update

算法分析与设计/大作业/coin.py

@@ -0,0 +1,35 @@
+def coin_change_min_weight(v, w, y):
+
+    n = len(v)  # 硬币种类数
+
+    # 初始化动态规划表
+    F = [float('inf')] * (y + 1)
+    F[0] = 0
+
+    # 用于记录选择的硬币
+    choice = [[0] * n for _ in range(y + 1)]
+
+    # 动态规划过程
+    for j in range(1, y + 1):
+        for k in range(n):
+            if v[k] <= j:
+                if F[j - v[k]] + w[k] < F[j]:
+                    F[j] = F[j - v[k]] + w[k]
+                    choice[j] = choice[j - v[k]].copy()
+                    choice[j][k] += 1
+
+    return F[y], choice[y]
+
+# 测试代码
+if __name__ == "__main__":
+    v_test = [1, 4, 6, 8]  # 硬币面值
+    w_test = [1, 2, 4, 6]   # 硬币重量
+    y_test = 12             # 需要支付的总价值
+
+    min_weight, coin_used = coin_change_min_weight(v_test, w_test, y_test)
+
+    print(f"最小总重量: {min_weight}")
+    print("使用的硬币:")
+    for i, count in enumerate(coin_used):
+        if count > 0:
+            print(f"面值 {v_test[i]}: {count} 个")

算法分析与设计/大作业/dazuoye.tex

@@ -0,0 +1,177 @@
+\documentclass{article}
+\usepackage{fancyhdr}
+\usepackage{extramarks}
+\usepackage{amsmath}
+\usepackage{amsthm}
+\usepackage{amsfonts}
+\usepackage{tikz}
+\usepackage[UTF8, scheme=plain, punct=plain, zihao=false]{ctex}
+\usepackage[shortlabels]{enumitem}
+\usetikzlibrary{automata,positioning}
+\usepackage{algorithm}
+\usepackage{algpseudocode}
+\usepackage{float}
+\usepackage{lipsum}
+
+\renewcommand{\algorithmicrequire}{\textbf{Input:}}  % Use Input in the format of Algorithm
+\renewcommand{\algorithmicensure}{\textbf{Output:}} % Use Output in the format of Algorithm
+\newcommand{\True}{\textbf{true}}
+\newcommand{\False}{\textbf{false}}
+\usepackage{graphics}
+\usepackage{epsfig}
+\usepackage{indentfirst}
+
+% 算法跨页
+\makeatletter
+\newenvironment{breakablealgorithm}
+  {\begin{center}
+     \refstepcounter{algorithm}% New algorithm
+     \hrule height.8pt depth0pt \kern2pt% \@fs@pre for \@fs@ruled
+     \renewcommand{\caption}[2][\relax]{% Make a new \caption
+      {\raggedright\textbf{\ALG@name~\thealgorithm} ##2\par}%
+      \ifx \relax##1\relax % #1 is \relax
+         \addcontentsline{loa}{algorithm}{\numberline{\thealgorithm}##2}%
+      \else % #1 is not \relax
+         \addcontentsline{loa}{algorithm}{\numberline{\thealgorithm}##1}%
+      \fi
+      \kern2pt\hrule\kern2pt
+     }
+  }{\kern2pt\hrule\relax
+   \end{center}
+  }
+\makeatother
+
+\topmargin=-0.45in
+\evensidemargin=0in
+\oddsidemargin=0in
+\textwidth=6.5in
+\textheight=9.0in
+\headsep=0.25in
+
+\linespread{1.1}
+
+\pagestyle{fancy}
+\lhead{\hmwkAuthorName}
+\chead{\hmwkClass\ \hmwkTitle}
+\rhead{\firstxmark}
+\lfoot{\lastxmark}
+\cfoot{\thepage}
+
+\renewcommand\headrulewidth{0.4pt}
+\renewcommand\footrulewidth{0.4pt}
+
+\newcommand{\enterProblemHeader}[1]{
+    \nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
+    \nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
+}
+
+\newcommand{\exitProblemHeader}[1]{
+    \nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
+    \stepcounter{#1}
+    \nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
+}
+
+\setcounter{secnumdepth}{0}
+\newcounter{partCounter}
+\newcounter{homeworkProblemCounter}
+\setcounter{homeworkProblemCounter}{1}
+\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
+
+\newenvironment{homeworkProblem}[1][-1]{
+    \ifnum#1>0
+        \setcounter{homeworkProblemCounter}{#1}
+    \fi
+    \section{\huge{Problem} \arabic{homeworkProblemCounter}}
+    \setcounter{partCounter}{1}
+    \enterProblemHeader{homeworkProblemCounter}
+}{
+    \exitProblemHeader{homeworkProblemCounter}
+}
+
+% Homework Details
+\newcommand{\hmwkTitle}{第1次作业}
+\newcommand{\hmwkDueDate}{Oct. 29, 2020}
+\newcommand{\hmwkClass}{计算机算法设计与分析}
+\newcommand{\hmwkAuthorName}{张三 202180132...}
+
+\title{
+    \vspace{2in}
+    \textmd{\textbf{\Huge{\hmwkClass}
+    \\ \Large{\hmwkTitle}}}\\
+    \vspace{3in}
+}
+
+\author{\Large{\textbf{张三}}\\\textbf{2020280132.....}}
+\date{}
+
+\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
+
+\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
+
+% For derivatives
+\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
+
+% For partial derivatives
+\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
+
+% Integral dx
+\newcommand{\dx}{\mathrm{d}x}
+
+% Alias for the Solution section header
+\newcommand{\solution}{\textbf{\large Solution}}
+
+% Probability commands: Expectation, Variance, Covariance, Bias
+\newcommand{\E}{\mathrm{E}}
+\newcommand{\Var}{\mathrm{Var}}
+\newcommand{\Cov}{\mathrm{Cov}}
+\newcommand{\Bias}{\mathrm{Bias}}
+
+\begin{document}
+\alglanguage{pseudocode}
+\maketitle
+
+\pagebreak
+\begin{homeworkProblem}
+\textbf{\large{Assignment Problem4}}
+\\\solution\\
+\begin{breakablealgorithm}
+\begin{algorithmic}[1]
+  \caption{Algorithm}
+    \Require{$A,B$}\Comment{A is ..., B is ...}
+    \Ensure{$[C,D]$}\Comment{C is ..., D is ...}
+    \State Initialize $A=-1,B=-1$;
+    \State \Call{SubFunction}{A,B}
+    \If{A}
+        \State return C;
+    \EndIf
+\Function{SubFunction}{...}
+\Comment{$如果...就...$}
+\EndFunction
+\end{algorithmic}
+\end{breakablealgorithm}
+\subsection{Subproblem Reduced Graph}
+以...为例，查看图...。
+\subsection{Proof of the correctness}
+要证明的循环不变量：
+\begin{itemize}
+    \item Initialization：
+    \item Maintenance：
+    \item Termination:
+\end{itemize}
+
+\证明：
+\begin{itemize}
+    \item Initialization：
+    \item Maintenance：
+    \item Termination:
+\end{itemize}
+
+\subsection{Analysis of Complexity}
+    \textbf{时间复杂度}
+
+        $O(x)$。
+
+    \textbf{空间复杂度}
+\end{homeworkProblem}
+
+\end{document}

