[TUCTF 2022] 部分

Crypto

RSA

A sheep jumps over fence

More Effort

没作几个题，网站太慢了，经常题都不出。

Crypto

RSA

忘了记名字了，是个远程题，e=5不变，n,c在变化，显然是用中国剩余定理。然后再开5次幂。


n1=10482537034616502990056118030126164434407037176718331962667011326955491120566031634285535049167438082826079349272253648544730225816672178907764563795724843238117833575393198105750393276477056062212621581826908061516664802589223889385525044351279674490228268440747101433020011881495784429621597371307426622757779128699745573729906521433559644189341958800662044774541
c1=6087004369946053227982494913706901186274614577662910050380033779616661997035829135770453917032222963288513864576850581583066480871865034839084616393770636880369071627374934883683905306826847313587047467574498166973462816444236751014698316815611495234751053545484591628748764097916135305447550851423345959821240953597530179672401858302119597171603198037366269233719
n2 =12367732114126301913737674911174659623260421701922083115463688097892335835832137676235926929476836047205277037213556311855697316742401499464166580788816568220539558161216494644597740553321733102039206472053356162972506585408259141048898509380926861285805055391399828624294182942246152692738571838819392996502010209610177387578042322991636983388504268335500249420617
c2 =6598110242539995863074355056158570164049216335364925288931323757253535398103184489531242032575274289060700366841161402204146875959261549401532459608455604827722243085170385981035222200389913788409648676752548513449147184508753978498499016611762538512670708489158929810784104583789226504345306015964098627253554658284373096694929056015453847074887650556440052878630
n3 =10256038725288950855196758629933306580783576388418827903732634844615198789778181879287006217785874404464941165322121223311218355864125867810588164953430819701659125700907576823337578399226408327622582145531259253939397764804840588945160103734078760521203955556671490266695737131932977852304750132447539392473513180775946728590263319845517676125152164659427731699479
c3 = 7307506932852480055779150743698192577518060811090089847831839584489356780048260286294057804235251817041195453111379116685659989141101142713850347634848460781996991307627778136048623024426933507486743478530416521616605247651996711033755127847375528743357717132830144403674863976913789724872493255786744573370517927939711839563307201365321241992779597973050791722751
e=5
#sage
#v = crt([c1,c2,c3],[n1,n2,n3])
v =73448690174896106135676773887969898376313644525180577635638534667360398313545968245865457044768650498073919986296816184849746808491061728181876720693332170726810820658795124493571143780153343023652595257458464865059042345641702358771008787607355295320693998300553350404027525319624789400937306774517103831179992855723955092191381869960795275997575970901077630183000778125
from gmpy2 import iroot
m = iroot(v,5)[0]
#(mpz(149003617072514967705362966617328574884671140357033588603732115961358085245), True)
bytes.fromhex(hex(m)[2:])
b'TUCTF{0bl1g4t0ry_RSA_chall_l0l}'

A sheep jumps over fence

从名字上看是个栅栏密码，给了密文

0x56455a7059574e7459584a685a576c756557646f59565637633364765a57647664474e305a48427965584e4464476c6c62584e6c6257463165574e3059574e7a5647687a6247567a5a6d527a636d46796233427366513d3d

尾部的3D3D应该是个hex后的base,先转bytes再base64


>>> bytes.fromhex('56455a7059574e7459584a685a576c756557646f59565637633364765a57647664474e305a48427965584e4464476c6c62584
e6c6257463165574e3059574e7a5647687a6247567a5a6d527a636d46796233427366513d3d')
b'VEZpYWNtYXJhZWlueWdoYVV7c3dvZWdvdGN0ZHByeXNDdGllbXNlbWF1eWN0YWNzVGhzbGVzZmRzcmFyb3BsfQ=='
>>> from base64 import *
>>> b64decode('VEZpYWNtYXJhZWlueWdoYVV7c3dvZWdvdGN0ZHByeXNDdGllbXNlbWF1eWN0YWNzVGhzbGVzZmRzcmFyb3BsfQ==')
b'TFiacmaraeinyghaU{swoegotctdprysCtiemsemauyctacsThslesfdsraropl}'

