Editorial of A, B are available.

I was adding numbers one by one from postion $$$1$$$ to position $$$n$$$. Let's say we are adding currently element at position $$$i$$$ (there is $$$n-i+1$$$ possible values). In case some element can not be right for all other elements (rest $$$n-i$$$ elements have same $$$a_i$$$ value), then put that element on position $$$i$$$. Otherwise put smallest or second smallest element on position $$$i$$$.

Obviously we can not get smaller lex permutation than this one, but also it is possible that last two elements are bad in the permutation:

For example case:

$$$ n = 5$$$

$$$a = [2, 3, 1, 5, 4]$$$

My algorithm will give result: $$$[1, 3, 2, 4, 5]$$$ and it is incorrect result. But we can make it better, just pick last three elements and try every permutation — at least one of permatutions will be valid (that is reason because there is no answer just for $$$n = 2$$$).

There is an editorial available. editorial

Edit: Just saw that the english version is not available for all problems, sorry.

Here's a nice analogy: you can interpret the problem as follows- There's a $$$k\times n$$$ grid. First you have to distribute $$$a_i$$$ black hokey pucks among the $$$n$$$ cells of the $$$i$$$-th row (for each $$$i$$$). Then you have to distribute $$$n$$$ red hockey pucks on top of the $$$\sum a_i$$$ black pucks such that every column has exactly one red puck. Count how many ways are there to do that.

I still don't know how to solve C

editorial is out already .... on yesterday