这是俄罗斯方块的原代码 虽然不是太完善 但是已经很具构架

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MODBUS协议调试软件，做这的比较好用，有用的就下载-MODBUS protocol debugging software, make this comparison useful and useful on the download

DOS系统下真彩显示24KBMP图象的源程序-DOS system 24KBMP true color images show the source code

restoring除法器设计 经典算法了，可以仿真通过-divider restoring a classical algorithm design, simulation can be adopted

[输入] 图的顶点个数N，图中顶点之间的关系及起点A和终点B [输出] 若A到B无路径，则输出“There is no path” 否则输出A到B路径上个顶点 [存储结构] 图采用邻接矩阵的方式存储。 [算法的基本思想] 采用广度优先搜索的方法，从顶点A开始，依次访问与A邻接的顶点VA1,VA2,...,VAK, 访问遍之后,若没有访问B，则继续访问与VA1邻接的顶点VA11,VA12,...,VA1M，再访问与VA2邻接顶点...，如此下去，直至找到B，最先到达B点的路径，一定是边数最少的路径。实现时采用队列记录被访问过的顶点。每次访问与队头顶点相邻接的顶点，然后将队头顶点从队列中删去。若队空，则说明到不存在通路。在访问顶点过程中，每次把当前顶点的序号作为与其邻接的未访问的顶点的前驱顶点记录下来，以便输出时回溯。 #include int number //队列类型 typedef struct{ int q[20] -[imported] map of the number of vertices N, Vertex map of the relationship between the starting point and end point A and B [output] if A to B without path then exporting "There is no path" Otherwise output A to B on the path vertices [storage structure] plan adopted adjacency matrix of storage. [Algorithm for the basic idea] BFS way from the vertex A, A visit followed with adjacent vertices VA1, VA2 ,..., VAK, visiting times, if not visit B, continue to visit with the VA1 adjacent vertices VA11, VA12 ,..., VA1M. VA2 and then

一个银行业务的程序-A banking procedures

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linux下的俄罗斯方块，程序短小精悍，代码仅有不到100k，适合于初学C++语言想看看大程序结构的同行们。-Russia under linux box, procedures small and code with less than 100k, suitable for a novice C language to look at the program structure counterparts.

此代码为udp通信程序，主要是从客户端发送一数据，然后在在服务器段接收。-This code for the udp communication procedures, mainly from one client to send data, and then the server receives paragraph.

teach yourself C++ in 21 days 第五版-teach yourself C in the fifth edition of 21 days