Modern Clueless Child

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WriteUp来源

https://dunsp4rce.github.io/csictf-2020/crypto/2020/07/18/Modern-Clueless-Child.html

by anishbadhri

题目描述

I was surfing the crimson wave and oh my gosh I was totally bugging. I also tried out the lilac hair trend but it didn’t work out. That’s not to say you are any better, you are a snob and a half. But let’s get back to the main question here- Who am I? (You don’t know my name).
Ciphertext = "52f41f58f51f47f57f49f48f5df46f6ef53f43f57f6cf50f6df53f53f40f58f51f6ef42f56f43f41f5ef5cf4e" (hex) Key = "12123"

题目考点

解题思路

This question needs a number of observations.

First, it can be seen that f occurs after every 2 characters. Splitting the ciphertext on f yields an array of bytes.

It is known that the flag starts with csictf{. When represented in bytes, this results in an array, 63 73 69 63 74 66 7b. On taking an xor of this array with the first 7 elements of the ciphertext, we get a bytearray, 31 32 31 32 33 31 32. It can be seen that the unit digit of this bytearray is key.

Hence, taking each element of the key and prepending 3 before it gives an array of bytes. Taking the xor of this key with the ciphertext returns the flag.

import base64
cipher = "52f41f58f51f47f57f49f48f5df46f6ef53f43f57f6cf50f6df53f53f40f58f51f6ef42f56f43f41f5ef5cf4e".split('f')
key = ['3' + i for i in "12123"]
res = []
for i, n in enumerate(cipher):
  x = int(n, 16)
  y = int(key[i % len(key)], 16)
  res.append(hex(x ^ y)[2:])
res = "".join(res)
print(bytes.fromhex(res).decode())

Flag

csictf{you_are_a_basic_person}