DASCTFXFATE2022

2022-04-27

本次月赛是和校队的师傅一起参加的，学到不少东西。进行记录和复现。

MISC

SimpleFlow

分离zip文件，将第50组左右的tcp流里的base进行解密找到压缩命令（-p后边就是压缩包密码），打开压缩包得到flag。

熟悉的猫

关于kdbx文件，先了解一下参考资料。其为KeepPass的密码数据文件，所以用KeepPass来打开。但没有告诉密码，可以使用passware（windows密码恢复工具）恢复：

用KeepPass打开查看密码为 jbRw5PB2kFmor6IeYYil ，用这个解压压缩包得到png和txt。题目为”熟悉的猫”，加上图片特征，显然是猫脸变换，简单学习了下，有几个关键点：

1.通过对像素坐标多次乘上变换矩阵（乘一次就进行一次变换）得到变换以后的图像，即像素值不变，位置改变。

2.猫脸变换具有周期性，对变换k次的图像再进行周期T-k次变换后可恢复原图。更好的做法是对变换矩阵求逆矩阵，左乘k次逆即可恢复。

3.这样的变换类似于图像加密，变换次数为密钥key。

txt中有一串很长的k值，这里有两部分利用：

1.零宽解出字符 22*160。

2.k值很长，猜测是塔珀自指公式。

脚本改改参数即可：

import numpy as np
import matplotlib.pyplot as plt
from PIL import Image
def Tupper_self_referential_formula(k):
    aa = np.zeros((22, 160))
    def f(x, y):
        y += k
        a1 = 2 ** -(-22 * x - y % 22)
        a2 = (y // 22) // a1
        return 1 if a2 % 2 > 0.5 else 0
    for y in range(22):
        for x in range(160):
            aa[y, x] = f(x, y)
    return aa[:, ::-1]
k = 92898203278702907929705938676672021500394791427205757369123489204565300324859717082409892641951206664564991991489354661871425872649524078000948199832659815275909285198829276929014694628110159824930931595166203271443269827449505707655085842563682060910813942504507936625555735585913273575050118552353192682955310220323463465408645422334101446471078933149287336241772448338428740302833855616421538520769267636119285948674549756604384946996184385407505456168240123319785800909933214695711828013483981731933773017336944656397583872267126767778549745087854794302808950100966582558761224454242018467578959766617176016660101690140279961968740323327369347164623746391335756442566959352876706364265509834319910419399748338894746638758652286771979896573695823608678008814861640308571256880794312652055957150464513950305355055495262375870102898500643010471425931450046440860841589302890250456138060738689526283389256801969190204127358098408264204643882520969704221896973544620102494391269663693407573658064279947688509910028257209987991480259150865283245150325813888942058
aa = Tupper_self_referential_formula(k)
plt.figure(figsize=(15, 10))
plt.imshow(aa, origin='lower')
plt.savefig("tupper.png")
img = Image.open('tupper.png')
# 翻转
dst1 = img.transpose(Image.FLIP_LEFT_RIGHT).rotate(180)
plt.imshow(dst1)
plt.show()

得到的图片中可见参数33，121，144，做一个arnold置乱（a=121,b=144），但33貌似没有用，因为第一轮就可以得到flag了:

脚本：Arnold-Python

import numpy as np
import cv2
r = 800 # 图像宽高
a = 121
b = 144
img = cv2.imread("flag.png", 1)
p = np.zeros([r, r, 3], np.uint8)
def dearnold(img):
    for x in range(0, r):
        for y in range(0, r):
            xx = -((a * b + 1) * x - b * y) % r
            yy = -(-a * x + y) % r
            p[xx][yy] = img[x][y]
    return p
for i in range(0, 33):
    p = dearnold(img)
    img = p
    cv2.imwrite(str(i) + ".png",p)

冰墩墩

脚本解压：

import zipfile

zipfilename = 'BinDunDun.zip'
z = zipfile.ZipFile(zipfilename, 'r') # r表示解压
name_list = z.namelist() # 压缩包内的文件名，这里就是BinDunDun
for i in name_list:
    z.extract(i, path='zip\\')

有超级多个txt文件，每个txt中含有01序列和下一个txt文件名：

1	101000001001011 =>The txt you should view is m9312r95cr.txt # 01序列补成16位刚好是504b

经过观察发现需要把每个txt中的01序列补成16位。用脚本批量处理（套宝的）：

import re
flag = ''
next = 'zip/BinDunDun/start.txt'
i = 0

while len(next) != 0:
    try:
        # 正则过滤
        f = open(next).read()
        tmp = re.search('(.*) =>',f).group(0)
        flag += tmp[:-3].zfill(16)
        tmp = re.search('is (.*)',f).group(0)
        next = 'zip/BinDunDun/'+ tmp[3:]
        i += 1
        print(f,i,next)
    except:
        f2 = open('out.txt','w')
        f2.write(flag)
        exit(0)

二进制序列转hex，发现是zip文件：

存一下，解压得到pyc和jpg，pyc隐写得到payload：$BingD@nD@n_in_BeiJing_Winter_Olympics$

最后利用这个作为密码进行jphs05隐写seek得到base64：$REFTQ1RGe0dvb2RfSm9kX0dpdmVfVGhlX0ZGRkZMQGdfVG9fWW91IX0=$

解码得：$DASCTF{Good_Jod_Give_The_FFFFL@g_To_You!}$

CRYPTO

easy_real

签到。

from hashlib import *
from gmpy2 import *
from Crypto.Util.number import *

# e_hash = '37693cfc748049e45d87b8c7d8b9aacd'
# for i in range(100):
#     if e_hash == md5(str(i).encode()).hexdigest():
#         print(i)

e = 23
n = 4197356622576696564490569060686240088884187113566430134461945130770906825187894394672841467350797015940721560434743086405821584185286177962353341322088523
c = 3298176862697175389935722420143867000970906723110625484802850810634814647827572034913391972640399446415991848730984820839735665233943600223288991148186397
p = 64310413306776406422334034047152581900365687374336418863191177338901198608319
q = n//p

# 解crypto
phi = (p-1) * (q-1)
d = invert(e,phi)
cry = long_to_bytes(pow(c,d,n))
print(cry)
cry = b'ndios_;9kgE;WK8e;W?gWn<\\;k|nu'
flag = ''
for i in range(10):
    for j in cry:
        flag = flag + chr(j^i)
    print(flag)

# flag{W31coM3_C0m3_7o_f4T3ctf}