2016 年 ACM/ICPC 青岛区域赛 Problem C Pocky

昨晚乱入学弟的训练赛，想了一下这个题。推导的过程中，加深了对公理化的概率论的理解。\(\newcommand{\d}{\mathop{}\\!\mathrm{d}}\)

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解法一

考虑 $ d < L$ 的情形。
\begin{equation*}
P(X = 1) = \frac{d}{L}
\end{equation*}
\begin{align*}
P(X = 2) &= \int_0^{L - d} \frac{\d x}{L} \frac{d}{L - x} \\
&= \frac{d}{L}\ln\frac{L}{d}
\end{align*}
\begin{align*}
P(X = 3) &= \int_{0}^{L-d}\frac{\d x}{L}\int_{0}^{L-x-d}\frac{\d y}{L-x}\frac{d}{L-x-y} \\
&= \int_{0}^{L-d} \frac{\d x}{L} \frac{d}{L - x} \ln \frac{L - x}{d} \\
&= \frac{d}{L}\frac{1}{2}\ln^2\frac{L}{d}
\end{align*}
\begin{align}
P(X = 4) &= \int_0^{L - d}\frac{\d x}{L} \int_0^{L - x - d} \frac{\d y}{L-x}\int_0^{L-x-y-d}\frac{\d z}{L - x -y}\frac{d}{L - x - y - z} \notag\\
&= \int_0^{L - d}\frac{\d x}{L} \int_0^{L - x - d} \frac{\d y}{L-x} \frac{d}{L - x -y} \ln \frac{L - x -y}{d} \notag\\
&= \int_0^{L - d}\frac{\d x}{L} \frac{d}{L-x}\frac{1}{2}\ln^2\frac{L-x}{d} \label{Int:1}
\end{align}
令 \(u = \frac{L-x}{d}\) ，则 \(\d x = -d\d u\) ，有
\begin{align*}
\eqref{Int:1} &= \int_1^\frac Ld\frac dL\frac{\d u}{u}\frac 12\ln^2u \\
&= \int_1^\frac Ld\frac dL\frac 16\d\ln^3u \\
&= \frac 16\frac dL\ln^3\frac Ld
\end{align*}
不难推出
\begin{equation*}
P(X = n) = \frac dL\frac1{(n-1)!}\ln^{n-1}\frac Ld
\end{equation*}
所以
\begin{align*}
E(X) &= \sum_{ n \ge 1 } n P(X=n) \\
&= \frac dL \sum_{n \ge 1} \frac n{(n-1)!}\ln^{n-1}\frac Ld \\
&= \frac dL \sum_{n \ge 0} \frac{n+1}{n!} \ln^n\frac Ld \\
&= \frac dL (\ln\frac Ld + 1) \mathrm{e}^{\ln\frac Ld} \\
&= \ln\frac Ld + 1
\end{align*}
上式中的求和用到了 \((x+1)\mathrm{e}^x\) 的 Maclaurin 展开：
\begin{equation*}
(x+1)\mathrm{e}^x = \sum_{n\ge 0} \frac{n + 1}{n!} x^n
\end{equation*}

解法二

用 \(f(x)\) 表示绳长为 \(x\) 时切割次数的期望，则有

\[f(x) = \begin{cases} 0, &&\text{if $x\le d$;} \\\\ 1 + \int_0^x \frac{dy}{x}f(y), && \text{otherwise.} \end{cases} \]

考虑 \(x>d\) 的情形，此时有
\begin{align}
f(x) &= 1 + \int_0^x \frac{\d y}{x}f(y) \notag\\
&= 1 + \int_0^d \frac{\d y}{x}f(y) + \int_d^x \frac{\d y}{x}f(y) \notag\\
&= 1 + \int_d^x \frac{\d y}{x}f(y) \label{Int:2}
\end{align}
对 \eqref{Int:2} 式两边求导，得
\begin{align*}
f'(x) &= \frac{f(x)}x - \frac1{x^2}\int_dx\d yf(y) \\
&= \frac{f(x)}x - \frac1x(f(x) -1) \\
&= \frac1x
\end{align*}
又 $\lim\limits_{x\to d^+} f(x) = 1 $，得 $$ f(x) = \ln x + 1 - \ln d $$

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解法二来自 Huo Chen