Codeforces Round 889 (Div. 1) |
---|
Finished |
You have a set $$$S$$$ of $$$n$$$ distinct integers between $$$1$$$ and $$$m$$$.
Each second you do the following steps:
What is the expected number of seconds until $$$S$$$ is empty?
Output the answer modulo $$$1\,000\,000\,007$$$.
Formally, let $$$P = 1\,000\,000\,007$$$. It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{a}{b}$$$, where $$$a$$$ and $$$b$$$ are integers and $$$b \not \equiv 0 \pmod{P}$$$. Output the integer equal to $$$a \cdot b^{-1} \bmod P$$$. In other words, output an integer $$$z$$$ such that $$$0 \le z < P$$$ and $$$z \cdot b \equiv a \pmod{P}$$$.
The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \leq n \leq m \leq 500$$$) — the number of elements in the set $$$S$$$ and the upper bound on the value of the elements in $$$S$$$.
The second line contains $$$n$$$ integers $$$S_1,\,S_2,\,\dots,\,S_n$$$ ($$$1 \leq S_1 < S_2 < \ldots < S_n \leq m$$$) — the elements of the set $$$S$$$.
Output a single integer — the expected number of seconds until $$$S$$$ is empty, modulo $$$1\,000\,000\,007$$$.
2 3 1 3
750000009
5 10 1 2 3 4 5
300277731
5 10 2 3 6 8 9
695648216
1 100 1
100
For test 1, here is a list of all the possible scenarios and their probabilities:
Adding them up, we get $$$\frac{1}{2}\cdot 4 + \frac{1}{4} \cdot 4 + \frac{1}{4} \cdot 3 = \frac{15}{4}$$$. We see that $$$750000009 \cdot 4 \equiv 15 \pmod{1\,000\,000\,007}$$$.
Name |
---|