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+\documentclass[aspectratio=169,hyperref={pdfencoding=auto,psdextra}]{beamer}
+\usepackage[utf8]{inputenc}
+\usepackage{CJKutf8}
+\usepackage{ulem}
+\usepackage{graphicx}
+\usepackage{fancyvrb}
+\usetheme{Malmoe}
+\usecolortheme{default}
+\begin{CJK*}{UTF8}{gbsn}
+\title{「知道错了没」}
+\subtitle{——如何让队友认错}
+\author{Chris Xiong}
+\date{2017-07-21}
+\begin{document}
+	\frame{\titlepage}
+	\begin{frame}
+		\frametitle{「知道错了没」}
+		\framesubtitle{Outline}
+		\begin{itemize}
+			\item 采访
+			\item dp: d(ui)p(ai)
+			\item dp: Knapsack problem
+			\item 如何让队友认错之如何殴打队友
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{采访}
+		\framesubtitle{\sout{a.k.a. 教学质量检查}}
+		\begin{itemize}
+			\item 上次课讲的内容大家都听懂了吗？\pause
+			\item 什么？不懂？\pause
+			\item 那还记得上次讲的什么吗？\pause
+			\item A Water Problem \pause
+			\item 已知$$f(x+1) =
+				\begin{cases}
+				a & x=0
+				\\
+				b & x=1
+				\\
+				f(x)+f(x-1)+sin(\frac{\pi x}{2}) & otherwise
+				\end{cases}$$
+			对于给定的$a,b,n$，求$f(n)$。$n\leq 10^{18}$。
+			\item 给大家5分钟的思考时间。
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{采访}
+		\framesubtitle{怎么样，是不是不会啊？}
+		\begin{itemize}
+			\item 都怪宇宙智障。\\
+			\includegraphics[scale=0.75]{zz.png}\pause
+			\item 提示：\pause周期！！\pause
+			\item 还不会的话就去找宇宙智障。\\
+			\includegraphics[scale=0.75]{zz1.png}
+			\item （听说你想要表扬　厚颜无耻）
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{d(ui)p(ai)}
+		\begin{itemize}
+			\item WTF is duipai?\pause
+			\item Automated generation of test data and execution of several programs.\pause
+			\item And most importantly, compare their results.\pause
+			\item \sout{A nice way to waste time if you are stuck.}
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{A sample script for UNIX-like OS}
+		\begin{Verbatim}
+#!/bin/bash
+i=0
+while(true)
+do
+	./170312cgen > test.in
+	./170312ca < test.in > aa.out
+	./170312cb < test.in > bb.out
+	diff aa.out bb.out
+	if [ $? -ne 0 ]
+	then
+		break
+	fi
+	echo $i passed
+	let i++
+done
+		\end{Verbatim}
+\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{How to use it?}
+		\begin{itemize}
+			\item Modify the script to your needs.
+			\item Save it as a script, e.g.: "xxx.sh".
+			\item Give it the permission to execute.
+			Run \verb|chmod +x <your_script_name_here>| in a terminal.
+			\item Run it!
+			Type \verb|./<your_script_name_here>| in a terminal.
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{What does this script do?}
+		\begin{itemize}
+			\item Run the input generator.
+			\item Feed the generated input to the compared program A and gather results from it.
+			\item Do the same thing with program B.
+			\item Check the output. If they differ, terminate the script. Otherwise loop. 
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Explanation}
+		\begin{itemize}
+			\item
+				\begin{verbatim}
+					while(true)
+					do
+					done
+					break
+				\end{verbatim}
+			\item
+				\begin{verbatim}
+					> <
+				\end{verbatim}
+				redirection
+			\item
+				\begin{verbatim}
+				if
+				then
+				fi
+				$?
+				[, -ne
+				\end{verbatim}
+			\item Verification: \verb|diff| / custom program
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Alternative approaches}
+		\begin{itemize}
+			\item Write a \verb|C/C++| program instead of a shell script?
+			\item \verb|system()| in \verb|stdlib.h| (\verb|cstdlib|)
+			\item return value of \verb|system()|
+			\item Windows batch file:
+				\begin{itemize}
+					\item \verb|IF %ERRORLEVEL% EQU 0(GOTO :loop)|
+				\end{itemize}
+			\item \sout{Powershell}?
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Writing input generators}
+		\begin{itemize}
+			\item Random?
+			\item Constructed special cases?
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{I suck at this}
+		\begin{itemize}
+			\item Unbounded knapsack problem
+			\item Bounded knapsack problem
+				\begin{itemize}
+					\item 0/1 knapsack problem
+				\end{itemize}
+			\item NP-complete!
+			\item A No-Dynamic-Programming-At-All variant
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{The No-DP-At-All variant}
+		Fractional knapsack problem (a.k.a. Continuous knapsack problem)
+		\begin{itemize}
+			\item A knapsack of capacity $W$.
+			\item $N$ items, each having its weight $w_i$ and value per unit weight $v_i$.
