Another Div2 !! YAY!

Школа ЦПМ — лучшая школа России!

As a tester, I can assure you that all tasks are interesting and worth trying to solve.

Also GL to all participants :)

As a school cpm student i recommend you to participate >.<

As a top -1 contributor and tester of this round, I recommend you to participate in this round!

In my opinion, the round is high-quality and has pleasant tasks.

Good luck to all participants!

As a tester I highly recommend you writing this round!

As a tester, i declare, that this contest is by far the best. Good luck everyone!

As a tester, I have enjoyed solving problems, and I hope the participants will enjoy it too!

GL HF!!! (◕ω◕)

don't trust these guys

finally a non-interactive round <3

yo

I am sure that this rounds problems will be quite engaging and very interesting . We will love solving them. Lots of learning on the way!!

.

As a tester, I think everyone should write this quality round

Hoping for the best Div.2 round!

Will the contest have interactive problems??

Hope Um_nik can solve problem C.

Why is this not on the home page yet? This post would definitely decorate the home page with all these cute emojis!

Hope to solve A and B in this round

Good Luck for every parcipicant!

Omg! So many emojis!

Dahm the post is so wholesome

One of the most beautiful announcement ever!

Hopefully i will back my Specialist rank

emojiforces

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What is NP complete problems

I also love penguins UwU.

orz penguin round !!! 

orz penguin round!!! 

i am not in danger skyler

i am the danger

Another Div2 !! YAY!

emojis;)

Hope to see myself cyan!!

Seems like SpeedForces looking at the score distribution

Пацаны, и вам удачи на Открытой олимпиаде по программированию!

Emojis spoiled a very good blog :(

as always hope for interesting and SHORT STATEMENT problems!

Excited for a thematic twist! With problems inspired by your school, this round promises to be both nostalgic and challenging.

As best pop singer in Russia, i recommend you to visit my concerts (and ofc participate in this round)

No interactive?

hope there won't be any bizarre questions.

this post so cutee XD

Надеюсь задач с реализацией будет больше чем математических

Why the announcement looks like copy pasta of instagram's comment section

As a participant, I hope to get expert back, but from a div $$$2$$$ this time.

bro used all his recently "used" emojis from his phone

Legends say you'll turn gay after this round :v

this guys are cool i know

This round's surely gonna be WILD!

need to see authors' faces to know if this contest is reputable

no interactive?

Can you make it a dating contest(blog has some lovey dovey emojis ) aviralarpan3301 any suggestions?

Please upvote, thanks.

Don't use emojis again , please.

Hoping to improve Problem solving skills.

I hope to retrieve my rate :(

I hope to retrieve my rate :(

I wish the authors made us some easy C problems this time

I hope everyone gains some rating this round! (even if its impossible)

is it just me or have Div2 contests become harder ;-;

I wish the problem-setters specified that messages can be sent in any order in problem C.

It took me an hour and 3 WAs to realize that I was solving a much more difficult version of the problem.

C and D should be swapped

Problem c seems like binary search..

any hints?

D <<< C

Thanks!

C>D for me..

Please how to do B?

Can C be solved with greedy?

This may sound like an excuse, but I spent much time trying to login to facebook

Damn I got pranked hard by C xd

Can someone explain why n^2logn solution fails for C for the given submission? https://codeforces.me/contest/1935/submission/249797192

C > D

D is some high school mathematics and I worked it out within 10min after a lengthy, failed debugging session on C. Yes, I did read C and D at the same time. The reason why I didn't try D earlier was that I got an idea for C that works in O(n^2) but is very implementation-heavy, and eventually I did not get C.

In problem F, how do you create a valid set of edges in the $$$cost = (degree - 1)$$$ case?

The $$$cost = degree$$$ case seems much easier since for each [min_node, max_node] subtree range you can just sort the ranges in increasing order of min and then for each range except the first add an edge from [min_node — 1, min_node] ([min_node — 2, min_node] for the edge over $$$v$$$) to produce a valid tree.

