货物配送物流系统 1. 综合查询和统计 对客户、货物和运输线路的数量、收入和成本的综合查询和统计分析。 2. 流程和预警 系统提供分级的流程...

本代码是光子晶体光纤切面的光强分布仿真程序，对搞光子晶体的同行很有帮助。-the code is photonic crystal fiber plane, the intensity distribution simulation program to engage in peer photonic crystal helpful.

This software Upgrader Ali3101

本文详细介绍了IAR FOR avr 编译器的基础，为刚接触此编译器的人提供了快速掌握的方法。-This paper describes the IAR FOR avr compiler foundation for people new to this compiler provides a fast method of disposal.

dt900原码 casio盘点机源代码 可用作参考-dt900 inventory of the original code Casio machine can be used as a reference source

利用GeoSuff开发网络电子地图的详细设计书~~希望对大家有所启发-GeoSuff development of network use electronic maps detailed design document ~ ~ hope to inspire everyone!

51源程序,希望大家能喜欢，谢谢大家的支持

Delphi做的图书馆管理系统的相应毕业论文-library management system corresponding thesis

8051 based microcontroller project

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

S3C2410 linux指导书，很经典的，-S3C2410 linux指导书