2025年湾区杯网络安全大赛初赛&决赛 Writeup

from collections import defaultdict

# ========= 需要你手动填充的全局表 =========
# 在 IDA 或 Ghidra 里找到 byte_2020 和 byte_2120 的内容
byte_2020 = [0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01, 0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76, 0xCA, 0x82, 0xC9, 0x7D, 0xFA, 0x59, 0x47, 0xF0, 0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4, 0x72, 0xC0, 0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC, 0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15, 0x04, 0xC7, 0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2, 0xEB, 0x27, 0xB2, 0x75, 0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84, 0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB, 0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF, 0xD0, 0xEF, 0xAA, 0xFB, 0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 0x9F, 0xA8, 0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5, 0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2, 0xCD, 0x0C, 0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17, 0xC4, 0xA7, 0x7E, 0x3D, 0x64, 0x5D, 0x19, 0x73, 0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A, 0x90, 0x88, 0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB, 0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C, 0xC2, 0xD3, 0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79, 0xE7, 0xC8, 0x37, 0x6D, 0x8D, 0xD5, 0x4E, 0xA9, 0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A, 0xAE, 0x08, 0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6, 0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A, 0x70, 0x3E, 0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 0x61, 0x35, 0x57, 0xB9, 0x86, 0xC1, 0x1D, 0x9E, 0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9, 0x8E, 0x94, 0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF, 0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99, 0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16]
byte_2120 = [0x97, 0xD5, 0x60, 0x43, 0xB4, 0x10, 0x43, 0x73, 0x0F, 0xDA, 0x43, 0xCD, 0xD3, 0xE8, 0x73, 0x4A, 0x94, 0xC3, 0xCD, 0x71, 0xBD, 0xDC, 0x97, 0x1A]

# ========= 工具函数 =========
def rol(x, r):
    return ((x << r) & 0xFF) | (x >> (8 - r))

def ror(x, r):
    return ((x >> r) | (x << (8 - r))) & 0xFF

def sub_1313(a1):
    if a1 == 0:
        return 0
    v2 = 1
    v3 = 255
    v4 = a1
    while v3:
        if v3 & 1:
            v2 = (v4 * v2) % 257
        v4 = (v4 * v4) % 257
        v3 >>= 1
    return v2

def sub_13E1(a1):
    v1 = rol(a1 ^ 0x5A, 3)
    tmp = ((3 * (v1 >> 4)) & 0xF) * 16 | ((5 * (v1 & 0xF)) & 0xF)
    v3 = sub_1313(tmp & 0xFF)
    return byte_2020[ror(v3 & 0xFF, 2)]

def sub_1492(s):
    out = []
    for i, ch in enumerate(s):
        out.append(rol(ch, (i % 7) + 1))
    return out

# ========= 逆向匹配 =========
def recover_flag():
    flag = []
    for idx, target in enumerate(byte_2120):
        # 枚举可能输入字符
        found = None
        for c in range(32, 127):  # printable ascii
            buf = rol(c, (idx % 7) + 1)
            if sub_13E1(buf) == target:
                found = chr(c)
                break
        if not found:
            found = '?'  # 没找到说明范围要放宽
        flag.append(found)
    return "".join(flag)

if __name__ == "__main__":
    print("Recovered flag:", recover_flag())
# Bl@st1ng_1s_a_g00d_Way!!