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【wp】2022Sloth选拔赛
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上周末的Sloth选拔赛，趁着还有余温先来补补自出题的wp。

整个RE部分只有checkininin有一个解，大哭，逆向手岌岌可危（

仓库传送门：scnu-sloth/hsctf-2022-trial: Sloth 2022 选拔赛 题目及wp

Reverse

checkininin

拖进IDA里面，关键逻辑就这个（题目的三个in就取自这三个xor）

由hint给的HSCTF知识科普|第三期：Reverse逆向可以知道，xor有一种特性是两次xor相同的数以后可以得到其本身。

当s1[i] ^ v5[i] ^ v6[i] ^ v7[i] == s2[i]时，s1为正确的flag。

那在v5、v6、v7、s2都是已知的情况下，用s2[i] ^ v5[i] ^ v6[i] ^ v7[i]即可恢复出应该为flag的s1。

然后把IDA里的伪代码复制过来改改就能拿到flag（python也可，不过需要处理一下负数问题x

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#include <stdio.h>

int main(int a1, char **a2, char **a3)
{
  int i; // [rsp+Ch] [rbp-B4h]
  char v5[32]; // [rsp+10h] [rbp-B0h]
  char v6[32]; // [rsp+30h] [rbp-90h]
  char v7[32]; // [rsp+50h] [rbp-70h]
  char s2[33] = {0}; // [rsp+70h] [rbp-50h] BYREF
  // char s1[32]; // [rsp+90h] [rbp-30h] BYREF

  v5[0] = -119;
  v5[1] = -85;
  v5[2] = 0;
  v5[3] = 85;
  v5[4] = -23;
  v5[5] = -107;
  v5[6] = -58;
  v5[7] = -16;
  v5[8] = 124;
  v5[9] = -21;
  v5[10] = 28;
  v5[11] = 87;
  v5[12] = 43;
  v5[13] = 30;
  v5[14] = 106;
  v5[15] = -18;
  v5[16] = 80;
  v5[17] = 78;
  v5[18] = -55;
  v5[19] = 97;
  v5[20] = 124;
  v5[21] = -86;
  v5[22] = 98;
  v5[23] = 39;
  v5[24] = -74;
  v5[25] = 11;
  v5[26] = -31;
  v5[27] = 35;
  v5[28] = 74;
  v5[29] = 24;
  v5[30] = -61;
  v5[31] = 74;
  v6[0] = 108;
  v6[1] = -124;
  v6[2] = -80;
  v6[3] = 25;
  v6[4] = -16;
  v6[5] = 72;
  v6[6] = -5;
  v6[7] = -61;
  v6[8] = -115;
  v6[9] = -120;
  v6[10] = 122;
  v6[11] = 109;
  v6[12] = -37;
  v6[13] = -87;
  v6[14] = 92;
  v6[15] = 8;
  v6[16] = 10;
  v6[17] = 69;
  v6[18] = 0x80;
  v6[19] = -27;
  v6[20] = 10;
  v6[21] = 14;
  v6[22] = -74;
  v6[23] = -96;
  v6[24] = -53;
  v6[25] = 30;
  v6[26] = -4;
  v6[27] = 114;
  v6[28] = -115;
  v6[29] = 84;
  v6[30] = -42;
  v6[31] = 23;
  v7[0] = -96;
  v7[1] = 103;
  v7[2] = 68;
  v7[3] = -101;
  v7[4] = 121;
  v7[5] = -23;
  v7[6] = -65;
  v7[7] = 111;
  v7[8] = -96;
  v7[9] = -126;
  v7[10] = 46;
  v7[11] = -120;
  v7[12] = -123;
  v7[13] = 109;
  v7[14] = -11;
  v7[15] = 27;
  v7[16] = -95;
  v7[17] = -7;
  v7[18] = 30;
  v7[19] = 55;
  v7[20] = 127;
  v7[21] = -4;
  v7[22] = -69;
  v7[23] = 35;
  v7[24] = -74;
  v7[25] = 125;
  v7[26] = 7;
  v7[27] = 117;
  v7[28] = -113;
  v7[29] = -112;
  v7[30] = -13;
  v7[31] = 121;
  s2[0] = 45;
  s2[1] = 59;
  s2[2] = -105;
  s2[3] = -93;
  s2[4] = 6;
  s2[5] = 79;
  s2[6] = -35;
  s2[7] = 14;
  s2[8] = 98;
  s2[9] = -43;
  s2[10] = 4;
  s2[11] = -2;
  s2[12] = 44;
  s2[13] = -123;
  s2[14] = -100;
  s2[15] = -72;
  s2[16] = -95;
  s2[17] = -83;
  s2[18] = 8;
  s2[19] = -20;
  s2[20] = 81;
  s2[21] = 23;
  s2[22] = 61;
  s2[23] = -21;
  s2[24] = -103;
  s2[25] = 39;
  s2[26] = 72;
  s2[27] = 5;
  s2[28] = 105;
  s2[29] = -3;
  s2[30] = -71;
  s2[31] = 89;
  for ( i = 0; i <= 31; ++i )
  {
    s2[i] ^= v5[i];
    s2[i] ^= v6[i];
    s2[i] ^= v7[i];
  }
  printf("%s\n", s2);
  return 0;
}

