CTFshow-1024杯

1024_Trick

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from Crypto.Util.number import getPrime, bytes_to_long
from gmpy2 import gcd, invert
from secret import flag

def gen(x, y):
	while True:
		k = getPrime(1024)
		if gcd(x * y - x - y + 1, k) == 1 and gcd(x * y - x - y + 1, k + 1024) == 1:
			return k

p = getPrime(1024)
q = getPrime(1024)

d = gen(p, q)

e1 = invert(d, (p - 1) * (q - 1))
e2 = invert(d + 1024, (p - 1) * (q - 1))

n = p * q
c = pow(flag, e1, n)

out = open('output', 'w')
out.write(str(n)+'\n')
out.write(str(e1)+'\n')
out.write(str(e2)+'\n')
out.write(str(c)+'\n')

​ 分析：

​ exp：

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from Crypto.Util.number import *
import gmpy2

n = 28247956190796077548821036165832051095182478841343849341201099128819479182033196745888916043527789810068912335285845353736774484336851382817061906161464410813261565043154569983001168979661397673033159029064565590415202266364668996185387927547647414609049665701634943970073424680262167057713264162243044242199992525550717430218592937779319306804366853371496949090902097569447358748738601201092248706444400984549491240349808244847814292328867056531536176757195282237815281620345660712787511847864860157060561247341926472934445937054346872777130907776700747053879719058858665504682167737096909091658310703848714995507087
e1 = 14729696809459508716385569304221898763069627997586157384420106315856541102841676901689428248411442723122274238893709627729613555080244422116252212813081403190521848377387091820856797688344508117977167198575999802122072100976783815485227448685251402928453641606244138534312889639695831020684859444038879931443306689809413522781182634833029742866760422550903997471974167552388516014202843112478278655370138693696356621310668692751596233613748393311787483393850367208650431878268345462622307080790201460993914739555030933412223443727451306619192444521947087311524606746672804056939248534450817019868870802552074243882377
e2 = 10907651584541464409470494737289836122935666586931706055595902003883184928824627549246585826596861981965927482456455737073705999708471130440398909870554351524288791148142553288839156753336551721274021596567766812030470532876231205718107266170708315964229471470530690537533600851746463150957473253899821019421342552091338365357017748742821451767352420328276163005698272017204771807428933429417340619373269170581813206653190286582334483827342338454461627946086922006705360281421384249860374825149668002500611584152958000206147344275567898096476692898681393229095674001876869497249423498186867190630700665116613171171985
c = 14151986355027540236877811729876734276410473157789551074684191809631364271052357249861862424614207635605041553420473293592242919606316628442514124449901660763278200999279517540020047022360186952956318844301844675024175636253509419780669989143178706203706350138854046573467948881219084764605132312446701967740275650109903208220772228525117766500225573404421894020148747991820530966508849170593946661068452445685905577771836515980941955814291933511646362511812580324895180140431064196870852032763940873198931010750588123937360585160510295014722360134758829576304291022740315244570633962516612490451736612193602659033973

kphi = e1 - e2 -1024*e1*e2
d2 = gmpy2.invert(e1,kphi)
m = pow(c,d2,n)
print(long_to_bytes(m))
#flag{wh47_1F_1_H4V3_pr1v473_K3Y+X???}

1024_密码系统

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# -*- coding:utf-8 -*-
# Author:airrudder

from Crypto.Cipher import DES
from binascii import b2a_hex, a2b_hex
from secret import FLAG,KEY,HINT
from itertools import *
import random

def Encode_1024sys(data,key):
    data_list = [data[i:i+8] for i in range(0,len(data),8)]
    k1 = DES.new(key.encode(), DES.MODE_ECB)
    k2 = DES.new(key[::-1].encode(), DES.MODE_ECB)
    data_res = ''
    for i in range(0,len(data_list)):
        k = random.choice([k1,k2])
        c = k.encrypt(data_list[i].encode())
        data_res += b2a_hex(c).decode() #字符串转16进制
    return data_res

def Encode_1024(data,key):
    len_data=len(data)
    choices = cycle('1024')
    while len_data%8!=0: #pad
        data += next(choices)
        len_data=len(data)
    data_res = Encode_1024sys(data,key)
    data_out = hex(int(data_res,16)**12 + random.randint(10**1023,10**1024))[2:]
    return data_out

def main():
    menu = '''
1. Encode
2. Verify your FLAG
3. Exit
'''
    try:
        while True:
            print(menu)
            choice = input("> ")
            if choice == "1":
                msg=input("Please input your msg: ")
                data_out = Encode_1024(msg+FLAG,KEY)
                print("hex(Encode_1024(msg+flag)) is :",data_out)
            elif choice == "2":
                yourFLAG = input('Please input your FLAG: ')
                if yourFLAG == FLAG:
                    print(HINT)
                else:
                    print('1024, nonono..., come on!!!')
            elif choice == "3":
                print("Bye!")
                return
            else:
                print("Invalid choice!")
                continue
        
    except:
        print('error')
        
if __name__ == "__main__":
    main()

