0x00 前言

CTF 加解密合集：CTF 加解密合集

0x01 题目

给了一个秘钥，三个加密后的文件

0x02 Write Up

先获取n和e

# 导入公钥
with open(r"C:\Users\wdd\Downloads\flag\fujian\public.key", "rb") as f:
    key = RSA.import_key(f.read())
    n = key.n
    e = key.e

然后通过n进行在线分解

得到p和q
现在题目就变成了，已知p，q，e，c求m，脚本如下

from Crypto.PublicKey import RSA
import libnum
import gmpy2
# 导入公钥
with open(r"C:\Users\wdd\Downloads\flag\fujian\public.key", "rb") as f:
    key = RSA.import_key(f.read())
    n = key.n
    e = key.e
print(n)
with open(r"C:\Users\wdd\Downloads\flag\fujian\encrypted.message1", "rb") as f:
    c1 = libnum.s2n(f.read())
with open(r"C:\Users\wdd\Downloads\flag\fujian\encrypted.message2", "rb") as f:
    c2 = libnum.s2n(f.read())
with open(r"C:\Users\wdd\Downloads\flag\fujian\encrypted.message3", "rb") as f:
    c3 = libnum.s2n(f.read())

p = 302825536744096741518546212761194311477
q = 325045504186436346209877301320131277983
d = libnum.invmod(e, (p - 1) * (q - 1))
c = [c1, c2, c3]
flag = ''
for i in c:
    m = pow(i, d, n)
    m1 = str(libnum.n2s(int(m)))
    flag += (m1.split("x00")[1])[:-3]  # 以x00分段取后面那段，将 \n' 去掉
print(flag)

以上