登录
首页 » Windows_Unix » ECDSA_Verilog

ECDSA_Verilog

于 2009-09-01 发布 文件大小:3KB
0 190
下载积分: 1 下载次数: 1

代码说明:

说明:  椭圆曲线加解密算法的verilog实现,帮助初学者有效理解ECC算法。(Elliptic curve encryption and decryption algorithm verilog implementation, to help beginners understand the ECC algorithm is effective.)

文件列表:

ECDSA_Verilog.v

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

  • 简单的RSA范例
    简单的RSA加密范例-simple example of the RSA encryption
    2022-08-05 12:37:46下载
    积分:1
  • Delphi控件源码,包括对称钥匙控件和散列函数控件
    Delphi加密控件源码,包括对称钥匙加密控件和散列函数控件 -Delphi encryption controls the source code, controls including the symmetrical key encryption and disperses the row letter numerical control
    2022-06-11 22:19:37下载
    积分:1
  • DesSecurity
    C# DES 对称加密,通过向量和密钥控制加密内容,可加密,解密(C# DES symmetric encryption, and key vector control through the encrypted content can be encrypted and decrypted)
    2011-01-14 17:50:21下载
    积分:1
  • RC4 cipher
    最常见的流密码RC4的visual c++实现。希望对大家有帮助。-RC4 cipher
    2022-05-14 15:32:21下载
    积分:1
  • 这是MD5算法的 Asm 实现,代码效率高,
    这是MD5算法的 Asm 实现,代码效率高,-This is the MD5 algorithm Asm implementation, code efficiency,
    2022-02-02 04:43:01下载
    积分:1
  • 在Delphi开发快速CRC算法。
    Fast CRC Algorithmes develop in Delphi. Simple to uses and very fast to run
    2023-04-29 18:50:02下载
    积分:1
  • 硬件求平方根
    硬件求解平方根源代码加密 (硬件求解平方根的,将license添加到原有的MaxplusII或QuartusII的license中就可以直接使用,但源代码加密。altera提供 )(solving square root of the hardware encryption code (square root of the hardware solution will be added to the original license MaxplusII or Quartus II of the license which can be directly used, but the source code encryption. ALTERA provide))
    2004-10-05 11:06:50下载
    积分:1
  • NHD
    程序采用当前最先进的Rijndael算法作为核心算法,带有可变块长和可变密钥长度的迭代块密码。块长和密钥长度可以分别指定成 128、192 或 256 位。多线程操作,保证加密时程序界面流畅,加密速度大约9M每秒。(Procedures using the most advanced as a core algorithm Rijndael algorithm with variable block length and variable key length of the iterative block password. Block length and key length can be specified as 128,192 or 256. Multi-threaded operation, to ensure smooth program interface encryption, encryption speed of about 9M per second.)
    2009-07-16 18:38:49下载
    积分:1
  • 关于AES算法的释文档,是初学者的好帮手,有助于对AES算法的理...
    关于AES算法的解释文档,是初学者的好帮手,有助于对AES算法的理解-AES algorithm on the interpretation of documents, beginners are a good helper to help the understanding of the AES algorithm
    2022-09-16 13:15:03下载
    积分:1
  • java非对称的源代码(RSA),鉴于rsa的重要性和相关源代码的匮乏,经过整理。...
    java非对称加密的源代码(RSA),鉴于rsa加密的重要性和相关源代码的匮乏,经过整理。-asymmetric encryption source code (RSA), in view of encryption rsa the importance and relevance of the lack of source code, collated.
    2022-08-25 03:39:46下载
    积分:1
  • 696518资源总数
  • 105559会员总数
  • 1今日下载