​ 面向wp学习：

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假设flag为		flag{12345678}						#共14位
填充后			flag{12345678}10					#16位，明文分组为“flag{123”、“45678}10”
#flag前面的数据可控，假如填入的msg为“111”
flag会被填充到	111flag{12345678}1024102			#共24位，明文分组为“111flag{”、“12345678”、“}1024102”
#也就是说flag的最后一位“}”就被挤出了原来的分组，总长度增加。而且最后一个明文分组为“}1024102”。
#那么假如我在一开始填入的msg为“}1024102111”
flag会被填充到	}1024102111flag{12345678}1024102	#共32位，明文分组为“}1024102”、“111flag{”、“12345678”、“}1024102”
#这也就是说第一个明文分组和最后一个明文分组是一样的，所以ECB模式下，所对应的密文分组也一样。
#所以对输入的msg逐位添加爆破即可
爆破8位后，得到	2345678}1024102111flag{12345678}	#共32位
#到第9位时，就需要比较第一个明文分组和倒数第二个明文分组所对应的密文分组是否一样了
12345678}1024102111flag{12345678}1024102	#共40位，明文分组“12345678”、“}1024102”、“111flag{”、“12345678”、“}1024102”
#同样的，到(8k+1)位是，需要比较第一个明文分组和倒数第(k+1)位的明文分组

​ exp：

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from pwn import *
import gmpy2
import string
import random

def main():
   try:
      sh = remote('pwn.challenge.ctf.show',28175)
      print('\n[+] 开始爆破FLAG')
      FLAG = get_FLAG(sh)
      print('\n[+] 开始解密得到flag')
      flag = yunying(FLAG)
      print('flag{'+flag.lower()+'}')
   except:
      print_msg()
      exit(0)

def print_msg():
   print('='*50)
   print('Usage: python exp.py <ip> <port>')
   print('   Eg: python exp.py 127.0.0.1 10248')
   print('='*50)

#验证FLAG
def verify_FLAG(sh,FLAG):
   sh.recvuntil('> ')
   sh.sendline('2')
   sh.recvuntil('Please input your FLAG: ')
   sh.sendline(FLAG)
   data =  sh.recvline().decode()
   if 'nonono' not in data:
      print('\nHINT: '+data)
      return FLAG
   return 0

def get_FLAG(sh):
   FLAG = ''
   table = '1024'
   # 不知道FLAG被分成几组，这里先爆破5组
   for k in range(5):
      #DES分组，每组8字节
      for j in range(8):
         #由于key有两个，不知道到底使用的哪一个，所以多循环几次爆破即可
         while True:
            i = random.choice(['1','0','2','4'])
            sh.recvuntil('> ')
            sh.sendline('1')
            sh.recvuntil('Please input your msg: ')
            payload = i+FLAG+'1024102111'
            sh.sendline(payload)
            #接收到的data数据
            data = sh.recvline().decode().split(' : ')[1][:-1]
            #print(data)
            datadec = int(data,16)
            #开方后十进制表示的数据
            dec1024 = gmpy2.iroot(datadec,12)
            hex1024 = hex(dec1024[0])[2:]
            if (k==0 and hex1024[:16]==hex1024[-16:]) or hex1024[:16]==hex1024[-16*(k+1):-16*(k)]:
               FLAG = i+FLAG
               print('\rFLAG = '+FLAG,end='')
               if verify_FLAG(sh,FLAG):
                  return FLAG
               break
   return 0

#云影密码(01248密码)解密
def yunying(FLAG):
   flag = ''
   strs = string.ascii_uppercase
   flag_list = FLAG.split('0')
   for j in flag_list:
      tmp = 0
      for k in j:
         tmp += int(k)
      flag += strs[tmp-1]
   return flag

if __name__ == '__main__':
   main()

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