BUU-CTF basic rsa

1.题目代码：

import gmpy2
from Crypto.Util.number import *
from binascii import a2b_hex,b2a_hex

flag = "*****************"

p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551

e = 65533
n = p*q


c = pow(int(b2a_hex(flag),16),e,n)

print c

# 27565231154623519221597938803435789010285480123476977081867877272451638645710

2.复现

import gmpy2
import libnum
from Crypto.Util.number import *
from binascii import a2b_hex,b2a_hex

flag = "*****************"

p = 262248800182277040650192055439906580479
q = 262854994239322828547925595487519915551
c=27565231154623519221597938803435789010285480123476977081867877272451638645710
e = 65533
n = p*q
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(libnum.n2s(int(m)))
# flag{B4by_Rs4}