The optimal sequence of operations is very simple.

The optimal sequence of operations is adding $$$k-1$$$ 1-s into the set each time, at the same time decreasing $$$n$$$ by $$$k-1$$$. This implies that the answer is $$$\lceil \frac{n-1}{k-1}\rceil$$$.

"Most sequences" can be transformed into $$$[1]$$$. Conditions for a sequence to be un-transformable is stringent.

Find several simple substrings that make the string transformable.

We list some simple conditions for a string to be transformable:

• If 111 exists somewhere (as a substring) in the string, the string is always transformable.
• If 11 appears at least twice in the string, the string is always transformable.
• If the string both begins and ends with 1, it is always transformable.
• If the string begins or ends with 1 and 11 exists in the string, it is always transformable.

These can be found by simulating the operation for short strings on paper.

Contrarily, if a string does not meet any of the four items, it is always not transformable. This can be proved using induction (as an exercise).

The answer is a simple construction.

The maximum length for $$$2^k-1\ (k>1)$$$ is $$$k+1$$$.

TODO

Formulate the problem.

Some variables can't be large.

Suppose monster $$$i$$$ is killed in round $$$b_i$$$. Then, the total health decrement is the sum of $$$a_i\times b_i$$$. The "independent set" constraint means that for adjacent vertices, their $$$b$$$-s must be different.

Observation: $$$b_i$$$ does not exceed $$$\lfloor \log_2 n\rfloor+1$$$. In this problem, $$$b_i\le 19$$$ always holds for at least one optimal $$$a_i$$$.

By dp, we can find the answer in $$$O(n\log n)$$$ or $$$O(n\log^2 n)$$$, depending on whether you use prefix/suffix maximums to optimize the taking max part.

Bonus: Find a counterexample for $$$b_i\le 18$$$ when $$$n=300000$$$. (Pretest 2 is one case)

Formulate the problem on a cartesian tree.

Find a clever bruteforce. Calculate the time complexity carefully.

Build the cartesian tree of $$$a$$$. Let $$$lc_x,rc_x$$$ respectively be $$$x$$$'s left and right children.

Consider a vertex $$$x$$$. If we delete it, what would happen to the tree? Vertices that are on the outside of $$$x$$$'s subtree will not be affected. Vertices inside the subtree of $$$x$$$ will "merge". Actually, we can see that, only the right chain of $$$x$$$'s left child ($$$lc_x\to rc_{lc_x}\to rc_{rc_{lc_x}}\to \dots$$$) and the left chain of $$$x$$$'s right child will merge in a way like what we do in mergesort.

Now, if we merge the chains by bruteforce (use sorting or std::merge), the time complexity is $$$O(n)$$$! It's easy to see that each vertex will only be considered $$$O(1)$$$ times.

We can use dp to calculate the number of permutations of $$$1\sim n$$$ with $$$i$$$ prefix maximums and $$$j$$$ ascents, $$$f(n,i,j)$$$: consider where 1 is inserted, we will have a $$$O(n^3)$$$ dp that finds $$$f(n,i,j)$$$ for all suitable $$$(n,i,j)$$$-s.

For suffix maximums, (the number of permutations of $$$1\sim n$$$ with $$$i$$$ prefix maximums and $$$j$$$ ascents, $$$g(n,i,j)$$$, we can just reverse some dimension of $$$f$$$).

To calculate the answer, consider the position of $$$n$$$. Suppose it's $$$p$$$. The the answer is

Let $$$u(x,y)=\sum_{i}f(x,i,y)a_{i+1}$$$, $$$v(x,y)=\sum_{z}g(x,i,y)b_{i+1}$$$ (both of these are calculated in $$$O(n^3)$$$), then the answer is

By seeing $$$u$$$ and $$$v$$$ as 2D polynomials, this can be calculated with 2D FFT in $$$O(n^2\log n)$$$.