继续打印上述rsa-crt没打印完整的代码

好的，下面是完整的 RSA-CRT 代码：

# Compute n = p * q
n = p * q

# Compute phi(n) = (p - 1) * (q - 1)
phi_n = (p - 1) * (q - 1)

# Choose an integer e such that e and phi(n) are coprime
# Use a relatively small value for e to speed up the encryption process
e = 65537

# Compute d such that d * e = 1 (mod phi(n))
# d is the modular inverse of e modulo phi(n)
# We can use the extended Euclidean algorithm to compute d
(d, _) = extended_euclidean_algorithm(e, phi_n)

# Make sure d is positive
d = d % phi_n
if d < 0:
    d = d + phi_n

# Public key is (e, n)
public_key = (e, n)

# Private key is (d, p, q)
private_key = (d, p, q)

def rsa_crt_encrypt(m, public_key):
    e, n = public_key
    p, q = factorize(n)
    dp = d % (p - 1)
    dq = d % (q - 1)
    qinv = modular_inverse(q, p)

    # Compute c = m^e (mod n) using the Chinese Remainder Theorem
    c1 = pow(m, dp, p)
    c2 = pow(m, dq, q)
    h = (qinv * (c1 - c2)) % p
    c = c2 + h * q

    return c

def rsa_crt_decrypt(c, private_key):
    d, p, q = private_key
    dp = d % (p - 1)
    dq = d % (q - 1)
    qinv = modular_inverse(q, p)

    # Compute m = c^d (mod n) using the Chinese Remainder Theorem
    m1 = pow(c, dp, p)
    m2 = pow(c, dq, q)
    h = (qinv * (m1 - m2)) % p
    m = m2 + h * q

    return m

def rsa_crt_sign(m, private_key):
    d, p, q = private_key
    dp = d % (p - 1)
    dq = d % (q - 1)
    qinv = modular_inverse(q, p)

    # Compute s = m^d (mod n) using the Chinese Remainder Theorem
    s1 = pow(m, dp, p)
    s2 = pow(m, dq, q)
    h = (qinv * (s1 - s2)) % p
    s = s2 + h * q

    return s

def rsa_crt_verify(s, m, public_key):
    e, n = public_key
    p, q = factorize(