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建立链表的算法： 新节点链入表尾建立链表： 动态分配第一个节点变量； 用指针head 妥善保存链表第一个节点的地址值； 依次动态分配一个新节点变量； 将上一个节点的指针域指向新生成的节点； 循环3),4)步骤，直到所有节点均连入链表； 将最后一个节点的指针域赋值为0； (Establish the list of algorithms: The new node links footer build list: Dynamic allocation of the first node variable Keep the list head pointer of the first node address value In order to dynamically allocate a new node variable The domain of a node pointer to the new generation of nodes Cycle 3), 4) step, until all nodes are connected to the chain Pointer field will last node assigned to 0 )

采用Kruskal算法求最小生成树主要数据结构 edgeset GE 存放图中的所有边 int n,int e 存放图中的顶点数与边数 edgeset C 存放生成树中的边 vexlist gv 图中结点的顶点值 adjmatrix s 用来处理图中结点的查找与合并 int m1,m2 一条边上两顶点所属集合的序号 int k 最小生成树中的边数 int d //图中待扫描边元素的下标(For the Minimum Spanning Tree by Kruskal main data structures edgeset GE store all edges in the figure int n, int e store graph vertices and edges edgeset C storage spanning tree edges in the vexlist gv graph node vertex values ​ ​ adjmatrix s used to process graph node to find and merge int m1, m2 an edge belongs to the set number two vertices int k minimum spanning tree edges int d // figure to be Scan edge element subscript)

简单LR语法分析程序自动生成器的实现，可用作数据结构课程设计(LR (0) grammar analysis)

魔王语言：输入一串字符，可自动翻译成一串语言。(the program of language)

多项式求和，用链表实现的一元多项式求和程序，比较 靠谱(Polynomial sum, with the linked list implementation of a polynomial summation procedure, more reliable)

阿当姆斯显式和隐式求解方法，用四阶龙格库塔作为起始，然后运用四阶阿当姆斯算法求解初值问题。主要程序包含在test2.cpp中，方法简单易懂。编译环境VC2010(Adam James explicit and implicit method for solving fourth-order Runge-Kutta as a start, then use the fourth-order A Williams algorithm for solving initial value problem. The main program contains in test2.cpp method is simple and easy to understand. Compile environment VC2010)

X位赋0mintest是实现芯片功耗测试不确定位填充算法(X-bit chip is to achieve the power conferred 0mintest test uncertain bit stuffing algorithm)

　广度优先搜索及深度优先搜索，用C语言实现，比较使用(Breadth-first search and depth first search, using C language, compared with)

约瑟夫问题，通过C语言对约瑟夫问题进行了分析和讲解，从而解决问题(Joseph problems, by Joseph C language issues are analyzed and explained, in order to solve the problem)

给定n种物品和一背包。物品i的重量是wi>0,价值为vi>0,背包的容量为c。选择装入背包的物品，使得装入背包中物品的总价值最大。(Given n kinds of goods and a backpack. Weight of the item i is wi> 0, the value of vi> 0, knapsack capacity c. Select the items into the backpack, so into the backpack of items in the total value of the maximum.)