Hospital Management System Database Design

SSD7 考试2 EXAM2 的最完整答案，自己好不容易写的，呵呵~-SSD7 exam 2 EXAM2 the most complete answers to their own hard-won writing,呵呵~

JXARM9-2410 Flash烧写流程.pdf-JXARM9-2410 Flash programmer flow. Pdf

包含有软件需求开发与需求管理、高质量软件开发之道、软件测试、软件过程改进与CMM、需求开发与需求管理等5个文档，对与已经从事或者即将从事软件开发的人员将有很大的帮助。-includes software development and demand for demand management, high-quality software development generally, software testing, software process improvement and CMM, and the development of demand management needs five documents, and to have engaged in or about to engage in software development will be of great help.

Java代码编写的30条建议,很有参考价值-Java code prepared 30 recommendations, a good reference

peer to peer : systems and concepts

This a test project-This is a test project

此shopping购物程序小巧使用，对小型购物商店来说实用。-small shopping procedures used for small shops practical terms.

使用威尔配对,本文提出了一种基于身份的环签密方案,给出了具体的算法.该方案能够使消息的 发送者以一种完全匿名的方式发送消息,并且同时实现保密性和认证性两种功能.我们证明了在决策双线性Diffie- Hellman问题难解的假设下,新提出的方案对自适应选择密文攻击是安全的.与传统的先签名后加密的方案 相比,本方案中密文长度有了明显的降低,在低带宽的要求下更加可行.-Will the use of matching, this paper, a ring of identity-based signcryption program, given a specific algorithm. The program will enable the sender information in a completely anonymous way to send messages, and at the same time to achieve confidentiality and authentication of two functions. We prove that in the decision-making bilinear Diffie-Hellman problem difficult to resolve under the assumption that the new proposal for adaptive chosen ciphertext attack is safe. with the traditional first signed encrypted program compared ciphertext of the program significantly reduced the length, in the low-bandwidth request more feasible.

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

电话卡系统，能完整的实现电话卡充值，查询的系统-Calling card system that can realize the complete calling card recharge, query system