真·逆向签到题。

3ZVW

hints给了一个迷宫题的wiki，其实看代码也能看出来是迷宫。而且题目名字上下颠倒再倒着读不就是MAZE嘛~（逃

很明显只有sub_7D1()这一个check，稍微分析一下（考验IDA快捷键的熟练运用）

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__int64 __fastcall sub_7D1(_BYTE *a1)
{
  int v2; // eax
  char v3; // [rsp+1Ch] [rbp-14h]
  char v4; // [rsp+1Dh] [rbp-13h]
  unsigned __int8 v5; // [rsp+1Eh] [rbp-12h]
  unsigned __int8 i; // [rsp+1Fh] [rbp-11h]
  char v7[4][2]; // [rsp+20h] [rbp-10h]
  unsigned __int64 v8; // [rsp+28h] [rbp-8h]

  v8 = __readfsqword(0x28u);
  if ( memcmp(a1, "hsctf{", 6uLL) || a1[37] != '}' )
    return 0LL;
  v7[0][0] = 0;
  v7[0][1] = -1;
  v7[1][0] = -1;
  v7[1][1] = 0;
  v7[2][0] = 0;
  v7[2][1] = 1;
  v7[3][0] = 1;
  v7[3][1] = 0;
  v3 = 1;
  v4 = 1;
  for ( i = 6; i <= 0x24u; ++i )
  {
    v2 = (char)a1[i];
    if ( v2 == 'j' )
    {
      v5 = 0;
    }
    else if ( v2 > 'j' )
    {
      if ( v2 == 'k' )
      {
        v5 = 2;
      }
      else
      {
        if ( v2 != 'l' )
          return 0LL;
        v5 = 1;
      }
    }
    else
    {
      if ( v2 != 'h' )
        return 0LL;
      v5 = 3;
    }
    if ( ((maze[v3][v4] >> v5) & 1) == 0 )
      return 0LL;
    v3 += v7[3 - v5][0];
    v4 += v7[3 - v5][1];
    if ( (maze[v3][v4] & 0x10) != 0 )
      return 1LL;
  }
  return 0LL;
}

明显，v7是一个二维数组{{0, -1}, {-1, 0}, {0, 1}, {1, 0}}，是用来指定方向的；v5是一个临时值，由根据flag的字符赋值（flag中间值必为hjkl中的一个），根据maze的对应值判断能不能过check。

v3是x，v4是y，maze题型参考前年新生赛【wp】2020HSCTF_PC逆向 | c10udlnk_Log，有一点不同的是这一次maze不是存每个点是墙还是路，而是存四周的情况，比如v5=3（即当前值为h）时，就判断(maze[v3][v4] >> v5)（二进制的第3位）是不是1，是的话（二进制为1xxx）说明没有墙，可以通过；否的话（二进制为0xxx）说明有墙，不可通过。

二进制的第4位为1说明这个点是终点，第0/1/2/3位代表当前点往左/上/右/下的情况。

存四周的情况下，画图其实是不好画的（也能画，但是懒），考虑到是选拔赛难度所以把迷宫规格加到了18*18，exp可以写一个bfs自动走迷宫就好了。

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def getVal(arr, t):
    return arr[t[0]][t[1]]

def walkMaze(maze_data, lens, start, end):
    dirs = [(0,-1), (-1,0), (0,1), (1,0)]
    FLAG_CHAR = ['h', 'k', 'l', 'j']
    path = [start]
    step = [["" for _ in range(lens)] for _ in range(lens)]
    flag = ""
    while len(path) != 0:
        prev = path[0]
        del path[0]
        for i in range(4):
            cur = (prev[0]+dirs[i][0], prev[1]+dirs[i][1])
            if getVal(maze_data, prev)&(1<<(3-i)) != 0 and getVal(step, cur) == "":
                step[cur[0]][cur[1]] = getVal(step, prev) + FLAG_CHAR[i]
                path.append(cur)
    return getVal(step, end)

