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

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

Dahm the post is so wholesome

One of the most beautiful announcement ever!

emojiforces

What is NP complete problems

I also love penguins UwU.

orz penguin round !!! 

orz penguin round!!! 

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!

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

this post so cutee XD

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

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

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

Please upvote, thanks.

Hoping to improve Problem solving skills.

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..

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://mirror.codeforces.com/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!!

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 ?