DES encryption algorithm examples of a

文件验证，利用MD5算法做文件的HASH，得到统一结果判断文件内容的一致性-File Hash

MD4的源代码。该算法来源于应用密码学的随书附带光盘。已通过调试-MD4 source. The algorithm derived from the application of cryptography book with CD-ROM attached. Through debug

一种国际上通用的加密算法―DES算法的原理-a common encryption algorithms-DES algorithm Principle

6）jm.tar.gz是上述程序的程序包,解包后产生一个安装程序install-jm,运行 ./install-jm就能根据需求选择安装 退出安装程序,安装程序会被自行删除。 8） 以上各程序在RedHatLINUX7.2平台编译通过。 10）本程序仅供学习使用。-6) jm.tar.gz above is the procedure for packages, unpacks resulting from the installation of a procedure install-jm, run./Install- jm demand can choose to exit the setup program to install, the installer will be replaced. 8) The above procedure RedHatLINUX7.2 platform compile. 10) only to learn the procedures used.

sha密码检验的示例 希望对大家有 帮助-sha password test examples we hope to help

在原作者Jim Gillogly代码的基础上修改优化，并实现C++改写。实现了DES加密。当然，密钥可在源文件中修改，或提供修改的接口函数。-the originator Jim Gillogly code revision on the basis of optimization, and to achieve C rewritten. Implementation of DES encryption. Of course, the key in the source documents changes, or amendments to provide the interface function.

个人编写的简单gsm保密A5算法，思路有点凌乱， a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法a5算法

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). ]:

RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ，编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s

this code implement an Order Preserving Encryption Scheme (OPES).