maze = [0x03, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x09, 0x05, 0x03, 0x08, 0x01, 0x03, 0x0A, 0x0B, 0x0A, 0x0B, 0x0B, 0x0B, 0x09, 0x03, 0x0A, 0x0B, 0x0A, 0x09, 0x05, 0x05, 0x07, 0x0A, 0x0F, 0x0E, 0x08, 0x06, 0x08, 0x05, 0x04, 0x04, 0x06, 0x0F, 0x08, 0x04, 0x02, 0x0C, 0x05, 0x05, 0x05, 0x03, 0x0E, 0x09, 0x02, 0x0A, 0x0A, 0x0F, 0x0A, 0x0B, 0x08, 0x07, 0x0B, 0x0B, 0x0A, 0x08, 0x05, 0x05, 0x04, 0x07, 0x09, 0x07, 0x0B, 0x0B, 0x08, 0x04, 0x03, 0x0F, 0x08, 0x05, 0x05, 0x07, 0x0B, 0x08, 0x05, 0x05, 0x03, 0x0D, 0x04, 0x05, 0x05, 0x07, 0x08, 0x02, 0x0D, 0x06, 0x09, 0x05, 0x04, 0x05, 0x04, 0x01, 0x05, 0x05, 0x05, 0x04, 0x01, 0x04, 0x05, 0x06, 0x08, 0x02, 0x0F, 0x09, 0x04, 0x04, 0x02, 0x0F, 0x08, 0x05, 0x05, 0x05, 0x05, 0x01, 0x07, 0x08, 0x06, 0x0A, 0x09, 0x01, 0x04, 0x07, 0x0B, 0x0A, 0x08, 0x06, 0x0A, 0x0D, 0x05, 0x05, 0x07, 0x0F, 0x0D, 0x02, 0x09, 0x01, 0x06, 0x0D, 0x02, 0x0D, 0x06, 0x09, 0x02, 0x0B, 0x0B, 0x0C, 0x05, 0x05, 0x05, 0x05, 0x07, 0x0B, 0x0E, 0x0C, 0x01, 0x07, 0x09, 0x05, 0x01, 0x06, 0x09, 0x04, 0x04, 0x01, 0x05, 0x05, 0x05, 0x04, 0x04, 0x05, 0x01, 0x03, 0x0C, 0x04, 0x04, 0x04, 0x07, 0x08, 0x06, 0x0B, 0x0B, 0x0D, 0x05, 0x05, 0x06, 0x0B, 0x08, 0x07, 0x0F, 0x0F, 0x0A, 0x0B, 0x0A, 0x0B, 0x0E, 0x0B, 0x09, 0x05, 0x05, 0x05, 0x05, 0x05, 0x02, 0x0F, 0x08, 0x05, 0x04, 0x07, 0x09, 0x05, 0x02, 0x0E, 0x09, 0x05, 0x04, 0x04, 0x05, 0x05, 0x05, 0x05, 0x03, 0x0F, 0x09, 0x07, 0x08, 0x04, 0x04, 0x06, 0x09, 0x02, 0x0C, 0x06, 0x09, 0x03, 0x0D, 0x17, 0x05, 0x05, 0x05, 0x05, 0x04, 0x06, 0x0B, 0x0A, 0x08, 0x01, 0x07, 0x0A, 0x08, 0x01, 0x04, 0x04, 0x05, 0x05, 0x05, 0x05, 0x04, 0x07, 0x0B, 0x08, 0x07, 0x0A, 0x0B, 0x0D, 0x06, 0x0B, 0x0A, 0x0F, 0x09, 0x01, 0x04, 0x05, 0x05, 0x05, 0x02, 0x0C, 0x04, 0x02, 0x0C, 0x02, 0x0C, 0x04, 0x02, 0x0E, 0x08, 0x04, 0x06, 0x0E, 0x08, 0x04, 0x05, 0x06, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0A, 0x0C]
a = 18
for i in range(len(maze)):
    if maze[i] & 0x10 != 0:
        break
end = (i//a, i%a)
maze = list(zip(*[iter(maze)] * a))
flag = "hsctf{" + walkMaze(maze, a, (1, 1), end) + "}"
print(flag)

