看雪题库_Reverse_马到成功
# Pragyan CTF 2015: Fast and Furious**Category:** Android
**Points:** 100
**Solves** 69
**Description:**
> (fast.apk)
## Write-up
by (https://github.com/abpolym)
After decompiling the `fast.apk` application file, we detect a hardcoded string, `65544231587a52794d3138316458417a636c396d4e44553343673d3d`, in `Main.java`.
Converting this Hex string to an ASCII string, we get `eTB1XzRyM181dXAzcl9mNDU3Cg=`, which is a base64 encoded string.
Base64 Decoding this string results in the flag:
```bash
$ base64 --decode <<< 'eTB1XzRyM181dXAzcl9mNDU3Cg='
y0u_4r3_5up3r_f457
```
The flag is `y0u_4r3_5up3r_f457`.
## Other write-ups and resources
* <https://ctf-team.vulnhub.com/pragyan-ctf-2015-android/>
# 注册
public static String a(String paramString) {
StringBuilder stringBuilder = new StringBuilder();
for (byte b = 0; b < paramString.length(); b += 2)
stringBuilder.append((char)Integer.parseInt(paramString.substring(b, b + 2), 16));
return stringBuilder.toString();
}
public static void main(String[] args) {
System.out.println("Hello JAVA");
String str = a("65544231587a52794d3138316458417a636c396d4e44553343673d3d");
System.out.println(str);
}
返回值是：
eTB1XzRyM181dXAzcl9mNDU3Cg==
解密这串base64就是了
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