此篇用来记录我在jarivsOJ上的一些题解，只给解题思路，不放flag

Misc

0x01 You Need Python(300)

题目有两个文件，一个py文件，另一个是经过编码的key

文件key

首先看到提示key_is_here_but_do_you_know_rfc4042，baidu一下了解到是utf9编码，于是将这个文件解码，得到一串字符（md里对下划线要转义才能正常显示，偷个懒所以我这里贴出的字符是没转义过的，请不要直接拿去进行下一步的操作以免出错）

_____*((//++_-_%___)**((%(-))+_______+(%+_____+_______%+_-(_//(_%_)))))+__*(((________/)+_%+_____-(________//____))**((_____+_____)+_______+_________%))+_____(((________//+______%)+(_____-_))**((+_)+______-(______//)))+_____*((+______-(______//-_%%))**(_____+_____+_____))+__*(+_______-(//-_________%_____%__))**(_________-____+_______)+(+____)**(________%%+_+_)+(_-)*((____//____-_____%____%)+________)**(_____-(_______//_______+_________%)+_)+(_+(_______%_______)*+)_________+_______(((_________%_______)_+______-(________//________))_)+(_____/)(((____-+______)(_+_))**)+*((+_______-_)**_____)+___*(((+_-_/_+-_______%_____%__)*(-+______/+_______%_____))**__)+(//)*(((________%%+_+_)%_____)+_______-_)**+__*((______/(_____%))+_)((_________%_______)+_+)+_//+______+_________/___

对下划线计数，运算符正常运算可以得到5287002131074331513，一开始以为这就是真正的key，然后死活解不出来，后来尝试ascii解码得到I_4m-k3y，这才是正确的key

文件flag.py

marshal.loads是反序列化操作，用uncompyle2即可解决。

脚本如下

#参考链接：https://www.aliyun.com/jiaocheng/475933.htmlimport uncompyle2import marshal, zlib, base64co = marshal.loads(zlib.decompress(base64.b64decode('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')))f = open('test','w')uncompyle2.uncompyle('2.7.3',co,f)

拿到源码之后就很好做了，运行源码可以知道我们需要输入key和flag，然后经过encrypt函数的加密和cipherText比较，而key我们已经拿到了，简单分析一下encrypt函数可以发现，连爆破都不用，直接赋值就好了。。

解题脚本如下

#!/usr/bin/python# -*- coding: utf-8 -*-__Author__ = "LB@10.0.0.55"import utf9import hashlibfrom libnum import n2sfrom string import printabledef get_key():    a = open('key','r').read()    print type(a)    key = utf9.utf9decode(a)    print key    exp = ''    num = 0    for i in key:        if i == '_':            num += 1        elif num != 0:            exp += str(num) + i            num = 0        else:            exp += i    exp += str(num)    return n2s(eval(exp))def sha1(string):    return hashlib.sha1(string).hexdigest()def calc(strSHA1):    r = 0    for i in strSHA1:        r += int('0x%s' % i, 16)    return rdef decrypt(key):    print key    keySHA1 = sha1(key)    print keySHA1    intSHA1 = calc(keySHA1)    print intSHA1    r = ''    flag = '-185-147-211-221-164-217-188-169-205-174-211-225-191-234-148-199-198-253-175-157-222-135-240-229-201-154-178-187-244-183-212-222-164'    flag = flag.split('-')[1:]    flag = [ eval('-'+i) for i in flag ]    print len(flag),flag    for i in range(33):        r += chr(-int('0x%s' % keySHA1[i % 40], 16)+intSHA1+flag[i])        intSHA1 = calc(sha1(r[:i + 1])[:20] + sha1(str(intSHA1))[:20])    print len(r),rdef main():        key = get_key()    decrypt(str(key))    main()

Crypto

Reverse

Pwn

作者： LB919

出处：http://www.cnblogs.com/L1B0/

如有转载，荣幸之至！请随手标明出处；