（其实还有个revenge版的，这次没人做出3ZVW就没放，留到下一场比赛啦~

5rn4

题外话，对国密熟悉的朋友们应该会知道描述指的是SM4（后来hint也给了）。

因为选拔赛难度，想着用来防AK所以加了个SEH（结构化异常处理）+花指令，果然防住了呢（。

花指令的说明见RE套路 - 关于逆向常客花指令 | c10udlnk_Log。这里是有点考汇编功底的（我错了，下次还敢

从没办法解析到的汇编往上看，可以看到inc edi这条是不太正常的。

edi只在前面存了一个栈上的地址，这边自增的话就错位了（当然也不排除这种可能。最不正常的应该是再下一句inc ebp，ebp存着栈底地址，理论上是不能动的。

可以把这部分试一下怎么拆分能把剩下的全部转成可读代码，然后把没用的花指令patch掉，也可以用动态调试发现接下来应该走的代码（不过这里有很多次循环SEH，用于反调试，比较阴间）。

然后把这个0xff给nop掉再整个转函数就可以生成反编译代码了。

sub_9F1032可以猜到是printf，sub_9F1262猜是scanf。

实际上就是一个sm4的实现（可以对比SM4的实现，如果看不出来就对比源码参考链接https://blog.csdn.net/weixin_42700740/article/details/102667012），密钥是`746573746f766572`，密文在byte_9FB000，可以说是一个密码题极简模板。

就是有一个魔改密文被藏在异常处理里了：

也可以看交叉引用发现（如果想到出题人这么阴间的话）

所以写exp：

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// sm4 code from https://blog.csdn.net/weixin_42700740/article/details/102667012

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

/******************************定义系统参数FK的取值****************************************/
const unsigned long TBL_SYS_PARAMS[4] = {
    0xa3b1bac6,
    0x56aa3350,
    0x677d9197,
    0xb27022dc
};

/******************************定义固定参数CK的取值****************************************/
const unsigned long TBL_FIX_PARAMS[32] = {
    0x00070e15,0x1c232a31,0x383f464d,0x545b6269,
    0x70777e85,0x8c939aa1,0xa8afb6bd,0xc4cbd2d9,
    0xe0e7eef5,0xfc030a11,0x181f262d,0x343b4249,
    0x50575e65,0x6c737a81,0x888f969d,0xa4abb2b9,
    0xc0c7ced5,0xdce3eaf1,0xf8ff060d,0x141b2229,
    0x30373e45,0x4c535a61,0x686f767d,0x848b9299,
    0xa0a7aeb5,0xbcc3cad1,0xd8dfe6ed,0xf4fb0209,
    0x10171e25,0x2c333a41,0x484f565d,0x646b7279
};

/******************************SBox参数列表****************************************/
const unsigned char TBL_SBOX[256] = {
    0xd6,0x90,0xe9,0xfe,0xcc,0xe1,0x3d,0xb7,0x16,0xb6,0x14,0xc2,0x28,0xfb,0x2c,0x05,
    0x2b,0x67,0x9a,0x76,0x2a,0xbe,0x04,0xc3,0xaa,0x44,0x13,0x26,0x49,0x86,0x06,0x99,
    0x9c,0x42,0x50,0xf4,0x91,0xef,0x98,0x7a,0x33,0x54,0x0b,0x43,0xed,0xcf,0xac,0x62,
    0xe4,0xb3,0x1c,0xa9,0xc9,0x08,0xe8,0x95,0x80,0xdf,0x94,0xfa,0x75,0x8f,0x3f,0xa6,
    0x47,0x07,0xa7,0xfc,0xf3,0x73,0x17,0xba,0x83,0x59,0x3c,0x19,0xe6,0x85,0x4f,0xa8,
    0x68,0x6b,0x81,0xb2,0x71,0x64,0xda,0x8b,0xf8,0xeb,0x0f,0x4b,0x70,0x56,0x9d,0x35,
    0x1e,0x24,0x0e,0x5e,0x63,0x58,0xd1,0xa2,0x25,0x22,0x7c,0x3b,0x01,0x21,0x78,0x87,
    0xd4,0x00,0x46,0x57,0x9f,0xd3,0x27,0x52,0x4c,0x36,0x02,0xe7,0xa0,0xc4,0xc8,0x9e,
    0xea,0xbf,0x8a,0xd2,0x40,0xc7,0x38,0xb5,0xa3,0xf7,0xf2,0xce,0xf9,0x61,0x15,0xa1,
    0xe0,0xae,0x5d,0xa4,0x9b,0x34,0x1a,0x55,0xad,0x93,0x32,0x30,0xf5,0x8c,0xb1,0xe3,
    0x1d,0xf6,0xe2,0x2e,0x82,0x66,0xca,0x60,0xc0,0x29,0x23,0xab,0x0d,0x53,0x4e,0x6f,
    0xd5,0xdb,0x37,0x45,0xde,0xfd,0x8e,0x2f,0x03,0xff,0x6a,0x72,0x6d,0x6c,0x5b,0x51,
    0x8d,0x1b,0xaf,0x92,0xbb,0xdd,0xbc,0x7f,0x11,0xd9,0x5c,0x41,0x1f,0x10,0x5a,0xd8,
    0x0a,0xc1,0x31,0x88,0xa5,0xcd,0x7b,0xbd,0x2d,0x74,0xd0,0x12,0xb8,0xe5,0xb4,0xb0,
    0x89,0x69,0x97,0x4a,0x0c,0x96,0x77,0x7e,0x65,0xb9,0xf1,0x09,0xc5,0x6e,0xc6,0x84,
    0x18,0xf0,0x7d,0xec,0x3a,0xdc,0x4d,0x20,0x79,0xee,0x5f,0x3e,0xd7,0xcb,0x39,0x48
};

