buu-re-findit

本文最后更新于:2023年11月8日 中午

拿到手是一个apk文件,利用Android Killer打开后查看mainActivity里面的相关信息,可以看到一串十六进制数字:0x70,0x76,0x6b,0x71,0x7b,0x6d,0x31,0x36,0x34,0x36,0x37,0x35,0x32,0x36,0x32,0x30,0x33,0x33,0x6c,0x34,0x6d,0x34,0x39,0x6c,0x6e,0x70,0x37,0x70,0x39,0x6d,0x6e,0x6b,0x32 ,0x38,0x6b,0x37,0x35,0x7d。写了一个C语言的小程序来把这串数字转换为字符串

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#include<iostream>
#include<vector>

using namespace std;

int main()
{
vector<int> flag={0x70,0x76,0x6b,0x71,0x7b,0x6d,0x31,0x36,0x34,0x36,0x37,0x35,0x32,0x36,0x32,0x30,0x33,0x33,0x6c,0x34,0x6d,0x34,0x39,0x6c,0x6e,0x70,0x37,0x70,0x39,0x6d,0x6e,0x6b,0x32 ,0x38,0x6b,0x37,0x35,0x7d};
char s[100];
for(int i =0; i<flag.size();i++)
{
s[i]=char(flag[i]);
}
cout<<s<<endl;
return 0;
}

得到结果pvkq{m164675262033l4m49lnp7p9mnk28k75},看符号判断应该是被加密过,使用凯撒解密得到结果flag{c164675262033b4c49bdf7f9cda28a75}


buu-re-findit
https://www.0error.net/2021/03/3d2fcb0e.html
作者
Jiajun Chen
发布于
2021年3月10日
许可协议