title: buu-re-简单注册器
    tags:

    • CTF
    • RE
    • buu
      abbrlink: 55aa4ca9
      date: 2021-03-10 20:59:15

    拿到题目后，是一个APK文件，使用jdx反编译后在mainActivity里面发现了一段核心代码

    1.                 if (flag == 1) {
    2.                     char[] x = "dd2940c04462b4dd7c450528835cca15".toCharArray();
    3.                     x[2] = (char) ((x[2] + x[3]) - 50);
    4.                     x[4] = (char) ((x[2] + x[5]) - 48);
    5.                     x[30] = (char) ((x[31] + x[9]) - 48);
    6.                     x[14] = (char) ((x[27] + x[28]) - 97);
    7.                     for (int i = 0; i < 16; i++) {
    8.                         char a = x[31 - i];
    9.                         x[31 - i] = x[i];
    10.                         x[i] = a;
    11.                     }
    12.                     textview.setText("flag{" + String.valueOf(x) + "}");
    13.                     return;
    14.                 }

    然后将此段用C程序写出来即可得到答案

    1. #include<iostream>
    3. using namespace std;
    5. int main()
    6. {
    7.     char x[] = "dd2940c04462b4dd7c450528835cca15";
    8.                     x[2] = (char) ((x[2] + x[3]) - 50);
    9.                     x[4] = (char) ((x[2] + x[5]) - 48);
    10.                     x[30] = (char) ((x[31] + x[9]) - 48);
    11.                     x[14] = (char) ((x[27] + x[28]) - 97);
    12.                     for (int i = 0; i < 16; i++) {
    13.                         char a = x[31 - i];
    14.                         x[31 - i] = x[i];
    15.                         x[i] = a;
    16.                     }
    17.     cout<<x<<endl;
    18.     return 0;
    19. }

    求得flag：59acc538825054c7de4b26440c0999dd