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本文介绍: 文档被加密查看图片的属性在备注可以看到解压密码解密成功在选项里面把隐藏的文本显示出来可以看到ffag。

word_sercet

文档被加密

查看图片的属性在备注可以看到解压密码

解密成功

在选项里面把隐藏的文本显示出来

可以看到ffag

easy_encode

得到一个bmp二维码

使用qr research

得到的密文直接放瑞士军刀

base32解码+base64解码+hex解码

dir_pcap

直接搜索flag

发现flag.zip，zip_pass.txt

像是密码

从流量包下载 zip_pass.txt

导出分组字节流

flag.zip被加密了

用密码3.6*3.6进入得到flag

ezsm

# -*- coding: utf-8 -*-
import binascii
from gmssl import sm4
from secret import flag

def encode(key, data):
    sm4_a = sm4.CryptSM4()
    sm4_a.set_key(key.encode(), sm4.SM4_ENCRYPT)
    ciphertext = sm4_a.crypt_ecb(str(data).encode()).hex()
    return ciphertext

if __name__ == '__main__':
    key = '4765?df?0170?44?'
    ciphertext = encode(key,flag)
    print("密文:",ciphertext)

#c49f4552b22f27969c07d9371d1aa093b54f97ccd44261a5fc92cd3461a38d68d20218a51686a3f9d0cc50679e36cd4f

key中存在4个不确定位，通过代码尝试所有字符，暴力检索flag关键字即可

from gmssl import sm4
from Crypto.Util.number import *

def encode(key, data):
    sm4_a = sm4.CryptSM4()
    sm4_a.set_key(key.encode(), sm4.SM4_ENCRYPT)
    ciphertext = sm4_a.crypt_ecb(str(data).encode()).hex()
    return ciphertext

def decode(key, data):
    sm4_a = sm4.CryptSM4()
    sm4_a.set_key(key.encode(), sm4.SM4_DECRYPT)
    text = sm4_a.crypt_ecb(long_to_bytes(data))
    return text

if __name__ == '__main__':
    l = [0,1,2,3,4,5,6,7,8,9,'a','b','c','d','f']
    for i in l:
        for n in l:
            for m in l:
                for j in l:
                    key = f'4765{i}df{n}0170{m}44{j}'
                    ciphertext = 0xc49f4552b22f27969c07d9371d1aa093b54f97ccd44261a5fc92cd3461a38d68d20218a51686a3f9d0cc50679e36cd4f
                    text = decode(key,ciphertext)
                    if b'flag{' in text:
                        print(text)