实验内容
1、对称密码技术的实现
（1）熟悉相应的DES加密/解密函数的使用，在单机完成对特定文件（test-1.txt）的单重DES加密/解密；
（2）针对DES密钥特点（8字节，64比特位，其中真正起作用的为56比特位），设计一次一密的密钥产生机制（即每次会话所使用的密钥都是临时随机产生的）；
2、非对称密码技术的实现
（1）熟悉RSA相关函数（公私钥生成函数，加密/解密函数）；
（2）利用公钥加密特定文件（test-1.txt），并用私钥进行解密；反过来，再利用私钥加密特定文件，并用公钥进行解密。对比解密后的文件与原始文件是否一致。
3、信息摘要技术的实现
（1）熟悉MD5算法的实现函数；
（2）利用MD5实现函数，对特定文件（test-1.txt）进行信息摘要的计算，得到信息摘要M1；尝试改变特定文件中的内容（比如改变其中某个字母），再利用MD5计算一次信息摘要，看同之前得到的信息摘要是否相同；
设计要求
1、JAVA语言编程实现
2、单机上完成
3、加密内容为一文本文件（从服务器下载），从服务器上下载相关软件包（jce-1_2_2和帮助文件，主要用于RSA密钥生成和加/解密,如果安装了JDK1.6版本以上，则不需要jce支持）后，请先阅读帮助再进行编程。
DES/RSA/MD5
The implementation of symmetric cryptography
(1) familiar with the use of the corresponding DES encryption / decryption function, complete single DES encryption / decryption of a particular file (test-1.txt) in a single machine;
(2) for DES key features (8 bytes, 64 bits, which really works for 56 bits), the design of a secret key generation mechanism (i.e. using the session key is randomly generated temporary); (2017-10-11, Java, 16KB, 下载10次)
完成分布式对称密钥分配
说明：作为通讯的双方A和B已经共享一个会话主密钥Km。在完成会话密钥Ks的分配后，A利用Ks对特定文件（test-1.txt）进行加密，并发送给B；B利用Ks对密文进行解密，并对比解密后的明文信息同原来的文件是否一致。
要求：
（1）JAVA语言编程实现，基于DES完成对称密码技术的加/解密（JAVA有库函数支持）；
（2）通信内容为一文本文件（test-1.txt）；
（3）基于网络完成密钥分配工作（作为通信的双方A、B应部署于两台主机上）；
（4）所分配的对称密钥皆为一次一密，随机产生；
（5）具体密钥分配过程请以课本中分布式密钥分配方案为参考，完成从步骤1到步骤3的全过程；
Complete distributed symmetric key distribution Description: As both A and B have shared a communication session master key Km. After the completion of the distribution of the session key Ks, A specific file using Ks (test-1.txt) encrypts and sends B B using Ks to decrypt the ciphertext, and the decrypted plaintext comparison with the original whether the files are consistent. Requirements: (1) JAVA programming language, based on DES symmetric cryptographic techniques to complete the encryption/decryption (JAVA there are library functions support) (2) the contents of a text file to communicate (test-1.txt) (3) based on the network completion of key distribution work (both as a communications A, B should be deployed on two hosts) (4) are all assigned to the symmetric key one-time pad, randomly generated (5) the specific key distribution process please textbooks distributed key distribution scheme as a reference, complete step 1 to step 3 in the whole process (2014-10-23, Java, 15KB, 下载63次)
完成基于非对称密码技术的对称密钥分配工作
说明：
（1）在完成上述基于公钥证书的非对称密钥分配工作的基础上，根据教材上基于非对称密码技术的对称密钥分配方案，完成对称密钥Ks的分配工作。
（2）A用Ks对一图象文件（test_pic.bmp)进行加密并发送给B，B收到后解密，对比原图象看是否一致，并记录整个加密/解密的所用时间T2；
（3）对比T1和T2；
要求：
（1）JAVA语言编程实现，基于RSA完成非对称密码技术的加/解密，基于DES完成对称密码技术的加/解密（JAVA有库函数支持）；
（2）通信内容为一图象文件（test_pic.bmp)；
（3）基于网络完成密钥分配工作（作为通信的双方A、B应部署于两台主机上）；
（4）所分配的对称密钥皆为一次一密，随机产生；
（5）具体密钥分配过程请以课本中基于非对称密码技术的对称密钥分配方案为参考，完成从步骤1到步骤4的全过程；
Completed based on asymmetric cryptography symmetric key distribution work
Description:
( 1 ) Upon completion of the above-mentioned asymmetric public key certificate-based key distribution based on the work , according to the materials on asymmetric cryptography-based symmetric key distribution scheme , the completion of the distribution of symmetric key Ks work .
