teraqqq's blog

By teraqqq, history, 13 months ago, translation, In English

Hello Codeforces! This Saturday, the student final of BSUIR Open XIII will take place, accompanied by rounds Codeforces Round 1021 (Div. 1) and Codeforces Round 1021 (Div. 2) at Apr/26/2025 11:35 (Moscow time), featuring problems from this competition. Note the unusual time of the round.

This year, the problems for the rounds and the olympiad were prepared by teraqqq, FairyWinx, Kirill22, and wilcot. Each round will contain 6 problems, and you will have 3 hours to solve them. Notably, wilcot has been contributing problems to BSUIR Open for multiple years and has helped involve new authors in this event. We believe this effort has resulted in a diverse and exciting student final, part of which has been adapted into this Codeforces round.

We extend our special thanks to MikeMirzayanov, geranazavr555, and KAN for the fantastic Polygon and Codeforces platforms! We also thank:

Special thanks to the people who made BSUIR Open XIII possible:

  • Velikovich Vladimir, Shved Elizaveta, Vnuk Olga, Yurchenko Olga, Gorokh Andrey, Makarevich Darya, and Sevryukov Stepan for organizing the championship;
  • Udovin wilcot Ivan, Markovets Roman, Baitasov Rinat, Romashevsky German, Butoma Vitaly, Lyubashenko Andrey, and Sitnikov Alexey for preparing the semifinal problems;
  • Yefimchyk Alexander and Adarov Dmitry for technical support of the championship.

For yet another year, the BSUIR Open championship has taken place within the walls of BSUIR. Each time, the organizers bring something new, unique, and unforgettable. Thanks to them, BSUIR Open continues to delight participants with its friendly atmosphere, engaging problems, and wonderful prizes!

UPD. The score distribution is:

Div 2: $$$500 + 1250 + 1500 + 2250 + 2750 + 3250$$$

Div 1: $$$500 + 1000 + 1500 + 2000 + 3000 + 3500$$$

UPD2: Editorial

  • Vote: I like it
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  • Vote: I do not like it

| Write comment?
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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

As a tester, I observe that the problems were prepared by a tsundere and a chuunibyou, which certainly does make a combo for preparing rounds.

P.S. look at their profile pictures :)

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    13 months ago, hide # ^ |
     
    Vote: I like it -13 Vote: I do not like it

    Kurisu is the best waifu in the world. I've only met girls like her a few times in my life. But she's truly the type I'd feel comfortable living with, and it's girls like her who inspire my creativity.

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      13 months ago, hide # ^ |
       
      Vote: I like it +4 Vote: I do not like it

      You don't love Kurisu, you only love kirill22...

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        13 months ago, hide # ^ |
        Rev. 2  
        Vote: I like it +34 Vote: I do not like it

        I love Kirill22 as a friend, a brother, and a family member. Though I wrote above about romantic relationships, this is something different.

        Shoutout to kirill22 for being the MVP drinking partner. When I’m staring down 30 shots, he doesn’t hesitate to take 30 himself—that’s loyalty.

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    13 months ago, hide # ^ |
     
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    Yeah.Tsundere is peak.Mid-2th Disease is peak

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13 months ago, hide # |
 
Vote: I like it +15 Vote: I do not like it

I wish we could have a

^-^

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13 months ago, hide # |
 
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When the score distribution will be updated?

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    13 months ago, hide # ^ |
     
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    Since there are only 6 problems and 3 full hours to solve them!!, I'm assuming the scores will differ by a big margin with each-other.

    It will be one of the breainstorming rounds.

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      13 months ago, hide # ^ |
       
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      Yeah. Hope to solve ABC. Though C is going to super tough i think.

      I assuming score distribution for ABC will be like : 500 — 1000 — 1750.

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Please make a note about unusual contest starting time also.

I'm waiting for score distribution too.

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Score distribution?

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13 months ago, hide # |
 
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Will round be rated?

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13 months ago, hide # |
 
Vote: I like it +26 Vote: I do not like it

Been training myself to wake up at 4:30 am for next 2 rounds. Can't wait to drop back to purple.

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    13 months ago, hide # ^ |
     
    Vote: I like it -33 Vote: I do not like it

    It's normal for the challenge to feel tough, especially waking up at 4:30 AM.
    Even if you drop back to purple for a while, you'll come back stronger.
    Focus on the journey, not just the result — the grind will pay off!