//4字节无符号数组转无符号long型
void four_uCh2uLong(unsigned char* in, unsigned long* out)
{
    int i = 0;
    *out = 0;
    for (i = 0; i < 4; i++)
        *out = ((unsigned long)in[i] << (24 - i * 8)) ^ *out;
}

//无符号long型转4字节无符号数组
void uLong2four_uCh(unsigned long in, unsigned char* out)
{
    int i = 0;
    //从32位unsigned long的高位开始取
    for (i = 0; i < 4; i++)
        *(out + i) = (unsigned long)(in >> (24 - i * 8));
}

//左移，保留丢弃位放置尾部
unsigned long move(unsigned long data, int length)
{
    unsigned long result = 0;
    result = (data << length) ^ (data >> (32 - length));

    return result;
}

//秘钥处理函数,先使用Sbox进行非线性变化，再将线性变换L置换为L'
unsigned long func_key(unsigned long input)
{
    int i = 0;
    unsigned long ulTmp = 0;
    unsigned char ucIndexList[4] = { 0 };
    unsigned char ucSboxValueList[4] = { 0 };
    uLong2four_uCh(input, ucIndexList);
    for (i = 0; i < 4; i++)
    {
        ucSboxValueList[i] = TBL_SBOX[ucIndexList[i]];
    }
    four_uCh2uLong(ucSboxValueList, &ulTmp);
    ulTmp = ulTmp ^ move(ulTmp, 13) ^ move(ulTmp, 23);

    return ulTmp;
}

//加解密数据处理函数,先使用Sbox进行非线性变化，再进行线性变换L
unsigned long func_data(unsigned long input)
{
    int i = 0;
    unsigned long ulTmp = 0;
    unsigned char ucIndexList[4] = { 0 };
    unsigned char ucSboxValueList[4] = { 0 };
    uLong2four_uCh(input, ucIndexList);
    for (i = 0; i < 4; i++)
    {
        ucSboxValueList[i] = TBL_SBOX[ucIndexList[i]];
    }
    four_uCh2uLong(ucSboxValueList, &ulTmp);
    ulTmp = ulTmp ^ move(ulTmp, 2) ^ move(ulTmp, 10) ^ move(ulTmp, 18) ^ move(ulTmp, 24);

    return ulTmp;
}

//解密函数（与加密函数基本一致，只是秘钥使用的顺序不同，即把钥匙反着用就是解密）
//len:数据长度 key:密钥 input:输入的加密后数据 output:输出的解密后数据
void decode_fun(unsigned char len, unsigned char* key, unsigned char* input, unsigned char* output)
{
    int i = 0, j = 0;
    unsigned long ulKeyTmpList[4] = { 0 };//存储密钥的unsigned long数据
    unsigned long ulKeyList[36] = { 0 };  //用于密钥扩展算法与系统参数FK运算后的结果存储
    unsigned long ulDataList[36] = { 0 }; //用于存放加密数据

