题目：
from Crypto.Util.number import *
from secret import flag
p = getPrime(64)
q = getPrime(64)
assert p % 4 == 3
assert q % 4 == 3
n = p * q
e = 2
m = bytes_to_long(flag)
c = pow(m,e,n)
print('n =', n)
print('c =', c)
# n = 201354090531918389422241515534761536573
# c = 20442989381348880630046435751193745753容易知道这是一道Rabin加密题，首先通过网站（factordb.com）暴力破解N，得到P和Q，再根据Rabin解密模板进行破解
在根据中国剩余定理
import gmpy2
import libnum
from Cryptodome.Util.number import *
p= 13934102561950901579
q= 14450452739004884887
n= p*q
c= 20442989381348880630046435751193745753
e= 2
mp = pow(c, (p + 1) // 4, p)
mq = pow(c, (q + 1) // 4, q)
inv_p = gmpy2.invert(p, q)
inv_q = gmpy2.invert(q, p)
a = (inv_p * p * mq + inv_q * q * mp) % n
b = n - int(a)
c = (inv_p * p * mq - inv_q * q * mp) % n
d = n - int(c)
# 因为rabin 加密有四种结果，全部列出。
aa = [a, b, c, d]
for i in aa:
print(long_to_bytes(int(i)))
print(libnum.n2s(int(i)))
得到结果
