词法编译程序，不是太长

GSM_TEST软件，可通过对GSM900模块进行在线测试！

Turbo c 2.0开发,可以对硬盘的主引导扇区的某些信息进行加密,从而对这个硬盘其到加密作用!!这样就不能进入系统了-Turbo c 2.0, the drive can master boot some of the information encrypted on the hard drive so as to its encryption role! ! This can not enter the system

一个简单人事管理系统，主要介绍ASP.NET和数据库以及网站等相关方面知识的应用-A simple personnel management system that focuses on ASP.NET and databases, as well as websites related to the application of knowledge

用C++实现m着色问题-achieve m C Coloring

QC-LDPC是准循环奇偶校验码，在编码和解码的的过程中更加简单，能够避免生成出现较多的1，QC-LDPC的校验矩阵的构造采用了PEG算法，并且带有了消除4环和6环的功能，构造方法简单。

是游戏初学者的良师益友，我很喜欢，新手必不可少的书籍及参考资料-The mentor is the game for beginners, I like, and newcomers essential reading and reference materials

text parser Please read your package and describe it at least 40 bytes in English. System will automatically delete the directory of debug and release, so please do not put files on these two directory.

c# 画面分割软件 监控系统使用  nvr dvr

windows 编程基础知识，讲解清楚透彻，windows初学者有极大的帮助！-basic knowledge of windows programming, on a clear and thorough, windows have great help for beginners!

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