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      13 months ago, hide # ^ |
       
      Vote: I like it +3 Vote: I do not like it

      Brother you become pink in just 2 month and you solved almost 500 problems in a month orz!. Any tips do you like to share ☠️

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        13 months ago, hide # ^ |
         
        Vote: I like it -45 Vote: I do not like it

        Thank you for your kind words!
        I’ve been focusing on consistency and targeted practice—solving problems from various topics and revisiting weak areas regularly.
        Virtual contests and upsolving helped a lot with speed and accuracy.
        I also try to analyze editorials after contests, and I keep track of mistakes to avoid repeating them.
        Staying disciplined and setting daily goals made a huge difference.
        Wishing you all the best on your CP journey!

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          13 months ago, hide # ^ |
           
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          Bro couldnt even resist using an LLM to generate his response.

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        13 months ago, hide # ^ |
         
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        Just learn OCaml and get free rating.

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13 months ago, hide # |
 
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Sorry to ask, but does the scoring distribution indicate the toughness of the problem? I've never really paid attention to this.

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    13 months ago, hide # ^ |
     
    Vote: I like it +5 Vote: I do not like it

    Generally a large score gap between problems indicates the difference in difficulty is also large

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rated?

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13 months ago, hide # |
Rev. 2  
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This is the first time I see the starting time of contest is 6 hours earlier than usual contest

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13 months ago, hide # |
 
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ABC would seal the deal for many of us.

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13 months ago, hide # |
 
Vote: I like it +17 Vote: I do not like it

6 problems, 3 hours. seems hard contest

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13 months ago, hide # |
 
Vote: I like it +7 Vote: I do not like it

The score distribution suggests that this contest will be good for fast solvers. Many users will solve three problems, and the fourth one will be critical, as it has a 2250 score, so those who solve the first three fast will be in a better rank.

Best of luck to all(me included).

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13 months ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

just 15k people?

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13 months ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Why no extra registration?

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13 months ago, hide # |
 
Vote: I like it +4 Vote: I do not like it

the questions were terribly worded!!

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13 months ago, hide # |
 
Vote: I like it +25 Vote: I do not like it

wat the fox with the statement of problem B in div 2? terrible

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13 months ago, hide # |
 
Vote: I like it +30 Vote: I do not like it

bro i am not able to understand B. The explanation is really bad...

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13 months ago, hide # |
 
Vote: I like it +22 Vote: I do not like it

i want to know if it is allowed to use chatgpt to provide a clear problem meaning during the contest, so that i wont waste so much time reading in B and C

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    13 months ago, hide # ^ |
     
    Vote: I like it -10 Vote: I do not like it

    Bro you solved B in 9 min over an hour ago and now you are asking about using chatgpt for "clear explanation" ? sounds sus to me

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      13 months ago, hide # ^ |
      Rev. 3  
      Vote: I like it +9 Vote: I do not like it

      cause B is easy for me, the hardest point for me is my first language is not English and I cant read problems well

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        13 months ago, hide # ^ |
         
        Vote: I like it -10 Vote: I do not like it

        But you pulled it off in 9 min and me here can't get to understand B despite english being my second language T_T

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13 months ago, hide # |
 
Vote: I like it +14 Vote: I do not like it

problem b is L

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13 months ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

I can't understand the problem statement, it's a terrible experience.

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13 months ago, hide # |
Rev. 4  
Vote: I like it +3 Vote: I do not like it

Is it only me or the questions are terribly written?

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13 months ago, hide # |
Rev. 2  
Vote: I like it +2 Vote: I do not like it

I understood B after 60 minute.... Dont know how to solve! :(( after asking them 4 questions too. I am crying

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13 months ago, hide # |
 
Vote: I like it +20 Vote: I do not like it

I think my english comprehension skill have been dragged through the mud today.

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13 months ago, hide # |
 
Vote: I like it +42 Vote: I do not like it

RIP people from div. 2

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Also post a separate editorial on how to comprehend these questions.

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13 months ago, hide # |
 
Vote: I like it +12 Vote: I do not like it

Уничтожить учетную запись DocinhoS is cheater. skipped: https://mirror.codeforces.com/submissions/DocinhoS/contest/2070 Strange submission from edu round: https://mirror.codeforces.com/contest/2043/submission/298277798 and there are more like this. Suspicious submission times today. Ban account.

