RSA3027的python加解密实现

RSA3027是一个非常大的RSA密钥，其加解密过程需要使用高精度算法。以下是一个Python实现的RSA3027加解密示例：

import math

# RSA parameters
p = 1830293723692707533460203820354173158250326337871530100473137939399205415045329862394080467869376930295655298439781276747912307643621603657882219318585461398021
q = 1569729998666316092466934639524662644867260999971388196840836113328811052659167141660747783089619511421593587530989260501982412467205624811666199042694291743
n = p*q
e = 65537
d = 4477818688885519094344649795711719498160730313711265070728673732430275706891144308693516166876778497062689957894011236235100893175995658411256606509976309644739649303513183032009111334700590872535197922366069632582910305017509629886009101700211327279353578306396525024133195328322535933046390224072140083057

# Convert a string to a list of integers
def str_to_int(message):
    return [ord(c) for c in message]

# Convert a list of integers to a string
def int_to_str(int_list):
    return ''.join([chr(i) for i in int_list])

# Modular exponentiation
def mod_pow(base, exponent, modulus):
    result = 1
    while exponent > 0:
        if exponent % 2 == 1:
            result = (result * base) % modulus
        exponent = exponent // 2
        base = (base * base) % modulus
    return result

# Encrypt a message using RSA
def rsa_encrypt(message):
    int_message = str_to_int(message)
    encrypted = [mod_pow(m, e, n) for m in int_message]
    return encrypted

# Decrypt a message using RSA
def rsa_decrypt(int_list):
    decrypted = [mod_pow(c, d, n) for c in int_list]
    message = int_to_str(decrypted)
    return message

# Example usage
message = "Hello, RSA3027!"
encrypted = rsa_encrypt(message)
decrypted = rsa_decrypt(encrypted)
print("Original message: ", message)
print("Encrypted message: ", encrypted)
print("Decrypted message: ", decrypted)

需要注意的是，由于RSA3027的密钥非常大，加解密过程非常耗时。你可能需要等待一段时间才能得到结果。