算法分析与设计/大作业/lightest_coin.m

@@ -0,0 +1,45 @@
+% 测试代码
+v_test = [1, 4, 6, 8];  % 硬币面值
+w_test = [1, 2, 4, 6];   % 硬币重量
+y_test = 12;             % 需要支付的总价值
+[minWeight, coinUsed] = coinChangeMinWeight(v_test, w_test, y_test);
+fprintf('最小总重量: %d\n', minWeight);
+fprintf('使用的硬币:\n');
+for i = 1:length(v_test)
+    if coinUsed(i) > 0
+        fprintf('面值 %d: %d 个\n', v_test(i), coinUsed(i));
+    end
+end
+function [minWeight, coinUsed] = coinChangeMinWeight(v, w, y)
+    n = length(v);  % 硬币种类数
+    % 初始化动态规划表
+    F = inf(n+1, y+1);
+    F(:, 1) = 0;  % 修正：将第一列（总价值为0）初始化为0
+    % 用于记录选择的硬币
+    choice = zeros(n+1, y+1);
+    % 动态规划过程
+    for k = 2:n+1
+        for j = 1:y+1
+            % 不选择第k种硬币
+            F(k, j) = F(k-1, j);
+
+            % 如果可以选择第k种硬币
+            if j > v(k-1)
+                if F(k, j-v(k-1)) + w(k-1) < F(k, j)
+                    F(k, j) = F(k, j-v(k-1)) + w(k-1);
+                    choice(k, j) = 1;
+                end
+            end
+        end
+    end
+    % 回溯找出使用的硬币
+    minWeight = F(n+1, y+1);
+    coinUsed = zeros(1, n);
+    j = y+1;
+    for k = n+1:-1:2
+        while choice(k, j) == 1
+            coinUsed(k-1) = coinUsed(k-1) + 1;
+            j = j - v(k-1);
+        end
+    end
+end

算法分析与设计/大作业/作业要求.md

@@ -0,0 +1,2 @@
+设有n种不同面值的硬币，第i中硬币的币值是 v_i，重量是w_i=1,2,,,,n。且现在购买总价值为y的某些商品，需要用这些硬币付款，若每种钱币使用的个数不限，问如何选择付款的方式使得付出的总重量最轻？给出该问题的数学描述，并设计一个求解该问题的算法，给出算法的伪码描述，以及实现代码。假设问题的输入实例是：\nu_1=1,\quad\nu_2=4,\quad\nu_3=6,\quad\nu_4=8\text{ ; }w_1=1,\quad w_2=2,\quad w_3=4,\quad w_4=6,\quad y=12
+给出算法所获得的付款方式。

算法分析与设计/大作业/大作业latex/homework.aux

@@ -0,0 +1,33 @@
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