题目链接：https://adworld.xctf.org.cn/task/answer?type=reverse&number=4&grade=0&id=5074
1.准备
打开测试用例
首先分析程序
得到是win32程序
2.第一种方法
2.1 分析代码
使用IDA打开，找到main函数，F5得到C代码
得知主要是main_0这个函数，打开
1 "If m of the Nth lamp is 1,it's on ,if not it's off\n"
2 "At first all the lights were closed\n");
3 sub_45A7BE("Now you can input n to change its state\n");
4 sub_45A7BE(
5 "But you should pay attention to one thing,if you change the state of the Nth lamp,the state of (N-1)th and (N+1)th w"
6 "ill be changed too\n");
7 sub_45A7BE("When all lamps are on,flag will appear\n");
8 sub_45A7BE("Now,input n \n");
9 while ( 1 )
10 {
11 while ( 1 )
12 {
13 sub_45A7BE("input n,n(1-8)\n");
14 sub_459418();
15 sub_45A7BE("n=");
16 sub_4596D4("%d", &v1);
17 sub_45A7BE("\n");
18 if ( v1 >= 0 && v1 <= 8 )
19 break;
20 sub_45A7BE("sorry,n error,try again\n");
21 }
22 if ( v1 )
23 {
24 sub_4576D6(v1 - 1);
25 }
26 else
27 {
28 for ( i = 0; i < 8; ++i )
29 {
30 if ( (unsigned int)i >= 9 )
31 j____report_rangecheckfailure();
32 byte_532E28[i] = 0;
33 }
34 }
35 j__system("CLS");
36 sub_458054();
37 if ( byte_532E28[0] == 1
38 && byte_532E28[1] == 1
39 && byte_532E28[2] == 1
40 && byte_532E28[3] == 1
41 && byte_532E28[4] == 1
42 && byte_532E28[5] == 1
43 && byte_532E28[6] == 1
44 && byte_532E28[7] == 1 )
45 {
46 sub_457AB4();
47 }
48 }
49 }
通过
if ( byte_532E28[0] == 1
&& byte_532E28[1] == 1
&& byte_532E28[2] == 1
&& byte_532E28[3] == 1
&& byte_532E28[4] == 1
&& byte_532E28[5] == 1
&& byte_532E28[6] == 1
&& byte_532E28[7] == 1 )
判断出我们需要将这八个数组的值都变为1，也就是8盏灯都闭合。
2.2 分析规则
打开sub_4576D6(v1 - 1);函数
bool __cdecl sub_45E640(int a1)
{
bool result; // al
if ( a1 )
{
if ( a1 == 7 )
{
byte_532E28[7] = byte_532E28[7] == 0;
byte_532E27[7] = byte_532E27[7] == 0;
result = 1;
byte_532E28[0] = byte_532E28[0] == 0;
}
else
{
byte_532E28[a1] = byte_532E28[a1] == 0;
byte_532E27[a1] = byte_532E27[a1] == 0;
result = byte_532E29[a1] == 0;
byte_532E29[a1] = result;
}
}
else
{
byte_532E28[0] = byte_532E28[0] == 0;
byte_532E29[0] = byte_532E29[0] == 0;
result = 1;
byte_532E28[7] = byte_532E28[7] == 0;
}
return result;
}
分析得到按键与电路闭合的关系：
- 按1--闭合1,2,8
- 按8--闭合1,7,8
- 按i(除1,8)--闭合i-1,i,i+1
2.3 暴力破解
由此并结合C，汇编代码我们写出暴力破解的C++代码：
#include <iostream>
#include <vector>
using namespace std;
#define for(a,b,c) for(int a = b; a < c; ++a)
#define N 8
vector<int> flag(8,-1);
void func(int *arr){
