diff --git a/.idea/acm-cheat-sheet.iml b/.idea/acm-cheat-sheet.iml new file mode 100644 index 0000000..e28549d --- /dev/null +++ b/.idea/acm-cheat-sheet.iml @@ -0,0 +1,8 @@ + + + + + + + + \ No newline at end of file diff --git a/.idea/compiler.xml b/.idea/compiler.xml new file mode 100644 index 0000000..96cc43e --- /dev/null +++ b/.idea/compiler.xml @@ -0,0 +1,22 @@ + + + + + + + + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/.idea/copyright/profiles_settings.xml b/.idea/copyright/profiles_settings.xml new file mode 100644 index 0000000..e7bedf3 --- /dev/null +++ b/.idea/copyright/profiles_settings.xml @@ -0,0 +1,3 @@ + + + \ No newline at end of file diff --git a/.idea/encodings.xml b/.idea/encodings.xml new file mode 100644 index 0000000..d821048 --- /dev/null +++ b/.idea/encodings.xml @@ -0,0 +1,4 @@ + + + + \ No newline at end of file diff --git a/.idea/misc.xml b/.idea/misc.xml new file mode 100644 index 0000000..699f6ac --- /dev/null +++ b/.idea/misc.xml @@ -0,0 +1,4 @@ + + + + \ No newline at end of file diff --git a/.idea/modules.xml b/.idea/modules.xml new file mode 100644 index 0000000..c1dec64 --- /dev/null +++ b/.idea/modules.xml @@ -0,0 +1,8 @@ + + + + + + + + \ No newline at end of file diff --git a/.idea/scopes/scope_settings.xml b/.idea/scopes/scope_settings.xml new file mode 100644 index 0000000..922003b --- /dev/null +++ b/.idea/scopes/scope_settings.xml @@ -0,0 +1,5 @@ + + + + \ No newline at end of file diff --git a/.idea/vcs.xml b/.idea/vcs.xml new file mode 100644 index 0000000..94a25f7 --- /dev/null +++ b/.idea/vcs.xml @@ -0,0 +1,6 @@ + + + + + + \ No newline at end of file diff --git a/C++/abstract.tex b/C++/abstract.tex index 939fb10..e5fa328 100644 --- a/C++/abstract.tex +++ b/C++/abstract.tex @@ -9,6 +9,8 @@ \subsubsection{内容简介} 全书的代码，使用“纯C + STL”的风格。本书中的代码规范，跟在公司中的工程规范略有不同，为了使代码短（方便迅速实现）: + +test file \begindot \item 所有代码都是单一文件。这是因为一般OJ网站，提交代码的时候只有一个文本框，如果还是 按照标准做法，比如分为头文件.h和源代码.cpp，无法在网站上提交； diff --git 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Job i, 1 ≤ i ≤ N, +needs ti units of processing time. If job i is finished by time T, where T is a given +deadline, then a profit pi is earned; otherwise, a penalty qi +is imposed. (Both pi and qi are positive integers.) We want to select a subset S of jobs such that +(i) ∑i∈S ti ≤ T, and (ii) f(S) = ∑i∈S pi −∑i /∈S qi is maximum. +Show how to find such a set of jobs using dynamic programming. +*/ +int const N=3; +int const T=6; +//profit +int p[N]={12,3,4}; +//penalty +int q[N]={1,4,2}; +//time consuming +int t[N]={1,4,3}; +//net profit table +int f[N+1][T+1]; +//pointer to index which job should