CTF-BUUCTF-CRPTO-[HDCTF2019]bbbbbbrsa

1、题目有两个文件

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

 

txt文件内容是：

 p = 177077389675257695042507998165006460849
n = 37421829509887796274897162249367329400988647145613325367337968063341372726061
c = ==gMzYDNzIjMxUTNyIzNzIjMyYTM4MDM0gTMwEjNzgTM2UTN4cjNwIjN2QzM5ADMwIDNyMTO4UzM2cTM5kDN2MTOyUTO5YDM0czM3MjM

2、这个题关键是两点，一个是求c，看上去是base32，实际上是（from base64 import b64encode as b32encode ）base64，而且是字符串首尾反转过来的，再反转回去就可以。

另一个是求e，e的值在50000-70000，爆破e，代码如下

#!/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)

 3、因为每个符合条件的e会产生一个m值 ，我们知道CTF的flag值 ，都 会有flag、CTF、{、}等字符，设置条件后，m的值也不是很多，手工查找一下，就可以找到flag