I tried to avoid the clash with Opencups but I didn't want ARCs to be too late in Japan, so that's the reason for the tight schedule. Though I can't change the contest time now, perhaps I'll have a margin of 5 mins next time, if not 15mins.

This is Atcoder and problem A so the solution would most likely be greedy.

May you provide which line contains your objective function? (or something related to that)

here is my solution of Prob B, it is different like my old binary search before $$$O(n\log n)$$$.

In code,my thoughts are finding the non-decreasing subsequence,but in fact I am looking for strictly ascending subsequence. and then is pass.

This maybe help you

I am wondering what would be his real rating if he do all the problems that he can in each contest!

Nice idea.

Can't A be solved in $$$\mathcal O(1)$$$?

It was also pretty classic.

I can explain a different solution that works in $$$O(2^KNK)$$$. Let us call the sequence $$$1,2,\cdots, K$$$ a "good sequence". If I choose $$$K$$$ different numbers at indices $$$x_1 < x_2 < \cdots < x_K$$$, I want to create a good sequence starting at $$$i$$$. Then, the optimal way of doing this can be thought of as performing two steps:

1. "Compressing" the indices $$$x_1, x_2, \cdots, x_K$$$ into the indices $$$i,i+1,\cdots, i+K-1$$$ by the means of adjacent swaps, while preserving the order. So if I start with $$$2,*,*,1,*$$$ and I want to make the good sequence start at $$$i=2$$$, I would make the sequence $$$*,2,1,*,*$$$. The cost of doing this is clearly $$$\sum_{k=1}^K |x_k - (i+k-1)|$$$.
2. Once I do this, I need to sort the sequence. The minimum number of adjacent swaps to do this is the inversion count, i.e. $$$inv(a_{x_1},a_{x_2}, \cdots, a_{x_k})$$$.

This gives us the same cost from the editorial. My goal is to find the minimum cost of having a good sequence starting at any $$$i \in [1,n]$$$. To do this, I want to use DP, but the roadblock here is that I'd need to consider numbers coming from both the left and the right of $$$i$$$. Instead, I'll split this into two parts — the prefix and the suffix. Let $$$dp_{pr}[i][S]$$$ be the cost of getting a contiguous sorted sequence of numbers in $$$S$$$, such that all the numbers are from indices $$$\leq i$$$, ending at $$$i$$$. Similiarly, let $$$dp_{su}[i][S]$$$ be the cost of getting a contiguous sorted sequence of numbers in the set $$$S$$$, such that all the numbers are from indices $$$\geq i$$$, starting at $$$i$$$. Then, we can get the cost of a good sequence starting at $$$i-|S|$$$ from the formula:

where $$$[k]$$$ is the set of the first $$$k$$$ natural numbers. $$$dp_{pr}[i][S]$$$ will cover the cost of "compressing" the prefix as well the inversion count within the prefix, $$$dp_{su}[i][[k] \setminus S]$$$ will cover the cost of doing this for the suffix, and $$$inv(S,[k] \setminus S)$$$ will cover the inversion count across the prefix and suffix. Hence, the minimal cost would be the minimum of $$$f(i,S)$$$ over all $$$i$$$ and $$$S \subseteq [k]$$$.

We can find the DPs in $$$O(2^KNK)$$$ and $$$inv$$$ in $$$O(1)$$$ after $$$O(K2^K)$$$ precalculation. The transitions are fairly trivial, you can see them in my submission.

If the first element of two pairs are equal, we can choose at most one of them. So we have to sort them somehow that doesnt conflict with the LIS part. Now the best way is to let the one with larger second element, come first