Description: rsa算法演示，运用数论知识和经典的RSA算法，可生成超长的密钥，对文件经行加密，解密的演示。当时的开发环境是JCreator 3.50。-rsa algorithm demo, using knowledge of number theory and classical RSA algorithm, can generate very long keys, the document by line encryption, decryption of the presentation. At that time, the development environment is JCreator 3.50. Platform: |
Size: 4096 |
Author:zgyang |
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Description: 利用C\C++实现RSA算法的加、解密运算。
具体包括：
1）利用扩展的Euclid计算 a mod n 的乘法逆元；
2）Miller-Rabin素性测试算法对一个给定的大数进行测试；
3）实现的运算，并计算；
4）利用Euler定理手工计算，并与3）计算的结果对比；
5）实现RSA算法。并对 I LOVE NANJING UNIVERSITY OF AERONAUTICS AND ASTRONAUTICS 加解密。说明：为了方便实现，分组可以小一点，比如两个字母一组。
字母及其数字编码 字母及其数字编码
空格 00 N 14
A 01 O 15
B 02 P 16
C 03 Q 17
D 04 R 18
E 05 S 19
F 06 T 20
G 07 U 21
H 08 V 22
I 09 W 23
J 10 X 24
K 11 Y 25
L 12 Z 26
M 13 -Use of C \ C++ implements the RSA algorithm encryption and decryption operations.
These include:
1) using the extended Euclid calculate a mod n multiplicative inverse
2) Miller-Rabin primality testing algorithm for a given test large numbers
3) to achieve the operation, and the calculation
4) the use of Euler Theorem manual calculation, and compared with the results of the calculation 3)
5) implement the RSA algorithm. And I LOVE NANJING UNIVERSITY OF AERONAUTICS AND ASTRONAUTICS encryption and decryption. Description: In order to facilitate the achievement of the packet may be smaller, for example, a group of two letters.
Alphabet letters and their digital encoding and digital encoding
Spaces 00 N 14
A 01 O 15
B 02 P 16
C 03 Q 17
D 04 R 18
E 05 S 19
F 06 T 20
G 07 U 21
H 08 V 22
I 09 W 23
J 10 X 24
K 11 Y 25 Platform: |
Size: 1024 |
Author:刘洋 |
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