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7_Rsa
- RSA公钥加密算法基于大整数因式分解困难这样的事实。 选择两个素数，p，q。（一般p,q选择很大的数） 然后计算 z=p*q f=(p-1)(q-1) 选择一个n，使gcd(n,f)=1（gcd代表greatest common divider,一般n也选择一个素数）， n和z就作为公钥。 选择一个s,0<s<f,满足n*s ％ f=1，s就作为私钥。-RSA public key encryption algorithm based on the integer fa
BigInteger_src
- C# BigInteger class. BigInteger.cs is a csharp program. It is the BIgInteger class. It has methods: abs() , FermatLittleTest(int confidence) ,gcd(BigInteger bi) , genCoPrime(int bits, Random rand) , genPseudoPrime(int bits, int confidence, Random ran
gudianmima
- 古典密码中，主要的思想为移位算法及置换算法。 1.移位密码 密钥K为整数，且取值空间为0到25；加密函数：x = x + k (mod 26)；解密函数：x = x - k (mod 26)。当K=3时，为凯撒密码。 2.仿射密码 密钥对由a、b组成，整数a满足 gcd(a, 26) = 1，整数b的取值空间为0到25；加密函数：x = ax + b(mod 26)；解密函数：x = a*y - a*b (mod 26)。当a=1,b=3时，为凯撒密码。 3.维吉
GCD
- 求sa和tb的和使等于a和b的最大公约数-For sa and tb, and so equal to a and b of the common denominator
recurs
- /* Divide and Conquer */ /* Find minimum and maximum from a given series of numbers */ /* by using above said policy */ gcd fibonacci-/* Divide and Conquer*/ /* Find minimum and maximum from a given series of numbers*/ /* by using
BasicRSA_latest.tar
- RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman i
least_common.tar
- a linux program that enables to find the least common multiple of two integers. by means of this lcm, you can easiky find the gcd.
RSA
- 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 experi
gcd
- oujilide suanfa de kuozhansuanfade shxiamzj jdsa lfjdas f-Euclid suandafa a ldjfdkuodahznadnfa,dfna dsfjdas fjdas
mimaxue
- 内含背包程序，RSA程序，中国剩余定理程序和GCD程序的源代码-Includes backpack program, RSA program, Chinese remainder theorem programs and GCD program' s source code
Shell_Scripting
- Sample programs of shell scr ipting. finding the maximum no. finding the gcd of a no. etc.
gcd
- The greatest common divisor of integers x and y is the largest integer that evenly divides both x and y. Write a recursive function gcd that returns the greatest common divisor of x and y, define recursively as follows. It y is equal to 0, then gcd(x
num.cpp
- 数论mod有关算法 丢翻图方程 线性同余方程解 原始根 gcd exgcd 中国剩余定理 需要高精度库gmpc-gcd exgcd diophantic equation linear congruence Chinese reminder square root primitive root
456
- RSA算法的C语言实现 1.密钥的产生 (1)选两个安全的大素数p和q。 (2)计算n=p×q，φ(n)=(p-1)(q-1)，其中φ(n)是n的欧拉函数值。 (3)选一整数e，满足1<e<φ(n)，且gcd(φ(n),e)=1。 (4)计算d，满足de≡1 modφ(n)，即d是e在模φ(n)下的乘法逆元，因e与φ(n)互素，由模运算可知，它的乘法逆元一定存在。 (5)以{e,n}为公开钥，{d,n}为秘密钥。 2.加密 加密时首先将明文M比特串分组
RC42003
- this about GCD function -this is about GCD function
gcd
- DIDD Interface module for Eicon active cards.
gcd
- Example configuration for a single TS1003 tilt switch that rotates around one axis in 4 steps and emitts the current tilt via two GPIOs.
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