for(i,0,N){
int n = arr[i];
if(n == 0){
flag[0] *= -1;
flag[1] *= -1;
flag[7] *= -1;
}else{
if(n == 7){
flag[0] *= -1;
flag[6] *= -1;
flag[7] *= -1;
}else{
flag[n] *= -1;
flag[n-1] *= -1;
flag[n+1] *= -1;
}
}
}
}
bool Judge(){
for(i,0,8)
if(flag[i] == -1)
return false;
return true;
}
int main(void)
{
int array[N] = {0};
for(i,0,8)
for(j,0,8)
for(k,0,8)
for(m,0,8)
for(n,0,8)
for(p,0,8)
for(q,0,8)
for(t,0,8){
array[0] = i;
array[1] = j;
array[2] = k;
array[3] = m;
array[4] = n;
array[5] = p;
array[6] = q;
array[7] = t;
func(array);
if(Judge()){
cout << "success:" << i+1 << j+1 << k+1 << m+1 << n+1 << p+1 << q+1 << t+1 << endl;
system("PAUSE");
}else{
for(x,0,8)
fill(flag.begin(), flag.end(), -1);
}
}
cout << "over!";
system("PAUSE");
return 0;
}
2.4 get flag！
输入到程序中
3.第二种方法
3.1 分析代码
通过分析代码，我们很容易获知sub_457AB4()就是输出flag的函数
if ( byte_532E28[0] == 1
&& byte_532E28[1] == 1
&& byte_532E28[2] == 1
&& byte_532E28[3] == 1
&& byte_532E28[4] == 1
&& byte_532E28[5] == 1
&& byte_532E28[6] == 1
&& byte_532E28[7] == 1 )
{
sub_457AB4();
}
int sub_45E940()
{
char v1; // [esp+0h] [ebp-164h]
signed int i; // [esp+D0h] [ebp-94h]
char v3; // [esp+DCh] [ebp-88h]
char v4; // [esp+DDh] [ebp-87h]
char v5; // [esp+DEh] [ebp-86h]
char v6; // [esp+DFh] [ebp-85h]
char v7; // [esp+E0h] [ebp-84h]
char v8; // [esp+E1h] [ebp-83h]
char v9; // [esp+E2h] [ebp-82h]
char v10; // [esp+E3h] [ebp-81h]
char v11; // [esp+E4h] [ebp-80h]
char v12; // [esp+E5h] [ebp-7Fh]
char v13; // [esp+E6h] [ebp-7Eh]
char v14; // [esp+E7h] [ebp-7Dh]
char v15; // [esp+E8h] [ebp-7Ch]
char v16; // [esp+E9h] [ebp-7Bh]
char v17; // [esp+EAh] [ebp-7Ah]
char v18; // [esp+EBh] [ebp-79h]
char v19; // [esp+ECh] [ebp-78h]
char v20; // [esp+EDh] [ebp-77h]
char v21; // [esp+EEh] [ebp-76h]
char v22; // [esp+EFh] [ebp-75h]
char v23; // [esp+F0h] [ebp-74h]
char v24; // [esp+F1h] [ebp-73h]
char v25; // [esp+F2h] [ebp-72h]
char v26; // [esp+F3h] [ebp-71h]
char v27; // [esp+F4h] [ebp-70h]
char v28; // [esp+F5h] [ebp-6Fh]
char v29; // [esp+F6h] [ebp-6Eh]
char v30; // [esp+F7h] [ebp-6Dh]
char v31; // [esp+F8h] [ebp-6Ch]
char v32; // [esp+F9h] [ebp-6Bh]
char v33; // [esp+FAh] [ebp-6Ah]
char v34; // [esp+FBh] [ebp-69h]
char v35; // [esp+FCh] [ebp-68h]
char v36; // [esp+FDh] [ebp-67h]
char v37; // [esp+FEh] [ebp-66h]
char v38; // [esp+FFh] [ebp-65h]
char v39; // [esp+100h] [ebp-64h]
