资源列表
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
ecdsa-src-vc6
- 使用openssl实现的椭圆密码学签名、认证的简单例子，vc6.0。-Elliptical cryptography signature authentication using openssl achieve simple example, vc6.0.
EAX
- EAX AES implementation
RSA
- this a RSA encrypt and decrypt program using java-this is a RSA encrypt and decrypt program using java
LZW
- LZW algorithm and program in Java
js_base64_gb2312
- javascr ipt base64 中文加密 解密 与c++ c# 结果一致-javascr ipt base64 encode decode
Vigenere
- iMPLEMENTATION OF vIGNERE CRYPTO sYSTEME
rsa
- RSA algorithm including long integer type, some test algorithm, large prime number generator and the general theory of numbers algorithm
crypt19-master-(1)
- Ruby code of some algoritms. RSA CSC and auther. Upload github -Ruby code of some algoritms. RSA CSC and auther. Upload github
SampleDBLogin
- sample login method made in visual basic
VITERBI
- viterbi编码算法verilog实现-viterbi encoder, developed by verilog language
DES_-KEY-dan-contoh-gambar
- Encrypt a picture using DES