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Test Crypt API
- CryptAPI测试程序源代码，包含获取证书链、枚举证书、获取设置证书属性、生产证书请求等源代码。-CryptAPI test program source code, including access to the certificate chain, enumeration certificate, certificate of property get set, production certificate request。
aes1
- aes的一种加密的C语言代码，只有加密的，目前还没弄到解密的-An encrypted aes C language code, only encrypted, decryption is currently not get
CA
- 提供两个说明文档，一个是函数查询手册（ Reference Manual），另一个是使用说明（User’s Manual） 你也可以把文档User’s Manual中的程序拿过来用用试试。逐渐熟悉一下一些常见的函数的使用方法。 -Documentation provided by two, one function query Manual (Reference Manual), the other is the instructions for use (User' s M
watermarking
- CT based encryption 1. Open contourlet 5 and run aaa.m for image decomposition and for denoising 2. Open tsss and A_hidetext.m and set break point at line 109 and run 3. After selecting image ind text press F5 4. Now run H_finalextra
jCryption-2.0
- In short words jCryption is a javascr ipt HTML-Form encryption plugin, which encrypts the POST/GET-Data that will be sent when you submit a form.
Escape.java
- Java Escape算法编码示例，包括解码类文件源码，解码说明：本方法保证 不论参数s是否经过escape()编码，均能得到正确的“解码”结果。-Java Escape arithmetic coding examples, including the decoding class source files, decode Note: This method ensures that regardless of whether the parameter s through the escap
rsa
- 取机器吗并加密解密get machine s serial code,crypt and decrypt-get machine s serial code,crypt and decrypt
Demo_RSA
- 通过java的keyPairGenerator得到秘钥来实现加密-KeyPairGenerator get through the java key encryption
read_cert_name
- 这是一个从usbkey中读取用户证书的例子，运行后将获取usbkey中的证书名称与颁发者信息，使用usbkey封装的CSP服务实现，编译环境VC6.0。 要配合usbkey使用。根据使用的key型号修改代码中宏CSP_NAME的值。（各厂商的sdk中都会有相应的说明）-This is an example of a read user certificate the usbkey, after the operation to get the name of the certificate
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
androidOTP
- 用来生成动态口令，基于时间同步的方式的获得6位随机的动态密码。-Used to generate dynamic password, get six dynamic random password based time synchronization mode