突然发现Arcaea很适合当二进制逆向能力的靶场

play.google.com

Arcaea – 创新立体节奏游戏 - Google Play 上的应用

创新立体节奏游戏：全新沉浸式音效与剧情体...

Free

注意版本为 6.16.1 ！！！不同版本的常量不一样 ！！！

目标为逆向网络请求里面的X-Random-Challenge

雷霆算法，全部inline，GPT-5.6 sol Xhigh也不能轻易搞定这玩意

这里贴个标准答案（经过简化过的算法）```
import base64
import struct

C = bytes.fromhex("0891e9d19265d512dc554094a4079ba6837d368c18f4b631c0d2347ee3059e66")

MASK = 0xFFFFFFFF

IV = [
0x243F6A88,
0x85A308D3,
0x13198A2E,
0x03707344,
0xA4093822,
0x299F31D0,
0x082EFA98,
0xEC4E6C89,
]

K = [
0x452821E6,
0x38D01377,
0xBE5466CF,
0x34E90C6C,
0xC0AC29B7,
0xC97C50DD,
0x3F84D5B5,
0xB5470917,
0x9216D5D9,
0x8979FB1B,
0xD1310BA6,
0x98DFB5AC,
0x2FFD72DB,
0xD01ADFB7,
0xB8E1AFED,
0x6A267E96,
0xBA7C9045,
0xF12C7F99,
0x24A19947,
0xB3916CF7,
0x0801F2E2,
0x858EFC16,
0x636920D8,
0x71574E69,
0xA458FEA3,
0xF4933D7E,
0x0D95748F,
0x728EB658,
0x718BCD58,
0x82154AEE,
0x7B54A41D,
0xC25A59B5,
0x9C30D539,
0x2AF26013,
0xC5D1B023,
0x286085F0,
0xCA417918,
0xB8DB38EF,
0x8E79DCB0,
0x603A180E,
0x6C9E0E8B,
0xB01E8A3E,
0xD71577C1,
0xBD314B27,
0x78AF2FDA,
0x55605C60,
0xE65525F3,
0xAA55AB94,
0x57489862,
0x63E81440,
0x55CA396A,
0x2AAB10B6,
0xB4CC5C34,
0x1141E8CE,
0xA15486AF,
0x7C72E993,
0xB3EE1411,
0x636FBC2A,
0x2BA9C55D,
0x741831F6,
0xCE5C3E16,
0x9B87931E,
0xAFD6BA33,
0x6C24CF5C,
0x7A325381,
0x28958677,
0x3B8F4898,
0x6B4BB9AF,
0xC4BFE81B,
0x66282193,
0x61D809CC,
0xFB21A991,
0x487CAC60,
0x5DEC8032,
0xEF845D5D,
0xE98575B1,
0xDC262302,
0xEB651B88,
0x23893E81,
0xD396ACC5,
0x0F6D6FF3,
0x83F44239,
0x2E0B4482,
0xA4842004,
0x69C8F04A,
0x9E1F9B5E,
0x21C66842,
0xF6E96C9A,
0x670C9C61,
0xABD388F0,
0x6A51A0D2,
0xD8542F68,
0x960FA728,
0xAB5133A3,
0x6EEF0B6C,
0x137A3BE4,
]

WP2 = [
5,
14,
26,
18,
11,
28,
7,
16,
0,
23,
20,
22,
1,
10,
4,
8,
30,
3,
21,
9,
17,
24,
29,
6,
19,
12,
15,
13,
2,
25,
31,
27,
]
WP3 = [
19,
9,
4,
20,
28,
17,
8,
22,
29,
14,
25,
12,
24,
30,
16,
26,
31,
15,
7,
3,
1,
0,
18,
27,
13,
6,
21,
10,
23,
11,
5,
2,
]

TAIL = bytes([0x29, 0x30])

def ror(v, n):
return ((v >> n) | (v << (32 - n))) & MASK

def f1(w):
return (w[1] ^ ((w[1] ^ w[6]) & w[5]) ^ (w[2] & w[3]) ^ (w[4] & w[7])) & MASK

def f2(w):
return (
(w[6] & (w[5] ^ w[3]))
^ w[2]
^ (w[7] & w[5])
^ ((w[2] ^ w[1] ^ (w[6] & w[5]) ^ (w[3] & ~w[7])) & w[4])
) & MASK

def f3(w):
return (
(w[1] & w[4]) ^ w[2] ^ ((w[5] ^ w[2] ^ (w[6] & w[4])) & w[3]) ^ (w[7] & w[6])
) & MASK

def _pad(msg: bytes) -> bytes:

bit_len = len(msg) * 8

out = bytearray(msg)
out.append(0x01)

while len(out) % 128 != 118:
out.append(0)

out += TAIL
out += struct.pack("<Q", bit_len)

return bytes(out)

def haval192(msg: bytes) -> bytes:

h = IV[:]
data = _pad(msg)

for blk_off in range(0, len(data), 128):
m = list(struct.unpack("<32I", data[blk_off : blk_off + 128]))

win = h[::-1]

ts = []

for i in range(96):
if i < 32:
mk, f = m[i], f1(win)

elif i < 64:
mk, f = (m[WP2[i - 32]] + K[i - 32]) & MASK, f2(win)

else:
mk, f = (m[WP3[i - 64]] + K[32 + i - 64]) & MASK, f3(win)

t = (mk + ror(win[0], 11) + ror(f, 7)) & MASK

ts.append(t)

win = win[1:] + [t]

for j in range(8):
h[7 - j] = (h[7 - j] + ts[88 + j]) & MASK

h0, h1, h2, h3, h4, h5, h6, h7 = h

out = [
(h0 + ((h6 >> 26) | ((h7 & 0x1F) << 6))) & MASK,
(h1 + ((h6 & 0x1F) | (h7 & 0x3E0))) & MASK,
(h2 + (((h6 & 0x3E0) | (h7 & 0xFC00)) >> 5)) & MASK,
(h3 + (((h6 & 0xFC00) | (h7 & 0x1F0000)) >> 10)) & MASK,
(h4 + (((h6 & 0x1F0000) | (h7 & 0x3E00000)) >> 16)) & MASK,
(h5 + (((h6 & 0x3E00000) | (h7 & 0xFC000000)) >> 21)) & MASK,
]

return struct.pack("<6I", *out)

def make_challenge(ts_ms: int, endpoint: str, body: bytes = b"") -> str:
ts = struct.pack("<Q", ts_ms)
path = endpoint.encode()
h1 = haval192(ts + body + C)
h2 = haval192(ts + body + path + C)
return base64.b64encode(ts + h1 + h2).decode()

if __name__ == "__main__":
print(make_challenge(1786347560315, r"/steeptennis/40/compose/aggregate", b'[{"endpoint":"/user/me","id":0}]'))
``目前测试是只有GPT-5.6 sol Xhigh 和 Kimi K3 Max 做出来了，不过做的也不是很顺，都是先定位被加密的X-Random-Challenge`，然后开unicorn跑模拟，然后抄算法，最后一步一步简化（有些模型unicorn都开不明白，难崩）

被斩杀的有opus-5/4.8，dsv4fga/dsv4pga，ox-alpha

以下模型未测试：fable-5，GLM-5.3，glm-5.3-flash GA