    /*开始生成子秘钥*/
    four_uCh2uLong(key, &(ulKeyTmpList[0]));
    four_uCh2uLong(key + 4, &(ulKeyTmpList[1]));
    four_uCh2uLong(key + 8, &(ulKeyTmpList[2]));
    four_uCh2uLong(key + 12, &(ulKeyTmpList[3]));

    ulKeyList[0] = ulKeyTmpList[0] ^ TBL_SYS_PARAMS[0];
    ulKeyList[1] = ulKeyTmpList[1] ^ TBL_SYS_PARAMS[1];
    ulKeyList[2] = ulKeyTmpList[2] ^ TBL_SYS_PARAMS[2];
    ulKeyList[3] = ulKeyTmpList[3] ^ TBL_SYS_PARAMS[3];

    for (i = 0; i < 32; i++)             //32次循环迭代运算
    {
        //5-36为32个子秘钥
        ulKeyList[i + 4] = ulKeyList[i] ^ func_key(ulKeyList[i + 1] ^ ulKeyList[i + 2] ^ ulKeyList[i + 3] ^ TBL_FIX_PARAMS[i]);
    }
    /*生成32轮32位长子秘钥结束*/

    for (j = 0; j < len / 16; j++)  //进行循环加密,并将加密后数据保存
    {
        /*开始处理解密数据*/
        four_uCh2uLong(input + 16 * j, &(ulDataList[0]));
        four_uCh2uLong(input + 16 * j + 4, &(ulDataList[1]));
        four_uCh2uLong(input + 16 * j + 8, &(ulDataList[2]));
        four_uCh2uLong(input + 16 * j + 12, &(ulDataList[3]));

        //解密
        for (i = 0; i < 32; i++)
        {
            ulDataList[i + 4] = ulDataList[i] ^ func_data(ulDataList[i + 1] ^ ulDataList[i + 2] ^ ulDataList[i + 3] ^ ulKeyList[35 - i]);//与加密唯一不同的就是轮密钥的使用顺序
        }
        /*将解密后数据输出*/
        uLong2four_uCh(ulDataList[35], output + 16 * j);
        uLong2four_uCh(ulDataList[34], output + 16 * j + 4);
        uLong2four_uCh(ulDataList[33], output + 16 * j + 8);
        uLong2four_uCh(ulDataList[32], output + 16 * j + 12);
    }
}

unsigned char byte_40B000[32] = {
    0xF6, 0x66, 0xAE, 0xF4, 0xFE, 0x17, 0xAE, 0x18, 0xAE, 0x68, 0xC7, 0x81, 0x18, 0x86, 0xC1, 0x9B, 
    0xF5, 0xEA, 0xE6, 0x22, 0x8F, 0x42, 0xCF, 0xF0, 0xD0, 0x15, 0x2A, 0xB7, 0xDC, 0xB3, 0x68, 0x4F
};

int main() {
    unsigned char i, len;
    unsigned char rslt[32] = { 0 };
    unsigned char key[20] = "746573746f766572";
    for (int i = 0; i <= 512; i++) { // 做了513次交换，其实就是做了一次交换，魔改密文
        byte_40B000[23] ^= byte_40B000[24];
        byte_40B000[24] ^= byte_40B000[23];
        byte_40B000[23] ^= byte_40B000[24];
    }
    decode_fun(32, key, byte_40B000, rslt);
    for (int i = 0; i < 32; i++) {
        printf("%c", rslt[i]);
    }
    return 0;
}

Misc

TapIt

附件里有一个Mikuuuuuu.mp4，给了一个演唱工具，明显是让你用这个演唱工具唱出跟mp4一样的旋律。

flag格式给了一个正则r'[0-9a-v]{30}'，意思是长度为30，字符为0-9或者a-v的某一个字符。

演唱工具魔改自HFIProgramming/mikutap: A Mainland China Friendly and independent version extracted from https://aidn.jp/mikutap，逆一下mikutap.js（不要问为什么逆这个，其他看起来就是导入的js，只有这个是用户自己写的）。

画面呈现存放在数组m中：

声音则是存在data/main/main.json中，通过base64解码得到音频。

而通过测试或者逆向代码能发现，0-9a-v可以呈现32种不同的声音，16种不同的画面类型，也就是说有两个字符的画面是一种类型的，但是声音呈现是唯一的。那么就可以通过画面+声音筛选拿到flag：l0vemikufor3verdoul1ke7hi5misc。

值得一提的是，原工具中每一次打开时同一个字符对应的画面类型是不同的，但是在同一次打开中同一个字符对应的画面类型是相同的（shuffle操作只在初始化的时候有做），但是本题工具中为了方便解题把这句shuffle给删掉了，所以无论多少次打开都是一样的。

本文作者: c10udlnk
本文链接: https://c10udlnk.top/p/wpFor-2022SlothTrial/
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