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13 months ago, hide # |
 
Vote: I like it +17 Vote: I do not like it

problem b is cooked... still can't understand what it is trying to say...

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13 months ago, hide # |
 
Vote: I like it +13 Vote: I do not like it

It took me an eternity to understand the language of div2 B problem but just a minute to write the code for it : (

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13 months ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

Problem B was hell. Took 1 hour to understand and 3 min to code.

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13 months ago, hide # |
 
Vote: I like it +24 Vote: I do not like it

LanguageForces

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13 months ago, hide # |
Rev. 2  
Vote: I like it +9 Vote: I do not like it

It is so challenging to understand problems B & C. Problem statements are not clear and example explanations are poor. It took me a significant amount of time to comprehend how the samples produce the specified results.

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

UnderstandingForces ... spent time understand what is happening in B

that also made me overthink and difficult in C to understand

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oh no.. I found problem E tempting and wasted all my time there ...and I didn't attempt D

I thought E was some diophantine equation kind of thing, but couldn't fully solve it :P

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13 months ago, hide # |
 
Vote: I like it +24 Vote: I do not like it

text-understanding-speed forces

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13 months ago, hide # |
Rev. 2  
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In problem Div2C, how its possible for Vadim to guarantee success in testcase-3?

Testcase-3
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    13 months ago, hide # ^ |
    Rev. 4  
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    B = bus, T = telescoping bridge

    First day 2 student: 3 = B, 4 = B Second day 2 student: 3 = T, 4 = B day 3 student: 4 = T, 5 = B First day 4 student: 5 = T, 6 = T Second day 4 student: 5 = T, 6 = B

    If day 4 is B, we win (with one of the day 2 students). What if day 4 is T? Then we only need Day 5 = B to win with day 3 student. What if day 5 is T? Then we win no matter what day 6 is (with one of the day 4 students)

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    13 months ago, hide # ^ |
     
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    say the two choices for a day are 0 or 1. on day 2, give two predictions 01, and 11. if any one of these is true, good, else both are false, both being false means, on day 4 it's 0. so on day 3 give prediction 01. if it's true, good, else it means, that it is 0 on day 5. so on day 4, give two predictions 01 and 00. atleast one of them will be true, if all previous predictions were false.

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    13 months ago, hide # ^ |
     
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    at day 2 give predictions like that:

    AB

    BB

    at day three one of the predictions are true so far, for the day 4, both of the predictions are B. so if at the day 4, B happens you've got at least one right prediction. but B may not happen so you give prediction like that in day 3 ( to predict day 4 and 5 ):

    AB

    then at day 4 you can specify these predictions:

    BA

    BB

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    13 months ago, hide # ^ |
     
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    so 2 affects (3,4) ... 3 affects(4,5) ... 4 affect (5,6)

    bet (a,b) -> (x,y) means a takes x-th way and b takes y-th way .. where x,y can be 0 or 1

    day 2 -> so I can bet (3,4) -> (0,1) and (3,4) -> (1,1) .. note that i bet 4 as 1 in both cases ... so if I win here we are done else I know for sure that 4 can't be 1

    day 4 -> same thing I can do for (5,6) as it also has two occurrence .. and I will have a definitive answer for 5 if I loose...

    so now I have correct answer for (4,5) which I can bet on day 3

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    13 months ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    On day 2 bet 00 and 10 if day 3 is a zero he wins

    day 3 bet 10 since any 0 is a win the only lose is 11

    day 4 bet 10 11 we cover all possible loses thus we win

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    13 months ago, hide # ^ |
     
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    on day 2 he makes 2 bets say he makes the following bets

    2-a bus 2-b bridge

    these are for the transportations of day 3 atleast one of them will be correct on day 3 he makes one bet with one person for the happenings of day 4 and 5

    lets say he makes a bets saying day 4 has bus transportation with they person of day 3 then he will make a bet of bridge with the 2 people from day 2 for the transportation of day 4.

    so atleast one person is sure to clear day 4 bet in the worst case it will be the person from day 3.

    now on day 4 he can make 2 bets so lets consider day 5 now we know that the guy from day 3 has made it to day 5(or else someone from day has already cleared the required on day 4)

    if he makes a bet of bus on day 5 with the person from day 3 he will make a bet of bridge with the two people from day 4.

    so if its a bus on day 5 the bet with person from day 3 was correct and we are done or else in the worst case its a bridge and the two people from day 4 clear day5 and enter day6

    now there are 2 people and a single day so he can make two different bets for that day and win

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13 months ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Div 2 problem B was a nightmare. I was using a static version of this website and did not see the post on problem B until I logged on to the main page, costing 1 hour and 30 minutes of contest time to be lost. I hope that next time the authors can put a little more time into making these problems, and clearly explain the ambiguity on the problems before the contest starts, thanks.

