Sorry about that. This was some kind of unintentional scheduling snafu, MikeMirzayanov somehow missed the existence of this TC round when scheduling the CF round, probably because some data wasn't up to date somewhere. We try to look avoid such clashes but sometimes mistakes happen.

Does TC still pay for editorialist (and how much)? Who should I contact for?

Let $$$D_k$$$ be the sum of product of $$$C_i$$$ taken $$$k$$$ at a time. $$$D$$$ can be found in $$$O(n \log^2 n)$$$ by finding $$$\prod (x + C_i)$$$.

Now, by symmetry, for any subset $$$S$$$ of $$$[n]$$$ of size k, $$$E[ \prod_{i \in S} x_i] = E [ \prod_{i=1}^{k} x_i] $$$. So, expanding the product, we have:

$$$ E [ \prod_{i=1}^{n} (C_i + x_i)] = \sum_{k=0}^{n} D_{n-k} E [ \prod_{i=1}^{k} x_i] $$$

So, we have to find sum of $$$\prod_{i=1}^{k} x_i$$$ over all n-tuples $$$x$$$ summing to m. One way to solve this is using generating functions ( by finding coefficient of $$$x^m$$$ in $$$(\sum i x ^ i) ^ k (\sum x ^ i) ^ {n - k} = (\frac{x}{(1-x)^2})^k (1-x)^{k-n})$$$). Another way is this:

Consider a line segment of length $$$m$$$ to be divided into $$$n$$$ integral parts. For each of the first $$$k$$$ segments you have to cut it in two parts at an integer point not coinciding with its left end. The number of ways to do this is our answer (as we have $$$x_i$$$ choices to cut a segment of length $$$x_i$$$). Equivalently, you have $$$n + k$$$ segments, with $$$k$$$ of them which must be non zero. This gives $$$\binom{n + m - 1}{m - k}$$$.

So, you have to find $$$\displaystyle \dfrac{\sum D_{n-k} \binom{n + m - 1}{m - k}}{ \binom{n+m-1}{n-1}}$$$

Yeah, it is...That's why it fails the system test :)