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libI-2.1pre6
- Arithmetic for integers of almost unlimited size for C and C++. Developed and copyrighted by Ralf Dentzer-Arithmetic for integers of almost unlimit ed size for C and C. Developed and copyrighted by Ralf Dentzer
cryptography.tar
- Cryptography and Network Security, 4th Ed.-Stallings ebook. free to download
The_Art_of_Error_Correcting_codes_ed2
- The Art of Error Correcting codes ed 2
ed
- 一个加密机与解密机实例源码 ,VC编写,测试过,功能是能够实现的,对有这方面需要的人来说很实用,而且代码不是太复杂,可以作为加解密编程学习。注:杀毒软件可能会报,但不碍事。-An encryption and decryption machine machine source code examples, VC prepared, tested, functions can be achieved on the needs of people in this regard is very pra
R4NR4_Utilities_Editor
- Rockey Editor can be used to edit and test Rockey, it is the best tool to learn Rockey
up.c
- decrypt a pice of malware that had part xor-ed with a constant
Audio-Power-Amplifier-Design-Handbook--4th-Ed.---
- Audio Power Amplifi er Design Handbook Douglas Self MA, MSc
rsn
- rsn算法RSA算法的描述 1、选取长度相等的两个大素数p和q,计算其乘积: n = pq 然后随机选取加密密钥e,使e和(p–1)(q–1)互素。 最后用欧几里德扩展算法计算解密密钥d,以满足 ed = 1(mod(p–1) ( q–1)) 即 d = e–1 mod((p–1)(q–1)) e和n是公钥,d是私钥 2、加密公式如下: ci = mi^e(mod n) 3、解密时,取每一密文分组ci并计算: mi = ci^d
SoftKey.Solutions.SENTINEL.Emulator.2007.FIXED-ED
- HASP-Sentinel Dongles Dumpper By Edge
rsa
- 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