RSA算法的简单实现

我们的网络安全试验，需用任一语言实现rsa算法，我就用python实现了。

用python来实现，相对于c来说，代码较短，但是速度明显没有c的快。

RSA算法描述见：wiki

代码如下：

#!/usr/bin/env python
#-*-coding: utf-8-*-
# author: "SK (skxiaonan@gmail.com)"
# date: "Date: 2009/05/22 09:10:10"
import math
def isPrime(number):
	i=2
	sqrtNumber=int(math.sqrt(number))
	for i in range(2, sqrtNumber+1):
		if number%i == 0:
			return False
		i = i+1
	return True

if __name__=="__main__":
	print "*"*77
	Flag = False
	while Flag == False:
		p = int(raw_input("Please input a prime(P): "))
		Flag = isPrime(p)
		if Flag == False:
			print "What you input is not a prime!"
	print "The P is: ", p
	
	Flag = False
	while Flag == False:
		q = int(raw_input("Please input a prime(Q): "))
		if p == q:
			continue
		Flag = isPrime(q)
		if Flag == False:
			print "What you input is not a prime!"
	print "The Q is: ", q
	n = p*q
	print "The N is: ", n
	t = (p-1)*(q-1)
	print "The T is: ", t
	
	print "*"*77
	Flag = False
	while Flag == False:
		e = int(raw_input("Please input a number(E): "))
		if (e<1 or e>t):
			continue
		d=0
		while (((e*d)%t) != 1):
			d+=1
		Flag = True
	print "The E is: ", e
	print "The D is: ", d
	print "The Public Key(E, N) is:", e, n
	print "The Private Key(D, N) is:", d, n

	print "*"*77
	Flag = False
	while Flag == False:
		plainText = int(raw_input("Please input a plaintext: "))
		if (plainText < n):
			Flag = True
	print "The plaintext is: ", plainText
	print "Encrypt"+"."*7
	cipherText = (plainText**e)%n 
	print "cipherText is: ", cipherText	
	print "Decrypt"+"."*7
	plain = (cipherText**d)%n
	print "The plain is: ", plain

	print "*"*77
	if plainText == plain:
		print "RSA Test success."
	else:
		print "RSA Test unsuccess!"