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小型CA系统.zip
- 前些天在网上看到了一些关于OPENSSL的介绍，觉得很有意思，于是做了一个程序，基本实现了数字证书的制作、SSL安全通讯、加解密操作等功能，秉承OPENSSL开放的原则，拿出来共享，主要实现写在了两个DLL中, The first days on-line saw some about the OPENSSL introduction, have thought very interestingly, thereupon has made a procedure, basically has
minidwep
- minidwep-gtk,一个aircrack-ng的图形前端gui， 全中文，比spoonwep好用 [attach]95802[/attach] 遵守规定，点确定 [attach]95803[/attach] 看看 左边有没有你的网卡，如果没有，说明软件不支持 你的 ... 想要的东西出来了 -minidwep-gtk, a graphical front-end aircrack-ng gui, all Chinese, easy to use than sp
TEA
- 本加密程序基于著名的TEA加密算法,密钥长度为128位,安全性高于DES算法(56位). 该算法安全、简洁、高效,加密解密速度极快,占用内存极少，非常适合于嵌入式系统 的加密解密，也能胜任大系统的安全性能要求。算法能防止常见的密码学攻击。 本程序中待加密的明文、加密生成的密文皆按64位数据分组进行加解密。 密钥长度128位，与AES、IDEA相当，从而确保足够的安全性。-The encryption program is based on the famous TEA
netbar
- 计件工资管理系统。挺好的一个系统，可以查询，添加……进行一些最基本的功能-Piece-rate management system. A very good system that can query, add ... ... to carry out some of the most fundamental function of
DTCSP
- 三代卡DTCSP编程手册，绝对难找，对加密卡的操作进行了详细的说明-Three generations of cards DTCSP programming manual, definitely hard to find, on encryption card operation carried out a detailed descr iption of
qiujumima
- 要穷举组合的算法，比如密码词典，算法有不够完善的地方大家指出来一起学习研讨-Algorithm to exhaustive combinations, such as the password dictionary, algorithms are not perfect enough where we pointed out that the study of the issues together
NOX5
- 感谢您选择龙脉科技加密产品！本文档帮助您快速完成加密保护工作。 以下建议旨在帮助您快速开展后面的工作： 1请认真阅读密码单中重要信息。 2将锁插入USB接口中，待指示灯点亮后继续下面的步骤。 3请在sample目录中找到您熟悉的开发语言或环境的例子，运行例程，参考这些例程了解加密锁API的使用方法。 4参考过程中，如果遇到陌生的概念或参数，请在help目录中查看帮助文档。 5熟悉了API的使用方法之后，在tools目录下运行设号工具，设置锁的相关 参数，请牢记超级密码
gtea
- Based on the key material generated 3584 Round table of the sample plug and a length of 256 x kb. Key schedule includes pre-encryption round key encryption procedure. Encryption is technology-dependent data permutations, sampling subkey for subblock
g_tea
- Based on the key material generated 3584 Round table of the sample plug and a length of 256 x kb. Key schedule includes pre-encryption round key encryption procedure. Encryption is technology-dependent data permutations, sampling subkey for subblock
1587-anqn
- 用于查看文件被谁加锁了，这个代码是很久以前的，但感觉还是很有用，于是就奉献出来了。-Whom used to view the file locking, and this code is a long time ago, but the feeling is still very useful, so he dedicated out.
KanqLic.dll
- 生成对应注册码的系统信息码 可在winxp系统下进行，无须解压-Generated code corresponding to your key system information can be carried out in the winxp system, without extracting
model_field_11[1].cache
- MD5伪造工具(不要做坏事噢) MD5碰撞测试程序并提供源代码. 利用该程序可伪造出有相同MD5的不同程序.-MD5 crash test program and provide the source code. Use this program can be forged out different programs with the same MD5.
PRAN
- 在使用C#编程，但伪随机数加密一直不知道是怎么实现的，今天研究了一下，看了一个使用伪随机数加密用户名密码的例子，觉得简单易懂，索性把源代码也共享出来，帮助更多想实现C#伪随机数加密的朋友。在演示窗口，输入密码的时候，会适时显示出经过加密的密码，挺有参考价值。-In the use of C# programming, but pseudo-random number encrypted did not know how to achieve today study a little, read
bfsh
- 加密算法blowfish的实现源码，本包中包涵有blowfish的三种语言的实现源码C、C++、java；都是很经典的代码，特别拿出来和大家分享-Blowfish encryption algorithm to achieve the source of the package bear with blowfish realize the source of the three languages C, C++, java are very classic code
Idea
- Overview: This code does the following: - print out all encryption and decryption subkeys which are used in the encryption and decryption process - encrypts plaintext message - decrypts ciphertext message - shows detailed, roun
Caesar-code
- software implementation of a cryptosystem, carrying out encoding and decoding system, Caesar cipher -software implementation of a cryptosystem, carrying out encoding and decoding system, Caesar cipher
cybercrypt
- It encrypts files into a file with COMPRESSION!!!. It also decrypts files out of a file.-It encrypts files into a file with COMPRESSION!!!. It also decrypts files out of a file.
diffeehellmanPROGRAM
- Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the answer. A handy standard no
RC4-Prog
- Example 1. To find 1537 x 4248 modulo 10, you could multiply out and take the last digit, but a better way would be to replace 1537 by 7 and 4248 by 8 to start, find 7 x 8 = 56, and then take 56 mod 10 to get 6 as the answer. A handy standard no
ChaosEncryptionJavaAnbeirar
- Modern symmetrical encryption systems use a starting “key” that is up to 256 bits long (for us civilians anyway) and combine this with an initialization-vector (also up to 256 bits long). By applying the vector to the key through some scary mathemati