be scheduled +int P[N+1][T+1]; + +void findMaxNetProfit(){ + //non recursive implementation + for(int i=0;il){ + f[k][l]=f[k-1][l]-q[i]; + } + else{ + f[k][l]=max(f[k-1][l]-q[i],f[k-1][l-t[i]]+p[i]); + if(f[k][l]==f[k-1][l-t[i]]+p[i]){ + P[k][l]=1; + } + } + } + } + cout<<"net profit is "<deadline){ + return findMaxNetProfit_rec(job_n-1,deadline)-q[job_n-1]; + } + else{ + int a=findMaxNetProfit_rec(job_n-1,deadline)-q[job_n-1]; + int b=findMaxNetProfit_rec(job_n-1,deadline-t[job_n-1])+p[job_n-1]; + if(b>a) + P[job_n][deadline]=1; + return max(a,b); + } + +} +void printJobSet(){ + int l=T; + for(int i=N;i>=1;i--){ + if(P[i][l]==1){ + cout<<"job "< +using namespace std; +//find the largest j which makes f[j]<=s[i] +int find_opt(double [], double [],int); +//return the maximum sum of the selected intervals +//non recurssive version +double max_select_nonrec(double [],double [],int); +//recurssive version +double max_select_rec(double [],double [],int); +//void print selected intervals +void print_interval(int *,int); + +int find_opt(double s[],double f[],int i){ + for(int j=i-1;j>=0;j--){ + if(f[j]<=s[i]){ + return j; + } + } + return 0; +} + +double max_select_nonrec(double s[], double f[],int n){ + double * T=new double[n]; + int * P=new int[n]; + for(int i=0;i=T[i-1]){ + T[i]=temp; + P[i]=j; + } + else{ + T[i]=T[i-1]; + P[i]=-1; + } + + } + //print + print_interval(P,n); + return T[n-1]; +} +void print_interval(int * P, int n){ + int i=n-1; + while(P[i]!=0){ + if(P[i]>0){ + cout< +using namespace std; +/* +Author: Jiankai Sun +Date: 2014/10/14 +*/ +/* +Let A=a1a2...am and B=b1b2...bn be two strings of characers. we +want to transform A into B using following operations: +delete a character +and a character +change a character +write a dynamic programming algorithm that finds the minimum number +of operations needed to transform A into B +*/ +int min_ope(char * c1,const int m, char *c2,const int n){ + + int F[m+1][n+1];//define operation matrix + + //boundary condition + for(int i=0;i<=m;i++){ + F[i][0]=i; + } + for(int i=0;i<=n;i++){ + F[0][i]=i; + } + for(int i=0;iT(i-1)} + +T[i]=T(p(i))+$f_i$-$s_i$ + +P[i]=p(i) +\Else + +T[i]=T[i-1] + +P[i]=-1 + +\EndIf +\EndFor +\State Return T[n] +\end{algorithmic} +\end{algorithm} +\item we use $P[1,2,...,n]$ to print selected intervals + +\begin{algorithm}[H] +\caption{function printInterval() } +\begin{algorithmic}[1] +\State global $P[1,...,n]$ +\State i=n +\While{P[i] $\neq$ 0} +\If{P[i] $>$ 0} + +print i + + +i=P[i] + +\EndIf +\EndWhile + +\State print i + +\end{algorithmic} +\end{algorithm} +\item We use $O(nlog(n))$ to sort. For each $i$ we