2022DASCTF Apr X FATE CTF部分wp_python

文章目录

- Misc
- - - SimpleFlow
    - 冰墩墩
- crypto
- - - easy_real

Misc SimpleFlow

打开数据包，追踪TCP流，发现是蚁剑的流量

在流50、51发现flag.txt和flag.zip

解密流50的数据
发现是使用了zip将flag.txt打包成flag.zip，并且密码为PaSsZiPWorD

将数据包丢进Kali，使用binwalk分离，得到压缩包
使用PaSsZiPWorD解压
得到flag

DASCTF{f3f32f434eddbc6e6b5043373af95ae8}

冰墩墩

解压压缩包，发现里面有10W+个txt
在里面发现start.txt

start.txt:
101000001001011 =>The txt you should view is m9312r95cr.txt
m9312r95cr.txt:
1100000100 =>The txt you should view is 4oyjbqwl59.txt
将上述二进制数据补全至16位，如下：

0101000001001011 --->504b
0000001100000100 --->34(补全前面的0，所以应当为0304)

504b0304多么熟悉的16进制数据啊，到这里我们就大概清楚这题的初步思路：正则按顺序提取txt文档中的2进制数据，再转成16进制生成zip压缩包

接下来就是搓脚本了
ps:在python中，转换过来的16进制数据小于17时会忽略前一位的0，所以需要在前面加一个0，如3替换为03

#coding:utf-8
import re
from binascii import *

next_txt = "start.txt"
path = r"E:\学习资料\CTF\fatectf\BinDunDun"
zip_data = ""
while True:
    if next_txt =='end.txt':
        break
    else:
        f = open(path+"\"+next_txt).read()
        bin_data = re.findall("(.*) =>",f)[0]
        full_bin_data = bin_data.zfill(16)
        zip_data +=full_bin_data
        next_txt = re.findall("is (.*)",f)[0]
        print(next_txt)
print(zip_data)
hex_file_data =""
for i in range(0,len(zip_data),8):
    hex_data = hex(int(zip_data[i:i+8],2))
    hex_data = hex_data.replace("0x","")
    if len(hex_data) == 1:
        hex_data = '0'+hex_data
    hex_file_data +=hex_data
with open("bdd_flag.zip",'wb') as f:
    f.write(unhexlify(hex_file_data))

打开压缩包，一个pyc文件和一个没有后缀的文件
作为misc题目，先尝试一下pyc的剑龙隐写，发现一串的字符串

BingD@nD@n_in_BeiJing_Winter_Olympics

难道说，这就是我们心心念念的flag？

开开心心去提交flag，啪的一下提交，很快嗷
flag错误？？？
出题人我劝你耗子尾汁

010 打开另一个没有后缀的文件，发现是一个jpg图片，感觉是jpg图片隐写了，上面那个字符串应该是密码
使用stegdetect检测，没检测出来

亚雷🐎，玩毛

采用笨方法，jpg隐写工具一个个去试咯
最后使用jphs05成功解密，得到一串base64字符串

REFTQ1RGe0dvb2RfSm9kX0dpdmVfVGhlX0ZGRkZMQGdfVG9fWW91IX0=

解密得到flag

DASCTF{Good_Jod_Give_The_FFFFL@g_To_You!}

crypto easy_real

题目：

import random
import hashlib

flag = 'xxxxxxxxxxxxxxxxxxxx'
key = random.randint(1,10)
for i in range(len(flag)):
	crypto += chr(ord(flag[i])^key)
m = crypto的ascii十六进制
e = random.randint(1,100)
print(hashlib.md5(e))
p = 64310413306776406422334034047152581900365687374336418863191177338901198608319
q = xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
n = p*q
c = pow(m,e,n)
print(n)
print(c)
#37693cfc748049e45d87b8c7d8b9aacd   23
#4197356622576696564490569060686240088884187113566430134461945130770906825187894394672841467350797015940721560434743086405821584185286177962353341322088523
#3298176862697175389935722420143867000970906723110625484802850810634814647827572034913391972640399446415991848730984820839735665233943600223288991148186397

取一个1~10的随机数作为Key，将每一位的flag与key做异或，得到的值作为明文再RSA加密一次。所以，求出m之后爆破key，再与key做异或即可得到flag
代码如下：

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

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

q = n//p
phi = (q-1)*(p-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
hex_m = long_to_bytes(m)
key = random.randint(1,10)
m1 = [hex(i) for i in hex_m]
m2 = ''
for key in range(1,11):
    flag = ''
    for i in m1:
        flag1 = int(i,16)^key
        flag +=chr(flag1)
    print(flag)