Hint: attempt to set every element in the answer to 1 at first, then decrement when you need, some small optimization may be needed

Cases $$$k = 1, 2$$$ are trivial. Notice that some element of $$$\{s - 1, s, s + 1\}$$$ should be divisible by $$$k$$$. Because $$$k \geq 3$$$ it must be exactly one element if solution exists. That way we found $$$b_1$$$. Solve recursively for others.

I have an easy intuitive solution for that.
Just take a var = 0 and check for decreasing powers of K(from n-1 to 0), whether adding/subtracting this to/from var will result in bringing var closer to s. Fill suitably 1 or -1 at this index in the answer array, and modify var.
After traversing all powers of K, if var != s, then print -2, else print the answer array.
My submission
Note why this works is because of the fact that K^N will always be greater than sum of K^(N-1) + K^(N-2) + .... + K + 1, so K^N will have the greatest hand in determining the closeness of var to required s.

Why multiset? You can use trie to get the maximum xor value.

We need to fix not only the starting element — $$$v$$$ but also the amount of it in the beginning — $$$c$$$. Then your formulas are almost correct

Try to fix the number of indices i such that b[i] is the prefix maximum, then, since we know that maximums will occur in increasing order and minimums in decreasing order, we only need to choose the indices where we will keep the maximums.

there is a caveat, though, you need to make sure that all places other than the maximums will surely be minimums to avoid overcounting; try to figure that out, it isn't very hard.

Yeah, your formula is correct