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Description: 本程序实现了用RSA加密算法加密、解密图片。本程序仅作为RSA原理理解，所以没有实现大数运算部分，RSA选取n为15~16位，加密图片不要选太大，否则会很慢。
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Size: 3737 |
Author: 涂进 |
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Description: RSA算法是第一个能同时用于加密和数字签名的算法，也易于理解和操作。RSA是被研究得最广泛的公钥算法，从提出到现在已近二十年，经历了各种攻击的考验，逐渐为人们接受，普遍认为是目前最优秀的公钥方案之一。RSA的安全性依赖于大数的因子分解，但并没有从理论上证明破译RSA的难度与大数分解难度等价。即RSA的重大缺陷是无法从理论上把握它的保密性能如何，而且密码学界多数人士倾向于因子分解不是NPC问题。 RSA的缺点主要有：A)产生密钥很麻烦，受到素数产生技术的限制，因而难以做到一次一密。B)分组长度太大，为保证安全性，n 至少也要 600 bits 以上，使运算代价很高，尤其是速度较慢，较对称密码算法慢几个数量级；且随着大数分解技术的发展，这个长度还在增加，不利于数据格式的标准化。目前，SET( Secure Electronic Transaction )协议中要求CA采用比特长的密钥，其他实体使用比特的密钥
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Size: 19520 |
Author: 陈泗勇 |
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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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Size: 1393 |
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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Size: 172963 |
Author: 李湘 |
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Description: RSA加密算法，界面操作，可以很方便的对文件的内容进行加密，而且为了能加密更长的内容，算法中采用了分段加密的手段。其算法步骤如下：1. 读取公钥e和n，作加密之用。 2. 从DecText中读取一大段明文，转化成明文字节数组Byte[]。 3. 再把Byte[]分成若干小段明文字节数组sectByte[]。 4. 使用e和n对所有的sectByte[]进行加密，并合成一大段密文,添加到EncText。 5. 读取下一大段明文，若明文为空，完成加密；否则转2-RSA encryption algorithm, interface operation, it is convenient to the contents of the documents for encryption, but in order to longer content encryption algorithm used in the sub-encryption tools. The algorithm steps are as follows : 1. Read public key e and n, for encryption purposes. 2. DecText from reading a big specifically, into an express byte array Byte []. 3. Byte [] then divided into pieces expressly sectByte byte [] array. 4. The use of e and n of all sectByte [] encryption and synthesis of a large dense text to be added EncText. 5. The next major reading of the express, if expressly for the space, complete encryption; Otherwise turn 2
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Size: 75988 |
Author: 何泽荣 |
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Description: RSA加长分段解密算法，界面操作。算法步骤如下：1. 读取私钥d和n，作解密之用。 2. 从EncText中读取一大段密文，再把该大段密文分成若干小段密文。 3. 使用d和n把所有小段密文进行解密成对应的小段密文sectByte[]，再合成一大段明文字节数组Byte[]，并转化成大段明文添加到DecText。 4. 读取下一大段密文，若密文为空，完成解密；否则转2-RSA decryption algorithm lengthened section, the interface operation. Algorithm steps are as follows : 1. Read private key d and n, used for decryption. 2. EncText from reading a big secret of the text, and then the big secret of the text is divided into a number of subparagraphs ciphertext. 3. The use of d and n all subparagraph declassified secret text into corresponding sections ciphertext sectByte [], then synthesis of a large expressly byte array Byte [], and conversion of the university to express DecText added. 4. The next major reading of dense text, if ciphertext empty, completed decryption; Otherwise turn 2
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Size: 33550 |
Author: 何泽荣 |
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Description: 1024位的大整数进行相乘（N方）取模，是RSA密钥算法的一部分。其中，我对十进制，二进制的高效转换部分非常满意-1024 for the large integer multiplication (N) The tray is RSA key part of the algorithm. Which, I decimal, binary conversion efficiency is very satisfactory
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Size: 57293 |
Author: 李春晖 |
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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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Size: 1024 |
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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Size: 342016 |
Author: 李湘 |
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Description: RSA加密算法，界面操作，可以很方便的对文件的内容进行加密，而且为了能加密更长的内容，算法中采用了分段加密的手段。其算法步骤如下：1. 读取公钥e和n，作加密之用。 2. 从DecText中读取一大段明文，转化成明文字节数组Byte[]。 3. 再把Byte[]分成若干小段明文字节数组sectByte[]。 4. 使用e和n对所有的sectByte[]进行加密，并合成一大段密文,添加到EncText。 5. 读取下一大段明文，若明文为空，完成加密；否则转2-RSA encryption algorithm, interface operation, it is convenient to the contents of the documents for encryption, but in order to longer content encryption algorithm used in the sub-encryption tools. The algorithm steps are as follows : 1. Read public key e and n, for encryption purposes. 2. DecText from reading a big specifically, into an express byte array Byte []. 3. Byte [] then divided into pieces expressly sectByte byte [] array. 4. The use of e and n of all sectByte [] encryption and synthesis of a large dense text to be added EncText. 5. The next major reading of the express, if expressly for the space, complete encryption; Otherwise turn 2
