Contest 2050 and Codeforces Round 718 (Div. 1 + Div. 2)
F
Reunion
Codeforces
3200
原题
← EG →
翻译中…

It is reported that the 2050 Conference will be held in Yunqi Town in Hangzhou from April 23 to 25, including theme forums, morning jogging, camping and so on.

翻译中…

The relationship between the $n$ volunteers of the 2050 Conference can be represented by a tree (a connected undirected graph with $n$ vertices and $n-1$ edges). The $n$ vertices of the tree corresponds to the $n$ volunteers and are numbered by $1,2,\ldots, n$.

翻译中…

We define the distance between two volunteers $i$ and $j$, dis$(i,j)$ as the number of edges on the shortest path from vertex $i$ to vertex $j$ on the tree. dis$(i,j)=0$ whenever $i=j$.

翻译中…

Some of the volunteers can attend the on-site reunion while others cannot. If for some volunteer $x$ and nonnegative integer $r$, all volunteers whose distance to $x$ is no more than $r$ can attend the on-site reunion, a forum with radius $r$ can take place. The level of the on-site reunion is defined as the maximum possible radius of any forum that can take place.

翻译中…

Assume that each volunteer can attend the on-site reunion with probability $\frac{1}{2}$ and these events are independent. Output the expected level of the on-site reunion. When no volunteer can attend, the level is defined as $-1$. When all volunteers can attend, the level is defined as $n$.

翻译中…

Input

翻译中…

The first line contains a single integer $n$ ($2\le n\le 300$) denoting the number of volunteers.

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Each of the next $n-1$ lines contains two integers $a$ and $b$ denoting an edge between vertex $a$ and vertex $b$.

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Output

翻译中…

Output the expected level modulo $998\,244\,353$.

翻译中…

Formally, let $M = 998\,244\,353$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x \lt M$ and $x \cdot q \equiv p \pmod{M}$.

翻译中…

Note

翻译中…

For the first example, the following table shows all possible outcomes. $yes$ means the volunteer can attend the on-site reunion and $no$ means he cannot attend. $$\begin{array}{cccc} 1 & 2 & 3 & level\\ yes & yes & yes & 3\\ yes & yes & no & 1\\ yes & no & yes & 0\\ yes & no & no & 0\\ no & yes & yes & 1\\ no & yes & no & 0\\ no & no & yes & 0\\ no & no & no & -1\\ \end{array}$$ The expected level is $\frac{3+1+1+(-1)}{2^3}=\frac{1}{2}$.

翻译中…

样例

Input #1
3
1 2
2 3
Output #1
499122177
InputCopy OutputCopy
Input #2
5
1 2
2 3
3 4
3 5
Output #2
249561089
InputCopy OutputCopy
Input #3
10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
Output #3
821796866
InputCopy OutputCopy
题目信息
题号
1517F
时间限制
3 s
内存限制
256 MB
标签
组合数学
动态规划
树

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。