first comment of this contest! first day of spring season! first ... hmm what else? let's add something)

As a tester, problems are very interesting :)

As a tester, I can assure that you will enjoy the problems.

As a tester, I enjoyed testing this round a lot! :)

.

As a tester, I can ensure you that the problems are quite engaging. You will love solving them. Lots of learning on the way!!

Hoping for the best Div.2 round!

[deleted]

First contest on the spring!

I hope the tasks will be as short and clear as the announcement of this contest!

Good Luck for EVERYone!

OMG purple round

Interactive problems? I don't even know how to solve them, does anyone has resource which I can refer

Thanks a lot for authour and tester for very good and high quality of problemset(at least A,B and C)

score distribution???

sounds like a tough contest tho

Codforces Round 931 (Div. 2) i think there is mistake

IIT Guwahati orz <3

interforces

I hope it's better than last round

aryanc403 testing back to back contests

Back to back cf contests. I like it !

As a second time tester, this is definetely one of the top 2 rounds I ever tested.

jdurie tester orz

jdurie :heart_eyes:

Jdurie

Hope C is not interactive :D

Give us the score distribution,precious! (Read in Gollum's voice.)

I want to become blue wish me luck

contest on the first day after 29th February. Let's see!

What happened to the chain of pictures of authors?

hope no bitmasks(

Bring it on, It's time for My Redemption.

One and only author harsh__h :)

Wow. Two contest in a row and two positive delta.

Can I be a newbie tester for some future contest. I want try what it's like to be a tester.

How can we hack others in interactive problems?

Problem setters pic where

with one of the problems having 2 versions One or more interactive problems XD XD

indian setter = mathforces

it was very enjoyable contest

Waiting for problem setters pics

There is just one author and people are asking him to post his selfie onto the announcement which will be shown on the home page lol. (no offense)

Amazing maths problem (propply)

Looks like recent Div 2 demotions in the current contest are considered “out of contest” even if their rating is 2100 by the start of the contest. Any reason why?

https://mirror.codeforces.com/contestRegistrants/1934/page/2?order=BY_RATING_DESC

whats wrong with last 2 rounds?

awful contest

Wtf, why C is interactive again? Is it so hard to make up NOT interactive problem for Div2?

No early screencast from Um_nik? :-(

can we stop with the interactive C please, welcome back pupil

russian translation for D1 is a bit strange. In english text it says: "This is the solo version of the problem". But word 'solo' is also used in russian text "Это сольная версия задачи". At first I thought it was somehow 'salty' version of problem D (salty is used to describe contention, bitterness, anger, or an otherwise-foul attitude). I got confused for a second, I even changed the language of the text to doublecheck.

I think the better way to say it both in english and in russian was "This is non-interactive version of the problem" / "Это неинтерактивная версия задачи"

Such a beatiful contest , I decided to give up just like my crush.

not able to solve problem B I should quit CP

Author probably thought he was proposing these problems for a math Olympiad

see y'all in green :|

Why is the limit on operations 63 in problem D when there is a solution with 2 operations?

Problem C seems to essentially be a duplicate of a problem which I wrote in a previous DMOJ round, but with one fewer step.

deleted

why c is interactive

can someone please explain me problem B

There is no hack phase this time?

Amazing interactive problems; I don`t know how to start.

Did anyone else get this annoying error for Problem C wrong output format Unexpected end of file — int32 expected (test case 2)?

Also, I technically never submitted a problem and passed test case 1, so is my participation unrated?

the problems were nice although I could only solve A , B

spent whole contest time for problem C

A and B problems Screencast here: https://www.youtube.com/live/iSkTzLR84pQ?si=EbJ_9lnTspSnm6Pw

D was easier than B.

249172627 can anyone please tell me why Div2.C giving such error?

I know this is a skill issue on my part, but still — CF should really give a interactor / tester like GCJ used to. I presume me and many others just YOLO submit interactive problems without testing at this point since it is a pain to validate compared to normal problems.

So bad...

Can someone please say, why I'm getting rt here? 249171286

I had a terrible experience in this round as well as the previous, both C and D1 are true mind benders.

Signing up AB -- 15mins, wrong idea for C -- 15mins, trying D1 and getting 3 WAs -- 45mins, correct idea for C -- 25mins, work on D1 again -- 15mins

C is a good problem but very painful to implement. Took so much of time for me.
Good round overall, thanks harsh__h and all testers!

What DO you Do in Interactive Problems???? Can someone pls explain
I followed the instructions but still getting Idleness Limit Exceeded
My Code

Thanks!

MathForces

-100 lets go

also what happened with my C

edit: lines intersect outside of boundary. Accepted now

Worst day ever

A: Sort a[i] and the answer is a[n]+a[n-1]-a[1]-a[2].

