Excited to see the problems! Misuki Kaey orz!

As a tester, I forgot the number of interactive problems there :)

Misuki orz!!

Interactive problem ❤

As a tester, I can confirm that the round is supercalifragilisticexpialidocious, too.

I came here from the author's X post

As a tester, I tested the round.

As a tester, I ate 2 bags of chips while testing.

Misuki orz

interactive problems..cool

Hope to reach -110 after this.

Who told them to say 1 or more interactive instead of 0 or more :)

Let's do our best

As a tester, I can confirm that the round is supercalifragilisticexpialidocious, too.

Good luck to everyone, and orz to Misuki!

Hope to reach 1300+ rating and solve ABC!

As a tester, I recommend solving AtCoder problems if you want to be as good as Misuki!

What does the score distribution mean?

As a tester.. i hope everyone get positive delta..

hope for new color

and hope for Salah7_a

Hope back to blue man :)

After getting down by the allegations of CHEATING, still standing to become RED

As a tester, i did nothing

I think this round will be great.

Hopefully, the color of my username will change

ill solve ABCDEF , ggez

As a participator, Thankyou for Interactive problem ❤️

As a participant all the best to all my fellow coders

.

I want to get +2500 in this contest

there is at least one interactive problem

it's so over

hope cadidate master today:3

Why did a game a long time ago change from scoring to not scoring, and the questions passed were also displayed as skipp

Why did a game a long time ago change from scoring to not scoring, and the questions passed were also displayed as skipp

Thanks to everyone who make this contest happen! Let's go!

ready for another opportunity to come back to cyan :)

Interactive Problems after a long time...

Hope to reach Specialist after this contest.

Misuki so strong!!!

Giving a contest after almost 2 months, hope I don't get destroyed..

Misuki

I really want to participate in it,but it's a pity that I don't have time...

I hope to solve A,B,C

so orzful for this contest <3

interactive problem will have two subtasks most likely

Taiwanese? I had to participate!

With a heavy heart and determined spirit, I'm back for higher +ve deltas

It can be a big challenge for my to solve interactive problems. But I have to try my best because I don't want to find my Rating going down again after the competition. It's really depressing!

Hope to slove A,B,C and D.

wish I could become 1200 rating after this contest!

Hope to accept A, B and C

As a tester, I did not test the problems.

Wish me to boost my rating in this round. And Good Luck Friends!

Going to participate after almost a year, let's see what happens.

Couldn't join the contest for 15 minutes. Am I the only one experiencing this?

Why is C of trees I am gonna be Pupil again

Poorly written B.

Absolutely LOVED C! Took me a lot of time, but that was SATISFYINGGGG

is it just me who thinks they should have mentioned the word cirular array in A to make things clear. I was breaking my head thinking how 8 8 7 6 3 8 7 6 3

can be convert to equal array

Could only solve one.

Got humbled by C so hard, speedran A+B in 10 mins feeling positive, just to get completely stuck on C with -7 at 15 mins left and no clue how to proceed

Stuck on WA 3 for D, can anyone share some small/tricky testcases?

finally changed colors!

Worst contest ever

problem C is cute problem I think, didn't realize binary_search can be used in such way until very last minute.

but I still cannot believe how 5000 people come up with solution in time. like how this problem have rating difficulty of 1500?

knew i was cooked as soon as i saw A

and i was, time to return to pupil ! yayy

need 2 more minutes for D :(

I am so dumb that 17k solved B and i didn't even understand the question until now :(
Someone please explain the problem statement.

C was fun. First time trying out an interactive problem.

It took me more time to understand problem B than it took to actually solve it

Huge praise on C. I'm glad that I didn't have to do the binary search myself today (and pushed the heavy job to the interactor).

Once again, D got leaked on a certain "YouTube channel" (as expected), with about 1k views in just one hour before the contest finished :)

The statement of B is very bad. Lots of cheating again (near 5k official solved C).

In Problem B, if there are 3 different operations, and you ask us to minimize just one of them, at least highlight it please?

My implementation skill is getting worse and worse.

Hints for D. Longest Max Min Subsequence

Consider the first index where an element appears for the last time. Since this is your last chance to take this element, you have to construct the best sequence that you can using the elements to the left. Hence, greedily take big and small elements to the left (while ensuring that as soon as you take an element, you invalidate all its left neighbors). Then, if you have taken an element, you can say that this element will never contribute to the last occurrence anymore.

