-
在微软CryptoAPI公共使用
Microsoft CryptoAPI中公私密钥对的使用、HASH算法、数字签名等技术-Microsoft CryptoAPI in the use of public-private key pair, HASH algorithm, digital signature technology
- 2022-12-19 04:50:03下载
- 积分：1
-
With c++ Achieve affine password, I was DevC++ In running, it should be in Visua...
用c++实现仿射密码，我是在DevC++中跑的，应该在VisualC++中也行-With c++ Achieve affine password, I was DevC++ In running, it should be in VisualC++ Also line
- 2022-08-02 19:19:58下载
- 积分：1
-
Delphi写的base64算法程序
Delphi写的base64算法程序--Base64 algorithm program writed by Delphi
- 2022-03-04 22:03:17下载
- 积分：1
-
凯撒加密，用java实现的非
凯撒加密，用java实现 非扩展算法，C=M*K，有加密和解密两步-Caesar encryption, using java to achieve non-expansion algorithm, C = M* K, have a two-step encryption and decryption
- 2023-07-05 22:50:02下载
- 积分：1
-
An easy language tutorial encryption ....................
一个易语言的加密教程-An easy language tutorial encryption ....................
- 2022-02-06 10:18:21下载
- 积分：1
-
这是RC5算法的实现
这是RC5算法的实现-This is the realization of RC5 Algorithm
- 2022-12-26 09:25:03下载
- 积分：1
-
enumcsp
EnumCSP - Enumerates Crypto Service Providers
Copyright (c) 2000-2004 G.Chapillon
All Rights Reserved.
Contributor:
Jupiter -EnumCSP- Enumerates Crypto Service Providers Copyright (c) 2000-2004 G. Chapillon
- 2022-04-08 01:32:04下载
- 积分：1
-
md5加密的bindshe
md5加密的bindshe-md5 encrypted bindshe
- 2022-08-07 16:33:31下载
- 积分：1
-
inplementation of AES vhdl
The use of a list of law, VHDL language based polyn...
inplementation of AES vhdl
The use of a list of law, VHDL language based polynomial-based finite field multiplier, for the AES algorithm
- 2022-07-01 03:45:19下载
- 积分：1
-
设f（x）= 1×X2 X5 X27，分别尝试写下面的移位寄存器…
设f(x)=1+x+x2+x5+x27，试分别写出实现下列移位寄存器的程序：
以f(x)为联接多项式的DSR；
以f(x)为联接多项式的LFSR。
可供选择的联接多项式：
f1(x)=1+x+x4+x6+x30；
f2(x)=1+ x3+ x31；
f3(x)=1+ x6+ x31；
f4(x)=1+ x7+ x31；
f5(x)=1+ x13+ x31；
-Let f (x) = 1 x x2 x5 x27, respectively try to write the following Shift Register : f (x) connected to the DSR polynomial; f (x) for connectivity polynomial LFSR. Connection options polynomial : f (x) = 1 x x4 x6 x30; F2 (x) = 1 x3 x31; f3 (x) = 1 x6 x31; f4 (x) = 1 x11 x31; f5 (x) = 1 x13 x31;
- 2022-04-07 10:24:13下载
- 积分：1
