Are the pretests strong?

Don't forget to notice the unusual start time!!!

I had been FST or hacked five times including just now When my predict score achieve 2000.That's true there were some bugs in my program, but I hope problem writer can make test data stronger, It's not funny When I suffer a disastrous decline

The start time is very kind to Chinese participants.Thank you! Orz

How will we be able to solve a 5hour contest in just 2 hours as it is a Camp contest ???

Codeforces is Best site ever made. I used it like 10 years starting from childhood. Thanks for all creators of this fantastic Website and to everyone who is reading this now !!Happy New Year!!I Wish all of you to solve problems (lvl higher than 3000) and reach Nutella this coming year :)

Раунд от создателей тик-тока пропускать нельзя.

КАК ЖЕ Я ЛЮЛБЛЮ ТИКТОК, ПЕЙТОНА САТОРИАСА И ОСТАЛЬНЫХ КОНЕЧНО Я БУДУ ПИСАТЬ УРА

Auto comment: topic has been updated by Endagorion (previous revision, new revision, compare).

Please make Christmas problems!

On the site https://programcamp.bytedance.com it's written that registration for the online contest ended yesterday. Is it possible to extend the registration process?

Because I think that many teams (including my team) learned about this contest from the announcement.

Look's like codeforces got rid of the mask already

For those of you who didn't notice it already, just hover over the Christmas lights near the codeforces logo :P.

Codeforces got in a festive mode. Nice to see this after many days of "Make CodeForces, not CoronaForces" and the mask in the logo.

What is the score distribution of the problems?

Unfortunately, I won't be able to use dark mode on codeforces for entire winters

What about the score distribution?

This competition is friendly for Chinese competitor! Thanks Endagorion!

As a tester, I recommend participating.

Happy to see, Codeforces is bringing up some different contest rather than the normal ones. Thanks to Endagorion

Seems like Endagorion missed to thank Mike for the platform :(

Btw Can you please change Good Bye 2020 -> Finally Good Bye 2020!!!? Given 2020 was one of the most difficult and sad phase world saw in past many years, I think many of us are waiting for 2020 to end and hoping to have a positive 2021.

Auto comment: topic has been updated by Endagorion (previous revision, new revision, compare).

Glad to see jqdai0815 appear again ^v^

ByteDance!! nb!!

wtf 2min 2problems?

Tourist about to break his own record!

Please don't discuss the problems until 3pm MSK (25 minutes from now)!

problem C wrong answer on pretest 5 :(

Looking at the scoreboard, you could've given us 30 more minutes instead of telling us to not to discuss for 30 minutes :(

just to inform you guys , there's Atcoder ABC 20 minutes from now

Div2 is the new Div1.

Can the problem setters spell BORING?

Only ~8000 participated when ~16000 have registered. Lol

glad to know I wasnt the only one to find it extraordinarily tough

A very nice contest!

Just a little hard XD

"If you just participated in ByteDance 2021 Online Qualification, you should not take part in this round! "

I hope they didn't

It's past 12:00 UTC now, please open the submissions and others' solutions now.

I've always liked Endagorion's style of problem setting. And I think that this was a great contest. But, I think the jump from C to D in division 2 was a bit too much or it would not have been if the contest was half an hour longer. Nevertheless, the problems were really great. Kudos to all the setters.

Anyone solved problem B using this? OEIS A241496

Is this going to be unrated?

Also: Any idea for div2 B?

For me, E was much, much easier than D, because the approach seems pretty straightforward, and D is some kind of ad-hoc which I believe is possible not to come up with at all.

Is it true that in D1D the invariant is — the number of zeros, ones, and a multiset of the balance array?

Do I need to know some property about inverse of a permutation in order to solve div1 C? If so,what is it?

for problem div2 B i search on oeis 1,4,4,12, from there i got the pattern.

69th place, nice!

Problems in div1 were fairly interesting, I'd say. My (and probably intended) solution for C is to store pairs (row, column, value), where "inverse" operations flip either "row" or "column" with "value" and the other four only increment/decrement "row" or "column".

Is Div2 D/Div1 B DP?

how to solve div 2 B plzzzzzz help???

I tried to solve problem Div2 D , glass half spilled using dynamic programming . Where at dp[i][j] is stored a pair {maximum water that can be stored in j glasses if k==j,maximum water that is left corresponding to first value in pair}.

Could some one tell what is wrong in this approach .

