Note that the time of abc and Round 1113 doesn't conflict.

I think so. I chose Unrated register.

WTF, I lost rank but gained rating.

for D: https://pastebin.com/eiPZMTX3

you sent this the moment the contest ended huh

vwxyz0 cn449

I was thinking I need 2 loops 1. {x,minn} 2. {x,maxx}

where minn and maxx are from last M'th round.

but then for these pairs i need to know if the pair is getting partially overlapped with all M pairs. (this I couldn't think of how to do fast) with lesser complexity

Let $$$f_1[i]$$$ denote the total number of finals played by player $$$i$$$, which is initialized by counting the occurrences of each player in the input pairs. For a pair of players $$$(x, y)$$$, the number of distinct finals covered by them is given by:

where common(x,y) represents the number of times players x and y have played together in a final. A pair of players is valid if and only if they cover at least m distinct finals:

where m is the total number of finals.

Approach

We process players in increasing order of their IDs from $1$ to $$$n$$$, maintaining an ordered set (PBDS) of pairs (f2[player], player) for all previously processed players.

For each player $$$y$$$:

1. Adjust for shared finals: If player $$$y$$$ has played a final together with player $$$x$$$ (where $$$x \gt y$$$, since we store the larger index as the first component and push $$$y$$$ to $$$G[x]$$$, or vice-versa based on if (x < y) swap(x, y)), their combined count needs to subtract the shared finals. Since we maintain $$$f_2$$$ arrays, we temporarily decrement $$$f_2[x]$$$ in the ordered set for all $$$x \in G[y]$$$.

2. Count valid pairs: We then query the ordered set to find how many previously processed players satisfy the condition:

using:

int C = ((int)(os.size())) - os.order_of_key({m - f1[y], -1});

Binary search on the answer.

If you are checking whether v is able,you can set $$$o=1$$$ and $$$x=\frac{-v}{1-v}$$$,then check there's a non-negative subarray with k 'o's or not.

maybe try ssize()?