Problem C was harder than D. Spent most of the time in C.(T_T)

Super Fast Editorial!!

D >>> C （°0°）

In C, did we have to find total time needed to read all messages optimally and then remove messages which take maximum time until time is >= l? or is this approach incorrect?

I had trouble solving B even after getting the correct logic.

I thought the array can be divided in 2 parts always if there was an answer. I calculated the MEX of the whole array. Now I divided the array in two parts of size i(left array) and n-i(right array), where 1<i<n . I updated the MEX of the left and right array in each loop and if they were equal printed the indices. Here is my code, can someone help :

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

void solve(int tt)
{
    ll n,x;

    cin>>n;
    ll a[n];

    map<ll,ll>mp;

    for(ll i=0;i<n;i++)
    {
        cin>>a[i];
        mp[a[i]]++;
    }

    ll mex=n;
    for(ll i=0;i<n;i++)
    {
        if(mp.find(i)==mp.end())
        {
            mex=i;
            break;
        }
    }

    ll mex2=0;
    for(ll i=0;i<n;i++)
    {
        if(a[i]==mex2)
        {
            mex2++;
        }

        mp[a[i]]--;
        if(mp[a[i]]==0 && a[i]<mex)
        {
            mex=a[i];
        }

        if(mex==mex2)
        {
            cout<<2<<endl;
            cout<<"1 "<<i+1<<endl;
            cout<<i+2<<" "<<n<<endl;
            return;
        }
    }
    cout<<-1<<endl;
}

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int T=1;
    cin >> T;
    for(int tt=1; tt<=T; tt++)
    {
        solve(tt);
    }
    return 0;
}

nvm idk how to prove shit

Too hard for C. I get stuck for more than half an hour, then I just solved it using two segment trees.

Update: no need for segment trees, because when you calculate the range from i to j, you only need to take k-st minimum value, no requirement it must contain the ends. (Since if it does not contains the end, all of them will also appear in the smallest range)

I think you should have swapped D and C. Nevertheless, tasks were very interesting

I think latest trend is difficulty of D is less than C in div2. And System testing just after contest. :)

is Div2 start getting harder ?

Thanks for fast editoral!

Thanks for the contest, enjoyed all the problems.

Me after solving B:

Common in recent contests :

1. Fast editorial
2. Fast system testing
3. difficulty of D is less than C.

By By Expert

Как я должен был понять, что в задаче С порядок сообщений можно менять... Об этом ни слова не написано в условии

I think D is easier than C problem.

[Deleted]

задачи огонь, особенно понравился Центр Помощи Мамам

why is this round listed as unrated in my profile even though this post says rated? do i need a rating before this contest?

Terrible luck with C and didn't even care to read D which was apparently easier today somehow :( Good Bye positive delta.

strong pretests? lmao. my second question passed on the 5 pretests and i got wrong answer on test 6

very very strong pretests! thanku very much

Good round overall, but problem C was a bit harder than usual (both the idea and implementation)

C is harder than D, but the score of C is lower than D.

Maybe I will never become a Candidate Master. :(

Great contest! Thank you.

My only comments are

• It was not very clear that $$$p$$$ doesn't have to be increasing in problem C
• Problem D is easier than usual

C is harder than D, but the score of C is lower than D.

Maybe I will never become a Candidate Master. :(

It was a great contest. Problem D was a bit easier than C (even had more accepts) which was great because it teaches us to not get stuck on a problem and always remember to read the next ones. (and yes I made that mistake too) Thanks for the effort to make this great contest!

Can anyone please elaborate the solution to Problem:C by using Dynamic Programming ..

DP[i][j] — Minimum time required to choose any subsequences of Messages in [0...i] such that the corresponding length is 'j'..

At the end , the final answer is to find the Maximum length (LEN) for which there exists atleast one 'i' such that it satisfies the inequality : DP[i][LEN] <=l .