need $O(n)$ to scan all the interval list and repeat $n$ times, So the time complexity is $O(n^2)$ +\item source code: max\_selected\_intervals.cpp +\end{enumerate} +\begin{question}{2} +Let $A=a_1a_2...a_m$ and $B=b_1b_2...b_n$ be two strings of characers. we +want to transform $A$ into $B$ using following operations: + +delete a character + +add a character + +change a character + +write a dynamic programming algorithm that finds the minimum number of operations needed to transform $A $into $B$ +\end{question} +\begin{enumerate} +\item Let F[i,j] denotes the minimum number of operations needed to transform $a_1a_2...a_i$ to $b_1b_2...b_j$. +\item Recurrence relation + +$ +F[i,j]=or +\begin{cases} +F[i-1,j-1]; if a_i=b_j\\ +min +\begin{cases} +F[i-1,j-1]+1; change\\ +F[i,j-1]+1; add\\ +F[i-1,j]+1; delete\\ +\end{cases} +otherwise +\end{cases} +$ + +\item Boundary conditions + +$F[0,k]=k$ for k in [0,n] + +$F[k,0]=k$ for k in [0,m] + + \item implement the non recursive algorithm, see algorithm \ref{algminnum} + +\begin{algorithm}[H] +\caption{function MinNumOper() } +\label{algminnum} +\begin{algorithmic}[1] +\State global $F[1...m;1..n]$, A[1...m], B[1...n] +\State Initialize $F[0,k]=k$ for k in [0,n] and $F[k,0]=k$ for k in [0,m] + \For{i $\leftarrow$ 1 to m} + \For {j $\leftarrow$ 1 to n} + \If {A[i]==B[j]} F[i,j]=F[i-1,j-1] + \Else + + $ +F[i,j]=min +\begin{cases} +F[i-1,j-1]+1; change\\ +F[i,j-1]+1; add\\ +F[i-1,j]+1; delete\\ +\end{cases} +$ + \EndIf + \EndFor +\EndFor + \State return F[m,n] +\end{algorithmic} +\end{algorithm} +\item in the main function, we can call the function $MinNumOper()$ to to find the minimum number of operations needed to transform A into B +\item time complexity of this algorithm is $\Theta(mn)$ +\item source code: min\_operation\_dp.cpp +\end{enumerate} +%-------------------------------------------------------------- +% You don't have to mess with anything below this line. +% 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!visited.count(new_word)){ + result.push_back(new_word); + visited.insert(new_word); + } + swap(c, new_word[i]); + } + } + return result; + }; + + current.push(start); + while(!current.empty()&&!found){ + ++level; + while(!current.empty()&&!found){ + const string str= current.front(); + current.pop(); + + const auto& new_states = state_extend(str); + for(const auto& state: new_states){ + next.push(state); + if(state_is_target(state)){ + found = true; + break; + } + } + } + swap(next, current); + } + if(found) return level+1; + else return 0; + } +}; + +int main() { + Solution sol; + unordered_set dict = {"hot", "dot", "dog", "lot", "log"}; + cout << sol.ladderLength("hit", "cog", dict) << endl; + return 0; +} + + + + diff --git a/code/chatper9/WordLadder/a.out b/code/chatper9/WordLadder/a.out new file mode 100755 index 0000000..24017bc Binary files /dev/null and b/code/chatper9/WordLadder/a.out differ diff --git a/index.rst b/index.rst new file mode 100644 index 0000000..a7acee3 --- /dev/null +++ b/index.rst @@ -0,0 +1,83 @@ +# ACM Cheat Sheet +----------------- +## PDF下载 +ACM Cheat Sheet(C).pdf + +ACM Cheat Sheet(C++).pdf + +C 文件夹下是C版，内容一摸一样，代码是用纯C写的， + +C++ 文件夹下是C++版，内容一摸一样，代码是用C++写的， + +Java 文件夹下是Java版，内容一摸一样，代码是用Java写的 + +## LaTeX模板 +本书使用的是陈硕开源的[模板](https://github.com/chenshuo/typeset)。该模板已经用于陈硕写的书[《Linux 多线程服务端编程：使用 muduo C++ 网络库》](http://chenshuo.com/book/)，已经出版。 + +这个模板制作精良，很有taste，感谢陈硕 :) + +## 在Windows下编译 +1. 安装Tex Live 2016 。把bin目录例如`D:\texlive\2016\bin\win32`加入PATH环境变量。 +1. 安装字体。这个LaTex模板总共使用了9个字体，下载地址 ，有的字体Windows自带了，有的字体Ubuntu自带了，但都不全，还是一次性安装完所有字体比较方便。 +1. 安装TeXstudio +1. (可选)启动Tex Live Manager，更新所有已安装的软件包。 +1. 配置TeXstudio。 + + 启动Texstudio，选择 `Options-->Configure Texstudio-->Commands`，XeLaTex 设置为 `xelatex -synctex=1 -interaction=nonstopmode %.tex`； + + 选择 `Options-->Configure Texstudio-->Build` + + Build & View 由默认的 PDF Chain 改为 Compile & View； + + Default Compiler 由默认的PdfLaTex 修改为 XeLaTex ； + + PDF Viewer 改为 “Internal PDF Viewer(windowed)”，这样预览时会弹出一个独立的窗口，这样比较方便。 + +1. 编译。用TeXstudio打开`ACM-cheat-sheet.tex`，点击界面上的绿色箭头就可以开始编译了。 + + 在下方的窗口可以看到TeXstudio正在使用的编译命令是`xelatex -synctex=1 -interaction=nonstopmode "ACM-cheat-sheet".tex` + +## 在Ubuntu下编译 +1. 安装Tex Live 2016 + + 1.1. 下载TexLive 2016 的ISO 光盘，地址 + + 1.2 mount 光盘，`sudo ./install-tl` 开始安装 + + 1.3 加入环境变量 + + sudo vi /etc/profile + export PATH=$PATH:/usr/local/texlive/2016/bin/x86_64-linux + export MANPATH=$MANPATH:/usr/local/texlive/2016/texmf-dist/doc/man + export INFPATH=$INFPATH:/usr/local/texlive/2016/texmf-dist/doc/info + +1. 安装字体。这个LaTex模板总共使用了9个字体，下载地址 ，有的字体Windows自带了，有的字体Ubuntu自带了，但都不全，还是一次性安装完所有字体比较方便。 +1. 安装TeXstudio +1. 配置TeXstudio。 + + 启动Texstudio，选择 `Options-->Configure Texstudio-->Commands`，XeLaTex 设置为 `xelatex -synctex=1 -interaction=nonstopmode %.tex`； + + 选择 `Options-->Configure Texstudio-->Build` + + Build & View 由默认的 PDF Chain 改为 Compile & View； + + Default Compiler 由默认的PdfLaTex 修改为 XeLaTex ； + + PDF Viewer 改为 “Internal PDF Viewer(windowed)”，这样预览时会弹出一个独立的窗口，这样比较方便。 + +1. 编译。用TeXstudio打开`ACM-cheat-sheet.tex`，点击界面上的绿色箭头就可以开始编译了。 + + 在下方的窗口可以看到TeXstudio正在使用的编译命令是`xelatex -synctex=1 -interaction=nonstopmode "ACM-cheat-sheet".tex` +1. 懒人版镜像。如果不想进行上面繁琐的安装过程，我做好了一个Ubuntu VMware虚拟机镜像，已经装好了TexLive 2016, TexStudio和字体(详细的安装日志见压缩包注释)，开箱即用，下载地址 。 + +## 如何贡献代码 +编译通过后，就具备了完整的LaTeX编译环境了。 + +本书模板已经写好了，基本上不需要很多LaTeX知识就可以动手了。 + +欢迎给本书添加内容或纠正错误，在自己本地编译成PDF，预览没问题后，就可以发pull request过来了。 + + +## 小密圈 + +![](参考资料/silicon-job.jpeg) diff --git a/new.txt b/new.txt new file mode 100644 index 0000000..9dd04cd --- /dev/null +++ b/new.txt @@ -0,0 +1 @@ +lifadfliguifan diff --git a/new2.txt b/new2.txt new file mode 100644 index 0000000..fa2be78 --- /dev/null +++ b/new2.txt @@ -0,0 +1 @@ +liguifan diff --git "a/\345\217\202\350\200\203\350\265\204\346\226\231/silicon-job.jpeg" "b/\345\217\202\350\200\203\350\265\204\346\226\231/silicon-job.jpeg" index 58dca6b..171bf32 100644 Binary files "a/\345\217\202\350\200\203\350\265\204\346\226\231/silicon-job.jpeg" and "b/\345\217\202\350\200\203\350\265\204\346\226\231/silicon-job.jpeg" differ diff --git "a/\345\217\202\350\200\203\350\265\204\346\226\231/\350\203\214\345\214\205\351\227\256\351\242\230\344\271\235\350\256\262.html" "b/\345\217\202\350\200\203\350\265\204\346\226\231/\350\203\214\345\214\205\351\227\256\351\242\230\344\271\235\350\256\262.html" new file mode 100644 index 0000000..144296a --- /dev/null +++ "b/\345\217\202\350\200\203\350\265\204\346\226\231/\350\203\214\345\214\205\351\227\256\351\242\230\344\271\235\350\256\262.html" @@ -0,0 +1,180 @@ + +version 1.1 build 20071115 + +前言 +目录 +第一讲 01背包问题 +第二讲 完全背包问题 +第三讲 多重背包问题 +第四讲 混合三种背包问题 +第五讲 二维费用的背包问题 +第六讲 分组的背包问题 +第七讲 有依赖的背包问题 +第八讲 泛化物品 +第九讲 背包问题问法的变化 +附录一：USACO中的背包问题 +附录二：背包问题的搜索解法 +联系方式 +致谢 +前言 +本篇文章是我(dd_engi)正在进行中的一个雄心勃勃的写作计划的一部分，这个计划的内容是写作一份较为完善的NOIP难度的动态规划总结，名为《解动态规划题的基本思考方式》。现在你看到的是这个写作计划最先发布的一部分。 + +背包问题是一个经典的动态规划模型。它既简单形象容易理解，又在某种程度上能够揭示动态规划的本质，故不少教材都把它作为动态规划部分的第一道例题，我也将它放在我的写作计划的第一部分。 + +读本文最重要的是思考。因为我的语言和写作方式向来不以易于理解为长，思路也偶有跳跃的地方，后面更有需要大量思考才能理解的比较抽象的内容。更重要的是：不大量思考，绝对不可能学好动态规划这一信息学奥赛中最精致的部分。 + +你现在看到的是本文的v1.1版，发布于2007年11月15日。我会长期维护这份文本，把大家的意见和建议融入其中，也会不断加入我在OI学习以及将来可能的ACM-ICPC的征程中得到的新的心得。但目前本文还没有一个固定的发布页面，想了解本文是否有更新版本发布，可以在OIBH论坛中以“背包问题九讲”为关键字搜索贴子，每次比较重大的版本更新都会在这个论坛里发贴公布。也可以用“背包问题九讲”为关键字在搜索引擎中搜索以得到最新版本。 + +目录 +第一讲 01背包问题 +这是最基本的背包问题，每个物品最多只能放一次。 + +第二讲 完全背包问题 +第二个基本的背包问题模型，每种物品可以放无限多次。 + +第三讲 多重背包问题 +每种物品有一个固定的次数上限。 + +第四讲 混合三种背包问题 +将前面三种简单的问题叠加成较复杂的问题。 + +第五讲 二维费用的背包问题 +一个简单的常见扩展。 + +第六讲 分组的背包问题 +一种题目类型，也是一个有用的模型。后两节的基础。 + +第七讲 有依赖的背包问题 +另一种给物品的选取加上限制的方法。 + +第八讲 泛化物品 +我自己关于背包问题的思考成果，有一点抽象。 + +第九讲 背包问题问法的变化 +试图触类旁通、举一反三。 + +附录一：USACO中的背包问题 +给出 USACO Training 上可供练习的背包问题列表，及简单的解答。 + +附录二：背包问题的搜索解法 +除动态规划外另一种背包问题的解法。 + +联系方式 +如果有任何意见和建议，特别是文章的错误和不足，或者希望为文章添加新的材料，可以通过http://kontactr.com/user/tianyi/这个网页联系我。 + +值得说明的是，如果有OI方面的问题，例如不明白自己的程序为什么错了或者索要某种算法的源代码，使用这个联系方式可能得不到及时解答。请在OIBH论坛发问。 + +致谢 +感谢以下名单： + +阿坦 +jason911 +donglixp +LeafDuo +他们每人都最先指出了本文曾经存在的某个并非无关紧要的错误。