[MRCTF2020]千层套路

[MRCTF2020]千层套路

[MRCTF2020]千层套路

题目地址 : [MRCTF2020]%E5%8D%83%E5%B1%82%E5%A5%97%E8%B7%AF

解压压缩包，发现是密码竟然是压缩包名称，再根据题目，tnnd又是套娃题

import zipfilefile_path = input('path : ')#压缩包路径
path = "/Users/mzq/Desktop/ctf/buuctf/misc/[MRCTF2020]千层套路/"
# path = file_path[0:len(file_path)-len(file_path.split('//')[-1])]
print(path)
n = 0
s2 = ""def extract():i = file_path.split('\')[-1]#递归解压压缩包名称print(i)str1 = ''for x in range(1000):ss = i[:i.find(".")]print(ss)zpf = zipfile.ZipFile( path+ss + ".zip")list = zpf.namelist()#print(list)print(x + 1)for f in list:#print(f)passwd = ss#passwd = f.split('.')[0]#压缩包密码#print(act(f, path, de('utf-8')))if 'zip' in f:print(path, passwd)i = str(f)extract()

最后解出来了一个qr.zip

qr.zip解压后得到qr.txt ，里面都是一些像素值，或者是坐标值

使用脚本拼图

from PIL import Imagefile=open(&#')
MAX=200picture&#w("RGB",(MAX,MAX))
for y in range(MAX):for x in range(MAX):string &#adline()picture.putpixel([x,y],eval(string)) #直接使用eval()可以转为元组
picture.show()

MRCTF{ta01uyout1nreet1n0usandtimes}