ctf 常用python脚本及在线网站
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ctf
脚本
图片爆破宽高等脚本在ctf misc 图片题知识点中
二进制转二维码
运行之前先改好图片边长，即MAX的值，例如二进制字符串长度为625(25*25)，这里就改成25
import PIL
from PIL import Image
MAX = 25 #图片边长
img = Image.new("RGB",(MAX,MAX))
str='1111111001110010001111111100000100001111010100000110111010011100010010111011011101010111100001011101101110101010101000101110110000010011000101010000011111111010101010101111111000000000100000110000000011000111011101101000110000001000010110010010010100010011110100001110111001100111101001010110010010011000001001100001001101000111100011111101110010100010110111110011011111101111000110110010010101101100100011110011111111111011100000000101100011000101001111111010010100101010001100000101010101010001100110111010001001111111100101011101000011001011110111101110100100110010010000110000010110000110110110011111111011010000101110101'
i = 0
for y in range (0,MAX):
for x in range (0,MAX):
if(str[i] == '1'):
img.putpixel([x,y],(0, 0, 0))
else:
img.putpixel([x,y],(255,255,255))
i = i+1
img.show()
img.save("flag.png")
base64
base64异或
import base64
s='TkVLTFdUQVpvUlNda1ZXRUpAZVldTltgJCQhLCAgGSknPjc='
s=base64.b64decode(s)
for i in range(256):
flag=""
k=0
for j in s:
res=j^(k+i)
flag+=chr(res)
k+=1
print(i,flag)
base64隐写加密(py2)
# -*- coding: cp936 -*-
import base64
flag = 'Tr0yBase64isF4n' # flag
bin_str = ''.join([bin(ord(c)).replace('0b', '').zfill(8) for c in flag])
base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
with open('0.txt', 'rb') as f0, open('1.txt', 'wb') as f1: # '0.txt'是明文, '1.txt'用于存放隐写后的base64
for line in f0.readlines():
rowstr = base64.b64encode(line.replace('\\n', ''))
equalnum = rowstr.count('=')
if equalnum and len(bin_str):
offset = int('0b' + bin_str[:equalnum * 2], 2)
char = rowstr[len(rowstr) - equalnum - 1]
rowstr = rowstr.replace(char, base64chars[base64chars.index(char) + offset])
bin_str = bin_str[equalnum * 2:]
f1.write(rowstr + '\\n')
base64隐写解密(py2)
# -*- coding: cp936 -*-
b64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
#https://tr0y.wang/2017/06/14/Base64steg/
with open('1.txt', 'rb') as f:
bin_str = ''
for line in f.readlines():
stegb64 = ''.join(line.split())
rowb64 = ''.join(stegb64.decode('base64').encode('base64').split())
offset = abs(b64chars.index(stegb64.replace('=', '')[-1]) - b64chars.index(rowb64.replace('=', '')[-1]))
equalnum = stegb64.count('=') # no equalnum no offset
if equalnum:
bin_str += bin(offset)[2:].zfill(equalnum * 2)
print ''.join([chr(int(bin_str[i:i + 8], 2)) for i in xrange(0, len(bin_str), 8)]) # 8位一组
1-5位的crc32碰撞
6位的碰撞已经有现成的工具了
# coding:utf-8
"""
Author：spaceman
"""
import binascii
import string
from time import sleep
def is_number(s):
try:
float(s)
return True
except ValueError:
pass
try:
import unicodedata
unicodedata.numeric(s)
return True
except (TypeError, ValueError):
pass
return False
# 进度条
def progress(percent=0, width=40):
left = width * percent // 95
right = width - left
print ('\\r[', '#' * left, ' ' * right, ']',f' percent:.0f%',sep='', end='', flush=True)
# 一位字节
def crc1(strs,dic):
strs = hex(int(strs,16))
rs = ''
for i in dic:
s = i
if hex(binascii.crc32(s.encode())) == strs:
rs += s
print (strs+' ： '+s)
return rs
# 两位字节
def crc2(strs,dic):
strs = hex(int(strs,16))
rs = ''
for i in dic:
for j in dic:
s = i + j
if hex(binascii.crc32(s.encode())) == strs:
rs += s
print (strs+' ： '+s)
return rs
# 三位字节
def crc3(strs,dic):
strs = hex(int(strs,16))
rs = ''
for i in dic:
for j in dic:
for k in dic:
s = i+j+k
if hex(binascii.crc32(s.encode())) == strs:
rs += s
print (strs+' ： '+s)
return rs
# 四位字节
def crc4(strs,dic):
strs = hex(int(strs,16))
rs = ''
it = 1
for i in dic:
for j in dic:
for k in dic:
for m in dic:
s = i+j+k+m
if hex(binascii.crc32(s.encode())) == strs:
rs += s
print ()
print (strs+' ： '+s)
print ('\\n')
progress(it)
sleep(0.1)
it += 1
return rs
# 五位字节
def crc5(strs,dic):
strs = hex(int(strs,16))
rs = ''
it = 1
for i in dic:
progress(it)
for j in dic:
for k in dic:
for m in dic:
for n in dic:
s = i+j+k+m+n
if hex(binascii.crc32(s.encode())) == strs:
rs += s
print ()
print (strs+' ： '+s)
print ('\\n')
sleep(0.1)
it += 1
return rs
# 计算碰撞 crc
def CrackCrc(crclist,length):
print ()
print ("正在计算...")
