RSA & what -BUUCTF-共模-base64隐写

题目

附件

这里给了相同的N，所以无疑是共模攻击，每个文件有多个c，因为之前遇到的也就只有c1，c2，这次多来几个还有点不会了，其实每个相对应组成c1，c2，最后的结果拼接一下就行了，我脚本能力较差，所以就分开跑了六祖c1，c2

import gmpy2 as gp
def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

n = 
c1 = 
c2 = 
e1 = 1697
e2 = 599
print(gp.gcd(c1,c2))
s = egcd(e1, e2)
s1 = s[1]
s2 = s[2]
if s1<0:
	s1 = - s1
	c1 = gp.invert(c1, n)
elif s2<0:
	s2 = - s2
	c2 = gp.invert(c2, n)

m = pow(c1,s1,n)*pow(c2,s2,n) % n
print(m)
print(hex(m)[2:])
print(bytes.fromhex(hex(m)[2:]))

##十进制转ASCII码
'''
m=str(m)
flag=""
i=0
while i < len(m):
    if m[i]=='1':
        c=chr(int(m[i:i+3]))
        i+=3
    else:
        c=chr(int(m[i:i+2]))
        i+=2
    flag+=c
print(flag)
'''

但是，当我跑出第一组结果时我就愣住了

这个看似base64，又不太像，最后看了大佬wp才知道，原来这是base64隐写，那就直接上脚本

from Crypto.Util.number import *
import base64
c = b'' 
def get_base64_diff_value(s1, s2):
    base64chars = b'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
    res = 0
    for i in range(len(s2)):
        if s1[i] != s2[i]:
            return abs(base64chars.index(s1[i]) - base64chars.index(s2[i]))
    return res
 
def solve_stego():
    line=b''
    bin_str=''
    for i in c:
        k=long_to_bytes(i)
        if k==b'\n':
            steg_line = line
            norm_line = base64.b64encode(base64.b64decode(line))
            diff = get_base64_diff_value(steg_line, norm_line)
            #print(diff)
            pads_num = steg_line.count(b'=')
            if diff:
                bin_str += bin(diff)[2:].zfill(pads_num * 2)
            else:
                bin_str += '0' * pads_num * 2
            print(goflag(bin_str))
            line=b''
            continue
        line+=k
 
def goflag(bin_str):
    res_str = ''
    for i in range(0, len(bin_str), 8):
        res_str += chr(int(bin_str[i:i + 8], 2))
    return res_str
 
 
if __name__ == '__main__':
    solve_stego()

结果如下：

成功拿到flag!!

小结：共模攻击原理需熟练！