BUUCTF 每日打卡 2021-7-16
引言
颓废了一天，水了
[UTCTF2020]hill
hill？山丘？变异栅栏？ 没有头绪，找wp 原来是希尔密码，没见过 主要就是要求一个nxn的密钥矩阵，然后解密即可 从wp中得知utflag被加密成wznqca，只能构造6个式子，还要模26，所以不可能是3x3的矩阵，只有可能是2x2的矩阵，4个未知数 即 \[
\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}20 & 5 & 0 \\19 & 11 & 6 \end{pmatrix}\equiv \begin{pmatrix}22 & 13 & 2 \\25 & 16 & 0 \end{pmatrix}mod\space 26
\] 其中\(a,b,c,d\)为密钥矩阵的四个未知数 得到模方程组 \[
\begin{cases}
20a+19b\equiv 22\space mod\space 26\\
20c+19d\equiv 25\space mod\space 26\\
5a+11b\equiv 13\space mod\space 26\\
5c+11d\equiv 16\space mod\space 26\\
6b\equiv 2\space mod\space 26\\
6d\equiv 0\space mod\space 26
\end{cases}
\] 显然，最后两个式子是最好下手的，但是我没有什么好的方法，猜测\(a,b,c,d\)应该不会超过100，干脆直接爆破，代码如下： 1
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26import random
for k in range(0, 100):
a = 0; b = 0; c = 0; d = 0
while (20*a+19*b) % 26 != 22 or (20*c+19*d) % 26 != 25:
d = random.randint(0, 100)
b = random.randint(0, 100)
a = 0
c = 0
while (6*d) % 26 != 0:
d += 1
while (5*c+11*d) % 26 != 16:
c += 1
while (6*b) % 26 != 2:
b += 1
while (5*a+11*b) % 26 != 13:
a += 1
print(a, b, c, d)
with open('result.txt', 'r') as f:
r = f.readlines()
with open('result.txt', 'a') as f:
if not (str([a, b, c, d]) + '\n') in r:
f.write(str([a, b, c, d]) + '\n')1
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16[1, 22, 11, 91]
[1, 22, 11, 39]
[1, 100, 11, 39]
[1, 100, 11, 91]
[1, 48, 11, 39]
[1, 48, 11, 65]
[1, 74, 11, 39]
[1, 22, 11, 13]
[1, 100, 11, 65]
[1, 74, 11, 65]
[1, 74, 11, 13]
[1, 74, 11, 91]
[1, 22, 11, 65]
[1, 100, 11, 13]
[1, 48, 11, 91]
[1, 48, 11, 13]
忽略了符号、数字和大小写，修改一下即可 结果为utflag{d4nger0us_c1pherText_qq}
[INSHack2017]rsa16m
题干描述： When you need really secure communications, you use RSA with a 4096 bit key. I want really really really secure communications to transmit the nuclear launch codes (yeah IoT is everywhere man) so I used RSA with a 16777216 bit key. Surely russians will not be able to factor that one !
平时为了保证安全RSA会采用4096位的密钥，但他给我整了个16777216位的 wtm直接找wp 
啊这 这还真想不到 代码如下：
1 | import gmpy2 |
结果为 
结语
希望继续坚持