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

how to solve D? I tried making a graph that only included cells belong to at least 1 path, but wasn't able to calculate the answer from the graph

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    13 months ago, hide # ^ |
     
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    It's DP right? we should be able to track how much flows from the previous known cell to the next known cell, then accumulate the fraction after everything... But I was not fully able to implement it in contest. Very interesting problem!

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      13 months ago, hide # ^ |
       
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      was thinking about DP but wasn't sure, especially since later cells can actually reduce the answer compared to previous cells. admittedly I am terrible at DP so I didn't think about it too much XD

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13 months ago, hide # |
 
Vote: I like it +31 Vote: I do not like it

Add a new tag named Reading.

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13 months ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

Bro what did I do in recent 3 hours :(

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13 months ago, hide # |
 
Vote: I like it +108 Vote: I do not like it

Wtf is 1C 1D?

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13 months ago, hide # |
 
Vote: I like it +26 Vote: I do not like it

if problem doesn't seem hard enough make problem statement harder to understand. simple :)

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

My solution for failed on 4th pretest..I am storing the frequency of every element and sorting them in ascending order then I am checking the frequencies of consecutive elements.. Am I missing any other cases here?

void solve(){
    int n;
    cin>>n;
    vector<int> a(n);
    for(int i=0;i<n;i++) cin>>a[i];

    map<int,int> mp;
    for(int i=0;i<n; i++){
        mp[a[i]] +=1;
    }

    vector<vector<int>> cnt;
    for(auto it: mp){
        cnt.push_back({it.first, it.second});
    }

    int m = cnt.size();

    for(int i=0;i<m;i++){
        if(cnt[i][1] > 3){
            cout<<"YES"<<endl;
            return;
        }

        if(cnt[i][1] >= 2){
            if(i+1 < m && cnt[i+1][0] == cnt[i][0]+1 && cnt[i+1][1] >=2){
                cout<<"YES"<<endl;
                return;
            }
            if(i+2<m && cnt[i+2][0] == cnt[i][0]+2 && cnt[i+2][1] >= 2){
                cout<<"YES"<<endl;
                return;
            }
        } 
    }

    cout<<"NO"<<endl;
}
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    13 months ago, hide # ^ |
    Rev. 2  
    Vote: I like it 0 Vote: I do not like it

    what about the case 1,1,2,3,4,5,6,7,7?

    means the frequency is like 2,1,1,1,1...,2!

    this should give yes!

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      13 months ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      How?

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        13 months ago, hide # ^ |
         
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        lets say

        B -> bus, J-> Jett test case: 1 1 2 3 4 5 5 for the freq 2,1,1,1,2

        lets say you pick BJ,JJ for the first two student. lets start the guessing from day 2. now if its wrong, it must mean the remaining possibilities are JB,BB. so the 2nd guess it must be B. so now for the next student where the freq = 1, you pick BJ now if its wrong then you know it was BB!

        now for the next student you pick BJ and if you are wrong it again means the last guess was incorrect so it was actually BB. or if pick BB and doesnt match with the student then actually it wasy BJ.

        now it goes on and on for the sequence of 1's in the frequency.

        now if there is 2 student after some 1's then you know for the last 2 students,

        you can choose the guess as BJ or BB if you know the first guess is B. it must match with one of them.

        so the solution is a sequence of 2-1-1-1-1...2 or 4 student one a single day. to cover all the 4 possibilities.

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          13 months ago, hide # ^ |
           
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          Can you tell me what will you exactly guess for 2 1 1 2?