可以明显看出来头部TUCTF{所在的位置，再用栅栏

TUCTF{thisisawelcomemessagefromdatasecurityandcryptographyclass}

More Effort

RSA题，只给了p没有q,q是通过算法实现，不过这个运算次数并不大，不需要用智力处理


#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Nov 23 22:26:57 2022
@author: weiping
"""
 
import os
import random
from Crypto.Util.number import *
import gmpy2
 
flag = b'xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx'
 
class RSA():
	def __init__(self):
		self.p = getPrime(512)
		self.s = 0
		for i in range(1, 18000000):
			self.s += pow(i, self.p-2, self.p)  #sum i^-1
		self.s = self.s % self.p
		self.q = gmpy2.next_prime(self.s)
		self.n = self.p * self.q
		self.phi = (self.p - 1) * (self.q - 1)
		self.e = 65537
		self.d = pow(self.e, -1, self.phi)
 
	def encrypt(self, m: int):
		return pow(m, self.e, self.n)
 
 
def main():
    rsa = RSA()
    print(f"p = {rsa.p}")
    print(f"e = {rsa.e}")
    c = rsa.encrypt(bytes_to_long(flag))
    print('c = ', c)
    '''
    p = 11545307730112922786664290405312669819594345207377186481347514368962838475959085036399074594822885814719354871659183685801279739518405830244888530641898849
    e = 65537
    c =  114894293598203268417380013863687165686775727976061560608696207173455730179934925684529986102237419507146768083815607566149240438056135058988227916482404733131796310418493418060300571541865427288945087911872630289527954636816219365941817260989104786329938318143577075200571833575709614521758701838099810751
    '''
    
 
if __name__ == "__main__":
	main()

照着原来的样子，重复一遍拿到q就OK了，只是时间需要一点点，毕竟不是智力方法，纯暴力。


from Crypto.Util.number import *
from gmpy2 import next_prime,invert
 
p = 11545307730112922786664290405312669819594345207377186481347514368962838475959085036399074594822885814719354871659183685801279739518405830244888530641898849
e = 65537
c =  114894293598203268417380013863687165686775727976061560608696207173455730179934925684529986102237419507146768083815607566149240438056135058988227916482404733131796310418493418060300571541865427288945087911872630289527954636816219365941817260989104786329938318143577075200571833575709614521758701838099810751
 
s = 0
for i in range(1, 18000000):
	s += pow(i, -1, p)  #sum i^-1
	s = s % p
 
q = next_prime(s)
phi = (p-1)*(q-1)
d = invert(e, phi)
m = pow(c,d,p*q)
print(long_to_bytes(m))
#TUCTF{syqow82pam_%shsjQF; ^7dagsWCpsp_#aes}

不过后来一想，这个题并不需要求q，因为m很小可以直接用p求


d = invert(e,p-1)
long_to_bytes(pow(c,d,p))
#TUCTF{syqow82pam_%shsjQF; ^7dagsWCpsp_#aes}

Umix the flag

前边是类似凯撒密码的偏移转码，后边是替换


import string
 
upperFlag = string.ascii_uppercase[:26]
lowerFlag = string.ascii_lowercase[:26]
MIN_LETTER = ord("a")
MIN_CAPLETTER = ord("A")
 
def mix(oneLetter,num):
 
    if(oneLetter.isupper()):
        word = ord(oneLetter)-MIN_CAPLETTER
        shift = ord(num)-MIN_CAPLETTER
        return upperFlag[(word + shift)%len(upperFlag)]
    if(oneLetter.islower()):
        word = ord(oneLetter)-MIN_LETTER
        shift = ord(num)-MIN_LETTER
        return lowerFlag[(word + shift)%len(upperFlag)]
 
def puzzled(puzzle):
    toSolveOne = ""
    for letter in puzzle:
    
