Reserve ctf SSE_KEYGENME VAX2学习

 最近的一场ctf比赛，让我去学习了这个

首先附上我查看的一个很重要的资料

打开ida查看代码

 

 

 

 发现主要判断函数再于这个check_login这个函数，然后点进去看看

loc_94E:
mov     [rbp+var_108], 1
mov     [rbp+var_107], 1
mov     [rbp+var_106], 1
mov     [rbp+var_105], 1
mov     [rbp+var_104], 1
mov     [rbp+var_103], 1
mov     [rbp+var_102], 1
mov     [rbp+var_101], 1
mov     [rbp+var_100], 1
mov     [rbp+var_FF], 1
mov     [rbp+var_FE], 1
mov     [rbp+var_FD], 1
mov     [rbp+var_FC], 1
mov     [rbp+var_FB], 1
mov     [rbp+var_FA], 1
mov     [rbp+var_F9], 1
movzx   edx, [rbp+var_100]
movzx   eax, [rbp+var_FF]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_FE]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_FD]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_FC]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_FB]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_FA]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_F9]
shl     rdx, 8
or      rdx, rax
movzx   eax, [rbp+var_108]
movzx   ecx, [rbp+var_107]
shl     rax, 8
or      rax, rcx
movzx   ecx, [rbp+var_106]
shl     rax, 8
or      rax, rcx
movzx   ecx, [rbp+var_105]
shl     rax, 8
or      rax, rcx
movzx   ecx, [rbp+var_104]
shl     rax, 8
or      rax, rcx
movzx   ecx, [rbp+var_103]
shl     rax, 8
or      rax, rcx
movzx   ecx, [rbp+var_102]
shl     rax, 8
or      rax, rcx
movzx   ecx, [rbp+var_101]
shl     rax, 8
or      rax, rcx
mov     qword ptr [rbp+var_130], rdx
mov     qword ptr [rbp+var_130+8], rax
vmovaps xmm0, [rbp+var_130]
vmovaps [rbp+var_C0], xmm0
lea     rax, p_box
mov     [rbp+var_C8], rax
mov     rax, [rbp+var_C8]
vlddqu  xmm0, xmmword ptr [rax]
vmovaps [rbp+var_B0], xmm0
lea     rax, x_box
mov     [rbp+var_D0], rax
mov     rax, [rbp+var_D0]
vlddqu  xmm0, xmmword ptr [rax]
vmovaps [rbp+var_A0], xmm0
vmovaps xmm0, [rbp+var_B0]
vmovaps [rbp+var_20], xmm0
vmovaps xmm0, [rbp+var_C0]
vmovaps [rbp+var_10], xmm0
vmovaps xmm0, [rbp+var_20]
vmovaps xmm1, [rbp+var_10]
vpsubb  xmm0, xmm0, xmm1
vmovaps [rbp+var_B0], xmm0
mov     rdx, [rbp+var_118]
mov     rax, [rbp+var_F8]
add     rax, rdx
mov     [rbp+var_D8], rax
mov     rax, [rbp+var_D8]
vlddqu  xmm0, xmmword ptr [rax]
vmovaps [rbp+var_90], xmm0
vmovaps xmm0, [rbp+var_90]
vmovaps [rbp+var_40], xmm0
vmovaps xmm0, [rbp+var_B0]
vmovaps [rbp+var_30], xmm0
vmovaps xmm1, [rbp+var_30]
vmovaps xmm0, [rbp+var_40]
vpshufb xmm0, xmm0, xmm1
vmovaps [rbp+var_80], xmm0
vmovaps xmm0, [rbp+var_80]
vmovaps [rbp+var_60], xmm0
vmovaps xmm0, [rbp+var_A0]
vmovaps [rbp+var_50], xmm0
vmovaps xmm1, [rbp+var_60]
vmovaps xmm0, [rbp+var_50]
vxorps  xmm0, xmm1, xmm0
vmovaps [rbp+var_80], xmm0
mov     rdx, [rbp+ptr]
mov     rax, [rbp+var_F8]
add     rax, rdx
mov     [rbp+var_E0], rax
vmovaps xmm0, [rbp+var_80]
vmovaps [rbp+var_70], xmm0
vmovaps xmm0, [rbp+var_70]
mov     rax, [rbp+var_E0]
vmovups xmmword ptr [rax], xmm0
add     [rbp+var_F8], 10h

发现了这一串函数，一开始我在查找vpshufb这个指令的时候以为是国密加密，所以去看了很久，然后发现其实这个是打乱你原本字节顺序的函数，而打乱是根据你的第一个操作数打乱的，所以通过代码不难发现每次的第一个操作数都是同一个，所以打乱顺序都一样，记录下来，在解密就可以轻松解决

    char s1[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 0 };
    char flag[] = { 0x43,0x51,0x43,0x36,0x40,0x52,0x21,0x55,0x24,0x42,0x5b,0x68,0x7d,0x67,0x1f,0x7b,0x5d,0x7e,0x4e,0x0e,0x58,0x4,0x22,0x40,0x1e,0x14,0x16,0x2c,0x20,0x22,0x26,0x34 };
    char temp1[33]{ 0 };
    for (int i = 0; i < 2; i++)
    {
        for (int j = 0; j < 16; j++)
        {
            temp1[i * 16 + j] = flag[i * 16 + j] ^ s1[j];
        }
        cout << temp1 << endl;
    }
    
    cout << endl;
    cout << endl;
    cout << endl;
    cout << endl;


    int arr1[] = { 4, 0, 7, 15, 2, 6, 1, 9, 14, 13, 10, 5, 12, 3, 11, 8 };
    char arr2[33] = { 0 };
    for (int i = 0; i < 2; i++)
    {
        for (int j = 0; j < 16; j++)
        {
            arr2[i * 16 + j] = temp1[i * 16+ arr1[j]];
        }
        cout << arr2<<endl;
    }