# 2017 XCTF\&NJCTF Easy Crack 100

2017XCTF南京站线上赛的题目，同时也是2017NJCTF的题目，可以在XCTF OJ上面找到，分值100

* <http://oj.xctf.org.cn/practice/defensetrain/6e65e388244c434ca9b6d7c3f1e5aa81/>

在Java层没有太多逻辑，对输入框进行监听，当有输入的时候调用native层的函数进行校验

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-1a7a6d2fcceb55e9dbb43479fd79ad2b48ef0cf5%2FF98999D6B1428310EACF038F94577943.png?alt=media)

IDA载入so，没有动态注册，直接分析`parseText()`函数

前面会调用Java层的`meaasgeMe()`函数

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-79bcd41dd111d1928f8de49e6078fcd7c490e738%2F3D63A01102C69B29CE19185049096419.png?alt=media)

而`messageMe"()`函数的功能是获取包名的最后一个串

```java
public String messageMe() {
    String result = "";
    int t = 51;
    String[] paname = getApplicationContext().getPackageName().split("\\.");
    for (char ch : paname[paname.length - 1].toCharArray()) {
        t ^= ch;
        result = result + ((char) t);
    }
    return result;
}
```

写个代码跑

```java
public class Main {
	
	public static void main(String[] args) throws Exception{
		String result = "";
        int t = 51;
        String[] paname = "com.njctf.mobile.easycrack".split("\\.");
        for (char ch : paname[paname.length - 1].toCharArray()) {
        	//System.out.println(ch);
            t ^= ch;
            result = result + ((char) t);
        }
        System.out.println(result);
	}
}
```

结果输出

```
V7D=^,M.E
```

然后这个传入native层，转为char数组

接着利用这个字符串对输入的字符串进行循环异或，就是把这个串循环复制自己，一个接一个，然后跟输入串长度一样，进行异或

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-54c46efa8b1aa006ac039099483e64e9a9d3976d%2F706389D1E6382F6CD566664EC5D635D4.png?alt=media)

异或后存在新malloc出来的空间，首地址指针`pvInput_new`

然后是一个key，这里有的同学如果没有经验的话可能会看不出来这是RC4加密的key

我看出这里是RC4加密的思路可以分享一下：

首先是这个key，如果你那里是三个很奇怪的数字，可以右键然后看`R`，就会和我这里一样的显示了

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-375666c271780983fd9a68ce0801bdf224ea2990%2F96A80408B7EBF211ECE2F5E56457FA95.png?alt=media)

加下来有两个`j_j_j__`开头的函数，如果没有看出这里是RC4加密的秘钥初始化的话可以跟进去

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-be5485a33df9bc7c25fc7781f07b82a584c7f2f6%2F6BF8448EF666E96DF16865D212403A86.png?alt=media)

跟到最里面的实现，RC4的秘钥初始化有比较明显的特征

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-b2f8b0643b72aea055f111bdff7b91f8331653d5%2F1EC36A71863125C73FBA54AF109349BB.png?alt=media)

然后是后面的加密过程

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-b4f724a7b36bcd0411a1a8b8bb3b21b43e413156%2F9EA8FB41B430B4AE83294280AC2331B8.png?alt=media)

跟进去，到最里面的实现过程，很明显的RC4加密

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-9fab1f17af4ee88ee15ac34ca2b63eaf7abc01ab%2F34CEAB44A73C163D8D9205FB24C82B62.png?alt=media)

最后的逻辑非常清楚，就是拷贝RC4加密后的数据然后跟一个字符串对比

![](https://2646860265-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FJh2iv6UDGq4O16NDjFK8%2Fuploads%2Fgit-blob-1f7cb9b23a633de17ac79f80ed0e4616b4275723%2FA9CD3A70E14735A3D304BC3BDFF8CEC8.png?alt=media)

所以到这里我们整理出流程

* 先输入数据，传入native层
* native层获取Java层`messageMe()`函数，根据包名获取到一个异或后的9字节字符串
* 循环使用异或后的字符串跟输入的字符串进行异或
* 将异或后的数据进行RC4加密，然后跟`compare`字符串对比，一样则校验通过

写个Python脚本来还原

```python
# coding = utf-8

import binascii

def main():
	key_rc4 = "I_am_the_key"
	rc4_encrypted = binascii.a2b_hex("C8E4EF0E4DCCA683088134F8635E970EEAD9E277F314869F7EF5198A2AA4")
	
	# Init rc4_key
	j = 0;
	s = range(256)
	for i in range(256):
		j = (j + s[i] + ord(key_rc4[i % len(key_rc4)])) % 256
		s[i], s[j] = s[j], s[i]
	i = 0
	j = 0
	
	# Rc4 decryption
	rc4_decrypted = []
	for data in rc4_encrypted:
		i = (i + 1) % 256
		j = (j + s[i]) % 256
		s[i], s[j] = s[j], s[i]
		rc4_decrypted.append(chr(ord(data) ^ s[(s[i] + s[j]) % 256]))
	rc4_decrypted =  "".join(rc4_decrypted)
	#print rc4_decrypted

	key_xor = "V7D=^,M.E"
	key_xor_len = len(key_xor)
	flag = []
	for i in range(len(rc4_decrypted)):
		 flag.append(chr(ord(rc4_decrypted[i]) ^ ord(key_xor[i % key_xor_len])))
	print "".join(flag)

if __name__ == '__main__':
	main()
```

运行结果

```
It_s_a_easyCrack_for_beginners
```