You can do anything, just don't give up!

I didn't solve the problem C

Not able get good rank again

oh c'mon, you are a lgm. Just solve all :)

not mathforces, just constructiveforces

I should find a matching green one aswell, given the amount of times I have dropped right after reaching cyan T-T

Pretty straightforward after trying to solve for a 1d array instead of matrix

There are $$$n^2-1$$$ possible differences $$$(1\dots n^2-1)$$$. And the chain $$$1, n^2, 2, n^2-1, \dots, \left\lfloor\dfrac{n^2}{2}\right\rfloor, \left\lfloor\dfrac{n^2}{2}\right\rfloor+1$$$ (chain ending correct modulo parity), contains all differences and has $$$n^2$$$ terms, for $$$n^2-1$$$ differences.

I code brute force as the max difference can be (n*n-1). Try to create all differences from (n*n-1) to 1.

I wasted time thinking that question is asking for only border elements.

Why do you think that is a silly thing to do? It trains observational skills.

Regardless, its not like patterns in grids of numbers are a meaningless or trivial thing to study. There are a ton of problems that can be represented as patterns in grids of numbers.. Doing regression in data analysis for example.

how did you solved B?

See this testcase:

1
3 5
2 3 5

I think C was a beautiful problem , though B was little annoying to me

Most of the good contests I participated in had at least one of them

some pattern based try for n = 4 it will be like

1->16->2->15

13->4->14->3

5->12->6->11

9->8->10->7

The whole point of them is to learn (standard) algorithms, not to solve interesting tasks. And I think that $$$A$$$, $$$B$$$, $$$F$$$ and $$$G$$$ are great for that ( educational ) matter.

It has to be on Binary Search but this mf B wasted an hour and my mind went blank coz of that so couldn't even think properly about C

Key observation is because each player plays each other the other players will have 1, 2, 3, 4... wins... If you can beat the player with exactly 1 more win than you while still maintaining the optimal number of wins than your place is (n-wins+2), else it is n-wins+1

Try thinking about $$$dp_{i,s} := \textrm{the number of possible suffixes of the array starting at } i\textrm{ having } a_{i+1} \textrm{ equal to } s$$$.

The transition will be something like $$$dp_{i,s} = dp_{i+1,a_{i+1}+s} + dp_{i+1, a_{i+1}-s}$$$ depending on whether $$$s = 0$$$.

I too want to know this, I got an intuition but couldn't mold it in any kind of solution.

I did it using binary search on the number of people we rank at least as low as. So to beat $$$k$$$ people, I pick the $$$k$$$ weakest opponents (in terms of $$$a_i$$$), if their sum of $$$a_i$$$ is less than $$$m$$$, we can rank $$$n-k+1$$$, there's also another small detail but this is mostly the idea.

Yes, we can binary search answer. When we check mid, we know that we can't lost the n — mid + 1 opponent. Let t = n — mid + 1. So we do it greedily to beat as many opponent as possible. Now there are 2 case : If the number we can beat >= t, we always win because player t can win at most t games. Else(number we can beat = t — 1) if we can beat player t, we can win too because now t win t — 1 games. 188457382

Yes.You can see my submission.

Yes

[E] You don't need a segment tree. Just do range sum using prefix arrays and iterate over all multiples of k for every k.

Do I check whether I can prepare for the k+1th opponent in the sorted array or in the orignal array?

Likewise

I didn't solve B, but I liked it. It tests one's observation ($$$n^2-1$$$ upper bound) and simplification (solve it for 1d array first) skills.

That is because you didn't find a key idea of it,so your code was very long.If you found that $$$1 \sim n^2-1$$$ must be able to express,you could know that $$$n^2$$$ must be adjacent to $$$1$$$,easily we could expand it into an easy solution.

Have you ever solved tasks from mathematical olympiads?

Thats how you solve it, xD