As an author, I can confirm that no problems are authored by AI!

As a bot, I can't participate in this contest.

As a tester, I was really looking forward to participating in this sponsored contest, only to realize that I had already tested it.

As the boyfriend of the girlfriend of the tester Caylex, may you good luck & have fun!

The round's number "985" is special for chinese, especially for our students.

Hope you all can perform well in the "special" round.

As a newbie, I hope that I learn more in this contest.

The score distribution is scaring. I hope I can reach master in the contest.

As a tester, I might lose a T-Shirt.

I wonder what the tourist will choose: the chance to win money with the risk of losing their rating, or doing nothing at all?

As an author, I prepared $$$998244353$$$ problems and wrote $$$998244353$$$ pieces of editorial in all (modulo $$$998244353$$$).

Love the score distribution, 750 for A problem means 1 WA only cost ~ 7% score...

It'll be less painful for stupidity submission sometimes.

As a tester, there will be 0.3 problems about penguins.

24 points to expert ! lesss gooooooooo

Bruh C is worth 1750!!! Thats actually too much. I'm assuming it'll be wayy more difficult than usual.

As a tester,I'm a bot.

As a tester, I really like some of these problems.

For Chinese high school students, "985" is a special number. Wish everyone to get into the university of their dreams!

As a tester and the friend of the author sszcdjr, I can confirm that the authors are all not AI, because I can't solve any of their problems! GL&HF!

Just out of curiosity, what is the relationship between the score of a question in a contest and its difficulty in the problem set (https://mirror.codeforces.com/problemset). I know higher score questions are more challenging, but after noticing 2 questions with score ≥ 5000 and no questions in the problem set with this difficulty it made me realize its not a identity mapping between the two (that is difficulty(score) != score)

Is there Chinese statement?

Finally got the Tourist title :)

How come the badges of the authors are not related to their PFP I thought in CF badges come from your CF PFP.

Please reschedule, it's my birthday

Want to see tourist's rating change!

_istil What's the behind your badage ?

light ray ? physics ?

btw like it

Good luck!

rainboy vs 5500 points problem

this contest is rated? ;-;

Hope to reach CM in this contest!

So this contest is rated?

I think it clashes with COCI :(

Loved the '0' interactive phrase .

Good luck for everyone!

I think, contest will be interesting without interactive problems, (because I can't really solve it)

The contest is about to start in an hour, but only 17000 people registered!? Is this a special feature of Div1+2?

Last Div2 30000+ people registered, and this one is a Div1+2, shouldn't there be more people?

As a tester,Hanghang007 hanghanghanghanged

this is the hardest div

rainboy being RAINBOY !!

As a participant, I read all the comments!

c any hints??

edit : got it thanks

I spent 2 hours on C, yet did not solve it.

Prolbem E. Why this is WA21? (I can't resubmit now, and in submission there is junk code.)

#include <bits/stdc++.h>

using namespace std;

const int N=1000000;
bool prime[N];

void prepare(){
    for(int i=1;i<N;i++) prime[i]=true;
    for(int i=2;i<N;i++){
        if(prime[i]){
            for(int j=i*2;j<N;j+=i){
                prime[j]=false;
            }
        }
    }
}

int solve(int n,vector <int> a){
    vector <int> pr;
    for(int i : a){
        if(prime[i]) pr.push_back(i);
    }
    if(pr.size()==0){
        return 2;
    }
    if(pr.size()>1){
        return -1;
    }
    int x=pr[0];
    bool good=true;
    for(int i : a){
        if(i==x) continue;
        if(i<2*x){
            good=false;
            break;
        }
        if(i>=2*x+6) continue;
        if((2*x)%3==0 && (i-2*x==1 || i-2*x==5)) good=false;
        if((2*x)%3==1 && (i-2*x==1 || i-2*x==3)) good=false;
        if((2*x)%3==2 && (i-2*x==1 || i-2*x==3 || i-2*x==5)) good=false;
    }
    if(good) return x;
    else return -1;
}

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

    prepare();

    int tt;
    cin>>tt;
    while(tt--){
        int n;
        cin>>n;
        vector <int> a(n);
        for(int &i : a) cin>>i;
        cout<<solve(n,a)<<'\n';
    }

    return 0;
}

wtf C, I'm so broken mentally, been doing cp for nearly 2 year still cannot solve C confidently, how this is even possible. god abandoned me

Why such a small number of participants? Only 7k+ solved A.

Edit: contest without subtasks feels so good.

How to do C?

Damn got cooked... someone explain C please?

Really enjoyed the problems. Great work authors!

Seems like some $$$O(3^n)$$$ passed H.Is it intended?

Looks like skipping C worked well for me, I solved A, B, D, E, F and then failed to solve C for about an hour lol

Any hints for F?

Things I could observe:

1. If all edges have same colour, answer is clearly yes.

2. If both $$$RR$$$ and $$$BB$$$ appear the answer is clearly no since there is at least one pair $$$(i, j)$$$ where all paths start with $$$R$$$ and end with $$$B$$$ and can thus never be palindrome.

3. For simple alternating colours ($$$RBRBRB\ldots$$$), the answer is yes for odd $$$n$$$ and no for even $$$n$$$.

4. The answer is no if an even length blocks of a colour appear more than once since there will be paths which are forced to start with $$$BB$$$ or $$$RR$$$ but can only have an odd number of the same colour at the end (example graph).

But I still have no clue how to generalize this to a solution.