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            13 months ago, hide # ^ |
            Rev. 2  
            Vote: I like it +1 Vote: I do not like it

            i suppose the testcase for this is

            1 1 2 3 4 4

            so 6 students.

            my guess for

            st1 -> BJ

            st2 -> JJ

            st3 -> BJ

            st4 -> BJ

            st5 -> BB

            st6 -> BJ

            so the guess at once seems like: if it dont match with st1 and st2 then for st3 it must match his first day. but his second day there is two possibilities. so i pick one. and if that doesnt match then for st4 the first day must match. again there is Two possibilities. but this time i've two guesses for a single day. say i pick both of them for different students. so there must be one win.

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13 months ago, hide # |
 
Vote: I like it +13 Vote: I do not like it

Worst Div2 B,C in recent times

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    13 months ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    why so negative... B not the most intuitive but can be solved in like 10 lines

    #include <bits/stdc++.h>
    using namespace std;
    
    
    int main(){
        int t, n, k;
        cin >> t;
        while(t--){
    
            cin >> n >> k;
            int b = n - k;
    
            vector<int> bars(n);
            for(int i = 0; i < n; i++){
                cin >> bars[i];
            }
            sort(bars.begin(), bars.end());
    
            int left_median = (b-1) / 2;
    
            cout << bars[n-1-left_median] - bars[left_median] + 1 << "\n";
        }
    }
    
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      13 months ago, hide # ^ |
       
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      can you provide the proof of this?, seeing the code will not help to understand.

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        13 months ago, hide # ^ |
         
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        My intuition is if you want your house to be central, you destroy the central neighbors...

        So the free land is the central/good location for ur house.

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13 months ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

Worst contest which I gave till now!

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13 months ago, hide # |
 
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In question C (Div 2) for input — 5 2 2 3 4 4

could some tell me what would be the output ?

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13 months ago, hide # |
 
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problem C was way easier than B. but it had many cases to consider, i cant believe i submitted 5 time.

got WA on i'th pretest on i'th submission for 1<=i<=4. and finally AC on 5'th. lmao

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    13 months ago, hide # ^ |
     
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    after taking my time on C.. I figure two kind of cases

    1. any day count >= 4
    2. any 2 days count > 1 .. but all days should be present between these two days.
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      13 months ago, hide # ^ |
       
      Vote: I like it +3 Vote: I do not like it

      exactly. but at first i thought there is only two possibilities:

      1. day count >= 4
      2. day count >= 2 && day+1 count >= 2

      i didnt know the second case could be expanded by day count == 1 in between...

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        13 months ago, hide # ^ |
         
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        I also did something similar and then submitted to see if I am correct but I was wrong ha ha ..

        then I figured out correct logic second time

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13 months ago, hide # |
 
Vote: I like it +18 Vote: I do not like it

What a bad description

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13 months ago, hide # |
Rev. 3  
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Insanely quick system testing

Also, does anyone think problem A is weird for only having 1 pretest

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13 months ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

This night contest and the last night contest were both so bad. Everytime I do one, I wish I just went back to sleep

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13 months ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Absolute bad problem description for B!! Please do something to fix this in future rounds.

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Why does sorting and calculating ranges based off medians if the all windows of size $$$n - k$$$ not work?

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13 months ago, hide # |
 
Vote: I like it +29 Vote: I do not like it

problem statements for both b and c were really vague, on top of all of that, 2d and 2e had a really large gap both difficulty-wise and knowledge-wise. talking to someone who solved 2d, I didn't even know some of the things(e.g. gaussian elimintaion). I had a solution for e using some magic with crt (i don't know if it even works), still geometry&crt for a 2e is too much in my opinion.

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13 months ago, hide # |
Rev. 2  
Vote: I like it +11 Vote: I do not like it

.

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13 months ago, hide # |
 
Vote: I like it +37 Vote: I do not like it

Do you really like geometry and linear algebra?

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13 months ago, hide # |
Rev. 2  
Vote: I like it -28 Vote: I do not like it

please all downvote this contest

Edit:Task achieved

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13 months ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

how to solve D ?

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    13 months ago, hide # ^ |
     
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    Separate the squares into connected components, where two squares are in the same component if there is an overlap in possible paths. If x1 != x2 and y1 != y2, you can just connect (x1, y2) and (y1, x2). Otherwise, there is only one possible path (we will call these "lines").

