So excited for Cyberpunk round and hope to get CM :D

As a tester, I confirm that problems are set in 2077. Hope you enjoy them!

Hoping that the problems will be easy.)

If theres a 2000 point problem, it should be increased to 2077 :D

Excited for my first contest :]

Hope that the streak of getting negative deltas breaks with this Cyberpunk contest and bring me big positive delta.

round will be rated for all participants with a rating below 2100? So if one's rating is 2100+ he can't get rating or something?

Down the line I hope codeforces round 2077 is written by you guys too

Let's hope that this round will be post patch 2.0 cyberpunk like :D

Im sure, My laptop cannot handle Cyberpunk!

And now you're going to leave with nothing but a sign? :(

wish my result gotta be the game NOW and not when it first came out. Gl to everybody tho

orz

will the warrior race get a positive delta??

will skip LeetCode Biweekly(35 leetcoins loss) for cyberpunk theme, lessgooo.... Hoping for gr8 contest and editorial too.

Hopefully, the problem statement will be short and precise just like the announcement!

Hope for not doing the Hat-trick of negative rating dealt by getting a positive rating dealt.....

#pray_for_me

Competing under the handle bartm0ss, I’m honored to join this cyberpunk contest :D

As a Cyberpunk 2077 fan, I am really excited for this

excited for this!

Lmao I am literally playing Cyberpunk currently

Hoping for a positive delta.

Hoping to get to pupil in this one :).

Just want to wish luck and enjoyment to everyone!

So excited for the cyberpunk theme!! Hope everyone performs well.

Damn, a true CP contest.

this photo is interesting

hope you all get positive delta this time for saving the world and me too!!

where is scores?

Why is it clashing with LC biweekly :(

score distribution??

I am gonna play Cyberpunk soundtrack while solving hehe

Wish me luck guys so that I can hit Pupil today <3

希望能解出4道题:)

my current rating is 839 . how many questions i have to solve in contest to reach 1200 rating.

Yeah I'm * out of competition

Why Rust is not allowed?

Gains don't stop!!!

What could be the difficulty range of problem C?

Today's sponsor: "All right, David. Let's go. To the top, then"

I love Cyberpunk 2077, one of the best stories in gaming in my opinion. Good luck to everyone who's participating :)

My First contest, lessgoooo

Almost the same as problem C was given this year at ONI (romanian informatics olympiad) link

Good contest, thank you

TOO CLASSIC D

for the first time i did Answer construction on a greedy problem.

I could not solve C :( . Let's see how much rating goes down

I hate C. How hard was it to find the d values :(

Why is it failed system tests on A for everyone? UPD: It's OK now

Problem F was very nice imo.

I enjoyed problems C and D .. I found them challenging and I was a little slow in solving these problems ..

also very happy to not make a silly WA submission for problem B .. lolzzz

Thanks for the nice problems..

E is cool. I couldn't code it in time, but I think it's cool.

Problem E appeared in last year's TAP (Argentinian Programming Tournament) Problem

hello, i found a hack testcase for my solution on problem D, but i couldn't figure out how to hack my own solution, any help on how to hack it?

edit: the incorrect solution still passes main tests.

edit2:

#include <bits/stdc++.h>

using namespace std;

#define ll long long

const int mod = 1000000007, N = 200005;


int main(){
cin.tie(0)->sync_with_stdio(0);
    cout<<1<<'\n';
    int n = 200000, m = 300000;
    cout<<n<<' '<<m<<'\n';
    for(int i = 1; i <= n; i++){
        cout<<1<<' ';
    }
    cout<<'\n';
    for(int i = 1; i < n; i++){
        cout<<i<<' '<<i + 1<<' '<<1<<'\n';
    }

    for(int i = n; i <= m; i++){
        cout<<i - n + 1<<' '<<i - n + 3<<' '<<1<<'\n';
    }
}

Problem E was interesting, but I didn't learn how to find Euler path :(

Can someone tell me why my solution of problem D fails?

I used Binary Search + Dijsktra

I found F very satisfying to solve. When I first saw it, I thought there was a good chance it would end up being bashy and/or involving lots of technical details, so I was pleasantly surprised to find that it just involved a couple of nice observations. Thanks for the round!

Nice contest, no ambiguous statement like other contests..Thank you..

good contest

What is the solution to C?

Why my F solution 321144387 got WA on 3?

The idea is:

f(x,y) is always less(or equal) than 2*min(x,y) and max(x,y).

So if we have a pair of numbers (x,y) which satisfy min(x,y)*2>max(x,y), then f(x,y) will equal to max(x,y) and we can ignore all the numbers less than max(x,y).

Thus, we can maintain a set, which don't have a single pair satisfy min(x,y)*2>max(x,y) in it, and each time we add a new number x into it, check if there have a number satisfy the condition mentioned above.

We get answer from the max of two parts: the maximum max(x,y) which satisfy min(x,y)*2>max(x,y) and maximum f(new element, numbers in set).

It can be shown that the number of the elemnts in the set always less than $$$\log 10^9$$$.

Sorry for my bad english :(

My F solution:

Consider sorted $$$a$$$ (non-decreasing order), and remove duplicate numbers:

for each $$$a_i$$$:

• if $$$a_i \lt 2a_{i-1}$$$, $$$\max(f(a_i,a[1,\ldots,i-1]))=a_i$$$, easy;

• if $$$a_i \ge 2a_{i-1}$$$, bruteforce $$$O(logA)$$$ such numbers.

I think it's absolutely correct, but I don't know how to TLE in #8.