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Paillier
Paillier加密算法的实现。产生512位大数,进行大数运算。1、产生密钥2、加密、3、解密。(Paillier encryption algorithm implementation. Produce 512 Tarsus, Tarsus operation performed. 1, 2 generates a key, encrypts, 3, decryption.)
- 2021-03-02 17:59:34下载
- 积分:1
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DES 加密 系统的 C 语言 实现的源代码
DES 加密 系统的 C 语言 实现的源代码-DES encryption system of the C language source code
- 2022-04-09 20:27:15下载
- 积分:1
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sha-256
sha256的签名算法,纯C实现,已验证正确,在控制台上直接运行即可,(sha256 signature algorithm, pure C implementation ,has been verified properly and can be run directly on the console.)
- 2021-04-28 16:08:43下载
- 积分:1
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c环境下的md5加密程序
c环境下的md5加密程序-c environment of md5 encrypted
- 2022-08-10 06:06:19下载
- 积分:1
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delphi file crypter full source code
delphi file crypter full source code
- 2022-08-13 03:30:27下载
- 积分:1
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use an iterative equation Xn Asin2 = (Xn
使用迭代方程Xn+1=Asin2(Xn-XB)生成混沌序列{Xi},需输入A和Xb值,不同的值产生不同的加密效果,但是仅在混沌区内的加密效果最好。在Microsoft Visual Studio .NET 2003上调试成功,用C++语言编写-use an iterative equation Xn Asin2 = (Xn- XB) generated chaotic sequence (Xi) need input A and isolate value, different values in different encryption effect, but only in the chaotic region of encryption is the best. In Microsoft Visual Studio.NET 2003 Debugger success C Language
- 2022-02-03 07:25:44下载
- 积分:1
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2的16次幂正整数d与n,编写计算d
2的16次幂正整数d与n,编写计算d-1 (mod n) 的程序;
2、对于三个不超过2的16次幂正整数a、e与n,编写计算ae (mod n) 的程序。
在上述程序基础上写出下列程序:
(1) 对给定的10000以内数判定其是否为素数;
(2) 进行ElGamal体制的加密与签名。
-two of 16 power-positive integer d and n, calculate the preparation of d-1 (mod n); 2. For not more than 2 3 16 Powers positive integers a, e and n, calculate the preparation ae (mod n) procedures. In the above procedure to write on the basis of the following procedures : (a) the right to 10,000 within a few judging whether a prime number; (2) ElGamal encryption and signature.
- 2022-02-14 21:21:03下载
- 积分:1
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FileEncrypt
这个是对文件进行加解密的项目,提供一些库函数,你可以直接调用(this source code s function is to encrypt and decrypt the file ,so ,of course ,it can
speed up your program)
- 2010-01-09 17:52:11下载
- 积分:1
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ASFDSAF
题目:某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:
每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。
1.程序分析:
2.程序源代码:(Topic: a company USES the public telephone relay data, data is four integers, in the relay in the process is encrypted, encryption rules are as follows:
Each Numbers add 5, then use and divided by the number of 10 instead of a remnant, then the first and fourth exchange, the second and third exchange.
1. The program analysis:
2. The source code:)
- 2012-05-26 20:07:28下载
- 积分:1
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Machine to make use of multi
机器采用多层感知器异或分类
- 2022-02-14 10:23:49下载
- 积分:1