char v40; // [esp+101h] [ebp-63h]
char v41; // [esp+102h] [ebp-62h]
char v42; // [esp+103h] [ebp-61h]
char v43; // [esp+104h] [ebp-60h]
char v44; // [esp+105h] [ebp-5Fh]
char v45; // [esp+106h] [ebp-5Eh]
char v46; // [esp+107h] [ebp-5Dh]
char v47; // [esp+108h] [ebp-5Ch]
char v48; // [esp+109h] [ebp-5Bh]
char v49; // [esp+10Ah] [ebp-5Ah]
char v50; // [esp+10Bh] [ebp-59h]
char v51; // [esp+10Ch] [ebp-58h]
char v52; // [esp+10Dh] [ebp-57h]
char v53; // [esp+10Eh] [ebp-56h]
char v54; // [esp+10Fh] [ebp-55h]
char v55; // [esp+110h] [ebp-54h]
char v56; // [esp+111h] [ebp-53h]
char v57; // [esp+112h] [ebp-52h]
char v58; // [esp+113h] [ebp-51h]
char v59; // [esp+114h] [ebp-50h]
char v60; // [esp+120h] [ebp-44h]
char v61; // [esp+121h] [ebp-43h]
char v62; // [esp+122h] [ebp-42h]
char v63; // [esp+123h] [ebp-41h]
char v64; // [esp+124h] [ebp-40h]
char v65; // [esp+125h] [ebp-3Fh]
char v66; // [esp+126h] [ebp-3Eh]
char v67; // [esp+127h] [ebp-3Dh]
char v68; // [esp+128h] [ebp-3Ch]
char v69; // [esp+129h] [ebp-3Bh]
char v70; // [esp+12Ah] [ebp-3Ah]
char v71; // [esp+12Bh] [ebp-39h]
char v72; // [esp+12Ch] [ebp-38h]
char v73; // [esp+12Dh] [ebp-37h]
char v74; // [esp+12Eh] [ebp-36h]
char v75; // [esp+12Fh] [ebp-35h]
char v76; // [esp+130h] [ebp-34h]
char v77; // [esp+131h] [ebp-33h]
char v78; // [esp+132h] [ebp-32h]
char v79; // [esp+133h] [ebp-31h]
char v80; // [esp+134h] [ebp-30h]
char v81; // [esp+135h] [ebp-2Fh]
char v82; // [esp+136h] [ebp-2Eh]
char v83; // [esp+137h] [ebp-2Dh]
char v84; // [esp+138h] [ebp-2Ch]
char v85; // [esp+139h] [ebp-2Bh]
char v86; // [esp+13Ah] [ebp-2Ah]
char v87; // [esp+13Bh] [ebp-29h]
char v88; // [esp+13Ch] [ebp-28h]
char v89; // [esp+13Dh] [ebp-27h]
char v90; // [esp+13Eh] [ebp-26h]
char v91; // [esp+13Fh] [ebp-25h]
char v92; // [esp+140h] [ebp-24h]
char v93; // [esp+141h] [ebp-23h]
char v94; // [esp+142h] [ebp-22h]
char v95; // [esp+143h] [ebp-21h]
char v96; // [esp+144h] [ebp-20h]
char v97; // [esp+145h] [ebp-1Fh]
char v98; // [esp+146h] [ebp-1Eh]
char v99; // [esp+147h] [ebp-1Dh]
char v100; // [esp+148h] [ebp-1Ch]
char v101; // [esp+149h] [ebp-1Bh]
char v102; // [esp+14Ah] [ebp-1Ah]
char v103; // [esp+14Bh] [ebp-19h]
char v104; // [esp+14Ch] [ebp-18h]
char v105; // [esp+14Dh] [ebp-17h]
char v106; // [esp+14Eh] [ebp-16h]
char v107; // [esp+14Fh] [ebp-15h]
char v108; // [esp+150h] [ebp-14h]
