buuctf re [GWCTF 2019]xxor

__int64 __fastcall main(int a1, char **a2, char **a3)
{
  int i; // [rsp+8h] [rbp-68h]
  int j; // [rsp+Ch] [rbp-64h]
  __int64 v6[6]; // [rsp+10h] [rbp-60h] BYREF
  __int64 v7[6]; // [rsp+40h] [rbp-30h] BYREF

  v7[5] = __readfsqword(0x28u);
  puts("Let us play a game?");
  puts("you have six chances to input");
  puts("Come on!");
  memset(v6, 0, 40);
  for ( i = 0; i <= 5; ++i )
  {
    printf("%s", "input: ");
    a2 = (char **)((char *)v6 + 4 * i);
    __isoc99_scanf("%d", a2);
  }
  memset(v7, 0, 40);
  for ( j = 0; j <= 2; ++j )
  {
    dword_601078 = v6[j];
    dword_60107C = HIDWORD(v6[j]);
    a2 = (char **)&unk_601060;
    sub_400686(&dword_601078, &unk_601060);
    LODWORD(v7[j]) = dword_601078;
    HIDWORD(v7[j]) = dword_60107C;
  }
  if ( (unsigned int)sub_400770(v7, a2) != 1 )
  {
    puts("NO NO NO~ ");
    exit(0);
  }
  puts("Congratulation!\n");
  puts("You seccess half\n");
  puts("Do not forget to change input to hex and combine~\n");
  puts("ByeBye");
  return 0LL;
}

查看 sub_400770

__int64 __fastcall sub_400770(_DWORD *a1)
{
  if ( a1[2] - a1[3] == 2225223423LL
    && a1[3] + a1[4] == 4201428739LL
    && a1[2] - a1[4] == 1121399208LL
    && *a1 == -548868226
    && a1[5] == -2064448480
    && a1[1] == 550153460 )
  {
    puts("good!");
    return 1LL;
  }
  else
  {
    puts("Wrong!");
    return 0LL;
  }
}

计算出 __int64 a[6] = { 3746099070, 550153460, 3774025685, 1548802262, 2652626477, 2230518816 };

打开sub_400686函数

__int64 __fastcall sub_400686(unsigned int *a1, _DWORD *a2)
{
  __int64 result; // rax
  unsigned int v3; // [rsp+1Ch] [rbp-24h]
  unsigned int v4; // [rsp+20h] [rbp-20h]
  int v5; // [rsp+24h] [rbp-1Ch]
  unsigned int i; // [rsp+28h] [rbp-18h]

  v3 = *a1;
  v4 = a1[1];
  v5 = 0;
  for ( i = 0; i <= 0x3F; ++i )
  {
    v5 += 1166789954;
    v3 += (v4 + v5 + 11) ^ ((v4 << 6) + *a2) ^ ((v4 >> 9) + a2[1]) ^ 0x20;
    v4 += (v3 + v5 + 20) ^ ((v3 << 6) + a2[2]) ^ ((v3 >> 9) + a2[3]) ^ 0x10;
  }
  *a1 = v3;
  result = v4;
  a1[1] = v4;
  return result;
}

这段是加密函数
这就是一个利用已知数组unk_601060对我们输入的整型数组进行异或操作
因此我们只需要将整个过程逆过来，for循环那段，你将异或过程看成一个整体就行，最后就能得到输入的整型数组。
&unk_601060的值其实就是{2，2，3，4}

#include 

#pragma warning(disable:4996)
using namespace std;

int main()
{
    __int64 a[6] = { 3746099070, 550153460, 3774025685, 1548802262, 2652626477, 2230518816 };
    unsigned int a2[4] = { 2,2,3,4 };
    unsigned int v3, v4;
    int v5;
    for (int j = 0; j <= 4; j += 2) {
        v3 = a[j];
        v4 = a[j + 1];
        v5 = 1166789954*0x40;
        for (int i = 0; i <= 0x3F; ++i) {
            v4 -= (v3 + v5 + 20) ^ ((v3 << 6) + a2[2]) ^ ((v3 >> 9) + a2[3]) ^ 0x10;
            v3 -= (v4 + v5 + 11) ^ ((v4 << 6) + *a2) ^ ((v4 >> 9) + a2[1]) ^ 0x20;
            v5 -= 1166789954;
        }
        a[j] = v3;
        a[j + 1] = v4;
    }

    /*将整型数组作为字符输出，注意计算机小端排序*/
    for (int i = 0; i < 6; ++i) {
        cout << *((char*)&a[i] + 2) << *((char*)&a[i] + 1) <<  * ((char*)&a[i]);
    }

    system("PAUSE");
    return 0;
}