        if (letter.isupper()):
            binary ="{0:015b}".format(ord(letter))
 
            toSolveOne += upperFlag[int(binary[:5],2)]
            toSolveOne += upperFlag[int(binary[5:10],2)]
            toSolveOne += upperFlag[int(binary[10:],2)]
 
        elif(letter.islower()):
            six = "{0:02x}".format(ord(letter))
            toSolveOne += lowerFlag[int(six[:1],16)]
            toSolveOne += lowerFlag[int(six[1:],16)]
        elif(letter == "_"):
            toSolveOne += "CTF"  
    return toSolveOne   
 
    
flag = "Figure it Out! :)"
numShift = "??"
mixed = ""
 
assert all([x in lowerFlag for x in numShift])
assert len(numShift) == 1
 
encoding = puzzled(flag)
print(encoding)
for count, alpha in enumerate(encoding):
    mixed += mix(alpha, numShift)
 
print(mixed)

但是偏移这块用的都是num减，这导致大小写减后模26后偏移不同。

密文比较特殊，看上去是盲文，提示是8点不是6点，但用的都是6点，是盲文的标记方法不同。

⠱⠁ ⠹⠣ ⠱⠹ ⠱⠁ ⠹⠣ ⠹⠳ ⠱⠁ ⠹⠣ ⠹⠪ ⠱⠁ ⠹⠣ ⠱⠩ ⠹⠣ ⠱⠩ ⠹⠱ ⠻⠁ ⠫⠩ ⠫⠡ ⠻⠻ ⠹⠣ ⠱⠩ ⠹⠱ ⠻⠁ ⠻⠪ ⠻⠁ ⠻⠱ ⠫⠡ ⠻⠻ ⠫⠡ ⠫⠩

先来手搓，得到密文

5a42545a42485a42495a42534253457a6361774253457a797a7561776163

然后再进行偏移转码


#a = bytes.fromhex('5a42545a42485a42495a42534253457a6361774253457a797a7561776163').decode()
a = 'ZBTZBHZBIZBSBSEzcawBSEzyzuawac'
for i in range(ord('a'), ord('z')+1):
    tv = ''
    for v in a:
        if v.isupper():
            tv += chr((ord(v) - i)%26 + ord('A'))
        else:
            tv += chr((ord(v) - i)%26 + ord('a'))
    print(tv)
#ACUACIACJACTCTFgjhdCTFgfgbhdhj

后边继续手搓


'''
ACU ACI ACJ ACT _ gj hd _ gfgbhdhj
0,2,20 0,2,8  0,2,9 0,2,19 _ 后边转后是16进制 69 73  65617379
THIS_is_easy 
TUCTF{THIS_is_easy}
'''

Program

Leisurely math

这类题型原来没见过，不知道叫什么，作完应该就是叫编程。

只有远端，连上去发现是个简单算式要求输入结果。编个循环eval处理。快到出flag的时候会出个exec(...) ，这时候eval会把程序办掉（这是说eval不安全，前几天看别人的都是用另一个安全的替代），知道是啥了就好办，检查到直接跳过就行了。


from pwn import *
from ast import literal_eval
 
p = remote('chals.tuctf.com', 30202)
context.log_level = 'debug'
 
while True:
    a = p.recvline()
    if b'exec' in a:
        a = p.recvline()
    b = eval(a)
    p.sendlineafter(b'Answer: ', str(b).encode())
    p.recvline()
 
