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KeyFile
- 用C#处理数字证书 使用System.Security.Cryptography.X509Certificate名称空间 System.Security.Cryptography.X509Certificate名称空间特别重要,因为它提供了开发人员处理数字签名的那些类. 提供了读取一个X.509证书的代码,和 使用不对称加密加密与解密数据的代码-C# using digital certificates System.Security.Cryptogra phy.X509C
1
- X.509de yingyong 和开发的.具体可以先学习一下.openssl的相关知识
rfc2459
- Internet X.509 公钥基础设施--证书和CRL简介-Internet X.509 Public Key Infrastructure -- Introduction certificate and CRL
RFC2511
- 本文描述了证书请求消息格式(CRMF)。它被用来向CA传递一个产生X.509证书请求 (可能通过RA)。请求消息一般包括公钥和有关的登记信息。-This paper describes a certificate request message format (CRMF). It was used to send a CA have X.509 certificate request (possibly through RA). Requests for general information,
nss-3.1.1.tar
- 支持SSL v2/v3, TLS, PKCS #5, PKCS #7, PKCS #11, PKCS #12, S/MIME, X.509v3证书等安全协议或标准的开发库编译用到NSPR-Support for SSL v2/v3, TLS, PKCS# 5, PKCS# 7, PKCS# 11, PKCS# 12, S/MIME, X.509v3 certificate security protocols or standards used to develop database to co
PlayfairCipher_codes
- Playfair Cipher 1.not even the large number of keys in a monoalphabetic cipher provides security 2.one approach to improving security was to encrypt multiple letters 3.the Playfair Cipher is an example 4.invented by Charles Wheatstone in 1854
CubicSpline
- Cubic spline algorithm approximating function y=f(x) - ((-2)*x*(2*x+3))/(x*x+4*x+5) in interval [-2 2]-Cubic spline algorithm approximating function y=f(x)- ((-2)*x*(2*x+3))/(x*x+4*x+5) in interval [-2 2]
4.x
- 基于U盘的强身份认证，附VB源码及dll接口说明-U disk-based strong authentication, with VB source code and dll interface descr iption
S-BGP_CMS-1.0
- S-BGP PKI makes use of X.509 version 3 public-key certificates, enabling verification of the identities and authorizations of BGP speakers, owners of AS numbers and of portions of the IP address space, repositories, repository administrators, and NOC
rfc2560
- 标准规范-X.509证书网络协议，供大家参考-X.509 Internet Public Key Infrastructure Online Certificate Status Protocol- OCSP
jj
- CORNACCHIA Algorithm * * GOAL: given d and p prime, find (x,y) such that x^2 + y^2 = d * p * * in this implementation, d = 1. This code does not accept d <> 1 !!! * see "A Course in Computational Algebraic Number Theory" by Henri
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
FTUnionControl
- 封装的一个ActiveX组件，里面使用的CAPI，有使用飞天诚信USBKEY中证书进行签名的过程，函数调用参看MSDN上说明。 -An ActiveX component package, which uses the CAPI, are used to FEITIAN USBKEY in the process, see MSDN on function call instructions.
aes-6.x-1.3.tar
- Simple aes algorithm
digital-certification-analysis
- 解析X.509规范编码的数字证书，利用开源的openssl库编程实现，方便易懂！-Resolution encoded X.509 standard digital certificates, use the openssl library open-source programming, easy to understand!
Elgamal
- ElGamal algorithm not only used for data encryption can be used for digital signatures, their safety depends on the calculation of a finite field discrete logarithm this problem. Key to the method. First of all, choose a prime number p, the two rando
There-are-three-types-of-cryptography-techniques.
- Caesar cipher is a simple substitution cipher where each letter in the plain text is replaced with a letter a fixed number of places down the aplhabet. EXAMPLE If the number of places to shift is 3, the letter A would be converted to ltter D. Usi
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