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Description: 使用RSA算法对一个数字进行加密和解密。可以自由指定p，q的值，并且当输入数字不是素数时，程序会给出提示，或自动指定一个素数。
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Author: 鹏鹏 |
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Description: RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ，编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s <= n, 通常取 s = 2^t), 则每一位数均小于 n, 然後分段编码...... 接下来, 计算 b == a^m mod n, (0 <= b < n), b 就是编码后的资料。解码的过程是, 计算 c == b^r mod pq (0 <= c < pq), 于是, 解码完毕-RSA algorithm : First, find a few 3, p, q, r, p, q is the two different prime number, with the r (p-1) (q-1) coprime several ... p, q, r it person_key number is three, and then find m, making r ^ m == a mod (p-1) (q-1 m ...................................... this must exist, because r (p-1) (q - 1) coprime, with the division algorithm can be a ..... Next, calculate n = pq ....... m, n is the number two public_key, coding process is that if a data and to treat it as a big Integer assuming a <n. ... If a> = n, a table will be rounding into s (s <= n, usually from 2 ^ s = t), each n is less than the median, and then sub-coding ..... . Next, calculate a == b ^ m mod n, (0 <= b <n), b is encoded information. Decoding is the process of calculating c == b ^ r mod pq (0 <= c &
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Author: 胡康康 |
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Description: /* RSA Demo 1.0 版 * 版权所有 (C) 2004 赵春生 * 2004.04.25 * http://timw.yeah.net * http://timw.126.com * 本程序调用Miracl ver 4.82大数运算库，详见其附带手册。 * P,Q,N,D,E使用RSATool2生成。 */ 编译提示： 一：将Project-Settings-Settings For(All Configuration)-C/C++中Category项的 Precompiled Headers设置成：Automatic use of precompiled headers（图1）。 二：将ms32.lib添加到工程中（图2）。 三：MIRACL是C库。 extern \"C\" { #include \"miracl.h\" #include \"mirdef.h\" } #pragma comment( lib, \"ms32.lib\" )-/ * RSA Demo version 1.0 * Copyright (C) 2004 Zhao Chunsheng 2004.04.25 * * * http://timw.126.com http://timw.yeah.net * The procedures called Miracl ver 4. The majority of computing for 82, as detailed in its fringe manual. * P, Q, N, D, E use RSATool2 generation. * / Compiler Tip : 1 : Project-Settings - Settings For (All Configuration) - C / C Category, of OO Headers set : Automatic use of precompiled headers (Figure 1). 2 : ms32.lib added to the project (Figure 2). 3 : MIRACL C library. Extern "C" (# include "miracl.h" # include "mirdef.h") # pragma comment (lib, "ms32.lib")
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Author: 李湘 |
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Description:
300多种加密算法
RSA
DES
ECC
Blowfish
RC4
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Author: guijiarui |
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Description: 密码学课程设计 对大家很有帮助
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Author: dhl004 |
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Description: RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ，编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s <= n, 通常取 s = 2^t), 则每一位数均小于 n, 然後分段编码...... 接下来, 计算 b == a^m mod n, (0 <= b < n), b 就是编码后的资料。解码的过程是, 计算 c == b^r mod pq (0 <= c < pq), 于是, 解码完毕-RSA algorithm : First, find a few 3, p, q, r, p, q is the two different prime number, with the r (p-1) (q-1) coprime several ... p, q, r it person_key number is three, and then find m, making r ^ m == a mod (p-1) (q-1 m ...................................... this must exist, because r (p-1) (q- 1) coprime, with the division algorithm can be a ..... Next, calculate n = pq ....... m, n is the number two public_key, coding process is that if a data and to treat it as a big Integer assuming a <n. ... If a> = n, a table will be rounding into s (s <= n, usually from 2 ^ s = t), each n is less than the median, and then sub-coding ..... . Next, calculate a == b ^ m mod n, (0 <= b <n), b is encoded information. Decoding is the process of calculating c == b ^ r mod pq (0 <= c &
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Author: 胡康康 |
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Description: /* RSA Demo 1.0 版 * 版权所有 (C) 2004 赵春生 * 2004.04.25 * http://timw.yeah.net * http://timw.126.com * 本程序调用Miracl ver 4.82大数运算库，详见其附带手册。 * P,Q,N,D,E使用RSATool2生成。 */ 编译提示： 一：将Project-Settings-Settings For(All Configuration)-C/C++中Category项的 Precompiled Headers设置成：Automatic use of precompiled headers（图1）。 二：将ms32.lib添加到工程中（图2）。 三：MIRACL是C库。 extern "C" { #include "miracl.h" #include "mirdef.h" } #pragma comment( lib, "ms32.lib" )-/* RSA Demo version 1.0* Copyright (C) 2004 Zhao Chunsheng 2004.04.25*** http://timw.126.com http://timw.yeah.net* The procedures called Miracl ver 4. The majority of computing for 82, as detailed in its fringe manual.* P, Q, N, D, E use RSATool2 generation.*/Compiler Tip : 1 : Project-Settings- Settings For (All Configuration)- C/C Category, of OO Headers set : Automatic use of precompiled headers (Figure 1). 2 : ms32.lib added to the project (Figure 2). 3 : MIRACL C library. Extern "C" (# include "miracl.h"# include "mirdef.h")# pragma comment (lib, "ms32.lib")
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Author: 李湘 |
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Description: 基于java的完整的RSA算法实现 /** * <p>Titre : RSA </p> * <p>Description : Encodage de donn閑s selon le protocole RSA </p> * <p>Copyright : Copyright (c) 2004</p> * @author Fran鏾is Bradette * @version 1.1 * version originale de Robert Sedgewick and Kevin Wayne.Copyright ? 2004 * pris sur le site http://www.cs.princeton.edu/introcs/104crypto/RSA.java.html * Modifier par Fran鏾is Bradette */-on the integrity of the RSA algorithm/*** lt; Pgt; Part : RSA lt;/Pgt;* Lt; Pgt; Description : Encodage de donn idle s selon le protocole RSA lt;/Pgt;* Lt; Pgt; Copyright : Copyright (c) 2004lt;/pgt;* @ author Fran is continuing Bradette** @ version 1.1 version originale de Robert Sedgewick and Kevin Wayne.Copyright 2004* pris sur le site http://www.cs.princeton.* Modifier edu/introcs/104crypto/RSA.java.html par Fran is continuing Bradette* /
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Author: 某男 |
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Description: RSA资料全集是我在学习RSA时收集的资料，它对我学习这种算法有着极大的帮助，不知对大家有帮助没有，希望对大家学习密码学有所帮助。-All of the information of RSA was collected when i was studying RSA.It gives me a great deal of help though i don t know how about you.I wish it also give you some help about the cryptology.
