.

i was told by the author to comment to boost engagement

as a tester, this was my first time testing a round :)

as a tester, I maybe should have stayed at specialist so we could have said that we had a div3-rated tester :|

Close your eyes when the cactus scene comes up.

yay another contest.

As a wacknana fornever I will gladly smurf in this round to take away rating from innocent newbies because I'm evil

finally a div 3 lets go..

As a tester, Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.

As a tester, I'm not paid to make this comment.

Damn, we makin' our way through lots of contests in the span of a few days

Going fire!

As a tester cry took my balls and I would like them back.

Can we expect another math-heavy contest?+

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This is going to my first ever contest at CF , (Wakanda ki Jai) ::))

Hey guys new to Codeforces and i am enjoying all of these

as a tester, I cheated and now I am in cry's basement.

Good luck to all participants!

Hopefully, the problem statements will be short, and I'll make it to Pupil in this contest, InshaAllah.

mibombooooooooooooooooooooo

mandatory tested comment !

so unqualified tester because of taking less than 5 rated rounds won't be included in official standing ?

As a participant, I'm excited to solve A, feel confident after B, and then spend the rest of the contest convincing myself that C is actually easy.

OOOOOOOOk!!!!!!!!!!!!!

Damn, the last 2 Div 2 contests were hell

I hope to make up for the rating i lost in this contest

I hope you cook well

Good luck

.

hoping to become a specialist this round. glhf everyone

.

It's awesome that we have so many contests in such a short period, with barely any long breaks between them :)

I hope there are no interactive problems.

Bro forgot 6 7 :sob:

It's a div3 contest, why are there no newbies, pupils, or specialists testing the contest?

I Hope to regain my lost ratings in this contest :-)

booooooiiiiiii is that 67 in the problem A pretest?

I can BARELY submit... the server is too late!!

crazy problems! Thanks

Nice problem set, but pls don't make Strict Alice formatting a reason people get WA.

thanks for the awesome div 3 bruh, was a booster solved my first two probs

yay I forgot long long on D and wasted 1 hour.

Such a well fabricated contest. The difficulty change for each problem was appropriate and balanced. Though I found E to be harder than F as the latter required just a minor observation that when & how can Bob win whereas in E, the counting in optimal Time Complexity is a dp-ish task in itself.

Anyways, this is just my opinion; F maybe harder because I can prove the necessity of my condition for parity of runs, but not its sufficiency. But after pondering post solving; I was able to come up with sufficiency as well using mirroring strategy by Bob.

Such a great contest without any interactive problem!

.#include <bits/stdc++.h> using namespace std;

int main(){ ios::sync_with_stdio(0); cin.tie(0);

int t;
cin >> t;
while(t--){
    int n;
    cin >> n;
    string s;
    cin >> s;
    int freq_1 = 0;
    int freq_2 = 0;

    int leng = s.size();
    for(int i = 0; i < leng; i++){
        if(s[i] == '1'){
            freq_1++;
            freq_2 = 0;
            if(freq_1 == 2){
                s.erase(i,1);
                i--;
                freq_1--;
            }
        }else{
            freq_2++;
            freq_1 = 0;
            if(freq_2 == 2){
                s.erase(i,1);
                i--;
                freq_2--;
            } 
        }
        leng = s.size();
    }
    leng = s.size();
    freq_1 = 0;
    freq_2 = 0;
    for(int i = 0; i < leng; i++){
        if(s[i] == '1'){
            freq_1++;
        }else{
            freq_2++;
        }
    }
    if(freq_1 % 2 != 0 && freq_2 % 2 != 0){
        cout << "2" << endl;
    }else{
        cout << "1" << endl;
    }
}
return 0;

}

this was my first solution for C

cant figure out why its wrong

I am so bad at trees....

For Problem F, my idea was that the only losing state is when the string has a single character left.

So, if the entire string has an odd number of inversions, or if we can remove a subsequence of length (n-1) (leaving only one character) such that the removed subsequence has an odd number of inversions, then Alice wins immediately.

Otherwise, I guessed that Bob wins.

Can someone provide a proof of why we don't need to consider states where 2, 3, or more characters are left? Why is it sufficient to check only whether Alice can reduce the game to a single-character string on her first move?

A mathematical proof or an intuitive explanation supporting (or disproving) this idea would be greatly appreciated.

Here is my implementation for reference:

:- https://codeforces.me/contest/2241/submission/380815649

Nice round, the tree problem is so fun to solve.

can anyone suggest me how to prepare trees and graphs?, whenever tree or graph related question comes up i get anxious and my mind goes blank

6.7k rank after solving A,B,C,E Was the contest too easy?? Because, I solved much easier 4 problems in another div 3 a few days ago and had a 4k rank

crazy times.

As a participant I want to ask how the fuck E has 2900+ solves

Good news: I went for the problems backwards (partially) in this ICPC-style contest.

Let me share a tragic experience with you all. Due to a network malfunction, I lost a precious 20 minutes and consequently fell behind about 2000 people.

:(

The carrot is showing -1 delta can I get +2 ToT

I really liked problems and one of the main reasons is Problem statements were short. that really helps me to focus more . BTW thanks for the contest :)

$$$B$$$ and $$$F$$$ are worst problems, rest were good.

How is E having more submissions that F? F was clearly easier.

Problem B uses the same idea as Problem C: ICPC India Online Prelims 2025 lol Enjoyed the problemset, great problems!

Great contest :D

such a good observation in B! wasted a lot of time precomputing good integers

nice touch with G had to use __int128.