(2) A use of an image file Ks (test_pic.bmp) is encrypted and sent to B, B to decrypt after receipt , compared to see if the same original image , and records the encryption/decryption used time T2
( 3 ) comparing T1 and T2
Requirements:
(1) JAVA programming language, based on the RSA asymmetric cryptographic techniques to complete encryption/decryption , based on the DES symmetric cryptographic techniques to complete the encryption/decryption (JAVA library functions are supported )
( 2) The content of the communication is an image file (test_pic.bmp)
( 3 ) complete network-based key distribution job ( as the (2013-11-05, Java, 15KB, 下载29次)
完成基于公钥证书的非对称密钥分配工作
说明：
（1）作为通讯的双方A和B，都有一个共同信赖的第三方CA，由CA为每个人生成公钥证书和相应的私钥并分发给A、B。
（2）A和B分别将自己的公钥证书提供给对方后，分别在本地验证对方公钥证书是否可靠（假定A、B已经获得CA的公钥），然后A用B的公钥加密一个数据X（整型，可自己定义，比如100）并发送给B，B用自己的私钥解密后得到数据X，然后按协商好的函数F（比如F(X)=2*X）计算得到F(X)，再利用A的公钥加密后发送给A；A在本地用自己的私钥解密后得到F(x)，在同本地计算得到的F(X)进行对比，如果一致，说明公钥分配成功。
（3）A用B的公钥对一图象文件（test_pic.bmp)进行加密并发送给B，B收到后解密，对比原图象看是否一致，并记录整个加密/解密的所用时间T1；
要求：
（1）JAVA语言编程实现，基于RSA完成非对称密码技术的加/解密（JAVA有库函数支持）；
（2）密钥长度不少于384位；
（3）基于网络完成密钥分配工作（作为通信的双方A、B应部署于两台主机上，CA可以部署在其中一台主机上或第三台主机上）；
Completed based on public key certificate allocation asymmetric key
Description:
( A ) as a communication between the parties A and B, there is a common trusted third-party CA, generated by the CA public key certificate for each person and the corresponding private key and distributed to A, B.
(2) A and B respectively own public key certificate to each other , respectively, the other public key certificate in the local authentication is reliable ( assuming A, B have the CA s public key ) , and A with B s public key to encrypt a data X ( integer, can define their own, such as 100 ) and sent to B, B to use their private key to decrypt the data obtained by X, and then press the negotiated function F ( eg F (X) = 2* X) is calculated F (X), and then encrypted using the public key of A to a A locally decrypted with the private key to obtain F (x), obtained in the same local computing F (X) for comparison , if the same , indicating that public key distribution success.
(3) A with B s pu (2013-11-05, Java, 5KB, 下载94次)
完成基于非对称密码技术的对称密钥分配工作
说明：
（1）在完成上述基于公钥证书的非对称密钥分配工作的基础上，根据教材上基于非对称密码技术的对称密钥分配方案，完成对称密钥Ks的分配工作。
（2）A用Ks对一图象文件（test_pic.bmp)进行加密并发送给B，B收到后解密，对比原图象看是否一致，并记录整个加密/解密的所用时间T2；
（3）对比T1和T2；
要求：
（1）JAVA语言编程实现，基于RSA完成非对称密码技术的加/解密，基于DES完成对称密码技术的加/解密（JAVA有库函数支持）；
（2）通信内容为一图象文件（test_pic.bmp)；
（3）基于网络完成密钥分配工作（作为通信的双方A、B应部署于两台主机上）；
（4）所分配的对称密钥皆为一次一密，随机产生；
（5）具体密钥分配过程请以课本中基于非对称密码技术的对称密钥分配方案为参考，完成从步骤1到步骤4的全过程；
Completed based on asymmetric cryptography symmetric key distribution work
Description:
( 1 ) Upon completion of the above-mentioned asymmetric public key certificate-based key distribution based on the work , according to the materials on asymmetric cryptography-based symmetric key distribution scheme , the completion of the distribution of symmetric key Ks work .