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Author: 王清华 |
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Description: 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 factorization great difficulties to the fact. Choice of two prime numbers, p, q. (General p, q very few options) and then calculate the z = p* q = f (p-1) (q-1) Choose a n, so gcd (n, f) = a (gcd representatives greatest common divider. n general also choose a prime number), n and z on as a public key. One's choice, 0
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Author: fasf |
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Description: 使用RSA算法对一个数字进行加密和解密。可以自由指定p，q的值，并且当输入数字不是素数时，程序会给出提示，或自动指定一个素数。
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Size: 114688 |
Author: 鹏鹏 |
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Description: Digital Signature Algorithm (DSA)是Schnorr和ElGamal签名算法的变种，被美国NIST作为DSS(DigitalSignature Standard)。算法中应用了下述参数：
p：L bits长的素数。L是64的倍数，范围是512到1024；
q：p - 1的160bits的素因子；
g：g = h^((p-1)/q) mod p，h满足h < p - 1, h^((p-1)/q) mod p > 1；
x：x < q，x为私钥 ；
y：y = g^x mod p ，( p, q, g, y )为公钥；
H( x )：One-Way Hash函数。DSS中选用SHA( Secure Hash Algorithm )。
p, q, g可由一组用户共享，但在实际应用中，使用公共模数可能会带来一定的威胁。签名及验证协议如下：
1. P产生随机数k，k < q；
2. P计算 r = ( g^k mod p ) mod q
s = ( k^(-1) (H(m) + xr)) mod q
签名结果是( m, r, s )。
3. 验证时计算 w = s^(-1)mod q
u1 = ( H( m ) * w ) mod q
u2 = ( r * w ) mod q
v = (( g^u1 * y^u2 ) mod p ) mod q
若v = r，则认为签名有效。
DSA是基于整数有限域离散对数难题的，其安全性与RSA相比差不多。DSA的一个重要特点是两个素数公开，这样，当使用别人的p和q时，即使不知道私钥，你也能确认它们是否是随机产生的，还是作了手脚。RSA算法却作不到。
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Author: wildkaede |
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Description: 1) 找出两个相异的大素数P和Q，令N＝P×Q，M＝（P－1）（Q－1）。
2) 找出与M互素的大数E，用欧氏算法计算出大数D，使D×E≡1 MOD M。
3) 丢弃P和Q，公开E，D和N。E和N即加密密钥，D和N即解密密钥。
-1) to identify two different large prime numbers P and Q, so N = P × Q, M = (P-1) (Q-1). 2) to identify and M large numbers coprime E, Euclidean algorithm used to calculate lump sum for the D, so that D × E ≡ 1 MOD M. 3) disposed of P and Q, the open E, D and N. E and N that encryption keys, D and N that the decryption key.
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Author: 阿达悟 |
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Description: RSA公钥加密的基本实现 bmp灰度图片加解密操作包括 RSA 的加减密算法； 素数检测算法；RSA 密钥生成算法； 应用该 RSA 密码体制加、解密； BMP 灰度图的算法； Pollard p-1 算法 ； Pollard r 算法 ； -RSA public key encryption to achieve the basic gray-scale picture bmp including RSA encryption and decryption operations of addition and subtraction Micronesia algorithm prime number detection algorithm RSA key generation algorithm application of the RSA cryptosystem, Encryption and Decryption BMP grayscale algorithm Pollard p-1 algorithm Pollard r algorithm
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Author: 魏汝垚 |
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Description: 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 in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key- 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 in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key
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Author: nb |
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Description: 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.
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Author: jhp627 |
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Description: RSA algorithem p=1024 bits
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Author: donweena |
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Description: rsa简单算法
输入p q e等加密元素，然后根据提示可进行解密-rsa Algorithm Analysis
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Author: 黄冉 |
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Description: RSA 原理：
选取两个不同的大素数p、q，并计算N=p*q
选取小素数d，并计算e，使d*e (p-1)(q-1)=1 -RSA works: select two different large prime numbers p, q, and compute N = p* q Select the small prime d, and calculate the e, so d* e (p-1) (q-1) = 1
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Author: 董然 |
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Description: 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.)
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Author: Appoint |
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