Rating:
# Starbucks | Reverse Engineering
### Solution
A quick look at the file tells us that this is compiled java file, so I found [decompiler online](http://www.javadecompilers.com/). After decompilation I have got this code:
```java
//
// Decompiled by Procyon v0.5.36
//
public class Challenge {
public static String f1(final String s) {
final StringBuilder b = new StringBuilder();
final char[] arr = s.toCharArray();
for (int i = 0; i < arr.length; ++i) {
b.append((char)(arr[i] + i));
}
return b.toString();
}
public static String f1_rev(final String s) {
final StringBuilder b = new StringBuilder();
final char[] arr = s.toCharArray();
for (int i = 0; i < arr.length; ++i) {
b.append((char)(arr[i] - i));
}
return b.toString();
}
public static String f2(final String s) {
final int half = s.length() / 2;
return String.valueOf(s.substring(half + 1)) + s.substring(0, half + 1);
}
public static String f3() {
return f1(f2("$aQ\"cNP `_\u001d[eULB@PA'thpj]"));
}
public static void main(final String[] args) {
System.out.println("You really thought finding the flag would be so easy?");
}
}
```
I modified code to execute 'f3()' function and ended up with:
```java
public static void main(final String[] args) {
System.out.println(f3());
}
```
Because I didn't have java compiler, I found this [online](https://www.online-java.com/)
### Flag
UMDCTF-{pyth0n_1s_b3tt3r}
#### Credits
- Writeup by [HuntClauss](https://ctftime.org/user/106464)
- Solved by [HuntClauss](https://ctftime.org/user/106464)
- WaletSec 2021
#### License
**CC BY 4.0** WaletSec + HuntClauss