(2) A use of an image file Ks (test_pic.bmp) is encrypted and sent to B, B to decrypt after receipt , compared to see if the same original image , and records the encryption/decryption used time T2
( 3 ) comparing T1 and T2
Requirements:
(1) JAVA programming language, based on the RSA asymmetric cryptographic techniques to complete encryption/decryption , based on the DES symmetric cryptographic techniques to complete the encryption/decryption (JAVA library functions are supported )
( 2) The content of the communication is an image file (test_pic.bmp)
( 3 ) complete network-based key distribution job ( as the (2013-11-05, Java, 44KB, 下载28次)
完成集中式对称密钥的分配工作
说明：作为通讯的双方A和B，都已经分别和KDC拥有会话密钥Ka和Kb，但A、B之间事先没有保密通道，需要依赖KDC为它们的会话分配临时密钥Ks。在完成会话密钥Ks的分配后，A利用Ks对特定文件（test-1.txt）进行加密，并发送给B；B利用Ks对密文进行解密，并对比解密后的明文信息同原来的文件是否一致。
要求：
（1）JAVA语言编程实现，基于DES完成对称密码技术的加/解密（JAVA有库函数支持）；
（2）通信内容为一文本文件（test-1.txt）；
（3）基于网络完成密钥分配工作（作为通信的双方A、B应部署于两台主机上，KDC可以部署在其中一台主机上或第三台主机上）；
（4）所分配的对称密钥皆为一次一密，随机产生；
（5）具体密钥分配过程请以课本中基于KDC的集中式密钥分配方案为参考，完成从步骤1到步骤5的全过程；
Complete centralized allocation symmetric key
Description: As a communication between the parties A and B, respectively, and the KDC already has a session key Ka and Kb, but the A, B no secret passage between the advance, they need to rely on the KDC allocate temporary session key Ks. Upon completion of the session key Ks allocation, A Ks on the use of a specific file (test-1.txt) is encrypted and sent to B B use Ks decrypt the ciphertext, and compare the decrypted plaintext message with the original files are consistent.
Requirements:
(1) JAVA programming language, based on the DES symmetric cryptographic techniques to complete the encryption/decryption (JAVA library functions are supported)
(2) the content of communications as a text file (test-1.txt)
(3) complete network-based key distribution job (as the communication between the parties A, B should be deployed on both hosts, KDC can be deployed on one host or a third host)
(4) are allocated once a secret symmetric key ra (2013-11-05, Java, 4KB, 下载29次)
如果CP1H没有密码保护，可以使用HOSTLINK C模式或FINS模式通讯协议用VB或VC编程轻易读出程序代码。大家都知道欧姆龙的CJ1M、CP1H可以设定UM读保护和任务读保护，在上载PLC程序时，需经过编译，PLC中的机器代码转变成梯形图。当CP1H设置了密码后，就禁止了读程序代码，通过修改CP1H时序设置值，修改时序来导致PLC的看门狗出错，这时CP1H减弱了对密码保护的限制，这时就可以载 PLC中的机器代码，先不经过反编译直接打包上传到电脑中。通过分析电脑中的PLC机器代码，找出先前通过串口监视软件得到的密码保护对应的机器代码。并修改其部分功能。然后软件开发者，用自己开发的下载工具把这些经过修改的PLC机器代码下载到另外一空白的 PLC 中。
如果CP1H没有密码保护，可以使用HOSTLINK C模式或FINS模式通讯协议用VB或VC编程轻易读出程序代码。大家都知道欧姆龙的CJ1M、CP1H可以设定UM读保护和任务读保护，在上载PLC程序时，需经过编译，PLC中的机器代码转变成梯形图。当CP1H设置了密码后，就禁止了读程序代码，通过修改CP1H时序设置值，修改时序来导致PLC的看门狗出错，这时CP1H减弱了对密码保护的限制，这时就可以载 PLC中的机器代码，先不经过反编译直接打包上传到电脑中。通过分析电脑中的PLC机器代码，找出先前通过串口监视软件得到的密码保护对应的机器代码。并修改其部分功能。然后软件开发者，用自己开发的下载工具把这些经过修改的PLC机器代码下载到另外一空白的 PLC 中。 (2013-08-10, Visual Basic, 60KB, 下载44次)
使用NET和CSharp编程，可以根据输入的模板格式和加密解密字符实现软件许可文件的自动生成
This article is about a license key generation program. As you know when you get a software package it usually requires a license key. The keys that are generated show various pieces of information, such as product, customer and version. With this program and library you can create the key and later read it in to validate that the correct key was used. This is used not only for selling commercial products but also in deploying enterprise applications. With a license key you can turn features on and off for each customer, have product feature expire at the end of a demo period, or any other type of enabling or disabling you desire. Since I wanted to put this code in CodeProject I adhered to using NUnit as my unit testing tool, NDoc as my documentation tool, and FxCop as my code analyzer. In this project, I created an interactive test tool to test the various features of the library. (2011-08-17, C#, 269KB, 下载11次)