Hence, you greedily take elements at each last occurrence of alive elements.

ayy just look at rank 1166 , tanish-jindal his D solution hahahah, just blatant cheating with stupid variable names

Watch out for these leaked solutions

#include <bits/stdc++.h>

using namespace std;

#define MAXN 1000
int q(int a, int b) {
    cout << "? " << a << " " << b << endl;
    fflush(stdout);
    int r;
    cin >> r;
    return r;
}

void function(int a, int b, vector<int>& vis, vector<pair<int, int>>& edges) {
    int x = q(a, b);
    if (x == a || x == b) {
        edges.push_back({a, b});
        vis[x] = 1;
        vis[b] = 1;
        return;
    }
    if (vis[a] && vis[x]) {
        function(x, b, vis, edges);
    } else {
        function(a, x, vis, edges);
        function(x, b, vis, edges);
    }
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;

        if (n == 2) {
            cout << "! 1 2" << endl;
            fflush(stdout);
            continue;
        }

        vector<int> vis(n + 1, 0);
        vector<pair<int, int>> edges;

        for (int i = 1; i < n; ++i) {
            for (int j = i + 1; j <= n; ++j) {
                if (vis[j]) continue;
                function(i, j, vis, edges);
            }
        }
        cout << "!";
        for (const auto& edge : edges) {
            cout << " " << edge.first << " " << edge.second;
        }
        cout << endl;
        fflush(stdout);
    }
    return 0;
}

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

int a[300001], c[300001], p[300001];
bool v[300001];
vector<int> t[300001];

void solve() {
    int n, pMin=1e9;
    cin >> n;
    for(int i=1; i<=n; ++i) {
        cin >> a[i];
        p[i] = -1;
        c[i] = 0;
        v[i] = false;
        t[i].clear();
    }
    for(int i=1; i<=n; ++i) {
        p[a[i]] = i;
        t[a[i]].push_back(i);
    }
    for(int i=1; i<=n; ++i)
        if(p[i] != -1) {
            c[p[i]]++;
            pMin = min(pMin, p[i]);
        }
    vector<int> r;
    for(int i=1; i<=n;) {
        if(i == pMin) {
            r.push_back(a[i]);
            c[p[a[i]]]--;
            while(c[pMin] == 0 && pMin <= n) pMin++;
            i++;
        } else {
            if(r.size() % 2 == 0) {
                int m=0;
                for(int j=i; j<=min(pMin, n); ++j) if(!v[j]) m = max(m, a[j]);
                for(int j=i; j<=min(pMin, n); ++j) if(a[j] == m) {
                    i = j + 1; break;
                }
                if(m == 0) break;
                r.push_back(m);
                c[p[m]]--;
                while(c[pMin] == 0 && pMin <= n) pMin++;
            } else {
                int m=1e9;
                for(int j=i; j<=min(pMin, n); ++j) if(!v[j]) m = min(m, a[j]);
                for(int j=i; j<=min(pMin, n); ++j) if(a[j] == m) {
                    i = j + 1; break;
                }
                if(m == 1e9) break;
                r.push_back(m);
                c[p[m]]--;
                while(c[pMin] == 0 && pMin <= n) pMin++;
            }
        }
        for(int idx: t[r.back()]) v[idx] = true;
    }
    cout << r.size() << endl;
    for(int x: r) cout << x << ' ';
    cout << endl;
}

signed main() {
    int T = 1; cin >> T; while(T--) solve();
}

#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,k,p,mem[4][501][501];
int sum1(int n,int k);
int dfs1(int n,int k);
int sum2(int n,int k);
int dfs2(int n,int k);
int sum1(int n,int k) {
    if(k==-1) return 0;
    if(mem[0][n][k]!=-1) return mem[0][n][k];
    return mem[0][n][k]=(sum1(n,k-1)+dfs1(n,k))%p;
}
int dfs1(int n,int k) {
    if(n==1||k==0) return 1;
    if(mem[1][n][k]!=-1) return mem[1][n][k];
    int s=0;
    for(int i=0;i<=k;++i) {
        s=(s+dfs1(n-1,i)*sum2(n-1,min(k-i,i-1)))%p;
        if(i+1<=k-i) s=(s+dfs2(n-1,i)*(sum1(n-1,k-i)-sum1(n-1,i)+p))%p;
    }
    return mem[1][n][k]=s;
}
int sum2(int n,int k) {
    if(k==-1) return 0;
    if(mem[2][n][k]!=-1) return mem[2][n][k];
    return mem[2][n][k]=(sum2(n,k-1)+dfs2(n,k))%p;
}
int dfs2(int n,int k) {
    if(n==1||k==0) return 1;
    if(mem[3][n][k]!=-1) return mem[3][n][k];
    int s=0;
    for(int i=0;i<=k;++i)
        s=(s+dfs2(n-1,i)*sum2(n-1,k-i))%p;
    return mem[3][n][k]=s;
}
void solve() {
    cin >> n >> k >> p;
    for(int i=0;i<=n;++i)
        for(int j=0;j<=k;++j)
            for(int l=0;l<=3;++l) mem[l][i][j]=-1;
    cout << dfs1(n,k) << endl;
}
signed main() {
    cin.tie(0);cout.tie(0);
    ios_base::sync_with_stdio(0);
    int T=1;cin >> T;while(T--) solve();
}

Deleted

B is the stupidest problem I have seen so far. Does providing enough test cases to understand the problem cost money? I hope you guys never set a question like this again. This kind of contest ruins my whole excitement toward contests.

Edit: Nvm, just my stupidity.

Worst Problem E1 & E2 I have seen, can't believe there exists a problem worse than "Let Me F**k you a Lesson". Downvoted.

How do I approach C?