EDstar

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:)

用C/C++实现加解密算法及其应用：AES 和基于AES 的CMAC(Encryption and decryption algorithm and its application with C/C++: AES and CMAC based on AES)

奇迹C1 C2 C3 C4协议加密与解密(MU C1 C2 C3 C4 encrypt and decrypt)

DES 算法描述DES 算法描述DES 算法描述(DES algorithm description)

DES加密的雪崩效应分析，按照作业要求做的。(DES encryption of the avalanche effect analysis, done in accordance with operational requirements.)

一个周期3的多表代替密码，第1个表由密钥字法产生（密钥字自拟），第2个表由洗牌法产生（注意，字母a~z与数字0~25一一对应），第三个表由公式法产生（数学公式自拟，注意它须是Z26上的一个一一变换）(a cycle of three multi-table instead of passwords, a table by key word method, (key word from the intended), Table No. 2 produced by shuffling method (note, a letter with a number of z ~ 0 ~ 25 he gathers), 3rd tables generated by the formula (from the proposed mathematical formula, it should pay attention to the Z26 is a transformation on January 1))

C network programming this source is vectorcross

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:)

rsa 实现代码，运用大数算法#include "afx.h" #include "Integer.h" #include <string> #include <iostream> using namespace std //构造大数对象并初始化为零 CInteger::CInteger() { m_nLength=1 for(int i=0 i<BI_MAXLEN i++)m_ulValue[i]=0 } //解构大数对象 CInteger::~CInteger() { } /*大数比较 调用方式：N.Cmp(A) 返回值：若N<A返回-1；若N=A返回0；若N>A返回1 */ int CInteger::Cmp(CInteger& A)(#include "afx.h" #include "Integer.h" #include <string> #include <iostream> using namespace std //构造大数对象并初始化为零 CInteger::CInteger() { m_nLength=1 for(int i=0 i<BI_MAXLEN i++)m_ulValue[i]=0 } //解构大数对象 CInteger::~CInteger() { } /*大数比较 调用方式：N.Cmp(A) 返回值：若N<A返回-1；若N=A返回0；若N>A返回1 */ int CInteger::Cmp(CInteger& A))

基于RSA算法的加密解密以及数字签名系统，实现加密解密签名的功能(RSA-based encryption and decryption algorithms and digital signature system to achieve the function of signature encryption and decryption )