自己做的一个学生成绩管理系统，用于对学生成绩进行查看，添加，删除等等。...

The system will automatically delete the debug and release directory, so please do not file on these two directory

A C source file printer with multiple options

往手机通讯录里添加图片的一个好程序，是微软智能手机学习的好例子-Mobile phone address book to add a good picture of the process, is Microsoft" s smart phone to learn a good example of

VC++ 云台摄像头控制程序源代码。控制云台和镜头。。

实用编程 实用编程-Practical Programming Prac tical Programming

2人五子棋游戏源，C写的，有兴趣的可以研究研究，希望大家一起讨论，献丑了-two games were the source of 331, C written in the interest of the studies can hope that we can discuss the shortcoming

孙鑫老师的VC++教程的光盘源代码，数写的很好，代码也很好，值得研究！第10章源程序！-SUN Xin teachers VC++ Tutorial CD-ROM source code, written in a few very good, code is also good, worthy of study! Chapter 10 source code!

One of the best document you may find to simplimfy and give the best results for code review.

还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不断地循环 e 次之后，很多情况下都可以循环出原文，只是计算量过余多一些罢了。不过有不少情况下，根本都无须循环 e 次，不过对于1024位的 n 级别来说，e 也是一个相当大的数值，所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考，如有不实，请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected

这是关于压缩感知的OMP算法，代码不仅包含OMP算法，还包括测量数M与重构成功概率的关系和稀疏度K与重构成功率的关系。