[summary] 2021unctf Campus (cry & MISC)

        After all, the first contact with regular competition questions , No experience , There are not many types of questions , In addition, there is a lack of basic knowledge , If you don't understand the principle, you will have a headache .

  The following is a summary of the idea of this problem solving ：( Refer to the official wp)

---

misc

• LPL:

1. Downloading an attachment is a zip Compressed package , Decompress directly to get .png and .zip file                                                  
2. picture win Can open (linux Cannot be opened ), No information , Look at the compressed package. You need a password (010 Observation is true encryption ), Unable to blast , Guess the password should be hidden in the picture
3. 010 Open picture discovery prompt , According to the official wp Methods , The lack of IDAT The module caused an error
4. CRC Error find the password according to the text ： Take the following one crc For example , Find out the password in order
5. Get the code ：EDGnb！！( Chinese exclamation point , It is recommended to copy directly from the text , To avoid being trapped )
6. decompression .zip obtain .png and flag.txt（ Never break, never build 1 Set - The documentary - HD exclusive online viewing -bilibili- Bili, Bili ）      Date and video link

      7. Open linked video , Find the comment corresponding to the date in the comment area ( By the way, I also saw the documentary ...)

ps： The password really didn't expect （ It is my brain hole that is out of date ）, Thought it was crc Types of questions , change crc The hexadecimal value of , Not paying attention to the text message , But in fact, the direction is wrong , according to LPL( League of heroes professional league ), Before contacting EDG take the crown , It should be possible according to crc The wrong text content is associated with , But I didn't think about that at all , Very good original title .

• telecommunication fraud ：

        Encrypted accounts qi]m^roVibdVbXUU`h 

 

1. According to the title Wan Jun , Commanders and the like may suggest that Caesar encrypts
2. But it should not be so simple , Look, there are some special characters in the account (^ This kind of ), It could be a variant of Caesar
3. turn ASCII Look at decimal numbers (113 105 93 109 94 114 111 86 105 98 100 86 98 88 85 85 96 104 10), First part comparison unctf(117 110 99 116 102 )
4. The former ascii Values are incremented one by one 4,5,6....... Finally, it is converted to a string

attach py Script ： obtain unctf{yauoreright}

c = "qi]m^roVibdVbXUU`h"     # contrast unctf perhaps flag See the displacement 
move = 4                     # Displacement 
flag = ""
for i in c:
    flag += chr(ord(i) + move)
    move += 1
print(flag)

• Handstand shampoo ：

1. Download it .txt file , Open is hexadecimal data , Look at the beginning and the end d9ff and ffd8(jpg Document mark ), Combined with the topic tips “ Handstand ”, Flashback the text ( Script , I use the online tool to flashback the text , Split and then flashback )

   f = open("key.txt", "rb")
   s = f.read()
   f.close()
   f = open("newtext.txt", "wb")
   f.write(s[::-1])
   f.close()
    
   

2. Combined with hint shampoo , take D8FF Change it to FFD8, Copy content to 010 in , Look at the text information and find base64 Encrypted content ( I don't know how to find it , Anyway, opening the picture is incomplete )
3. base64 decode

And Buddhism on Zen （ Japan ri Change to date yue, Although it looks the same , But if you can't solve it, you will be very angry , pit !）

Simple log audit ：

1. Download it juexue.log Log files ,notepad++ open , find base64 Code segment
2. Use your own plug-in base64 decode , obtain   'cat /CTF?YouShouJiuXing'
3. flag：unctf{CTF?YouShouJiuXing}（ I thought it was all in quotation marks , The results have not been submitted ... speechless ）

---

Crypto

• Explore medieval castles

1. Download the attachment to get Little note .txt, According to the prompt, the Emperor ( Caesar encryption ),64 position (bsae64), Double block ( The fence is encrypted )
2. First Caesar ran all the possible , Put it in base64 Decryption in ( Online tools are OK , I used it here CaptfEncoder)
3. Find the obvious structure ：UCFsbciet_inwnaoagNT{usrb_oXagadmwn}  ( Other solutions are garbled )
4. Online tools fence decryption ：UNCTF{subscribe_to_Xiangwandamowang}

• easy_rsa

1. Download it p,q,c,e(n=p*q), Script directly （py2）

import gmpy2
from Crypto.Util.number import long_to_bytes
 
q= 9961202707366965556741565662110710902919441271996809241009358666778850435448710324711706845973820669201482939820488174382325795134659313309606698334978471
p= 12525187149887628510447403881107442078833803097302579419605689530714690308437476207855511625840027119860834633695330551080761572835309850579517639206740101
c= 28587419802025513525354713621431206010395084854419372005671024739235625817936539010481222419824634956610184430308528941304950093228826213143262329902946812513518444587906469224383320964300417189270202019231856531012143472434842753891213128487132962453421971000901646523331476667655739056951415917218673801225
e = 65537

n=p*q
phi = (p-1)*(q-1)

d = gmpy2.invert(e, phi)  
m = pow(c, d, n)  
print (long_to_bytes(m)) 

• telecommunication fraud pro

 

1. According to the topic tips ： Four pits and seven rows , The Roman empire —> rot47( Caesar's displacement is 47 Bit encryption )
2. contrast ascii value (unctf), Difference between 64 position ,CyberChef On-line