1、题目信息
一眼看穿
查看代码
from Crypto.Util.number import *
from secret import flag
import gmpy2
flag=b''
r = getPrime(6)
a = 11001240791308496565411773845509754352597481464288272699325231395472137144610774645372812149675141360600469640492874223541765389441131365669731006263464699
p = gmpy2.next_prime(a - r)
q = gmpy2.next_prime(gmpy2.next_prime(a) + r)
n = p*q
def enc(flag, n):
m = bytes_to_long(flag)
return pow(m, 65537, n)
c = enc(flag, n)
print('n =', n)
print('c =', c)
# ('n =', mpz(121027298948349995679677982412648544403333177260975245569073983061538581058440163574922807151182889153495253964764966037308461724272151584478723275142858008261257709817963330011376266261119767294949088397671360123321149414700981035517299807126625758046100840667081332434968770862731073693976604061597575813313L))
# ('c =', mpz(42256117129723577554705402387775886393426604555611637074394963219097781224776058009003521565944180241032100329456702310737369381890041336312084091995865560402681403775751012856436207938771611177592600423563671217656908392901713661029126149486651409531213711103407037959788587839729511719756709763927616470267L))根据r的位数爆破即可
exp
from Crypto.Util.number import *
import gmpy2
n = 121027298948349995679677982412648544403333177260975245569073983061538581058440163574922807151182889153495253964764966037308461724272151584478723275142858008261257709817963330011376266261119767294949088397671360123321149414700981035517299807126625758046100840667081332434968770862731073693976604061597575813313
c = 42256117129723577554705402387775886393426604555611637074394963219097781224776058009003521565944180241032100329456702310737369381890041336312084091995865560402681403775751012856436207938771611177592600423563671217656908392901713661029126149486651409531213711103407037959788587839729511719756709763927616470267
a = 11001240791308496565411773845509754352597481464288272699325231395472137144610774645372812149675141360600469640492874223541765389441131365669731006263464699
for r in range(2**6):
p = gmpy2.next_prime(a - r)
q = gmpy2.next_prime(gmpy2.next_prime(a) + r)
if(n % p == 0):
d = inverse(65537,(p-1)*(q-1))
print(long_to_bytes(pow(c,d,n)))
exit()
#flag{621f7c4f-21de-8566-649e-5a883ce318dc}