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Size: 75776 |
Author: 何泽荣 |
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Description: RSA加长分段解密算法，界面操作。算法步骤如下：1. 读取私钥d和n，作解密之用。 2. 从EncText中读取一大段密文，再把该大段密文分成若干小段密文。 3. 使用d和n把所有小段密文进行解密成对应的小段密文sectByte[]，再合成一大段明文字节数组Byte[]，并转化成大段明文添加到DecText。 4. 读取下一大段密文，若密文为空，完成解密；否则转2-RSA decryption algorithm lengthened section, the interface operation. Algorithm steps are as follows : 1. Read private key d and n, used for decryption. 2. EncText from reading a big secret of the text, and then the big secret of the text is divided into a number of subparagraphs ciphertext. 3. The use of d and n all subparagraph declassified secret text into corresponding sections ciphertext sectByte [], then synthesis of a large expressly byte array Byte [], and conversion of the university to express DecText added. 4. The next major reading of dense text, if ciphertext empty, completed decryption; Otherwise turn 2
Platform: |
Size: 33792 |
Author: 何泽荣 |
Hits:
Description: 1024位的大整数进行相乘（N方）取模，是RSA密钥算法的一部分。其中，我对十进制，二进制的高效转换部分非常满意-1024 for the large integer multiplication (N) The tray is RSA key part of the algorithm. Which, I decimal, binary conversion efficiency is very satisfactory
Platform: |
Size: 3643392 |
Author: 李春晖 |
Hits:
Description: 本程序实现了用RSA加密算法加密、解密图片。本程序仅作为RSA原理理解，所以没有实现大数运算部分，RSA选取n为15~16位，加密图片不要选太大，否则会很慢。-This procedure achieved the RSA encryption algorithm with encryption, decryption picture. This procedure only as a principle of understanding of RSA, there is no part of the realization of large numbers computing, RSA select n to 15 ~ 16, encryption, image and not too much, otherwise they will be very slow.
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Size: 3072 |
Author: 涂进 |
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Description: RSA算法是第一个能同时用于加密和数字签名的算法，也易于理解和操作。RSA是被研究得最广泛的公钥算法，从提出到现在已近二十年，经历了各种攻击的考验，逐渐为人们接受，普遍认为是目前最优秀的公钥方案之一。RSA的安全性依赖于大数的因子分解，但并没有从理论上证明破译RSA的难度与大数分解难度等价。即RSA的重大缺陷是无法从理论上把握它的保密性能如何，而且密码学界多数人士倾向于因子分解不是NPC问题。 RSA的缺点主要有：A)产生密钥很麻烦，受到素数产生技术的限制，因而难以做到一次一密。B)分组长度太大，为保证安全性，n 至少也要 600 bits 以上，使运算代价很高，尤其是速度较慢，较对称密码算法慢几个数量级；且随着大数分解技术的发展，这个长度还在增加，不利于数据格式的标准化。目前，SET( Secure Electronic Transaction )协议中要求CA采用比特长的密钥，其他实体使用比特的密钥-err
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Size: 61440 |
Author: 陈泗勇 |
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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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Size: 6144 |
Author: 阿达悟 |
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Description: it is a rsa algorithm agent programme in cpp for ns2
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Size: 1024 |
Author: Vineet |
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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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Size: 81920 |
Author: jhp627 |
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Description: encrypt-decrypt with RSA algorithm. implementation encryption algorithm with RSA method. using public key and private key. first, generate public and private key with generate.php. you will get e,d, and n. for encrypt, use the e and n for private and public key. and using d and n as private and public key to decrypt your text
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Size: 8192 |
Author: wahyu |
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Description: 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.
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Size: 45056 |
Author: semmir |
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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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Size: 1108992 |
Author: Appoint |
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