B: We can believe that for some n0, we have dp[n]=dp[n-15]+1 for all n>=n0, and we can solve the problem by brute force for n<n0. (n0=300 can pass)

C: First query for (1, 1), let answer be d0, then there will be a mine on the line L0: x+y=2+d0. Then we query for the 2 endpoints of the intersection of L0 and the grid, let these points be A1,A2 and answer be d1,d2, then the circle (A1, d1) and (A2, d2) will have exactly 1 intersection with L0 in the grid, one of them will be the answer.

D1: If n is power of 2 we cannot do any operation. Otherwise, let r=floor(log2(n)). If 2**r<=m, then (m^n)<n, we can solve the problem in 1 operation. Otherwise, let t1=floor(log2(n-2**r)), t2=floor(log2(m)). If t2<t1, we can solve in 2 operation: n --> 2**r+m --> m. If t2==t1, we also have (m^n)<n and solve in 1 operation. If t2>t1, there's no solution.

D2: Let ppc(n) be the count of '1' in the binary representation of n. Then if ppc(n) is even, we always can do some operation to turn it into n1, n2 such that ppc(n1), ppc(n2) are odd. If ppc(n) is odd, for any n1, n2 such that (n1^n2)==n, one of ppc(n1) and ppc(n2) will be even. So If the initial ppc(n) is even we play first, otherwise we play second, and we can always let ppc(n) to be even on our turn. To win in 63 turns, we can let n1=2**r, n2=n-2**r, if the opponent choose n1 they will lose immediately, if they choose n2 the value of n will be halved.

I hate myself. Due to a series of stupid oversights I spent 35 minutes not seeing a single missing +1 in E, thinking that I verified all possible tests locally

Yes, it's interesting

Thank you, author, for a great contest! Both problem B and C were really cool! C is one of the best problems I solved so far. Sad i didn't have enough time(

I enjoyed thinking about problems D2 and E.

Thanks for the contest!

Amazing contest and Again amazing problem C. I think every contest need an interactive Problem.

What is the error in this code?

#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define double long double
#define endl '\n'
#define PI 3.1415926535897932384626433832795l
const int MAX_N = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const double EPS = 1e-9;
void solve()
{
    int n, m;
    cin >> n >> m;
    cout << "? 1 1" << endl;
    cout.flush();
    int d1;
    cin >> d1;
    if (d1 == 0)
    {
        cout << "! 1 1" << endl;
        cout.flush();
        return;
    }
    cout << "? 1 " << m << endl;
    cout.flush();
    int d2;
    cin >> d2;
    if (d2 == 0)
    {
        cout << "! 1 " << m << endl;
        cout.flush();
        return;
    }
    cout << "? " << n << " " << 1 << endl;
    cout.flush();
    int d3;
    cin >> d3;
    if (d3 == 0)
    {
        cout << "! " << n << " 1" << endl;
        cout.flush();
        return;
    }

    int x1 = ((d1 + d2 - n) + 1) / 2;
    int y1 = d1 - x1;
    int x2 = ((d1 + d3 - m) + 1) / 2;
    int y2 = d1 - x2;
    if (x1 + 1 > 0 && y1 + 1 > 0)
    {
        cout << "? " << y1 + 1 << " " << x1 + 1 << endl;
        cout.flush();
        int d4;
        cin >> d4;
        if (d4 == 0)
        {
            cout << "! " << y1 + 1 << " " << x1 + 1 << endl;
        }
        else
        {
            cout << "! " << y2 + 1 << " " << x2 + 1 << endl;
        }
        cout.flush();
    }
    else
    {
        cout << "? " << y2 + 1 << " " << x2 + 1 << endl;
        cout.flush();
        int d4;
        cin >> d4;
        if (d4 == 0)
        {
            cout << "! " << y2 + 1 << " " << x2 + 1 << endl;
        }
        else
        {
            cout << "! " << y1 + 1 << " " << x1 + 1 << endl;
        }
        cout.flush();
    }
}
int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

The code seems correct for C question, please help me find bug

B and C were very appropriate difficulty-wise. Good job.

Amazing problems and huge thanks to the author! Problems involving binary representations have always been one of my most feared type, and D1 and D2 really teached me a lot!

interactive forces ftw

Why am I getting Idleness Limit Exceeded in this Code?

fvck interactive

In problem D1 I tried to derive what those operations mean.

Let number $$$n$$$ has at least $$$2$$$ digits $$$1$$$. Then we can take prefix of these digits $$$1$$$, and either change them all to $$$0$$$ and the last to $$$1$$$, or change them all to $$$1$$$ and the last to $$$0$$$. All digits after the chosen prefix can be changed to any other without restrictions.

1000100010101010101
|   |   |
0   0   1
 000 000 [anything]

But it turned out, that the intended solution is "we can see that..."