulation for state $$$i$$$ is:

(where $j$ occurs before $$$i$$$ in the topological sort)

All states will be evaluated in $$$O(n^2)$$$.

The final answer is the value of the dp state corresponding to the the state $$$(n, m)$$$ divided by the total number of ways to go from $$$(0, 0)$$$ to $$$(n, m)$$$, which is just $$$\binom{n + m}{n}$$$.