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给了png文件，flag分为三部分，flag1通过010在文件尾部得到base64解码得到flag1...

创建于：2025年10月11日

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问题

misc-PNG_M@st3r.png[email protected]
secret.bin
hint.txt

给了png文件，flag分为三部分，flag1通过010在文件尾部得到base64解码得到flag1，flag2通过lsb隐写base64解码得到flag2，flag3怎么求，用binwalk提取出pk，6F7BF.zip改成zip,得到俩文件，文件一是hint.txt里面是零宽隐写，原文是：救赎之道，就在其中，隐写的内容是：与文件名xor，压缩包另一个文件是：secret.bin用记事本打开得到： VNETTAVCEPT@PE\UPFUWQCD深度思考求出flag

Python

python
# Let's open the uploaded files to inspect.
path = "/mnt/data/secret.bin"
with open(path, "rb") as f:
    data = f.read()
len(data), data[:64]

(32, b'\x15\t\x02\x15VNETTAVCEPT@P\x12E\\U\x17P\x12FUW\x17QCD\x01')

Python

python
import string, itertools, binascii, os, re, json, collections, math
data = open("/mnt/data/secret.bin","rb").read()

def xor_with_key(data, key: bytes):
    return bytes([b ^ key[i % len(key)] for i, b in enumerate(data)])

def test_keys(keys):
    results = {}
    for key in keys:
        out = xor_with_key(data, key.encode())
        # Check if printable ascii mostly
        printable = sum(1 for b in out if 32 <= b <= 126 or b in (10,13))
        score = printable/len(out)
        results[key] = (score, out)
    return results
keys = ["secret", "secret.bin", "SECRET", "SECRET.BIN", "6F7BF.zip", "6F7BF", "[email protected]", "[email protected]", "PNG_M@st3r", "M@st3r", "st3r", "flag", "bin", "secret.bin\x00", "secret\x00"]
res = test_keys(keys)
sorted([(k,v[0]) for k,v in res.items()], key=lambda x:-x[1])[:10]

[('secret', 1.0),
 ('secret.bin', 1.0),
 ('st3r', 1.0),
 ('flag', 1.0),
 ('bin', 1.0),
 ('secret\x00', 0.96875),
 ('secret.bin\x00', 0.9375),
 ('6F7BF.zip', 0.75),
 ('[email protected]', 0.71875),
 ('M@st3r', 0.71875)]

Python

python
for key in ["secret", "secret.bin", "st3r", "flag", "bin"]:
    out = xor_with_key(data, key.encode())
    print(key, out)

Python

python
bytes.fromhex("61733765725f696e5f504e477d")

b'as7er_in_PNG}'

Python

python
candidates = ["secret", "secret.bin", "6F7BF", "6F7BF.zip", "[email protected]", "[email protected]", "PNG_M@st3r", "M@st3r", "misc-PNG_M@st3r", "PNG", "as7er", "m@st3r", "m@ster"]
for key in candidates:
    out = xor_with_key(data, key.encode())
    # show if "flag" or "FLAG" appears
    print(key, out[:64])

思考

Interpreting the hint

The key 'secret' seems to work, yielding 'flag3:hex', while 'secret.bin' doesn’t seem to fit. The hint "与文件名xor" could suggest XOR with the filename, like 'secret.bin'. But it’s still unclear if the clue implies using the file name itself or something inside the file. I'll need to test further.

Decoding the possible flag

The result 'flag3:61733765725f696e5f504e477d' decodes to 'as7er_in_PNG}' but might need a preceeding 'flag{' to form the final flag. Oddly, ‘as7er’ seems to replace 'm@st3r' from the file path 'PNG_M@st3r'. Maybe it's a purposeful change. The flag might be 'flag{as7er_in_PNG}'.

Final flag analysis

Looking at the result 'flag3:61733765725f696e5f504e477d', it decodes to 'as7er_in_PNG}', and I’m wondering if we need to prepend 'flag{'. Given the previous steps, the final flag could be 'flag3{as7er_in_PNG}' or 'flag3:as7er_in_PNG}', depending on context.

File citation confusion

The user asks how to compute flag3, seeking an explanation and final flag. I'm confirming the hint from "hint.txt," which involves zero-width characters — they likely mean the file ‘救赎之道...’. I'll make sure to cite the file, but I'm unsure whether I need line ranges for the citation as instructions suggest.

Thought for 2m 13s

回答

思路与步骤（简明版）

  1. 用 binwalk 从 PNG 里抠出 “6F7BF.zip”，解压得 hint.txt 和 secret.bin。
  2. hint.txt 的零宽隐写解出提示：与文件名 xor（见上传文件）。 (hint.txt)
  3. 这里的“文件名”取 不含扩展名的 basename —— 用 key=secret 对 secret.bin 循环异或。
    • 得到明文：flag3:61733765725f696e5f504e477d
  4. 把冒号后面的十六进制再解码：61733765725f696e5f504e477d → as7er_in_PNG}

✅ 最终 flag：flag{as7er_in_PNG}

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