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光辉的确定有限自动机: 05-05 23:29:23 0x52, 0x7F, 0x7B, 0x73,...

创建于：2026年5月5日

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问题

光辉的确定有限自动机: 05-05 23:29:23
0x52, 0x7F, 0x7B, 0x73, 0x78, 0x65, 0x7F, 0x79, 0x78, 0x77, 0x7A, 0x49, 0x45, 0x7A, 0x77, 0x65, 0x7E, 0xB7

光辉的确定有限自动机: 05-05 23:29:38
逐字节映射

光辉的确定有限自动机: 05-05 23:29:47
用纯misc的方法能不能解

思考

undefined

Python

python
data=[0x52,0x7F,0x7B,0x73,0x78,0x65,0x7F,0x79,0x78,0x77,0x7A,0x49,0x45,0x7A,0x77,0x65,0x7E,0xB7]
print(len(data), bytes(data))
print('chars', ''.join(chr(b) if 32<=b<127 else '.' for b in data))
# xor single byte keys that produce printable last maybe
import string
printable=set(bytes(string.printable,'ascii'))
cands=[]
for k in range(256):
    out=bytes([b^k for b in data])
    if all(32<=c<127 for c in out[:-1]): # allow last?
        cands.append((k,out))
len(cands), cands[:20]


(52,
 [(2, b'P}yqzg}{zuxKGxug|\xb5'),
  (3, b'Q|xp{f|z{tyJFytf}\xb4'),
  (9, b'[vrzqlvpq~s@Ls~lw\xbe'),
  (10, b'Xuqyrousr}pCOp}ot\xbd'),
  (11, b'Ytpxsntrs|qBNq|nu\xbc'),
  (13, b'_rv~uhrtuzwDHwzhs\xba'),
  (14, b'\\qu}vkqwvytGKtykp\xb9'),
  (15, b']pt|wjpvwxuFJuxjq\xb8'),
  (16, b'BokchuoihgjYUjgun\xa7'),
  (17, b'CnjbitnhifkXTkfto\xa6'),
  (18, b'@miajwmkjeh[Whewl\xa5'),
  (19, b'Alh`kvljkdiZVidvm\xa4'),
  (20, b'Fkoglqkmlcn]Qncqj\xa3'),
  (21, b'Gjnfmpjlmbo\\Pobpk\xa2'),
  (22, b'Dimensional_Slash\xa1'),
  (23, b'Ehldorhno`m^Rm`ri\xa0'),
  (24, b'Jgck`}ga`obQ]bo}f\xaf'),
  (25, b'Kfbja|f`ancP\\cn|g\xae'),
  (27, b'Id`hc~dbclaR^al~e\xac'),
  (28, b'NcgodycedkfUYfkyb\xab')])

Python

python
data=[0x52,0x7F,0x7B,0x73,0x78,0x65,0x7F,0x79,0x78,0x77,0x7A,0x49,0x45,0x7A,0x77,0x65,0x7E,0xB7]
out=[(b & 0x7f)^0x16 for b in data]
print(bytes(out), ''.join(map(chr,out)))
for b,c in zip(data,out):
    print(f"{b:#04x} -> {b&0x7f:#04x} xor 0x16 = {c:#04x} {chr(c)!r}")

思考

Final answer formulation and citation consideration

The final answer can be in Chinese, and I think I don’t need citations if we’re relying mostly on reasoning, especially since the answer is based on computations. If I mention ASCII (like 0x21 being '!'), I could cite an ASCII table. But unless the computation is based on a web source, no need for detailed citations. The answer is certainly "yes," it can be solved with pure misc methods.

Thought for 1m 7s

回答

能，纯 misc 就能解，不需要真的去分析 DFA。

核心映射是：

python
chr((byte & 0x7f) ^ 0x16)

也就是每个字节先清掉最高位，再异或 0x16。

逐字节结果：

text
0x52 -> D
0x7F -> i
0x7B -> m
0x73 -> e
0x78 -> n
0x65 -> s
0x7F -> i
0x79 -> o
0x78 -> n
0x77 -> a
0x7A -> l
0x49 -> _
0x45 -> S
0x7A -> l
0x77 -> a
0x65 -> s
0x7E -> h
0xB7 -> !

解出来是：

text
Dimensional_Slash!

最后一个 0xB7 比较关键：
0xB7 & 0x7f = 0x37，然后 0x37 ^ 0x16 = 0x21，也就是 !。

如果题目需要 flag 格式，大概率填：

text
flag{Dimensional_Slash!}

所以这题的“逐字节映射”提示基本就是让你从单字节变换入手，暴力 XOR / 清高位即可。

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