XYCTF2025部分WP-Re

小结

emm,比赛到最后就剩我一个了，pwn爷来打的话能冲前60的，PYD逆向和VM逆向也要开始学起来了。

Re

WARMUP

vbs文件

参考这个

本质就是chr加密混淆

无魔改rc4，提数据痛苦

XYCTF{5f9f46c147645dd1e2c8044325d4f93c}

Dragon

bc文件逆向，学习过程见我blog，赛后会发

int __fastcall main(int argc, const char **argv, const char **envp)
{
    __int64 v4; // [rsp+30h] [rbp-50h]
    size_t v5; // [rsp+38h] [rbp-48h]
    char Str[80]; // [rsp+40h] [rbp-40h] BYREF
    _QWORD v7[13]; // [rsp+90h] [rbp+10h] BYREF
    int v8; // [rsp+FCh] [rbp+7Ch]

    _main(argc, argv, envp);
    v8 = 0;
    memcpy(v7, &unk_140004000, 0x60uLL);
    memset(Str, 0, 66uLL);
    printf("Input U flag:");
    scanf("%s", Str);
    v5 = 0LL;
    v4 = 0LL;
    while ( v5 < strlen(Str) >> 1 )
        {
            if ( calculate_crc64_direct((const unsigned __int8 *)&Str[v5], 2uLL) != v7[v4] )
            {
                printf("Error!");
                return 0;
            }
            v5 += 2LL;
            ++v4;
        }
    printf("Success");
    return 0;
}

进去后是crc校验，每两位为一组

__int64 __fastcall calculate_crc64_direct(const unsigned __int8 *a1, unsigned __int64 n2)
{
  unsigned __int64 j; // [rsp+0h] [rbp-28h]
  unsigned __int64 i; // [rsp+8h] [rbp-20h]
  __int64 v5; // [rsp+10h] [rbp-18h]

  v5 = -1LL;
  for ( i = 0LL; i < n2; ++i )
  {
    v5 ^= (unsigned __int64)a1[i] << 56;
    for ( j = 0LL; j < 8; ++j )
    {
      if ( v5 >= 0 )
        v5 *= 2LL;
      else
        v5 = (2 * v5) ^ 0x42F0E1EBA9EA3693LL;
    }
  }
  return ~v5;
}

密文

脚本：

def calculate_crc64_direct(data):
    crc = 0xFFFFFFFFFFFFFFFF  # 初始值 -1
    poly = 0x42F0E1EBA9EA3693  # CRC64/ECMA 多项式

    for byte in data:
        crc ^= (byte << 56) & 0xFFFFFFFFFFFFFFFF  # 高位异或
        for _ in range(8):  # 处理 8 位
            if (crc & 0x8000000000000000) != 0:  # 检查最高位
                crc = ((crc << 1) & 0xFFFFFFFFFFFFFFFF) ^ poly
            else:
                crc = (crc << 1) & 0xFFFFFFFFFFFFFFFF
    return (~crc) & 0xFFFFFFFFFFFFFFFF  # 取反并确保 64 位


# 给定的 96 字节密文
cipher_bytes = bytes([
    0x47, 0x7B, 0x9F, 0x41, 0x4E, 0xE3, 0x63, 0xDC, 0xC6, 0xBF, 0xB2, 0xE7, 0xD4, 0xF8, 0x1E, 0x3,
    0x9E, 0xD8, 0x5F, 0x62, 0xBC, 0x2F, 0xD6, 0x12, 0xE8, 0x55, 0x57, 0xCC, 0xE1, 0xB6, 0xE8, 0x83,
    0xCC, 0x65, 0xB6, 0x2A, 0xEB, 0xB1, 0x7B, 0xFC, 0x6B, 0xD9, 0x62, 0x2A, 0x1B, 0xCA, 0x82, 0x93,
    0x87, 0xC3, 0x73, 0x76, 0xA0, 0xF8, 0xFF, 0xB1, 0xE1, 0x5, 0x8E, 0x38, 0x27, 0x16, 0xA8, 0xD,
    0xB7, 0xAA, 0xD0, 0xE8, 0x1A, 0xE6, 0xF1, 0x9E, 0x45, 0x61, 0xF2, 0xE7, 0xD2, 0x3F, 0x78, 0x92,
    0xB, 0xE6, 0x6F, 0xF5, 0xA1, 0x7C, 0xC9, 0x63, 0xAB, 0x3A, 0xB7, 0x43, 0xB0, 0xA8, 0xD3, 0x9B
])

# 转换为 12 个 64 位整数（小端序）
v7 = []
for i in range(0, len(cipher_bytes), 8):
    chunk = cipher_bytes[i:i + 8]
    value = int.from_bytes(chunk, byteorder='little')
    v7.append(value)

# 枚举所有 2 字节组合，计算 CRC64 并匹配
flag = b""
for target in v7:
    found = False
    for i in range(0x10000):  # 0x0000 ~ 0xFFFF
        data = i.to_bytes(2, byteorder='little')
        crc = calculate_crc64_direct(data)
        if crc == target:
            flag += data
            print(f"Found: {data.hex()} -> {hex(crc)}")
            found = True
            break
    if not found:
        print(f"Failed for {hex(target)}")
        break

print("Flag:", flag.decode('latin-1'))
#flag{LLVM_1s_Fun_Ri9h7?}