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RSA算法实验报告和代码 1.选取两个素数p,q(不可相差悬殊) 2.计算n=pq,f(n)=(p-1)(q-1) 3.选取e,满足1<e<f(n),则gcd(e,f(n))=1 4.计算d,满足de=1 mod f(n)。一般d>=[n的四分之一方],(e,n)为公钥,(p,q,d)为私钥,将明文0,1序列分组,使每组十进制小于n。c=[m的e次方] mod n,m=[c的d次方] mod n。-RSA algorithm and code an experimental report. Choose two primes p, q (non-significant differences between) 2. Calculate n = pq, f (n) = (p-1) (q-1) 3. Select the e, to satisfy a <e<f(n),则gcd(e,f(n))=1 4.计算d,满足de=1 mod f(n)。一般d> = [n the fourth side], (e, n) for the public key, (p, q, d) for the private key will be expressly 0,1 sequence packet, so that each of the decimal is less than n. c = [m of the e-th power] mod n, m = [c of the d-th power] mod n.
Date : 2025-12-23 Size : 80kb User : jhp627

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用MFC编写的基于Julia集加密解密的算法,本程序能成功的绘制出2次,m次以及广义的Julia集,并且是在复数空间中,同时能实现无限制的放大,还可以选取任意大小空间截屏保存。接着利用Julia集对图像进行加密解密,并计算出他们的信噪比。全部功能都已实现,是本人的毕业设计。-Written with the MFC-based encryption and decryption algorithm for Julia sets, the program can successfully draw the two times, m times, as well as a broad Julia set, and is in a complex space, at the same time to achieve unlimited zoom, you can also select any size Save room for screen shots. Then the use of Julia sets of the image encryption and decryption, and calculate their signal to noise ratio. All of the features have been achieved, is my graduation project.
Date : 2025-12-23 Size : 4.44mb User : bends

C#加密解密程序及源代码,加密主要分两步进行,第一步选择文件,第二步随机产生对成加密钥匙Key和IV、使用发送者私钥签名随机密钥,使用接收者公钥加密密钥和签名、利用随机密钥使用DES算法分组加密数据、MD5摘要生成-->用发送者私钥签名;-Encryption and decryption procedures and source code in C#, encryption, mainly in two steps, the first step to select the file, the second step randomly generated as the encryption key Key and IV using the sender' s private key signature random key using the recipient public key encryption key key and signature, using the random key packet to encrypt data using the DES algorithm, the MD5 digest generation-> sender' s private key signature
Date : 2025-12-23 Size : 18kb User : 程亮

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SMS4算法的C语言实现。手动输入明文和密钥以及功能选择,按16进制输入和输出。-SMS4 algorithm in C language. Manually enter the plaintext and key functions to select, according to the hexadecimal input and output.
Date : 2025-12-23 Size : 1.17mb User : Chen Gillian

VC/C++源码,MD5,加密,文件多层MD5加密工具。本MD5加密工具速度快,适用于除纯英文类文件外的其他文件,如果加密或解密非纯英文类文件,建议使用此方法,加密或解密纯英文类文件,建议选择全部操作。-VC/C++ source, MD5, encryption, file multi-MD5 encryption tool. The MD5 encryption tool is fast, applicable to other documents in plain English class file, encrypt or decrypt non-plain English class file, it is recommended to use this method, encrypt or decrypt plain English class files, it is recommended that you select all.
Date : 2025-12-23 Size : 36kb User : 韩雅

Win版破解方法: 1.安装Unity后(联网破解不成功就断网试试),运行破解补丁。 2.点击Browse选择Unity所在路径 例如我的Unity安装路径 F:\Soft\unity5.0\Unity\Editor 3.点击Edit,选择Unity Pro,然后点ok 4.选择Unity版本 (注:Unity5.0以下版本全部选择4.0~4.5.3,Unity5.0以上版本选择Unity5.0.0f1(beta)),然后点Patch,破解完成。(0) Install Unity (DO NOT RUN IT). 1) Run patch as Administrator. 2) Select "Unity folder" by pressing "Browse" button. OR Copy the "Patch file" in "Unity folder". (default : "C:\Program Files (x86)\Unity\Editor") 3) Select Unity version by pressing "drop-down menu" (if needed). 4) Click the "Patch" button to start the patching process. Done.)
Date : 2025-12-23 Size : 144kb User : ARLTORIA

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1.问题描述 RSA密码系统可具体描述为:取两个大素数p和q,令n=pq,N=(p-1)(q-1),随机选择整数d,满足gcd(d,N)=1,ed=1 modN。 公开密钥:k1=(n,e) 私有密钥:k2=(p,q,d) 加密算法:对于待加密消息m,其对应的密文为c=E(m)=me(modn) 解密算法:D(c)=cd(modn) 2.基本要求 p,q,d,e参数选取合理,程序要求界面友好,自动化程度高。 4. 实现提示 要实现一个真实的RSA密码系统,主要考虑对大整数的处理。P和q是1024位的,n取2048位。(1. problem description The RSA cryptosystem can be specifically described as: take two large prime numbers P and Q, make n=pq, N= (p-1) (Q-1), select integer D randomly, and satisfy GCD (D, N) =1. Public key: k1= (n, e) Private key: k2= (P, Q, d) Encryption algorithm: for the encrypted message M, its corresponding ciphertext is c=E (m) =me (MODN) Decryption algorithm: D (c) =cd (MODN) 2. basic requirements P, Q, D, e parameters are selected reasonably, the program requires friendly interface and high degree of automation. 4. realization hints To implement a real RSA cryptosystem, the main consideration is to deal with large integers. P and Q are 1024 bits, and N takes 2048.)
Date : 2025-12-23 Size : 1.06mb User : Appoint
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