    For a connected component, you need to fill in K edges and you will have either K + 1 candidates or K candidates. If you have K + 1 candidates, you multiply res by K + 1 (since any deviation would result in the rest being fixed, and there are K + 1 spots for deviations). If you have K candidates, you multiply the result by 2 unless the component contains a line. If you ever have less candidates than edges, you can just return 0 since it is impossible.

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

in c div 2 problem I covered cases when
frequency is 4 then yes
or 2 (zero or any number of ones) 2 and all of them should be consecutive
what cases I am missing ?

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13 months ago, hide # |
 
Vote: I like it +7 Vote: I do not like it

D2B must be the worst worded statement I've seen.

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

How to solve div2 D ?

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13 months ago, hide # |
 
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How to solve Div2 D?

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

For C, don't use collections.Counter, (it won't pass)

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    13 months ago, hide # ^ |
     
    Vote: I like it +1 Vote: I do not like it

    mine passed with Counter

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    13 months ago, hide # ^ |
     
    Vote: I like it +1 Vote: I do not like it

    I got FST-ed once with Counter, and lesson learned.

    Not sure about this contest. But I'd rather use something else just to be safe.

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      13 months ago, hide # ^ |
       
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      what is wrong with python..

      i got FSTed in the last div3 because i used sets apparently

      now apparently even counter is no good

      recursion dp never works

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        13 months ago, hide # ^ |
        Rev. 2  
        Vote: I like it 0 Vote: I do not like it

        As a Pythoner:

        • Stop use collections.Counter and set.
        • If you have to use map, use custom xor random.
        • Never use recursion with O(n) and above. I do sometimes with O(log n) but really have to be careful. (Try to break recursion, it IS a skill and really helps in long term too.)
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        13 months ago, hide # ^ |
         
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        soo, it's related to how hashmaps work in Python,

        interestingly

        this is TLE:
        c = Counter(arr)
        

        and this passes

        arr.sort()
        c = Counter(arr)
        

        there is something with hash collisions, need to investigate a bit more.

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13 months ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

I think there was a problem in the problem Baggage Claim in python3, I don't think there was a way to solve it fast enough, If any one nows how, please let me know

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13 months ago, hide # |
 
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I think the Problem statement of B was either Wrongl Translated or was Incorrect !

bCOZ I Solved it in a way -> maximum houses he Can Buy while he can remove at most K bars having minimum f(x)

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13 months ago, hide # |
 
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I tried to solve div 2-D. I used a 2d dp, with 4 columns, storing the possible combination for p1,p3,p5 ...etc for up, down,left,right. I mean p1[0] stores possible combinations of p2 when it is placed on top of p1.

For example, if P3[0] can exist, that mean p4 can be placed on top of p3, then dp[p3][0] = dp[p1][0]+dp[p1][1]+ ... [3] But then to remove the commons, we subtract all previous value of dp, that could be present there. I know it sounds complicated, But for example, we add all dp[p7] to dp[p9][0] that means top of p9, so now we find what other grids share boundry with the top block of p9. Suppose left of that is p3, so we subtract dp[p3][right] from dp[p9][0] and so on. I think my logic is right, but I spent 1 hours writing it's code , but did simple mistakes with x and y coordinates, and I rage quit. Is my logic correct?

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13 months ago, hide # |
 
Vote: I like it +60 Vote: I do not like it

this guy looks very happy

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13 months ago, hide # |
 
Vote: I like it +7 Vote: I do not like it

Too hard 2D/1B.

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13 months ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Hello all. Will you be able to help me with the reason of TLE for https://mirror.codeforces.com/contest/2098/submission/317312937. Thank you.

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13 months ago, hide # |
 
Vote: I like it -22 Vote: I do not like it

Despite popular belief, I think this is a great contest with beatiful questions! Also, I would gain good rating so by definition, it must be a good contest!

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13 months ago, hide # |
 
Vote: I like it +41 Vote: I do not like it

Div2 B is the worst statement I have ever seen. We are supposed to guess the solution, not the problem.

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13 months ago, hide # |
 
Vote: I like it +37 Vote: I do not like it

The statement of div2 B is very poorly written and way too ambiguous on what we should take minimum over. I think this is caused by $$$f(x)$$$ implicitly depending on the selection of open bars.

