BUGKU (Take the maze)

首先进行查壳，没有壳。

随便输入，看程序执行信息。随意输入字符串，提示key error

放到IDA中打开，在左侧函数窗口中找到main0，F5反编译，进行分析。具体已在分析在图中标识。

关于main函数的逻辑是很容易理解的，但是接下来可就犯难了。我首先是打开45C748（对Str进行变换的函数）。如下图，看起来应该是在进行对Str变换之前，做一些初始化工作。

整个过程不难理解，但当我打开45DCD3函数时，就一头雾水了。

这没得分析（后来才知道是VM处理过的），于是打开OD输入012345678901234567891234，试试看（之前做过一个逆向题，印象深刻，那也是对字符串进行变换的函数，再IDA中查看，根本看不明白，拿到OD里面一跑，才知道，原来就是个base64编码啊。╮(╯▽╰)╭）经过变换，如下图所示。

我做过的逆向题不多，根据以往的经验，在IDA中看不懂，或者是不是很理解的情况下，在OD中跟进调试一下，往往会有所收获。所以就跟进了这个函数。结果，不会就是不会，一点办法都没有。╮（╯＿╰）╭。之后，也是在调试的时候偶然把上图的字符串（即0000000000::>>::'&**%'%#）当作输入，进行调试，结果发现返回的竟是012345678901234567891234。当时没有细想，没感觉有什么大不了的。后来才猛然间醒悟，如果知道了正确的输入，先输入进去，得到加密后的字符串，这不就是flag了吗？于是有了往下做下去的动力。

接着分析。

 再来打开其中的45CC4D函数。（其中的注释是后来看了别人的WP加的）。

之前做过类似的迷宫逆向题，知道得有一张“地图”，可找了半天也没发现。再者这个函数里面套函数，这种return，实在绕的头疼，最后还是看了别人的WP，找了找思路。原来得在if语句这里，写一个IDC内置脚本（第一次接触）。

这个脚本很容易理解。然后的话，四个函数里面的数组都是不一样的，然后一个个把Dword中的地址改掉，dump出“地图”来就行了。

auto i;
for(i=0;i<26*12;i++){
    if(Dword(0x540548+4*i)^Dword(0x540068+4*i))
    Message("0,");
    else Message("1,");
}

我用的python,它处理字符串不能够直接修改某个位置上的值，所以用C写会舒服一些。

up='0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0'.replace(',','')
down='1,1,0,1,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0'.replace(',','')
left='0,1,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1'.replace(',','')
right='0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0'.replace(',','')
tmp='~`-.'
for i in range(len(up)):
    if i%26==0:
        print '\n'
    if up[i]=='1':
        tmp=tmp.replace('~','U')
    if down[i]=='1':
        tmp=tmp.replace('`','D')
    if left[i]=='1':
        tmp=tmp.replace('-','L')
    if right[i]=='1':
        tmp=tmp.replace('.','R')
    print tmp+' ',
    tmp='~`-.'

最后在画图里面打开对应着地图，把路线画出来。

 

0-d
2-l
3-r
4-u
06260826062b0829072e0629

然后按照之前的想法，把06260826062b0829072e0629输进去，得到加密后的字符串，这个字符串再重新输入回去，成功，如下图。

给了张二维码，说是最终flag要加作者的名字。到此结束。

 最后感觉这个题，自己感觉还是挺有趣的，其实我如果之前接触过IDC的话，不看别人的WP也会自己做出来的。