ACTF新生赛2020_Splendid_MineCraft

[ACTF新生赛2020]Splendid_MineCraft

32位ida

首先是输入长度校验并且校验格式

dword+word = 4 + 2 = 6

看出ACTF{}里面是三段由_分割的xxxxxx

3 * 6 + 5 + 2 + 1 刚好26位

第一段

看到((int (__cdecl *)(int *))unk_4051D8)(&v17)猜测是一个函数

直接转换成代码

是SMC，写脚本

#include <idc.idc>
static main()
{
    auto addr = 0x4051FC;
    auto i = 0;
    for(i = 0; i <= 337; ++i) {
        PatchByte(addr+i, Byte(addr+i)^0x72);
    }
}

写脚本解密

v5 = "Welcome "
v6 = "3@1b;b"
for i in range(6):
    x = (ord(v5[i+1]) ^ ord(v6[i]))+ 0x23
    print(chr(x&0xff), end="")

yOu0y*

第二段

也是SMC，idc脚本

v9 动调可得为0x20

#include <idc.idc>
static main()
{
    auto addr = 0x405018;
    auto i = 256;
    for(i = 256; i < 496; ++i) {
        PatchByte(addr+i, Byte(addr+i)^0x20);
    }
}

从最后的判断可以看出，输入异或上(i^0x83)，将其作为表v6的索引与a2对比，a2则是内存中的一段数据，动调或直接计算偏移

写脚本

keyboard = [
0xF6, 0xA3, 0x5B, 0x9D, 0xE0, 0x95, 0x98, 0x68, 0x8C, 0x65,
  0xBB, 0x76, 0x89, 0xD4, 0x09, 0xFD, 0xF3, 0x5C, 0x3C, 0x4C,
  0x36, 0x8E, 0x4D, 0xC4, 0x80, 0x44, 0xD6, 0xA9, 0x01, 0x32,
  0x77, 0x29, 0x90, 0xBC, 0xC0, 0xA8, 0xD8, 0xF9, 0xE1, 0x1D,
  0xE4, 0x67, 0x7D, 0x2A, 0x2C, 0x59, 0x9E, 0x3D, 0x7A, 0x34,
  0x11, 0x43, 0x74, 0xD1, 0x62, 0x60, 0x02, 0x4B, 0xAE, 0x99,
  0x57, 0xC6, 0x73, 0xB0, 0x33, 0x18, 0x2B, 0xFE, 0xB9, 0x85,
  0xB6, 0xD9, 0xDE, 0x7B, 0xCF, 0x4F, 0xB3, 0xD5, 0x08, 0x7C,
  0x0A, 0x71, 0x12, 0x06, 0x37, 0xFF, 0x7F, 0xB7, 0x46, 0x42,
  0x25, 0xC9, 0xD0, 0x50, 0x52, 0xCE, 0xBD, 0x6C, 0xE5, 0x6F,
  0xA5, 0x15, 0xED, 0x64, 0xF0, 0x23, 0x35, 0xE7, 0x0C, 0x61,
  0xA4, 0xD7, 0x51, 0x75, 0x9A, 0xF2, 0x1E, 0xEB, 0x58, 0xF1,
  0x94, 0xC3, 0x2F, 0x56, 0xF7, 0xE6, 0x86, 0x47, 0xFB, 0x83,
  0x5E, 0xCC, 0x21, 0x4A, 0x24, 0x07, 0x1C, 0x8A, 0x5A, 0x17,
  0x1B, 0xDA, 0xEC, 0x38, 0x0E, 0x7E, 0xB4, 0x48, 0x88, 0xF4,
  0xB8, 0x27, 0x91, 0x00, 0x13, 0x97, 0xBE, 0x53, 0xC2, 0xE8,
  0xEA, 0x1A, 0xE9, 0x2D, 0x14, 0x0B, 0xBF, 0xB5, 0x40, 0x79,
  0xD2, 0x3E, 0x19, 0x5D, 0xF8, 0x69, 0x39, 0x5F, 0xDB, 0xFA,
  0xB2, 0x8B, 0x6E, 0xA2, 0xDF, 0x16, 0xE2, 0x63, 0xB1, 0x20,
  0xCB, 0xBA, 0xEE, 0x8D, 0xAA, 0xC8, 0xC7, 0xC5, 0x05, 0x66,
  0x6D, 0x3A, 0x45, 0x72, 0x0D, 0xCA, 0x84, 0x4E, 0xF5, 0x31,
  0x6B, 0x92, 0xDC, 0xDD, 0x9C, 0x3F, 0x55, 0x96, 0xA1, 0x9F,
  0xCD, 0x9B, 0xE3, 0xA0, 0xA7, 0xFC, 0xC1, 0x78, 0x10, 0x2E,
  0x82, 0x8F, 0x30, 0x54, 0x04, 0xAC, 0x41, 0x93, 0xD3, 0x3B,
  0xEF, 0x03, 0x81, 0x70, 0xA6, 0x1F, 0x22, 0x26, 0x28, 0x6A,
  0xAB, 0x87, 0xAD, 0x49, 0x0F, 0xAF
]
key = [0x30, 0x04, 0x04, 0x03, 0x30, 0x63, 0x90]
for i in range(6):
    x = keyboard.index(key[i])
    print(chr((x^(0x83+i))&0xff), end="" )

knowo3

第三段

能直接看到的东西差不多用完了，看汇编

调用Congratulations之前有个strcmp的判断

5mcsM<

ACTF{yOu0y*_knowo3_5mcsM<}