'''
[DEBUG] Received 0x1ff bytes:
    b'Correct!\n'
    b'exec(\'\\nimport os\\nscript_path = os.path.realpath( __file__ )\\nnew_program = ""\\nwith open( script_path, "r" ) as f:\\n    lines = f.readlines()\\n    for line in lines:\\n        for char in line:\\n            if char.isalpha():\\n                new_program += chr( ord( char ) + 1 )\\n            else:\\n                new_program += char\\nwith open( script_path, "w" ) as f:\\n    f.write( new_program )\\nos.system( "cls" )\\nos.system( "clear" )\\n\')\n'
    b'6993 - 5654 * 1086 + 6334 + 9502 - 954 + 126\n'
    b'Answer: '
[DEBUG] Sent 0x9 bytes:
    b'-6118243\n'
[DEBUG] Received 0x46 bytes:
    b'Correct!\n'
    b'5766 - 5967 - 4411 - 857 + 7792 + 5427 - 1369 + 7819\n'
    b'Answer: '
[DEBUG] Sent 0x6 bytes:
    b'14200\n'
[DEBUG] Received 0x52 bytes:
    b'Correct!\n'
    b'Here is your flag: TUCTF{7h4nk5_f0r_74k1n6_7h1n65_4_l177l3_5l0w_4268285}\n'
'''

Maze

第二道是个迷宫，走路的


'''
XOOOO###############
####OOOOOOOOOOOOOOO#
##################O#
##################O#
########OOOOOOOOOOO#
########OOOOOOO#####
##############O#####
##############OOO###
################O###
################O###
############OOOOO###
############OOOOOOOO
'''

没说要走到哪，手工试了一次是走到右下角。并且只有下左右3个方向可走，应该是要写遍历程序处理。其实10关和1000关没啥区别，可以也太漫长了，复杂度在加大，感觉题没个完，去了个超市回来才出。


from pwn import *
 
p = remote('chals.tuctf.com', 30204)
context.log_level = 'debug'
 
simple = '''
XOOOO###############
####OOOOOOOOOOOOOOO#
##################O#
##################O#
########OOOOOOOOOOO#
########OOOOOOO#####
##############O#####
##############OOO###
################O###
################O###
############OOOOO###
############OOOOOOOO
'''
ok = ''
width = 20
height = 11
def get_v(m,s, plc):
    global ok
    if ok != '':
        return False
    if plc == [height-1,width-1]:
        ok = s
        return True
    x,y = plc
    #print(1, s)
    if x1 and m[x+1][y] == 'O':
        ts = s+'V'
        tm = m.copy()
        tm[x][y] = '#'
        get_v(tm,ts,[x+1,y])
    #print(2, s)
    if y1 and m[x][y+1] == 'O':
        ts = s+'>'
        tm = m.copy()
        tm[x][y] = '#'
        get_v(tm,ts,[x,y+1])
    #print(3, s)
    if y>0 and m[x][y-1] == 'O':
        ts = s+'<'
        tm = m.copy()
        tm[x][y] = '#'
        get_v(tm,ts,[x,y-1])
    return 
 
def test():
    global ok
    a = simple.split('\n')[1:]
    m = [[0 for j in range(width)] for i in range(height)]
    for i in range(height):
        for j in range(width):
            m[i][j] = a[i][j]
    print(m)
    get_v(m,'',[0,0])
    print(ok)
 
#test()
 
def maze():
    global ok,width,height
    a = p.recvuntil(b'Move').decode().split('\n')[:-1]
    width = len(a[0])
    height = len(a)
    print(width,height, a)
    m = [[0 for i in range(width)] for j in range(height)]
    for i in range(height):
        for j in range(width):
            m[i][j] = a[i][j]
    
    ok = ''
    get_v(m,'',[0,0])
    print(f"{ok = }")
    for v in ok:
        p.sendlineafter(b': ', v)
 
p.recvuntil(b'down.\n')
while True:
    maze()
    p.recvuntil(b'level...\n\n')

p3 未完成

这个连上后跟第1个几乎一样，最后会出一堆英文表示的式子，写程序处理成算式，回答数字或者英文的都不对就卡住了。不知后事如何。


from pwn import *
from ast import literal_eval
import re
from num2words import num2words
 
