I don't really know how to answer that

As I have seen some previous ECPC contests, I think it will be 3-4 stars in difficulty.

Updated, 3 stars.

The contest will start 3 days later, you will see the problems when it starts.

Easier than SCPC, but not by much

I solved it by counting how many numbers are larger than the entered number(n), except the last one, let's call their count (CC) then did the following:

res = probability of getting the largest one from the first try (1/size(n)!)

find in how many ways I can choose 1 out of the larger numbers = C(CC,1) and multiply it by the probability of that which is (1/size(n)!)^2

find in how many ways I can choose 2 out of the larger numbers = C(CC,2) and multiply it by the probability of that which is (1/size(n)!)^3

Keep doing so until you reach C(CC,CC) * (1/size(n)!)^(CC+1) or whenever the result of that step is less than a small eps, but be careful eps should be really small! I got around 12 WA just because eps was 1e-10, when I changed it to 1e-14 I got AC.

Add up the whole thing and this will be the answer

I think there are other simpler solutions though.

nothing any editorials?!