char v109; // [esp+151h] [ebp-13h]
char v110; // [esp+152h] [ebp-12h]
char v111; // [esp+153h] [ebp-11h]
char v112; // [esp+154h] [ebp-10h]
char v113; // [esp+155h] [ebp-Fh]
char v114; // [esp+156h] [ebp-Eh]
char v115; // [esp+157h] [ebp-Dh]
char v116; // [esp+158h] [ebp-Ch]
sub_45A7BE((int)"done!!! the flag is ", v1);
v60 = 18;
v61 = 64;
v62 = 98;
v63 = 5;
v64 = 2;
v65 = 4;
v66 = 6;
v67 = 3;
v68 = 6;
v69 = 48;
v70 = 49;
v71 = 65;
v72 = 32;
v73 = 12;
v74 = 48;
v75 = 65;
v76 = 31;
v77 = 78;
v78 = 62;
v79 = 32;
v80 = 49;
v81 = 32;
v82 = 1;
v83 = 57;
v84 = 96;
v85 = 3;
v86 = 21;
v87 = 9;
v88 = 4;
v89 = 62;
v90 = 3;
v91 = 5;
v92 = 4;
v93 = 1;
v94 = 2;
v95 = 3;
v96 = 44;
v97 = 65;
v98 = 78;
v99 = 32;
v100 = 16;
v101 = 97;
v102 = 54;
v103 = 16;
v104 = 44;
v105 = 52;
v106 = 32;
v107 = 64;
v108 = 89;
v109 = 45;
v110 = 32;
v111 = 65;
v112 = 15;
v113 = 34;
v114 = 18;
v115 = 16;
v116 = 0;
v3 = 123;
v4 = 32;
v5 = 18;
v6 = 98;
v7 = 119;
v8 = 108;
v9 = 65;
v10 = 41;
v11 = 124;
v12 = 80;
v13 = 125;
v14 = 38;
v15 = 124;
v16 = 111;
v17 = 74;
v18 = 49;
v19 = 83;
v20 = 108;
v21 = 94;
v22 = 108;
v23 = 84;
v24 = 6;
v25 = 96;
v26 = 83;
v27 = 44;
v28 = 121;
v29 = 104;
v30 = 110;
v31 = 32;
v32 = 95;
v33 = 117;
v34 = 101;
v35 = 99;
v36 = 123;
v37 = 127;
v38 = 119;
v39 = 96;
v40 = 48;
v41 = 107;
v42 = 71;
v43 = 92;
v44 = 29;
v45 = 81;
v46 = 107;
v47 = 90;
v48 = 85;
v49 = 64;
v50 = 12;
v51 = 43;
v52 = 76;
v53 = 86;
v54 = 13;
v55 = 114;
v56 = 1;
v57 = 117;
v58 = 126;
v59 = 0;
for ( i = 0; i < 56; ++i )
{
*(&v3 + i) ^= *(&v60 + i);
*(&v3 + i) ^= 0x13u;
}
return sub_45A7BE((int)"%s\n", (unsigned int)&v3);
}
这实际上就是一段经过计算，输出flag的代码，for上面是已知条件，下面进行变换，我们可以转换为Python代码，输出flag
3.2 脚本获取flag
arr1 = [18, 64, 98, 5, 2, 4, 6, 3, 6, 48, 49, 65, 32, 12, 48, 65, 31, 78, 62, 32, 49, 32,
1, 57, 96, 3, 21, 9, 4, 62, 3, 5, 4, 1, 2, 3, 44, 65, 78, 32, 16, 97, 54, 16, 44,
52, 32, 64, 89, 45, 32, 65, 15, 34, 18, 16, 0]
arr2 = [123, 32, 18, 98, 119, 108, 65, 41, 124, 80, 125, 38, 124, 111, 74, 49,
83, 108, 94, 108, 84, 6, 96, 83, 44, 121, 104, 110, 32, 95, 117, 101, 99,
123, 127, 119, 96, 48, 107, 71, 92, 29, 81, 107, 90, 85, 64, 12, 43, 76, 86,
13, 114, 1, 117, 126, 0]
str = ''
for i in range(0, 56):
arr2[0 + i] ^= arr1[0 + i]
arr2[0 + i] ^= 0x13
str = str + chr(arr2[i]);
print(str)
3.3 get flag！
zsctf{T9is_tOpic_1s_v5ry_int7resting_b6t_others_are_n0t}