[HDCTF2019]bbbbbbrsa

[HDCTF2019]bbbbbbrsa

p = 177077389675257695042507998165006460849
n = 37421829509887796274897162249367329400988647145613325367337968063341372726061
c = ==gMzYDNzIjMxUTNyIzNzIjMyYTM4MDM0gTMwEjNzgTM2UTN4cjNwIjN2QzM5ADMwIDNyMTO4UzM2cTM5kDN2MTOyUTO5YDM0czM3MjM

from base64 import b64encode as b32encode
from gmpy2 import invert,gcd,iroot
from Crypto.Util.number import *
from binascii import a2b_hex,b2a_hex
import random

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

nbit = 128

p = getPrime(nbit)
q = getPrime(nbit)
n = p*q

print p
print n

phi = (p-1)*(q-1)

e = random.randint(50000,70000)

while True:
	if gcd(e,phi) == 1:
		break;
	else:
		e -= 1;

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

print b32encode(str(c))[::-1]

# 2373740699529364991763589324200093466206785561836101840381622237225512234632

运行脚本

#!/usr/bin/env python
# -*- coding:utf-8 -*-
from base64 import b64encode as b32encode
from base64 import b64decode
from gmpy2 import invert, gcd, iroot
from Crypto.Util.number import *
 
p = 177077389675257695042507998165006460849
n = 37421829509887796274897162249367329400988647145613325367337968063341372726061
c64 = '==gMzYDNzIjMxUTNyIzNzIjMyYTM4MDM0gTMwEjNzgTM2UTN4cjNwIjN2QzM5ADMwIDNyMTO4UzM2cTM5kDN2MTOyUTO5YDM0czM3MjM'
c = int ( b64decode ( str ( c64)[::-1] ) )
print ( c )
q = n // p
phi = (p - 1) * (q - 1)
for e in range ( 50000, 70000 ):
    if  gcd ( e, phi ) == 1:
        d = invert ( e, phi )
        m = pow ( c, d, n )
        flag=str(long_to_bytes(m))
        if 'flag' in flag or 'CTF' in flag or ("{" in flag and '}'in flag):
            print(flag)

在运行结果里面手动找一下 即可找到flag

flag{rs4_1s_s1mpl3!#}

标签：phi,nbit,gcd,HDCTF2019,flag,print,import,bbbbbbrsa
来源： https://blog.csdn.net/m0_52727862/article/details/117383992