南京邮电大学CTF密码学部分Writeup-白红宇

南京邮电大学CTF密码学部分Writeup

阅读量：5240 次

发布时间：2019-06-14

本文共 3505 字，大约阅读时间需要 11 分钟。

异性相吸

1.xor
2.hex2binary
3.len(bin(miwen))==len(bin(mingwen))

# -*- coding:utf-8 -*-file_de = open('decode1.txt')file_en = open('ene.txt')de = file_de.read()en = file_en.read()s = ''for i, j in zip(de, en):    s += chr(ord(i) ^ ord(j))print sfile_de.close()file_en.close()

MD5

明文:TASC?O3RJMV?WDJKX?ZM
MD5 密文：e9032???da???08????911513?0???a2

查了下 Python MD5的库，发现是内置了一个模块的：

暴力猜明文缺失字符。

import hashlibs1 = "TASC"s2 = "O3RJMV"s3 = "WDJKX"s4 = "ZM"dict = ['0','1','2','3','4','5','6','7','8','9','0','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']for i in range(36):    for j in range(36):        for k in range(36):            src = s1 + dict[i] + s2 + dict[j] + s3 + dict[k] + s4            pre = hashlib.md5()            pre.update(src)            md5 = str(pre.hexdigest())            if ('e9032' and 'da' and '911513') in md5 :                print md5,'  ',src

easy&base64&n次base64&mixed_base64

这几道题都是一个类型的……

import base64with open('string.txt') as f:        s = f.read().replace('\n','')while 1:        try:                s = base64.b16decode(s)        except:                try:                     s = base64.b32decode(s)                except:                        s = base64.b64decode(s)        if 'nctf' in s:                print s                break

KeyBoard

看键盘画图形，不过我没做对，想象力拙计……

骚年来一发吗

密文

iEJqak3pjIaZ0NzLiITLwWTqzqGAtW2oyOTq1A3pzqas

function encode($str){    $_o = strrev($str);    for($_0 = 0; $_0 < strlen($_o); $_0++){        $_c = substr($_o, $_0, 1);        $__ = ord($_c) + 1;        $_c = chr($__);        $_ = $_.$_c;    }    return str_rot13(strrev(base64_encode($_)));

没学过这种世界上最好的语言。去搜索了下 $ 貌似是“xx变量”的意思，暂且当成 CPP 来看了……然后直接搜一些函数语法：

substr() 返回字符串strrev(string)     反转字符串ord(string)   返回字符串中第一个字符的 ASCII 值chr(ascii)  从指定 ASCII 值返回字符(.)         PHP 并置运算符,将两个字符串连接起来

执行完函数后得到一个字符串 $_

然后将这个字符串 base64 再反转再 ROT13 。

,不过神奇的是，PHP 竟然自带 rot13 的函数。

解密的时候逆过来操作就好了。

我感到了智商缺陷，最开始在逆向操作 return那一行时从内层到外层写的……

import codecsimport base64with open('rot13decode.txt') as f:    s = f.read().replace('\n', '')flag = ""s = codecs.encode( s,"rot13" )s = s[::-1]s = base64.b64decode(s)for i in s:    s = s + chr(ord(i) - 1)s = s[::-1]print s

Vigenere

It is said that Vigenere cipher does not achieve the perfect secrecy actually :-)
Tips:
1.The encode pragram is given;
2.Do u no index of coincidence ？
3.The key is last 6 words of the plain text(with "nctf{}" when submitted, also without any interpunction)

code is here(without '
'):
F96DE8C227A259C87EE1DA2AED57C93FE5DA36ED4EC87EF2C63AAE5B9A7EFFD673BE4ACF7BE8923CAB1ECE7AF2
DA3DA44FCF7AE29235A24C963FF0DF3CA3599A70E5DA36BF1ECE77F8DC34BE129A6CF4D126BF5B9A7CFEDF3EB8
50D37CF0C63AA2509A76FF9227A55B9A6FE3D720A850D97AB1DD35ED5FCE6BF0D138A84CC931B1F121B44ECE70F
6C032BD56C33FF9D320ED5CDF7AFF9226BE5BDE3FF7DD21ED56CF71F5C036A94D963FF8D473A351CE3FE5DA3CB
84DDB71F5C17FED51DC3FE8D732BF4D963FF3C727ED4AC87EF5DB27A451D47EFD9230BF47CA6BFEC12ABE4ADF7
2E29224A84CDF3FF5D720A459D47AF59232A35A9A7AE7D33FB85FCE7AF5923AA31EDB3FF7D33ABF52C33FF0D673A
551D93FFCD33DA35BC831B1F43CBF1EDF67F0DF23A15B963FE5DA36ED68D378F4DC36BF5B9A7AFFD121B44ECE76
FEDC73BE5DD27AFCD773BA5FC93FE5DA3CB859D26BB1C63CED5CDF3FE2D730B84CDF3FF7DD21ED5ADF7CF0D63
6BE1EDB79E5D721ED57CE3FE6D320ED57D469F4DC27A85A963FF3C727ED49DF3FFFDD24ED55D470E69E73AC50DE
3FE5DA3ABE1EDF67F4C030A44DDF3FF5D73EA250C96BE3D327A84D963FE5DA32B91ED36BB1D132A31ED87AB1D021
A255DF71B1C436BF479A7AF0C13AA14794