f = (? * ?)

题目

附件下载有两个file1，file2，外加cipher和hint，因为hint写的e = 65537，推测是基于RSA出的题目，另外两个file是
file1

点击查看代码

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file2

点击查看代码

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cipher

点击查看代码

ve9MPTSrRrq89z+I5EMXZg1uBvHoFWBGuzxhSpIwu9XMxE4H2f2O3l+VBt4wR+MmPJlS9axvH9dCn1KqFUgOIzf4gbMq0MPtRRp+PvfUZWGrJLpxcTjsdml2SS5+My4NIY/VbvqgeH2qVA==

得知RSA加密的基础之上，我们有了e和c的值，我们猜测两个file应该是p，q值，所以我们进行解密看看。
观察给所给的文件，其中都为科学计数法的浮点数，并无实际意义，但是仔细观察不难发现每组数据只有两个值且对应科学计数法的指数为00，01，很明显的二进制数，将其转成十进制应该能得到p，q的值，从而解出，给个脚本

点击查看代码

from Crypto.Util.number import *
import base64

a = open(r"file1.txt", "r").readlines()
b = open(r"file2.txt", "r").readlines()
p = ""
for i in a:
    print(p)
    p += str(int(i[0]) % 2)
p = int(p, 2)
print(p)
q = ""
for i in b:
    q += str(int(i[0]) % 5)
q = int(q, 2)
e = 65537
c = 've9MPTSrRrq89z+I5EMXZg1uBvHoFWBGuzxhSpIwu9XMxE4H2f2O3l+VBt4wR+MmPJlS9axvH9dCn1KqFUgOIzf4gbMq0MPtRRp+PvfUZWGrJLpxcTjsdml2SS5+My4NIY/VbvqgeH2qVA=='
c = base64.b64decode(c)
c = bytes_to_long(c)
d=inverse(e,(p-1)*(q-1))
print(long_to_bytes(pow(c,d,p*q)))

这个是看大佬的脚本，他很巧妙的思想在于只看数据的第一位然后进行模操作，就很轻松得到了0，1值，输出flag$b'H&NCTF{Y0u_s@cce3d3d_in_finding_the_meaning_0f_these_d0cuments}'$