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
 
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
 
//cout<<"Case #"<<c<<": "<<ans<<"\n" ;

const ll N = 101;

pair<ld,ld> dp[N][N],c[N];

ld s,v,temp;

int main()
{
    IOS;

    ll  i,j,n;

    cin>>n;

    for(i=1;i<=n;++i)
        cin>>c[i].first>>c[i].second;


    for(i=1;i<=n;++i)
    {
        s += c[i].second;

        for(j=0;j<=i;++j)
        {
            if(j==0)
                dp[i][j]={0,s};
            else
            {
                dp[i][j] = {dp[i-1][j].first,dp[i-1][j].second+c[i].second};

                v = dp[i-1][j-1].second;
                temp = min(v/2.0,c[i].first-c[i].second);

                if((dp[i-1][j-1].first+c[i].second+temp)>dp[i][j].first)
                {
                    dp[i][j].first = dp[i-1][j-1].first+c[i].second+temp;
                    dp[i][j].second = (v-2*temp);
                }
                else if((dp[i-1][j-1].first+c[i].second+temp)==dp[i][j].first)
                {
                    if((v-2*temp)>dp[i][j].second)
                        dp[i][j].second = v-2*temp;
                }
            }
        }
    }

    for(i=1;i<=n;++i)
    {
        cout<<setprecision(11)<<fixed<<dp[n][i].first<<" ";
    }


    return 0;
}

UPD : I understood the mistake . I was not keeping track of space left which can be used to fill more.

Why do we have to wait, to submit? Could you turn on the practice mode?

Why would you include 2 problems that can easily be googled like div2 B and C? I mean if people thinking first of trying to cheat search them fast, they will get way higher in standings than someone that really tried to solve them on their own.

Div1A — 712 solved Div1B — 551 solved

Div2C — 1701 solved Div2D — 104 solved

Really weird difference in gap.

Tourist is ready to cross his own highest rating record.

when will ratings update

HOPE TO BE SPECIALIST (fingers crossed)

Did anybody else bruteforce Div.2B like me 101739165

Hi . Good job ! When will the ratings change ?! Thank you for your great time i really liked it and i'm not joking.

Did not liked Div2 this time. To much math, to less programming.

Solved Div2C with help of second last comment of this article

How to reduce the running time of this solution for problem B? Also I have tried only one part of the coordinate and multiply with 4 but same result TLE.

Yay ! Today expecting to become specialist for the first time :)

Is div 2 problem C similar to another problem?

It's so sad that C can be found here out some years ago :(

When will the changes in rating will be reflected ? Nothing specified in the announcement.

why the digits don't need to be sorted in problem A?

https://codeforces.me/contest/1459/submission/101721508

https://codeforces.me/contest/1459/submission/101718794

It is possible that I made amazingly silly mistake but for now I can't understand why did my Submission_TLED get a TLE ?

Is it still rated now?

I know DIV2 A was very easy just we to find for how many cars upper no is greater and same for lower written no's cards. but my code is ruturing something different even something wired here is my code

include<bits/stdc++.h>

using namespace std; int main( ) { #ifndef ONLINE_JUDGE freopen("input.text", "r", stdin); freopen("output.text", "w", stdout); #endif int T; cin>>T; while(T--){ int n; cin>>n; int a[n]; int b[n]; int r = 0 , c = 0; for ( int i = 0; i<n; i++){ cin>>a[i]; } for ( int j= 0; j<n ; j++){ cin>>b[j]; } for ( int k = 0; k<=n; k++){ if ( a[k] > b[k]){ r++; } else if ( a[k] < b[k]){ c++; } } if ( r > c){ cout<<"RED";

    }
    else if ( r < c){
       cout<<"BLUE";

    }
    else 
       cout<<"EQUAL";
    cout<<endl;
    }
}

Does anyone know how to solve problem-c

Can someone please tell me sources to learn fast googling techniques . It took me too much time to think for B and was unable to think C , Which would have been a matter of Minutes if I would have knew good googling techniques

Submissions not getting judged at the moment. Why?

How to solve Div2D ?

https://www.hackerearth.com/practice/data-structures/advanced-data-structures/suffix-trees/practice-problems/algorithm/employee-performance-1/description/

Problem C of today's contest can easily be solved using this already available problem

I have a talent at not being able to solve questions

2C is very similar to this problem G from ccpc Guilin site ccpc

Auto comment: topic has been updated by Endagorion (previous revision, new revision, compare).