urgvide

描述：足球俱乐部包括球员、普通行政人员。球员有姓名、年龄、能力、年薪、转会费、进球总数、服役年限等重要参数；普通行政人员有姓名、年龄、能力、年薪等参数。你需要给相关管理机构开发一个管理程序，实现对众多足球俱乐部管理的基本功能。具体功能包括： a) 俱乐部的增删改查，俱乐部的属性包括名称、现金、其下人员； b) 可以对所有球员的各项属性进行简单搜索，支持输入多个条件，条件之间用&&连接，例如： i. “年龄不大于25且服役年限大于5年且进球数大于100的球员”，查询表达式：!(@age>25) && @experience>5 && @kick>100 c) 支持球员在不同俱乐部间的买卖，买卖需符合以下条件：作为买方的俱乐部的现金>=球员的转会费，交易完成后，卖方将得到转会费。 要求： d) 必须采用面向对象的方式，使用类组织数据结构，并注意类的继承关系； e) 在题目要求基础上可以自行扩展功能，提供扩实用扩展功能者适当加分（加分不超过5分）。 (Description: Football club including players, general administrative staff. Player name, age, ability, salary, transfer fee, total number of goals, service life and other important parameters ordinary administrative staff name, age, ability, salary and other parameters. You need to give the relevant management agencies to develop a management program to achieve the number of football club management s basic functions. Specific features include: a) the club CRUD, the name of the club s property, including cash, under which personnel b) all players can perform a simple search of the property, supports input multiple conditions with && connection between conditions, for example: i. "Age is not greater than 25 and greater than 5 years service life and goals more than 100 players," query expression:! (@ age> 25) && @ experience> 5 && @ kick> 100 c) Support the players traded between different clubs, the sale subject to the following conditions: As a buyer s club cash> = pl)

实现链表的一些操作，方便进行遍历，查找等程序(Operation, convenient linked list traversal, the search procedure)

博弈树，是数据库课程中非常重要的一个编程程序。里面的编程思想非常适合初学者(Game tree, which is the database course a very important programming procedures. Inside the programming ideas for beginners)

任务：实现Dijkstra算法 工程名：Project_ShortestPath 要求：实现Dijkstra算法，算法主体写在CGraph类中；要求是从command窗口中输入源节点编号，并在command窗口中输出从源节点到所有目的节点的最短路长及最短路。注：要输出整条路径，而不仅仅是前一跳，也不允许逆序输出。 (Task: Dijkstra' s algorithm to achieve the project name: Project_ShortestPath requirements: achieve Dijkstra algorithm, the algorithm written in the body CGraph class requires input from the command window, the source node number, and in the command window, all of the output from the source node to the destination node shortest long and the shortest. Note: To export the entire path, not just before the jump, do not allow reverse output.)

智能车库管理系统 MFC 车辆的存放 计费 修改 会员管理 数据记录（DAQ数据库）(The MFC system of parkiing cars.)

单向链表的创建与操作 设单向链表中节点的数据域的数据类型为整型，编写函数实现以下操作： （1）实现单向链表的创建（包括初始化）与输出操作，节点的个数及节点的数据由用户输入。 （源代码：ds3-1.c） （2）查找给定的单链表中的第i个节点，并将其地址返回。若不存在第i个节点，则返回空地址。 （源代码：ds3-2.c） （3）查找给定的单链表中值为n的节点，并将其地址返回。若不存在值为n的节点，则返回空地址。同时，还应通过参数传回该节点的序号。 （源代码：ds3-3.c） （4）删除给定的单链表中的第i个节点，成功返回1，失败返回0。 （源代码：ds3-4.c） （5）删除给定的单链表中值为n的节点，成功返回1，失败返回0。 （源代码：ds3-5.c） （6）在给定的单链表的第i位上插入值为n的节点。 （源代码：ds3-6.c） （7）在给定单链表的值为m的节点的前面插入一个值为n的节点。 （源代码：ds3-7.c） (Creation and operation of a one-way linked list Set up a one-way linked list data type node integer data fields , write a function to achieve the following: ( 1 ) achieve the creation of a one-way linked list ( including initialization ) and output operation , the number of nodes and node data entered by the user . ( Source : ds3-1.c) ( 2 ) Find a single list given in the i-th node and returns its address . Without the presence of the i-th node , returns an empty address. ( Source : ds3-2.c) ( 3 ) Find a given node in a given value of n single list , and return address . Without the presence of the value of n nodes , returns an empty address. Meanwhile, the number should be returned by the parameters of the node . ( Source : ds3-3.c) ( 4 ) Delete the given singly linked list in the i-th node , the successful return 1, else return 0 . ( Source : ds3-4.c) ( 5 ) to delete a single node in the list is given n , the successful return 1, else return 0 . ( Source : ds3-5.c) ( 6 ) )

约瑟夫环（约瑟夫问题）是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。通常解决这类问题时我们把编号从0~n-1，最后结果+1即为原问题的解。 (Josephus (Josephus problem) is the application of a mathematical problem: Given n individuals (with numbers 1,2,3 ... n respectively) sitting around a round table. From the number of people gettin k, number of the m man out of the line he s the next person and a number of gettin number to m the man was out of the line and so the law is repeated until the round table were all out of the column. We numbered 0 ~ n-1, the final result is the original problem solution+1 usually solve these problems.)

本程序实现一元多项式合并。本程序输入方式为按项输入，每输入一个项，会先提示输入项的幂，再提示输入项的系数。(This procedure combined to achieve a dollar polynomial. This procedure according to entry input methods for the input, one for each input item, it will first prompt entry of the power, and then prompts entry of the coefficient.)

设计一个表达式求值的程序。该程序必须可以接受包含（，），+，-，*，/的中缀表达式，并求出结果。如果表达式正确，则输出表达式的结果；如果表达式非法，则不输出或输出其他字符。(Design an expression evaluation procedures. The program must accept (,), including +, -, *, / the infix expression, and the results obtained. If the expression is correct, the output of the expression is output; if the expression is illegal, no other characters are output or exported.)

为了点分 求分下载好麻烦 c++二叉排序树(In order to find points dotted download good trouble c++ binary sort tree)