自动关机（可设置时间，问题很多，可作为新手学习之用）.

参加运动会的 个学校编号为 。比赛分成 个男子项目和 个女子项目，项目编号分别为 和 。由于各项目参加人数差别较大，有些项目取前五名，得分顺序为7，5，3，2，1；还有些项目只取前三名，得分顺序为5，3，2。写一个统计程序产生各种成绩单和得分报表。 2、要求产生各学校的成绩单，内容包括各校所取得的每项成绩的项目号、名次（成绩）、姓名和得分；产生团体总分报表，内容包括校号、男子团体总分、女子团体总分和团体总分。 3、测试数据：对于 ， ， ，编号为奇数的项目取前五名，编号为偶数的项目取前三名，设计一组实例数据。 (The school serial number participating in Games is. Competition divides into men s event and women s event , the project serial number parts for the sum. The difference is bigger since every project participates in a number , some projects choose the first five , score order is 7 , 5 , 3 , 2 , 1 Still have some of projects taking the first three places , score only being 5 , 3 , 2 in proper order. Write a form for report counting procedure producing the various school report card and score. 2, demands the school report card producing every school , content includes every achievement project number , position in a name list (achievement) , full name and score per got by school Produce the group total points form for report , content including the school number , male person group total points , woman group total points and group total points. 3, testing data: Be that the odd number project chooses the first five to the serial number, the serial number is that the even number project c)

核心实例 POP3上电子邮件的检查,POP3协议，网络编程(POP3 email check)

outlook风格的界面例子(Outlook style interface example)

用动态规划法求两个字符串 A="xzyzzyx"和 B =”zxyyzxz“的最长公共子序列(For two strings A dynamic programming method = " xzyzzyx" and B = " zxyyzxz" the longest common subsequence)

[Introduction of the example] Jwt的使用Demo# 包含内容- asp.net core集成jwt权限验证- token的强制失效- 对api接口进行基于策略模式的权限验证[Screenshot of the example] [Core code]

残缺棋盘问题的VC++实现，用分而治之方法解决此问题(Defective chessboard problem VC++ Realize, using divide and conquer method to solve this problem)

3路寻迹测试程序 智能小车寻迹程序 红外对管(intelligent; intelligence; brain power; noopsyche dolly; wheelbarrow; handbarrow; sedan; handcart procedure; order; program; software; formality infrared)

说明：  1、 创建图类，存储结构使用邻接矩阵。 2、 输入图的节点数n（不超过10个）、边数m，节点分别用1-n代表。 3、 采用“起始节点，终止节点，权值”输入图的m条边，创建图。 4、 输出从节点1开始的BFS遍历，在遍历过程中，如有多个可以选择的节点，则优先选择编号较小的节点。 5、 输出从节点1开始的DFS遍历，在遍历过程中，如有多个可以选择的节点，则优先选择编号较小的节点。 6、 输出从第1节点到第n节点最短路径的长度，如果没有路经，输出0。(1. Create graph class and use adjacency matrix in storage structure. 2. The number of nodes n (no more than 10) and edge m in the input graph are represented by 1-n, respectively. 3. Create a graph by using the m-edges of the input graph of "Starting Node, Terminating Node, Weight Value". 4. Output BFS traversal starting from node 1. In the traversal process, if there are multiple nodes to choose, the node with smaller number will be preferred. 5. Output DFS traversal starting from node 1. In the traversal process, if there are multiple nodes to choose, the node with smaller number will be preferred. 6. Output the length of the shortest path from the first node to the n node. If there is no path, output 0.)

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this is a windows7 usb&dvd intalling tool source code.-this is a windows7 usb&dvd intalling tool source code.