I think so lol

Which new judge are you talking about?

D is a classic problem.......

Rubbish ABC

may i can help u after the contest xD

Use PBDS along with find_by_order().

You can use 2 stacks to maintain the median in a stream, similar to the Leetcode problem...

I think this problem has appeared in leetcode contest maybe. Anyways, you can solve this using two priority queues, maintain two priority queues, one for the smaller numbers than current median, and one for the larger numbers than current numbers. Let's say current median is 5, and we now add two numbers 2 and 3, now since both are less than 5, so we add them in the left pq, now to find the median, we take out the largest value from the left pq and throw it into the right one until the size of left is just 1 more than the right one, there the largest among the left one is median, similarily if right one has more elements, then we can throw smallest from the right to left until the size is fine.

My solution: https://pastebin.com/vVjwfwpF

This was my solution using the multiset

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

#define int long long

//For msb -> __lg(n)

/*
Problem is Easy

*/

/*
small observations--

*/

void solve() {
    int x,q;
    cin >> x >> q;

    multiset<int> a,b;



    while(q--){
        int A,B;
        cin >> A >> B;

        int e0 = min(A,B),e1 = max(A,B);

        if(x <= e1 && x >= e0){
            a.insert(e0);
            b.insert(e1);
        }else if(x < e0){
            a.insert(x);
            b.insert(e0);
            b.insert(e1);
            x = *b.begin();
            b.erase(b.begin());
        }else if(x > e1){
            a.insert(e0);
            a.insert(e1);
            b.insert(x);
            auto it = a.end();
            it--;
            x = *it;
            a.erase(it);
        }

        cout<<x<<'\n';
    }

}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}

true

Oh, Atcoder original proBlem Contest!

Well, it is possibly impossible to count the amount of origins of F... The are too many of them

OOOOOOOMMMMMMMMMMMGGGGGGGGGGGG! I'm Chinese, had an account on Luogu, but I'm sooooo foolish that I didn't even notice that! I wanna swear (but I can't). P.S. (To all non-Chinese readers) Luogu is in Chinese. If you don't understand the problems ask me. And writing blogs on it is forbidden unless you register your full name and scan the QR code. But there are some good problems in it (including CF ones). You can try to find it by searching https://www.luogu.com.cn/problem/CF(round id)(problem e.g. A~f).

Sadly, it is easier, atleast A-D. C was observational and D is standard

I think its almost identical to this G — Banned Substrings https://atcoder.jp/contests/abc305/tasks/abc305_g like damn I have the gist that it gonna be making dp based on automation transition of valid and invalid state then turn that dp into matrix to solve for large n and something, yeah sadly I did not solve it... I don't remember/understand much about ahocorasick automation sufficiently enough to use it in correct ways so I'm having mental breakdance mid compettition trying to remember how the hell I did solve that Banned Substrings before but damn... no luck. 🥀🥀🥀

You need to know how to get number of different solutions to a equation of form x1 + x2 +...xn = r, after that the question becomes pretty simple.

Hints:

Purely because Atcoder doesn't have a blog-writing function :(