攻防世界 reverse 进阶5-7

5.re-for-50-plz-50 tu-ctf-2016

流程很简单，异或比较

1 x=list('cbtcqLUBChERV[[Nh@_X^D]X_YPV[CJ')
2 y=0x37
3 z=''
4 for t in x:
5     z+=chr(ord(t)^y)
6 print(z)

View Code

TUCTF{but_really_whoisjohngalt}

6.key csaw-ctf-2016-quals

运行后打印完?W?h?a?t h?a?p?p?e?n? 便结束

修改指令跳过文件读取，输出=W=r=o=n=g=K=e=y=

关键点就是sub_4020c0函数

关注if比较处

动态调试可直接获得flag

这里的的v7其实是sub_4020c0函数中第三个参数

关注Memory，发现两次循环处理

脚本：

 1 x=[ 0x74, 0x68, 0x65, 0x6D, 0x69, 0x64, 0x61, 0x74, 0x68, 0x65,
 2   0x6D, 0x69, 0x64, 0x61, 0x74, 0x68, 0x65, 0x6D, 0x69, 0x64,
 3   0x6]
 4 y=[0x3E, 0x2D, 0x2D, 0x2D, 0x2D, 0x2B, 0x2B, 0x2B, 0x2B, 0x2E,
 5   0x2E, 0x2E, 0x2E, 0x3C, 0x3C, 0x3C, 0x3C, 0x2E]
 6 # print(len(x))
 7 # print(len(y))
 8 z=[]
 9 for i in range(18):
10     z.append((x[i]^y[i])+22+9)
11 
12 # print(' '.join(map(hex,z)))
13 print(''.join(map(chr,z)))

View Code

idg_cni~bjbfi|gsxb

7.simple-check-100 school-ctf-winter-2015

exe文件有坑，满是辛酸泪，分析elf文件

过掉check_key()函数在linux便可直接输出flag

分析：

 1   a1[0] = 0xE37EC854;
 2   a1[1] = 0x9A16C764;
 3   a1[2] = 0x326511CD;
 4   a1[3] = 0x43D3E32D;
 5   a1[4] = 0xD29DA992;
 6   a1[5] = 0xD32C6DE6;
 7   a1[6] = 0x6AFEBDB6;
 8   v14 = 0x13;
 9   v3 = alloca(32);
10   v15 = &v7;
11   printf("Key: ");
12   __isoc99_scanf((int)"%s", (int)v15, v5, v6, v7, v8, v9, (int)v10, v11, v12, a1[0], a1[1], a1[2], a1[3], a1[4], a1[5]);
13   if ( check_key((int)v15) )
14     interesting_function(a1);

 1 unsigned int *__cdecl interesting_function(int a1[7])
 2 {
 3   unsigned int *result; // eax
 4   unsigned int temp; // [esp+18h] [ebp-20h]
 5   int i; // [esp+1Ch] [ebp-1Ch]
 6   int j; // [esp+20h] [ebp-18h]
 7   int *__attribute__((__org_arrdim(0,7))) v5; // [esp+24h] [ebp-14h]
 8   char *ptr_temp; // [esp+28h] [ebp-10h]
 9   unsigned int v7; // [esp+2Ch] [ebp-Ch]
10 
11   v7 = __readgsdword(0x14u);
12   result = (unsigned int *)a1;
13   v5 = a1;
14   for ( i = 0; i <= 6; ++i )
15   {
16     temp = v5[i] ^ 0xDEADBEEF;
17     result = &temp;
18     ptr_temp = (char *)&temp;
19     for ( j = 3; j >= 0; --j )
20       result = (unsigned int *)putchar((char)(ptr_temp[j] ^ flag_data[i][j]));
21   }
22   return result;
23 }

脚本：

 1 win=[0x54, 0xB8, 0xFE, 0x61, 0x00, 0x13, 0x00, 0x00, 0x00, 0x61, 0x6A, 0xFE, 0xBD, 0xB6, 0xD3, 0x2C,
 2 0x6D, 0xE6, 0xD2, 0x9D, 0xA9, 0x92, 0x43, 0xD3, 0xE3, 0x2D, 0x32, 0x65, 0x11, 0xCD, 0x9A, 0x16,
 3 0xC7, 0x64, 0xE3, 0x7E, 0xC8, 0x30]#windows下调试获取的操作数据
 4 win2=[0x6A, 0xFE, 0xBD, 0xB6, 0xD3, 0x2C,
 5 0x6D, 0xE6, 0xD2, 0x9D, 0xA9, 0x92, 0x43, 0xD3, 0xE3, 0x2D, 0x32, 0x65, 0x11, 0xCD, 0x9A, 0x16,
 6 0xC7, 0x64, 0xE3, 0x7E, 0xC8, 0x54]
 7 win2=win2[::-1]
 8 
 9 flag_date=[0xDC, 0x17, 0xBF, 0x5B, 0xD4, 0x0A, 0xD2, 0x1B, 0x7D, 0xDA,
10   0xA7, 0x95, 0xB5, 0x32, 0x10, 0xF6, 0x1C, 0x65, 0x53, 0x53,
11   0x67, 0xBA, 0xEA, 0x6E, 0x78, 0x22, 0x72, 0xD3]
12 
13 
14 a1=[0 for i in range(7)]
15 a1[0] = 0xE37EC854;
16 a1[1] = 0x9A16C764;
17 a1[2] = 0x326511CD;
18 a1[3] = 0x43D3E32D;
19 a1[4] = 0xD29DA992;
20 a1[5] = 0xD32C6DE6;
21 a1[6] = 0x6AFEBDB6;
22 
23 x=[0xef,0xbe,0xad,0xde]
24 
25 s=''
26 for i in range(7):
27     temp=(a1[i]^0xDEADBEEF).to_bytes(4,'little')
28     for j in range(3,-1,-1):
29         s+=chr(temp[j]^flag_date[i*4+j])
30         print(chr(win[i * 4 + j] ^ flag_date[i * 4 + j] ^ x[j]), end='')
31 #         print(chr(v8[i*4+j]^flag_date[i*4+j]^x[j]),end='')
32 print()
33 print(s)

exe中栈内变量布局与linux不同，

exe文件：äìgÅ§;µ`n:ç,=c!hí±t
elf文件：flag_is_you_know_cracking!!!

flag_is_you_know_cracking!!!

攻防世界 reverse 进阶5-7

猜你喜欢