CTFshow Reverse 36D杯 签到 wp

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  __int64 v3; // rdx
  __int64 v4; // rcx
  int v6; // [rsp+4h] [rbp-8Ch]
  unsigned int v7; // [rsp+8h] [rbp-88h]
  int v8; // [rsp+Ch] [rbp-84h]
  int v9[31]; // [rsp+10h] [rbp-80h] BYREF
  char v10[4]; // [rsp+8Ch] [rbp-4h] BYREF
 
  v9[0] = 0x66;
  v9[1] = 109;
  v9[2] = 99;
  v9[3] = 98;
  v9[4] = 127;
  v9[5] = 58;
  v9[6] = 85;
  v9[7] = 106;
  v9[8] = 57;
  v9[9] = 82;
  v9[10] = 122;
  v9[11] = 55;
  v9[12] = 81;
  v9[13] = 19;
  v9[14] = 51;
  v9[15] = 35;
  v9[16] = 67;
  v9[17] = 70;
  v9[18] = 41;
  v9[19] = 61;
  v9[20] = 41;
  v9[21] = 32;
  v9[22] = 127;
  v9[23] = 28;
  v9[24] = 38;
  v9[25] = 77;
  v9[26] = 49;
  v9[27] = 20;
  v9[28] = 80;
  v9[29] = 94;
  v9[30] = -24;
  sub_4007F8(v10, 0LL, 4LL);
  v7 = 0;
  v6 = 0;
  a_puts((__int64)aFlag);
  do
  {
    v8 = a_getc();
    v6 |= v8 ^ v7 ^ (v7 + (v7 ^ v9[v7]));
    v4 = v7++;
  }
  while ( v8 && v8 != 10 && v8 != -1 );
  if ( v6 )
    a_write((__int64)aFailed, 0LL, v3, v4);
  else
    a_write((__int64)aCorrect, 0LL, v3, v4);
  return 0;
}

一上来的数字，应该是加密后的flag

aFlag为FLAG:说明这个函数是puts之类的输出函数，对应后边的aFailed就应该也是输出可能是printf，不重要

v8 = XXX  这里在循环里最后还比较-1和10说明是getc类的。

梳理完这个流程就比较明白了：flag输入每次一个字符，然与密文后作运算最后或到v6上，那么如果让v6为0就必需每个字符的运算结果都为0，也就得到了校验公式

v8 = v7 ^ (v7 + (v7 ^ v9[v7]))

对应解码程序

a = [102,109,99,98,127,58,85,106,57,82,122,55,81,19,51,35,67,70,41,61,41,32,127,28,38,77,49,20,80,94]
for i in range(len(a)):
    print(chr(i^(i+(i^a[i]))), end='')

最终得到flag{A_s1mpLe&E4sy_RE_i5Nt_1t}