_known = {
  'zero': 0,  'one': 1,  'two': 2,  'three': 3,  'four': 4,  'five': 5,
  'six': 6,  'seven': 7,  'eight': 8,  'nine': 9,  'ten': 10,  'eleven': 11,
  'twelve': 12,  'thirteen': 13,  'fourteen': 14,  'fifteen': 15,  'sixteen': 16,
  'seventeen': 17,  'eighteen': 18,  'nineteen': 19,  'twenty': 20,  'thirty': 30,
  'forty': 40,  'fifty': 50,  'sixty': 60,  'seventy': 70,  'eighty': 80,  'ninety': 90
  }
def spoken_word_to_number(n):
  n = n.lower().strip()
  if n in _known:
    return _known[n]
  else:
    inputWordArr = re.split('[ -]', n)
  assert len(inputWordArr) > 1 #all single words are known
  #Check the pathological case where hundred is at the end or thousand is at end
  if inputWordArr[-1] == 'hundred':
    inputWordArr.append('zero')
    inputWordArr.append('zero')
  if inputWordArr[-1] == 'thousand':
    inputWordArr.append('zero')
    inputWordArr.append('zero')
    inputWordArr.append('zero')
  if inputWordArr[0] == 'hundred':
    inputWordArr.insert(0, 'one')
  if inputWordArr[0] == 'thousand':
    inputWordArr.insert(0, 'one')
  inputWordArr = [word for word in inputWordArr if word not in ['and', 'minus', 'negative']]
  currentPosition = 'unit'
  prevPosition = None
  output = 0
  for word in reversed(inputWordArr):
    if currentPosition == 'unit':
      number = _known[word]
      output += number
      if number > 9:
        currentPosition = 'hundred'
      else:
        currentPosition = 'ten'
    elif currentPosition == 'ten':
      if word != 'hundred':
        number = _known[word]
        if number < 10:
          output += number*10
        else:
          output += number
      #else: nothing special
      currentPosition = 'hundred'
    elif currentPosition == 'hundred':
      if word not in [ 'hundred', 'thousand']:
        number = _known[word]
        output += number*100
        currentPosition = 'thousand'
      elif word == 'thousand':
        currentPosition = 'thousand'
      else:
        currentPosition = 'hundred'
    elif currentPosition == 'thousand':
      assert word != 'hundred'
      if word != 'thousand':
        number = _known[word]
        output += number*1000
    else:
      assert "Can't be here" == None
  return(output)
 
def aaa(a):
    arr = re.split('[ -]', a.replace(',',''))
    xx = {'plus':'+', 'minus':'*', 'negative':'-'}
    b = ''
    i = 0
    while i < len(arr)-1:
        if arr[i] in xx:
            b += xx[arr[i]]
            i+=1
        
        print(b)
       
        j = i 
        while j< len(arr):
            if arr[j] in xx:
                break 
            j+=1
        c = ' '.join(arr[i:j])
        print(c)
        b += str(spoken_word_to_number(c))
        i = j 
        print(b)
    return b
 
p = remote('chals.tuctf.com', 30200)
context.log_level = 'debug'
 
while True:
    a = p.recvline()
    while b'Correct!' in a or b'exec' in a or a == b'\n':
        a = p.recvline()
    open('log.txt', 'ab').write(a)
    if b'plus' in a or b'minus' in a :
        b = eval(aaa(a.decode().strip()))
        b = num2words(int(b)).encode()
    else:
        b = eval(a)
        b = str(int(b)).encode()
    p.sendlineafter(b'Answer: ', b)
    #p.recvline()

相关阅读:
网页排版和代码优化，求带我看看代码
 vue3热更新后导致组件执行两次的bug
一套轻量、安全的问卷系统基座，提供面向个人和企业的一站式产品级解决方案
 Linux知识点 -- Linux多线程（四）
大数据-玩转数据-Flink状态后端（下）
“动捕设备+飞兔渲染软件”，激发数字人短视频营销新动力
 数据分析与Excel
红海云签约广州无线电集团，头部国企人力资源数字化转型全面加速
 【脚本】生成已划分好训练集、验证集、测试集的数据集对应的train.txt、val.txt、test.txt【包含图像的绝对地址】
类继承关系梳理
原文地址：https://blog.csdn.net/weixin_52640415/article/details/128187318