攻防世界 reverse 新手练习区

1.re1 DUTCTF

IDA shift+F12 查看字符串

DUTCTF{We1c0met0DUTCTF}

2.game ZSCTF

zsctf{T9is_tOpic_1s_v5ry_int7resting_b6t_others_are_n0t}

3.Hello, CTF Pediy CTF 2018

CrackMeJustForFun

将16进制字符串转ascii字符串得到flag

4.open-source HackYou CTF

参数 51966 25 h4cky0u

flag为:

c0ffee

5.open-source 9447 CTF 2014

IDA查看字符串

9447{This_is_a_flag}

6.simple-unpack

脱壳搜索字符串

flag{Upx_1s_n0t_a_d3liv3r_c0mp4ny}

7.logmein RC3 CTF 2016

1 v8=":\"AL_RT^L*.?+6/46"
2 v7 = 'ebmarah'[::-1]
3 s=''
4 for i in range(len(v8)):
5     s+=chr(ord(v7[i%7])^ord(v8[i]))
6 
7 print(s)

RC3-2016-XORISGUD

8. no-strings-attached 9447 CTF 2014

 1 x=[58,  20,   0,   0,  54,  20,   0,   0,  55,  20,
 2     0,   0,  59,  20,   0,   0, 128,  20,   0,   0,
 3   122,  20,   0,   0, 113,  20,   0,   0, 120,  20,
 4     0,   0,  99,  20,   0,   0, 102,  20,   0,   0,
 5   115,  20,   0,   0, 103,  20,   0,   0,  98,  20,
 6     0,   0, 101,  20,   0,   0, 115,  20,   0,   0,
 7    96,  20,   0,   0, 107,  20,   0,   0, 113,  20,
 8     0,   0, 120,  20,   0,   0, 106,  20,   0,   0,
 9   115,  20,   0,   0, 112,  20,   0,   0, 100,  20,
10     0,   0, 120,  20,   0,   0, 110,  20,   0,   0,
11   112,  20,   0,   0, 112,  20,   0,   0, 100,  20,
12     0,   0, 112,  20,   0,   0, 100,  20,   0,   0,
13   110,  20,   0,   0, 123,  20,   0,   0, 118,  20,
14     0,   0, 120,  20,   0,   0, 106,  20,   0,   0,
15   115,  20,   0,   0, 123,  20,   0,   0, 128,  20,
16     0,   0,   0,   0,   0,   0]
17 y=[]
18 for t in x:
19     if t!=20 and t!=0:
20         y.append(t)
21 i=0
22 while i<len(y):
23     for j in range(1,6):
24         y[i]-=j
25         y[i]=chr(y[i])
26         i+=1
27         if i==len(y):
28             break
29 print(''.join(y))

View Code

9447{you_are_an_international_mystery}

9.python-trade NJUPT CTF 2017

使用EasyPythonDecompiler.exe

1 import base64
2 
3 buf = base64.b64decode('XlNkVmtUI1MgXWBZXCFeKY+AaXNt')
4 flag = ''
5 for i in buf:
6     i -= 16
7     i ^= 32
8     flag += chr(i)
9 print(flag)

nctf{d3c0mpil1n9_PyC}

10.getit SharifCTF 2016

主要逻辑：

 1 int __cdecl main(int argc, const char **argv, const char **envp)
 2 {
 3   char v3; // al
 4   __int64 v5; // [rsp+0h] [rbp-40h]
 5   int i; // [rsp+4h] [rbp-3Ch]
 6   FILE *stream; // [rsp+8h] [rbp-38h]
 7   char filename[8]; // [rsp+10h] [rbp-30h]
 8   unsigned __int64 v9; // [rsp+28h] [rbp-18h]
 9 
10   v9 = __readfsqword(0x28u);
11   LODWORD(v5) = 0;
12   while ( (signed int)v5 < strlen(s) )          // s=c61b68366edeb7bdce3c6820314b7498
13   {
14     if ( v5 & 1 )
15       v3 = 1;
16     else
17       v3 = -1;
18     t[(signed int)v5 + 10] = s[(signed int)v5] + v3;
19     LODWORD(v5) = v5 + 1;
20   }
21   strcpy(filename, "/tmp/flag.txt");
22   stream = fopen(filename, "w");
23   fprintf(stream, "%s\n", u, v5);
24   for ( i = 0; i < strlen(t); ++i )
25   {
26     fseek(stream, p[i], 0);
27     fputc(t[p[i]], stream);
28     fseek(stream, 0LL, 0);
29     fprintf(stream, "%s\n", u);
30   }
31   fclose(stream);
32   remove(filename);
33   return 0;
34 }

View Code

解题脚本：

 1 s='c61b68366edeb7bdce3c6820314b7498'
 2 t=list('SharifCTF{????????????????????????????????}')
 3 v5 = 0;
 4 while v5 < len(s):  #c61b68366edeb7bdce3c6820314b7498
 5 
 6     if (v5 & 1):
 7         v3 = 1;
 8     else:
 9         v3 = -1;
10 
11     t[v5 + 10] = chr(ord(s[v5])+ v3) ;
12     v5= v5 + 1;
13 
14 print(''.join(t))

View Code

SharifCTF{b70c59275fcfa8aebf2d5911223c6589}

来源：oschina

链接：https://my.oschina.net/u/4380330/blog/3413617

标签

python

upx