谢谢你们如此仔细地阅读拙作并弥补我的疏漏。 + +感谢 XiaQ，它针对本文的第一个beta版发表了用词严厉的六条建议，虽然我只认同并采纳了其中的两条。在所有读者几乎一边倒的赞扬将我包围的当时，你的贴子是我的一剂清醒剂，让我能清醒起来并用更严厉的眼光审视自己的作品。 + +sfita 提供了P01中的“一个常数优化”。 + +当然，还有用各种方式对我表示鼓励和支持的几乎无法计数的同学。不管是当面赞扬，或是在论坛上回复我的贴子，不管是发来热情洋溢的邮件，或是在即时聊天的窗口里竖起大拇指，你们的鼓励和支持是支撑我的写作计划的强大动力，也鞭策着我不断提高自身水平，谢谢你们！ + +最后，感谢 Emacs 这一世界最强大的编辑器的所有贡献者，感谢它的插件 EmacsMuse 的开发者们，本文的所有编辑工作都借助这两个卓越的自由软件完成。谢谢你们——自由软件社群——为社会提供了如此有生产力的工具。我深深钦佩你们身上体现出的自由软件的精神，没有你们的感召，我不能完成本文。在你们的影响下，采用自由文档的方式发布本文档，也是我对自由社会事业的微薄努力。 + +首页 + +Copyright (c) 2007 Tianyi Cui + +Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation. + +背包问题九讲 + +version 1.1 build 20071115 + +前言 +目录 +第一讲 01背包问题 +第二讲 完全背包问题 +第三讲 多重背包问题 +第四讲 混合三种背包问题 +第五讲 二维费用的背包问题 +第六讲 分组的背包问题 +第七讲 有依赖的背包问题 +第八讲 泛化物品 +第九讲 背包问题问法的变化 +附录一：USACO中的背包问题 +附录二：背包问题的搜索解法 +联系方式 +致谢 +前言 +本篇文章是我(dd_engi)正在进行中的一个雄心勃勃的写作计划的一部分，这个计划的内容是写作一份较为完善的NOIP难度的动态规划总结，名为《解动态规划题的基本思考方式》。现在你看到的是这个写作计划最先发布的一部分。 + +背包问题是一个经典的动态规划模型。它既简单形象容易理解，又在某种程度上能够揭示动态规划的本质，故不少教材都把它作为动态规划部分的第一道例题，我也将它放在我的写作计划的第一部分。 + +读本文最重要的是思考。因为我的语言和写作方式向来不以易于理解为长，思路也偶有跳跃的地方，后面更有需要大量思考才能理解的比较抽象的内容。更重要的是：不大量思考，绝对不可能学好动态规划这一信息学奥赛中最精致的部分。 + +你现在看到的是本文的v1.1版，发布于2007年11月15日。我会长期维护这份文本，把大家的意见和建议融入其中，也会不断加入我在OI学习以及将来可能的ACM-ICPC的征程中得到的新的心得。但目前本文还没有一个固定的发布页面，想了解本文是否有更新版本发布，可以在OIBH论坛中以“背包问题九讲”为关键字搜索贴子，每次比较重大的版本更新都会在这个论坛里发贴公布。也可以用“背包问题九讲”为关键字在搜索引擎中搜索以得到最新版本。 + +目录 +第一讲 01背包问题 +这是最基本的背包问题，每个物品最多只能放一次。 + +第二讲 完全背包问题 +第二个基本的背包问题模型，每种物品可以放无限多次。 + +第三讲 多重背包问题 +每种物品有一个固定的次数上限。 + +第四讲 混合三种背包问题 +将前面三种简单的问题叠加成较复杂的问题。 + +第五讲 二维费用的背包问题 +一个简单的常见扩展。 + +第六讲 分组的背包问题 +一种题目类型，也是一个有用的模型。后两节的基础。 + +第七讲 有依赖的背包问题 +另一种给物品的选取加上限制的方法。 + +第八讲 泛化物品 +我自己关于背包问题的思考成果，有一点抽象。 + +第九讲 背包问题问法的变化 +试图触类旁通、举一反三。 + +附录一：USACO中的背包问题 +给出 USACO Training 上可供练习的背包问题列表，及简单的解答。 + +附录二：背包问题的搜索解法 +除动态规划外另一种背包问题的解法。 + +联系方式 +如果有任何意见和建议，特别是文章的错误和不足，或者希望为文章添加新的材料，可以通过http://kontactr.com/user/tianyi/这个网页联系我。 + +值得说明的是，如果有OI方面的问题，例如不明白自己的程序为什么错了或者索要某种算法的源代码，使用这个联系方式可能得不到及时解答。请在OIBH论坛发问。 + +致谢 +感谢以下名单： + +阿坦 +jason911 +donglixp +LeafDuo +他们每人都最先指出了本文曾经存在的某个并非无关紧要的错误。谢谢你们如此仔细地阅读拙作并弥补我的疏漏。 + +感谢 XiaQ，它针对本文的第一个beta版发表了用词严厉的六条建议，虽然我只认同并采纳了其中的两条。在所有读者几乎一边倒的赞扬将我包围的当时，你的贴子是我的一剂清醒剂，让我能清醒起来并用更严厉的眼光审视自己的作品。 + +sfita 提供了P01中的“一个常数优化”。 + +当然，还有用各种方式对我表示鼓励和支持的几乎无法计数的同学。不管是当面赞扬，或是在论坛上回复我的贴子，不管是发来热情洋溢的邮件，或是在即时聊天的窗口里竖起大拇指，你们的鼓励和支持是支撑我的写作计划的强大动力，也鞭策着我不断提高自身水平，谢谢你们！ + +最后，感谢 Emacs 这一世界最强大的编辑器的所有贡献者，感谢它的插件 EmacsMuse 的开发者们，本文的所有编辑工作都借助这两个卓越的自由软件完成。谢谢你们——自由软件社群——为社会提供了如此有生产力的工具。我深深钦佩你们身上体现出的自由软件的精神，没有你们的感召，我不能完成本文。在你们的影响下，采用自由文档的方式发布本文档，也是我对自由社会事业的微薄努力。 + +首页 + +Copyright (c) 2007 Tianyi Cui + +Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation. + +> \ No newline at end of file