+			\item Select an amount $x_i$ of each item so that
+				the total weight doesn't exceed the capacity (
+				$\displaystyle\sum_{i}^{}x_i\leq W$
+			) and maximizing the total value
+				$\displaystyle\sum_{i}^{}x_i \times v_i$, where $x_i \in \mathbb{R}, x_i \geq 0$.
+			\pause
+			\item Greedy.
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		\begin{itemize}
+			\item Still a knapsack of capacity $W$.
+			\item Still $N$ items, each having its weight $w_i$
+				and value $v_i$.
+			\item For each item, determine whether to put it in
+				the knapsack so that the total weight doesn't
+				exceed the capacity and the total value is maximum.
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		A brute-force solution:
+		\begin{Verbatim}
+def dfs(i,remaining_capacity):
+	if(i==0): return 0;
+	if(remaining_capacity<0):
+		return -inf;
+	r1=dfs(i-1,remaining_capacity);
+	r2=dfs(i-1,remaining_capacity-w[i])+v[i];
+	return max(r1,r2);
+		\end{Verbatim}
+		\begin{itemize}
+			\item Call dfs(N,W) for answer.
+			\item Each non-trivial invocation of dfs branch into two paths.
+			\item Time complexity: $O(2^N)$.
+			\item A minor optimization: replace the second condition
+				statement with
+\verb|if(remaining_capacity<w[i]): return dfs(i-1,remaining_capacity);|
+		\end{itemize}
+\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		A effective optimization: memoization.
+		\begin{Verbatim}
+f=[[-1 for i in range(N)] for j in range(W)]
+def dfs(i,remaining_capacity):
+	if(i==0): return 0;
+	if(remaining_capacity<w[i]):
+		return dfs(i-1,remaining_capacity);
+	if(f[i][remaining_capacity]!=-1):
+		return f[i][remaining_capacity];
+	f[i][remaining_capacity]=max(
+			dfs(i-1,remaining_capacity),
+			dfs(i-1,remaining_capacity-w[i])+v[i]);
+	return f[i][remaining_capacity];
+		\end{Verbatim}
+\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		\begin{itemize}
+			\item For each parameter tuple of dfs, the function may only branch once.
+			\item Time complexity: $O(NW)$.\\
+				\tiny It's a pseudo-polynomial algorithm, so the knapsack problem is still NP-complete.
+				\normalsize
+			\pause
+			\item Why does this work?
+			\pause
+			\item Once the result for a specific parameter tuple has been calculated, will it change any further?
+			\item Non-aftereffect property.
+			\pause
+			\item Recursion? Phooey! That will be a lot of context switches!
+			\pause
+			\item Time for some black magic!
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		The iteration version.
+		\begin{Verbatim}
+f=[[0 for i in xrange(N)] for j in xrange(W)]
+for i in xrange(1,N):
+	for j in xrange(0,W):
+		f[i][j]=max(f[i-1][j],
+				f[i-1][j-w[i]]+v[i] if j>=w[i] else 0);
+		\end{Verbatim}
+\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		\begin{itemize}
+			\item Recall that in the memoization version, in order to calculate results for $f[i,remaining\_capacity]$
+				we must already have at least two results for $f[i-1,x]$.
+			\item Why don't we calculate all $f[i-1,x]$ before calculating $f[i,x]$?
+				\pause
+			\item Got the maximum value now! Want the list of selected items?
+				\pause
+			\item Traceback.
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		\begin{itemize}
+			\item When we are at $i=x$ of the outer loop, all values in $f[y],y<x-1$ are no longer used.
+			\item If we don't need to traceback, can we save a bit of memory?
+				\pause
+			\item Yes! Just throw them away!
+			\begin{Verbatim}
+f=[0 for i in xrange(W)]
+for i in xrange(1,N):
+	for j in xrange(W,w[i],-1):
+			f[j]=max(f[j],f[j-w[i]]+v[i]);
+			\end{Verbatim}
+		\end{itemize}
+\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{0/1 knapsack problem}
+		\begin{itemize}
+			\item How does this work?\\
+				\pause
+			($g[i][j]$ denotes the original $f[i][j]$ from the two dimensional iterative solution.)
+			\item When we are at $j=y$ of the inner loop, $f[0..y]$ are values from $g[i-1]$ and
+				$f[y+1..W]$ contains values from $g[i]$.
+			\item Why reverse the inner loop?
+				\pause
+			\item Because we still need the values with smaller $remaining\_capacity$ from the last iteration!
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{Unbounded knapsack problem}
+		\begin{itemize}
+			\item Same as the 0/1 knapsack problem, but each item has unlimited copies.
+				\pause
+			\item Converting to 0/1 knapsack problem?
+				\pause
+			\item Imitating Binary. We can obtaining any multiplicity of items from a combination of 1x, 2x, 4x, 8x, ... of that item.
+				\pause
+			\item Any other solutions?
