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@FeLvi

# SECCON 2019Online CTF Writeup coffee_break

More than 1 year has passed since last update.

2日目の朝8時に起きて「SECCON予選そういえば今日までじゃん」となったので解きました。

coffee_break以外にもBeeeeeeeeeerなども解きましたが解説がしんどいので他のつよつよの方に譲ります……

# coffee_break (crypto)

encrypt.py
```import sys
from Crypto.Cipher import AES
import base64

def encrypt(key, text):
s = ''
for i in range(len(text)):
s += chr((((ord(text[i]) - 0x20) + (ord(key[i % len(key)]) - 0x20)) % (0x7e - 0x20 + 1)) + 0x20)
return s

key1 = "SECCON"
key2 = "seccon2019"
text = sys.argv[1]

enc1 = encrypt(key1, text)
cipher = AES.new(key2 + chr(0x00) * (16 - (len(key2) % 16)), AES.MODE_ECB)
p = 16 - (len(enc1) % 16)
enc2 = cipher.encrypt(enc1 + chr(p) * p)
print(base64.b64encode(enc2).decode('ascii'))

# ここまで元のコード

# 元のコードの下に追記

def decrypt(key, s):
text = ""
for i in range(len(s)):
temp = ord(s[i]) - ord(key[i % len(key)]) + 0x20
while(temp < 0x20):
temp += 95
text += chr(temp)
return text

enctext = "FyRyZNBO2MG6ncd3hEkC/yeYKUseI/CxYoZiIeV2fe/Jmtwx+WbWmU1gtMX9m905"
enc2 = cipher.decrypt(base64.b64decode(enctext)).decode("ascii")
print(decrypt(key1, enc2))
```

encrypt.pyは二段階で暗号化していて、
enc1の1段階目でを文字をずらし、
enc2の2段階目でAES暗号化をしているだけです。

それを逆順にたどっていけば元の文字列が出てきます。

encrypt.py
```enc2 = cipher.encrypt(enc1 + chr(p) * p)
```

なので}以降は無視して送信しましょう。

flag.txt
```SECCON{Success_Decryption_Yeah_Yeah_SECCON}?AA56
```
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