登录
首页 » 加密解密 » MD5加密 MD5加密

MD5加密 MD5加密

于 2022-10-26 发布 文件大小:8.19 kB
0 222
下载积分: 2 下载次数: 1

代码说明:

MD5加密 MD5加密-MD5 encryption MD5 encryption MD5 encryption

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • This algorithm is used VC++ Compiled using RSA algorithm to encrypt any document...
    此算法是用VC++编译的,用RSA算法对任何文件加密,简单易懂。-This algorithm is used VC++ Compiled using RSA algorithm to encrypt any document, easy-to-read.
    2022-01-25 14:37:37下载
    积分:1
  • AES has been certified by the AES DELPHI, the better, we all give it a try and f...
    已经通过AES认证的AES的DELPHI实现,比较好,大家自己试试看,搜的-AES has been certified by the AES DELPHI, the better, we all give it a try and found the
    2022-03-19 00:08:16下载
    积分:1
  • MD5(支持Unicode码)
    .rar中包括两个文件MD5.cpp和MD5.h,在MD5的基础上改进让其支持Unicode的宽字符,可以对CString进行32位的加密,使用vs2010进行测试过,在解决方案“头文件”“源文件”分别使用添加“已有项”,将.h和.cpp文件加入工程,可能会有两三个警告,但不影响程序运行
    2023-03-02 17:25:03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
  • 一个简单实现RC5算法的例子!仅供参考!
    一个简单实现RC5加密算法的例子!仅供参考!-Realize a simple example of RC5 encryption algorithm! For reference purposes only!
    2022-02-14 04:31:46下载
    积分:1
  • 128192和256位钥的码Twofish的VHDL实现。
    VHDL implementation of the twofish cipher for 128,192 and 256 bit keys. The implementation is in library-like form All needed components up to, including the round/key schedule circuits are implemented, giving the flexibility to be combined in different architectures (iterative, rolled out/pipelined etc). Manual in English is included with more details about how to use the components and/or how to optimize some of them. All testbenches are provided (tables, variable key/text, ECB/CBC monte carlo) for 128, 192 and 256 bit key sizes, along with their respective vector files.-VHDL implementation of the twofish cipher for 128,192 and 256 bit keys. The implementation is in library-like form All needed components up to, including the round/key schedule circuits are implemented, giving the flexibility to be combined in different architectures (iterative, rolled out/pipelined etc). Manual in English is included with more details about how to use the components and/or how to optimize some
    2022-07-08 23:33:44下载
    积分:1
  • 对证书的基本操作,如读取证书DN,证书有效期等
    对证书的基本操作,如读取证书DN,证书有效期等-Basic operation of the certificate, such as reading the certificate DN, the certificate is valid, etc.
    2022-01-25 17:20:30下载
    积分:1
  • DES。网
    des .net
    2023-01-15 19:30:03下载
    积分:1
  • 码――码技术剖析与实战应用中的一个例子
    加密与解码――密码技术剖析与实战应用中的一个例子-encryption and decryption-- Password Technology Analysis and real application of an example
    2022-03-24 10:53:59下载
    积分:1
  • 对于这个问题的实现、刚刚给过一个程序,突然想起另外一种简单一些的方法,一并给出[笔者长期从事移动通信系统的无线链路调制与调、物理层实现方面的工作。在移动通信G...
    对于这个问题的实现、刚刚给过一个程序,突然想起另外一种简单一些的方法,一并给出[笔者长期从事移动通信系统的无线链路调制与解调、物理层实现方面的工作。在移动通信GSM系统中,我们进行语音或者业务信道解调时,都会遇到CRC的求解。通常在硬件DSP实现时,特别是40位CRC求解时候,由于生成多项式有41项,DSP最大一次能处理40位,所以使用单个寄存器会遇到一些困难,那么以下这个程序将会解决这一困难(这是针对定点DSP、C55xx的编程实现方法)。]: -For the realization of this issue, just to have a program, suddenly reminded of another simpler way to be given [the author has long been engaged in mobile communication system modulation and demodulation of wireless links, the physical layer to achieve work. GSM in the mobile communications system, we have a voice or channel demodulation operations, it will encounter the solution of CRC. DSP hardware usually realize, especially when solving the 40 CRC, as the generation polynomial has 41, DSP can handle the largest 40, so using a single register may encounter some difficulties, then following this process will solve this One difficulty (which is for fixed-point DSP, C55xx realize programming methods). ]:
    2022-03-14 13:54:53下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载