Flexnet-Trusted-Storage-based_PR_TS_V11.9.1

FlexNet Publisher 11.9.1 Trusted Storage-based Licensing Programming Reference

密码学BM算法的C++实现。可计算100阶m序列的生成多项式(Cryptography BM algorithm in C++. Computable 100 m sequence generator polynomial order)

sha256计算代码，非常好用的代码，大家可以用这个代码直接计算哈希值，sha256实现 依次执行 SHA256Init、 SHA256Update SHA256Final 三个函数即可得到sha256结果(sha256 sourse code，Is very usefull)

c++ 写的加密算法，可以加密解密，采用的算法是比较长用的(c write the encryption algorithm, encryption can be decrypted using the algorithm is relatively long with the)

该文件能用Visual C++6.0打开编译实现加密解密功能，并能通过少量的修改改(修改处程序源码中已标明)镶嵌到其他的程序源码中实现加密解密功能。 可直接使用。 (The file can use Visual C++6.0 to open compiled encryption and decryption functions through a few modifications reform (modify at program source code has been marked) inlaid into the other program source code to achieve encryption and decryption functions. Can be used directly.)

说明：  编制程序，将输入的一行字符以加密的形式输出，然后将其解密，解密的字符串序列与输入的正文相比较，吻合时输出解密的正文，否则解密失败。 加密时，将每个字符的ascii码依次反复加上“4962873”中的数字，并在32（‘’）~122（‘z’）之间作模运算。解密与加密的顺序相反。例如输入正文“the result of 3 and 2 is not 8” 结果为： (Programming, the input line character output in an encrypted form, and then decrypted, the decrypted text sequence of characters compared with the input, output to decrypt the body of the agreement, otherwise the decryption failed. Encryption, ascii code of each character in turn repeatedly with &quot 4962873&quot in the figures, and 32 (' ' )~ 122 (' z' ) as a model between the operations. Decryption and encryption in reverse order. For example, enter the text &quot the result of 3 and 2 is not 8&quot results:)

自己写的一份RSA算法。可正常执行。控制台程序。(Write your own copy of the RSA algorithm. Can be performed normally. Console program.)

RSA公钥加密程序，可实现密钥对生成，加解密任意字符串功能。MFC界面。(RSA public key encryption program, enabling the key pair generation, encryption and decryption functions of arbitrary strings. MFC interface.)

利用C语言实现的cmac算法，该cmac基于AES。程序清晰有注释便于查看(The use of C language cmac algorithm, the cmac based AES. There are clear and easy to view program notes)

本程序可以实现AES16,24,32位任意长度的密码与任意长度的明文的加密与解密，并附有一实例(This procedure can achieve the AES16 24,32 password of any length and any length of plaintext encryption and decryption, with an example)