Reverse

Reverse-ez_apk

用IDA进行打开，找到main函数

def decrypt(encrypted_text, decryption_key):

key_bytes = decryption_key.encode()

characters = list(encrypted_text)

decrypted_characters = [''] * len(characters)

for index, character in enumerate(characters):

if character in ['_', '{', '}']:

decrypted_characters[index] = character

elif 'a' <= character <= 'z':

# 应用解密逻辑

decrypted_characters[index] = chr(((ord(character) - ord('a') - (key_bytes[index % len(key_bytes)] - ord('a'))) % 26 + ord('a')))

else:

# 遇到非预期字符则停止解密

break

return ''.join(decrypted_characters)

# 解密密钥

decryption_key = "aptxcony"

# 密文

cipher_text = "fatd{sm_cgrmvc_ylvhokhuk_gxsgffc_wtec}"

# 解密结果

decrypted_result = decrypt(cipher_text, decryption_key)

print(decrypted_result)

加密处理，进行构造脚本

Reverse-jiqimao

用kali打开，进行反编译exe

发现不可以打开，那我进行反编译

找到一段字符

fVJXNjE0ODBpM2RrZmNSVzYxNDgwaTNka01BSlVPe25oc20=

这是一段base64编码，进行解密

凯撒偏移解码

}KP61480b3wdyvKP61480b3wdFTCNH{galf

再进行字符串逆序

Reverse-fenxi

找到一道类似的题目，进行参考一下[https://blog.csdn.net/weixin_52640415/article/details/123877043](https://blog.csdn.net/weixin_52640415/article/details/123877043" \t "_blank)

if (p[3]+p[5] != pow((I-------I).A,2) * (I-----I)+(I---I)) error(); // (3*3) ** 2 * 2 + 1 == 163

if (p[3]+p[4] != pow((o-----------o

| !

! !

! !

o-----------o).A,2)+(I---I)) error(); // (5*3) ** 2 + 1 == 226

if (p[4]*p[5] != (pow((o-------------o

| !

! !

! !

o-------------o).A,2)-(I---I))(I-----I)pow(I-------I,2)) error(); // ((63) ** 2 - 1)2* 3**2 = 5814

if (p[7]+p[8] != (o-----------o

|L </font>