+		\end{itemize}
+	\end{frame}
+	\begin{frame}[fragile]
+		\frametitle{Knapsack problem}
+		\framesubtitle{Unbounded knapsack problem}
+		Another solution:
+		\begin{Verbatim}
+f=[0 for i in xrange(W)]
+for i in xrange(1,N):
+	for j in xrange(w[i],W):
+			f[j]=max(f[j],f[j-w[i]]+v[i]);
+		\end{Verbatim}
+		Wait... Isn't this our final solution for the 0/1 knapsack problem?
+\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{Unbounded knapsack problem}
+		\begin{itemize}
+			\item Not exactly! Note that the inner loop now iterate from $w[i]$ to $W$.
+			\item Why?
+				\pause
+			\item Let's revisit the reason to iterate in reverse order in 0/1 knapsack problem:
+			\item We still need the values with smaller $remaining\_capacity$ from the last iteration.
+				\pause
+			\item Why do we need \textit{those values}, instead of the shiney new values we just obtained?
+				\pause
+			\item Because these values do not take the current item into consideration, effectively ensuring
+				that every item can be used at most once.
+			\item But now we have unlimited copies of each item!
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{Bounded knapsack problem}
+		\begin{itemize}
+			\item Same as the 0/1 knapsack problem, but each item has $C_i$ copies.
+			\item POJ 1276
+				\pause
+			\item Still solve by converting to a 0/1 knapsack problem.
+				\pause
+			\item How to limit the maximum number of copies?
+				\pause
+			\item By modifying the largest group so that if all groups are selected,
+				the sum of multiplicity equals to $C_i$.
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{Knapsack problem}
+		\framesubtitle{Bounded knapsack problem}
+		Another "stupid" solution that can also be applied to the unbounded knapsack problem:
+		\begin{itemize}
+			\item For each item, we have $C_i+1$ choices.
+			\item We just iterate through these choices to update $f[][]$.
+			\item This solution runs for $O(W\Sigma C_i)$.
+			\item However it can be further optimized to $O(NW)$ using some advanced DP optimization technics.
+				\pause
+			\item We are not covering that here today.
+			\item More about knapsack problems:\\
+				\url{https://github.com/tianyicui/pack}
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{如何殴打队友}
+		\begin{center}
+		\Huge{以下内容仅供娱乐}
+		\end{center}
+	\end{frame}
+	\begin{frame}
+		\frametitle{如何殴打队友}
+		\framesubtitle{——何时应当考虑殴打队友？}
+		\begin{itemize}
+			\item 当队友占3个小时键盘什么都没写出来时
+			\item 当队友开一题WA一题时
+			\item 当队友开始表演口技时
+			\item 当队友热身赛开可乐洒了一地时
+			\item 当队友看到树就想重心分解时
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{如何殴打队友}
+		\framesubtitle{——正题}
+		\begin{itemize}
+			\item 像现在这么殴打（宣传光辉事迹）
+			\item 比赛时不准碰键盘
+			\item 表演口技时录音
+			\item WA一题灌一瓶可乐，不准洒
+			（大家可以算一下光这张图就要喝多少瓶）\\
+			\includegraphics[scale=0.25]{zz2.png}\\
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{如何殴打队友}
+		\framesubtitle{——殴打队友时需要注意的地方}
+		\begin{itemize}
+			\item 注意殴打的度——虽然原则上是越重越好，但是如果你的队友是个卜力星人，殴打太重会导致其发射大量宇宙射线，导致「伤敌800，自损1000」的尴尬情形。
+			\item 殴打方式要适当。比如其在表演口技不应该使用灌可乐的手法，因为容易洒一地。
+			\item 适可而止。如果感觉队友能A题了就让其施展一发（没A就接着灌）。
+		\end{itemize}
+	\end{frame}
+	\begin{frame}
+		\frametitle{如何殴打队友}
+		\framesubtitle{Bonus: 利用宇宙射线}
+		如果你发现你的队友会发射宇宙射线，那么它可能是可以被利用的。可利用的宇宙射线的发射者是会认错的。这里有一个正面例子和一个反面例子：
+		\begin{itemize}
+			\item 黄焖蓉　：发射射线导致临近的队伍接连两次CE。
+			\item 宇宙智障：发射射线导致队友高数全部忘光。
+		\end{itemize}
+		如你所见，第一类射线是可以加以利用的；而第二类射线则是「射别人一个也射不中，射自己人一射一个准」的。大家要尽量做好对第二类射线的防护工作。关于这个问题我们下次再说（如果还有下次机会的话）。
+	\end{frame}
+	\begin{frame}
+		\frametitle{如何殴打队友}
+		\framesubtitle{So... what's the point?}
+		\begin{itemize}
+			\item 合理利用时间
+			\item 卡题时的处理方式
+			\item 队内的合作
+			\item 其他队伍的影响
+		\end{itemize}
+	\end{frame}
+\end{CJK*}
+\end{document}
+
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