I think the statement should have been written like this:

Let's call the indices of open bars $$$S \subset \{ 1, 2, \ldots, n \}$$$, such that $$$|S| \ge n - k$$$. Then define $$$f(S, x)$$$ as follows:

$$$\displaystyle f(S, x) = \sum_{i \in S} | x - a_i |. $$$

Find the number of $$$x$$$ 's ($$$1 \le x \le 10^9$$$) that satisfy

$$$\displaystyle f(S, x) = \min_{1 \le y \le 10^9} f(S, y) $$$

for some possible $$$S$$$.

Note: The original statement might mislead people into counting $$$x$$$ 's that satisfy

$$$\displaystyle \min_{S} f(S, x) = \min_{y, \ S} f(S, y), $$$

which is different from what's originally intended.

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13 months ago, hide # |
 
Vote: I like it +2 Vote: I do not like it

I think the concept of problem B and C was pretty good. But the author Couldn't make the statement clear and also the worst test case explanations ever. If the explanation were good then it could be better to understand the problems. __

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13 months ago, hide # |
Rev. 2  
Vote: I like it -32 Vote: I do not like it

I think this is the first contest announcement to ever have more than 250 downvotes. And the number is still skyrocketing.

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13 months ago, hide # |
Rev. 2  
Vote: I like it +3 Vote: I do not like it

is it unrated? if not how much more time it would take to update ratings?

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13 months ago, hide # |
 
Vote: I like it +14 Vote: I do not like it

Honestly I don't get it. Why people are downvoting announcement and editorial this much? Is it really just because of div2B statement?

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13 months ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

For people like me Div2 B is itself medium to hard difficulty and you guys make it more difficult by making such problem statement which is hard to comprehend. It should be atleast easy to understand what the problem is saying right?

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13 months ago, hide # |
 
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When you realize they have another round next week with same set of authors

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13 months ago, hide # |
 
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I personally enjoyed the problems even though I could only solve $$$1$$$ and D1B has a really cool solution. I'm unsure with how bad the statements are with D2B because it wasn't in my set. Upvoted!

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13 months ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

I love how thought is the main player in 2B & 2C. And knowing from the score distribution that I wouldn't get past C, after the first hour I stopped and went out with my friends. Still the highest rank I've ever got in a Div. 2. So, upvoted =))

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13 months ago, hide # |
Rev. 3  
Vote: I like it +76 Vote: I do not like it

I think the statements were really hard to understand. I do hope that the preparation is improved. I comment on Div.1 problems:

A: The story is a bit hard to imagine. The chronological order (bet vs information) is not written. It is not written that Vadim can decide the students' bet.

C: I think two irrelevant paragraphs are too much.

F: I cannot think of a worse choice than lost buggages to represent flows.... The story is too strange to tell "found" means disappear. You should not create a flow problem and then try to hide by writing a random story. Also, in the Input section, "the minimum number of lost pieces of luggage that will be found on the $$$i$$$-th day in each airport." is simply wrong. For future readers: it means there is a edge from $$$(i,j)$$$ to $$$(i+1,j)$$$ with capacity $$$b_{i,j}$$$.

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13 months ago, hide # |
 
Vote: I like it +54 Vote: I do not like it

I cannot understand what div2B is asking at all. Glad the round was not combined

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    13 months ago, hide # ^ |
    Rev. 8  
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    The problem's saying.. You can buy a house(let's say house numbered 'm') if you chose any number x, where x <= k, and close x bars, and then calculate the distance from your house to all remaining(opened) bars, will the combined distance be minimum among all the existing houses between 1st and 10^9th house? If it's minimum, then the m'th house is valid. Now we need to count the number of valid houses.

    Lets take the 3rd test case for example.. arr = [6,7,9] and k = 1.

    (.) Now, if we don't close any bars, then 7'th house is valid.
    (.) If we close the 6th bar, then house number 7,8,9 are valid.
    (.) If we close the 7th bar, then house number 6,7,8,9 are valid.
    (.) If we close the 9th bar, then house number 6 and 7 are valid.

    we can't close more than 1 bar at once(since k = 1) Hence, we can say that there are "4" valid houses, house no. 6,7,8 and 9.

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13 months ago, hide # |
 
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Thanks for the contest. The problems were really new and educational to me :)

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13 months ago, hide # |
 
Vote: I like it +16 Vote: I do not like it

Why their contest of 3rd may got cancelled? Do anyone know?

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13 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Why today's round was canceled? I gave up my trip plan for this!