print ()
dic = ''' !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\\\]^_`abcdefghijklmnopqrstuvwxyz|~''' # 碰撞需要的字符字典
dic = dic[::-1]
text = ''
for i in crclist:
if length == '1':
text += crc1(i,dic)
if length == '2':
text += crc2(i,dic)
if length == '3':
text += crc3(i,dic)
if length == '4':
text += crc4(i,dic)
if length == '5':
text += crc5(i,dic)
print ('\\n')
if text == '':
print ("碰撞失败,无结果")
exit()
print ("字符顺序组合：",end=' ')
print ()
print (text)
print ()
input("回车确认结束程序...")
# 主函数
print ('''
##############################
###### Author：spaceman ######
### Thank you for your use ###
##############################
''')
listcrc = [] # 用于存储crc值
length = (input("请输入文本字节大小(1-5)：")) # 即文本内容大小，如文本内容为flag，大小即为4
if is_number(length) == False or length not in ("1,2,3,4,5"):
exit("非指定数字，退出")
print ()
while 1:
crc = input('请输入crc值(例如:d1f4eb9a,输入n完成输入):')
if crc == 'n':
break
crc = '0x'+crc
if len(crc) != 10:
print ("rcr长度不对,请重新输入")
continue
listcrc.append(crc)
CrackCrc(listcrc,length)
将十进制数写入文件中
x=[55,122,188,175,175]
f=b''
for i in x:
f += i.to_bytes(1,'big')
out=open("1.7z","wb")
out.write(f)
usb流量
键盘流量
先tshark导出数据
┌──(volcano㉿kali)-[~/桌面]
└─$ tshark -r whereiskey.pcapng -T fields -e usb.capdata | sed ‘/^\\s*$/d’ > out.txt
然后每两个数字之间加一个冒号
f=open('1.txt','r')
fi=open('out.txt','w')
while 1:
a=f.readline().strip()
if a:
if len(a)==16: # 鼠标流量的话len改为8
out=''
for i in range(0,len(a),2):
if i+2 != len(a):
out+=a[i]+a[i+1]+":"
else:
out+=a[i]+a[i+1]
fi.write(out)
fi.write('\\n')
else:
break
fi.close()
再跑脚本
normalKeys = "04":"a", "05":"b", "06":"c", "07":"d", "08":"e", "09":"f", "0a":"g", "0b":"h", "0c":"i", "0d":"j", "0e":"k", "0f":"l", "10":"m", "11":"n", "12":"o", "13":"p", "14":"q", "15":"r", "16":"s", "17":"t", "18":"u", "19":"v", "1a":"w", "1b":"x", "1c":"y", "1d":"z","1e":"1", "1f":"2", "20":"3", "21":"4", "22":"5", "23":"6","24":"7","25":"8","26":"9","27":"0","28":"<RET>","29":"<ESC>","2a":"<DEL>", "2b":"\\t","2c":"<SPACE>","2d":"-","2e":"=","2f":"[","30":"]","31":"\\\\","32":"<NON>","33":";","34":"'","35":"<GA>","36":",","37":".","38":"/","39":"<CAP>","3a":"<F1>","3b":"<F2>", "3c":"<F3>","3d":"<F4>","3e":"<F5>","3f":"<F6>","40":"<F7>","41":"<F8>","42":"<F9>","43":"<F10>","44":"<F11>","45":"<F12>"
shiftKeys = "04":"A", "05":"B", "06":"C", "07":"D", "08":"E", "09":"F", "0a":"G", "0b":"H", "0c":"I", "0d":"J", "0e":"K", "0f":"L", "10":"M", "11":"N", "12":"O", "13":"P", "14":"Q", "15":实验吧CTF密码学Writeup-奇妙的音乐Writeup