xuenixiang_2020_re_Reverse6

roger

载入OD

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xuenixiang_2020_re_Reverse6

随意输入1111111111111111111

xuenixiang_2020_re_Reverse6

xuenixiang_2020_re_Reverse6

发现进行了base64加密

再向下单步 发现、

xuenixiang_2020_re_Reverse6

结合IDA看下

xuenixiang_2020_re_Reverse6

int sub_4156E0()

{

size_t v0;// eax@6

constchar*v1;// eax@6

size_t v2;// eax@9

char v4;// [sp+0h] [bp-188h]@6

char v5;// [sp+Ch] [bp-17Ch]@1

size_t v6;// [sp+10h] [bp-178h]@3

size_t j;// [sp+DCh] [bp-ACh]@6

size_t i;// [sp+E8h] [bp-A0h]@1

char Dest[108];// [sp+F4h] [bp-94h]@5

char Str;// [sp+160h] [bp-28h]@6

char v11;// [sp+17Ch] [bp-Ch]@6

unsignedint v12;// [sp+184h] [bp-4h]@1

int savedregs;// [sp+188h] [bp+0h]@1

 

memset(&v5,0xCCu,0x17Cu);

v12 =(unsignedint)&savedregs ^ __security_cookie;

for( i =0;(signedint)i <100;++i )

{

v6 = i;

if( i >=0x64)

sub_411154();

Dest[v6]=0;

}

sub_41132F("please enter the flag:", v4);

sub_411375("%20s",(unsignedint)&Str);

v0 = j_strlen(&Str);

v1 =(constchar*)sub_4110BE(&Str, v0,&v11);

strncpy(Dest, v1,'(');

sub_411127();

i = j_strlen(Dest);

for( j =0;(signedint)j <(signedint)i;++j )

Dest[j]+= j;

v2 = j_strlen(Dest);

strncmp(Dest, Str2, v2);

if( sub_411127())

sub_41132F("wrong flag!\n", v4);

else

sub_41132F("rigth flag!\n", v4);

sub_41126C(&savedregs,&dword_415890);

sub_411280();

return sub_411127();

}

 

分析可知:将输入的串Str1先进行base64加密 再与串Str2比较 若相等 则输出"right flag"

由此，我们只需将Str2也就是"e3nifIH9b_C@n@dH"进行解密即可



Python脚本：

import base64

s ="e3nifIH9b_C@n@dH"

flag =""

for i in range(len(s)):

flag += chr(ord(s<i>)- i)

flag = base64.b64decode(flag)

print(flag)
</i>

xuenixiang_2020_re_Reverse6

所以得到flag{i_l0ve_you}