我选择约瑟夫问题（附带一个合数段的问题这是我们第一次的考试题）做我最后一个上传的代码。因为这是我付出最多的一个，我的第一个程序，当时全班就我一个弄了出来现在想起...

vb编写的云台镜头控制。主要通过pc机的串口通讯。-vb prepared pan and camera control. Pc machine, mainly through the serial port communications.

只要你添加本章的内容到你希望联机的游戏上，再利用第七章编程的思想，任何简单的游戏，你都可以轻松地实现联机。-this chapter as long as you are adding to the content you want online games, we can use Chapter VII of programming ideas, any simple game, you can easily realize on-line.

网络实验：模拟网桥试验-network experiment : Analog Bridge Test

MDIPicSize控件 上传资料内有详细功能，自己查阅-MDIPicSize

在VC++环境下完全使用API实现的俄罗斯方块-VC++ API implementation of the Tetris

自己写的JAVA贪吃蛇程序,相当有参考价值-wrote JAVA huge memory process, have considerable reference value

功能：快进，播放，暂停，后退，调节声音大小，添加播放文件，-Features: fast forward, play, pause, rewind, adjust the size of the sound, add play the file,

[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

046测试~！！ ！！ ！！！ -046 test ~! ! ! ! ! ! ! ! ! ! ! 11

fsk modulation Frequency shift keying (FSK) is the most common form of digital modulation in the high-frequency radio spectrum, and has important applications in telephone circuits. This article provides a general tutorial on FSK in its many forms. Both modulation and demodulation schemes will be discussed