编程实现RSA算法。包括：生成公钥(e, n)和私钥d，对明文m加密，对密文m解密。
注：实际应用中，512比特的n 已经不够安全，所以建议公司用1024比特的n，及其重要的场合用2048比特的 n。所以大家要选择大整数n。
Programming RSA algorithm. Include: Creation of a public key (e, n) and private key d, m the plaintext encryption, decryption of ciphertext m. Note: The actual application, the 512-bit n is not secure, it is recommended that companies with 1024-bit n, and the important occasion with a 2048-bit n. Therefore, we should choose a large integer n. (2011-03-23, Visual C++, 44KB, 下载158次)
使用SUN JCE 1.6 的授权文件,按照里面的提示安装后就可正常使用SUN JCE 提供的安全编程功能。
JDK 1.6 "Unlimited Strength Jurisdiction Policy Files" (2009-02-26, Java, 9KB, 下载78次)
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). ]: (2008-07-17, Others, 3KB, 下载6次)
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）。
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). (2008-07-17, Others, 3KB, 下载11次)
指令动态加密法。这儿讲述的是用单条指令加密法，再用 int 1 单步中断解下一条指令的第一字节，由于用另外程序解密时无法预知指令长, 所以不能用编程的方法解密，只能用手工一条一条地解。具体实现见注释，这种加密法的麻烦只处就是加密时也要一句一句来。
directive dynamic encryption method. Here is on the order of the single-encryption method, and then a single step int slips off an interruption of the first byte instructions, As with the other procedures can not predict when declassification instructions long, so I can not use the method of decryption program. We can only use one hand to a solution. See Notes to realize that this encryption method only trouble is that the encryption but also for a one to. (2007-03-16, Asm, 2KB, 下载5次)
本文根据发表在CUJ Expert Forum上的两篇文章编译而成。C/C++ User s Journal是目前最出色的C/C++语言专业杂志，特别是在C++ Report闭刊之后，CUJ的地位更加突出。CUJ Expert Forum是CUJ主办的网上技术专栏，汇集2000年10月以来C++社群中顶尖专家的技术短文，并免费公开发布，精彩纷呈，是每一个C/C++学习者不可错过的资料。由Jim Hyslop和Herb Sutter主持的Conversation系列，是CUJ Expert Forum每期必备的精品专栏，以风趣幽默的对话形式讲解C++高级技术，在C++社群内得到广泛赞誉。译者特别挑选两篇设计模式方面的文章，介绍给大家。设计模式方面的经典著作是GoF的Design Patterns。但是那本书有一个缺点，不好懂。从风格上讲，该书与其说是为学习者而写作的教程范本，还不如说是给学术界人士看的学术报告，严谨有余，生动不足。这一点包括该书作者和象Bjarne Stroustrup这样的大师都从不讳言。实际上Design Pattern并非一定是晦涩难懂的，通过生动的例子，一个中等水平的C++学习者完全可以掌握基本用法，在自己的编程实践中使用，得到立竿见影的功效。这两篇文章就是很好的例证。本文翻译在保证技术完整性的前提下作了不少删节和修改，以便使文章显得更紧凑。
Based on the published in CUJ Expert Forum on the two articles from compiler. C/C User's Journal is the most excellent C/C language magazine, especially in the C Report to be closed, the status of CUJ more prominent. CUJ CUJ Expert Forum is sponsored by the online technology column Survey October 2000 C community has been the top technical expert appeared, and free public release, brilliant, Each is a C/C learners should not miss the information. By Jim Hyslop and Herb Sutter auspices of the Conversation series, CUJ Expert Forum is an essential quality each column a humorous forms of dialogue on the High C technology, the C community has been widely acclaimed. Translator selected two design patterns of articles introducing to you. Design patterns of the classics is GoF Design Pat (2006-12-10, C/C++, 131KB, 下载31次)
很多人都想学学做cracker，好象破了一个程序很风光的，人人都很佩服。 可是做一个cracker其实很累，需花费大量的时间，而且经常会碰壁，三五天毫无进展是极为平常的事情。 做cracker的五个必要条件是: 1. 知识 2. 经验 3. 感觉 4. 耐心 5. 运气 。 如果你刚开始学crack，也许你遇到不少麻烦，并且有点想退却，其实你不要着急，只要你认真学习，成功就在眼前。没有人是生来就什么都会的，如果你有问题，就大胆的去问你周围的人，多来我主页论坛请教。计算机水平不高怎么办？没关系，我也不是学计算机的，我自己的专业与电脑不沾边。只要努力学习就能成功。 所谓知识只要你肯学就可以了，刚入门时如你没汇编知识是不行的，你要掌握一下这门编程语言，能看懂就能上手，但是你想很顺手的话，除了把汇编掌握好，还有编程的基本功夫，保护模式等技术。 经验是跟你破解软件时间，掌握程度有关，接触多了，拿到一软件应该知道用哪种法比较省事，比较有把握。
many people would like to learn to do cracker, as if breaking a procedure very proud that everyone admire. But doing a cracker actually very tired, and need to spend a lot of time, but often run into snags, March 5 days with no progress is extremely commonplace. Do cracker necessary conditions for the five are : 1. Knowledge 2. Experience 3. Feeling 4. Patience 5. Luck. If you have just started to learn the crack, you may encounter a lot of trouble, and want to retreat a bit, actually, you should not worry, if you seriously study and success is at hand. Nobody is born on what would, if you have problems, boldly ask the people around you, the more I ask Home Forum. High-level computer how to do? Okay, I was not a computer, I own a computer professional and ask. If learning can be successful (2005-05-16, C#, 1912KB, 下载57次)
使用的是API编程,可格式化、校验和读写特殊扇区。可用作Windows下的磁盘加密。本函数还有以下两个缺点以待改进： 1.本函数还只能读能读 A: 和 B:,即只能对软盘操作 2.不能改变磁盘扇区大小，只能是标准的 512 个字节。 参数说明: command 操作: 0 重置磁盘 2 读扇区 3 写扇区 4 校验磁道 5 格式化磁道 8 得到设备参数 (int 1EH) drive 驱动器 A:=0 B:=1 head 磁头号,范围 0 - 1 track 磁道号,范围 0 - 84 ( 80 - 84 为特殊磁道，通常用来加密 ) sector 扇区号,范围 0 - 255 ( 19 - 255 为非标准扇区编号，通常用来加密) nsectors 每次读或写的扇区数,不能超出每磁道的最大扇区数 buffer 数据写入或读出的缓冲区，大小为 512 个字节 返回值 ( 同 Int 13H ): 0x0 成功 0x1 无效的命令 0x3 磁盘被写保护 0x4 扇区没有找到 0xa 发现坏扇区 0x80 磁盘没有准备好
using the API programming, formatting and testing literacy and special sectors. Windows can be used as a disk encryption. This function has the following two shortcomings to be improved : 1. This function can only be read, read A : and B : that only the right to operate two floppy disks. No change disk sector size, the standard is only 512 bytes. Parameter Description : command operations : 0 Reading 2 replacement disk write Sector Sector 3 4 5 checksum track to be formatted track eight equipment parameters (int 1EH) drive drive A : = 0 B : = 1 head head, range 0-1 track track, the range of 0-84 (80-84 for special track, normally used for encryption) sector sector, the scope of 0-255 (19-255 for a non-standard sector numbers, commonly used encryption) nsectors each read or write in the num (2005-04-23, Visual C++, 38KB, 下载131次)
S-DES的编程实现 S-DES的加密使用8位明文和10位密钥做为输入，产生8位的密文做为输出。加解密使用相同的密钥。要求:采用CBC模式对文本加解密 1.对话界面:选择加密或解密,输入密钥,在目录中选择明文或密文文件(内容为二进制,.txt文件,),提示加密或解密完成 2.输出:产生密文文本或明文文本 3.提供:说明文档,源码,可执行程序,通过加密实验,给出运行结果. 4.提交分析雪崩效应文档:相同密钥下明文改变任一位时密文的位差异情况 相同明文下密钥改变任一位时密文的位差异情况
S-DES Programming S-DES encryption using 8 and 10 expressly as a key input, have eight of dense text as output. Encryption and decryption using the same key. Requirements : CBC mode to a declassified version increases. Dialogue interface : Select encryption or decryption key input in the directory option expressly or dense text document (for binary content,. Txt file,), suggesting encryption or decryption completed 2. Output : produce dense text or text the text expressly 3. provide : documentation, source code, executables, through encryption experiment is running results. 4. analytical avalanche effect Document : Key under the same as an explicit change the text at the close of the differences in the same place where expressly as a key change at the potentiometer ciphertext the situation (2004-12-12, DOS, 2KB, 下载36次)