0x0 拯救鲁班七号
载入jeb,在Mainactivity里面可以看到程序将输入放到checkPass函数中验证
protected void onCreate(Bundle arg3) {
super.onCreate(arg3);
this.setContentView(2130903040);
this.findViewById(2131165184).setOnClickListener(new View$OnClickListener() {
public void onClick(View arg4) {
MainActivity.this.checkPass(MainActivity.this.findViewById(2131165185).getText().toString().trim());
}
});
}在这个checkPass函数里面会调用CheckUtil的checkPass函数验证是否正确，此函数为humen.so文件中的函数，将humen放到ida中进行分析。无用代码比较多，下面只将关键代码发出来。
length = strlen(input);
while ( 1 )
{
v23 += 2;
if ( length <= v23 )
break;
v7 = input[v23 - 2];
input[v23 - 2] = input[v23 - 1];
v8 = 0;
input[v23 - 1] = v7;
if ( length > 4 )
{
for ( i = 4; ; i += 4 )
{
v10 = (char *)&input[v8];
v9 = v10;
v11 = *v10;
v10 += 4;
*v9 = *v10;
*v10 = v11;
v8 = i;
if ( length <= i + 4 )
break;
}
}
} v14 = (&t_ptr)[v4 - 12208];
v15 = 0;
if ( *input == *v14 )
{
do
v16 = v14[v15++ + 1];
while ( input[v15] == v16 );
}程序将输入的字符串奇数和偶数做交换并且前八位的0,4,8,位分别与4,8,12,做交换，最后与t进行比较，直接附上脚本。
stt="S!@#@1FD23154A34"
st=list(stt)
v6=len(st)
for j in range(14,0,-2):
if v6>4:
for i in range(12,1,-4):
st[i-4],st[i]=st[i],st[i-4]
st[j-1],st[j-2]=st[j-2],st[j-1]
flag=''
for i in range(len(st)):
flag+=st[i]
print flag0X1 The Marauder’s Map
依照惯例，拖进jeb分析。发现和数据库有关，最近刚好了解了一下安卓连接数据库，只能说好巧。
package com.example.icontest;
public class ReadSe {
private boolean a;
static {
System.loadLibrary("test");
}
public ReadSe() {
super();
this.a = false;
}
public final boolean a(String arg3, String arg4) {
boolean v0 = false;
if(arg3 != null && arg3.length() > 0 && arg4 != null && arg4.length() > 0 && (arg4.equals(this.readbin(arg3)))) {
v0 = true;
}
return v0;
}
private native String readbin(String arg1) {
}
}代码简单明了，将readbin函数的返回值与arg4进行比较，arg4为user数据库中读到的id为2的生日列。
public a() {
super();
this.a = "";
this.b = "dGVzdA==";
this.c = "WWVhaH4h";
this.d = "dXNlcnM=";
this.e = "Mg==";
}
public final String a(Context arg10) {
String v0 = a.a(this.b);
String v1 = a.a(this.d);
a.a(this.c);
String v6 = a.a(this.e);
SQLiteDatabase v0_1 = new b(arg10, v0, null, 1).getReadableDatabase();
Cursor v1_1 = v0_1.query(v1, new String[]{"userid", "age", "birthday", "id"}, "id=?", new String[]{v6}, null, null, null);
while(v1_1.moveToNext()) {
v1_1.getString(v1_1.getColumnIndex("userid"));
v1_1.getString(v1_1.getColumnIndex("age"));
this.a = v1_1.getString(v1_1.getColumnIndex("birthday"));
}
v0_1.close();
return this.a;
}
private static String a(String arg3) {
return new String(new String(Base64.decode(arg3.getBytes(), 0)).toCharArray());
}
}数据库可以通过sqlitebrowser软件查看。
然后，我们分析一下readbin函数，
关键代码在sub_1220()中
s = (char *)a1;
v7 = strlen(a1);
v5 = 0;
v6 = 0;
src = (char *)operator new[](2 * v7 + 1);
do
{
v1 = (unsigned __int8)s[v5];
src[v6] = sub_1078(~(_BYTE)v1 & 0xF);
v2 = v6 + 1;
src[v2] = sub_1078((v1 >> 4) ^ 0xE);
++v5;
v6 = v2 + 1;
}
while ( v5 < v7 );
src[2 * v7] = 0;
strncpy(s, src, 2 * v7 + 1);
return s;代码如上，只能说算法还是挺简单的
将输入的每一位经过运算化为src的两位，然后返回得到的字符串。
逻辑大概就是这样，flag可以经过爆破得出，直接上脚本。
s="9838e888496bfda98afdbb98a9b9a9d9cdfa29"
st=list(s)
def subl(a):
v1=0
if(a>9 or a<0):
if(a<=9 or a>15):
v1=255
else:
v1=a+87
else:
v1=a+48
return v1
print len(s)
key=""
for i in range(0,38,2):
for j in range(32,128):
if subl(~j&0xF)==ord(st[i]) and subl((j>>4)^0xE)==ord(st[i+1]):
key+=chr(j)
break;
print key0x2取证密码
输入在GetString.encrypt函数里面进行验证，encrypt为XTU.so文件的函数，代码如下
v3 = a1;
_JNIEnv::NewStringUTF(a1, "yInS567!bcNOUV8vwCDefXYZadoPQRGx13ghTpqrsHklm2EFtuJKLzMijAB094W");
_JNIEnv::NewStringUTF(v3, "Welc0meT0XTUCTF");
v4 = (const char *)_JNIEnv::GetStringUTFChars((int)v3);
str = (const char *)_JNIEnv::GetStringUTFChars((int)v3);
input = (char *)_JNIEnv::GetStringUTFChars((int)v3);
v7 = j_j_strlen(v4);
v8 = j_j_strlen(str);
temp = (char *)j_operator new[](v7 + 1);
v10 = (char *)j_operator new[](v8 + 1);
v11 = j_j_strlen(input);
v15 = (char *)j_operator new[](v11 + 1);
j_j_memcpy(&dest, &unk_2018, 0x3Cu);
j_j_strcpy(temp, v4);
j_j_strcpy(v10, str);
j_j_strcpy(v15, input);
for ( i = 0; i < j_j_strlen(v4); ++i )
temp[i] = v10[*((_DWORD *)&dest + i)];
v13 = 0;
while ( (unsigned __int8)v15[v13] == (unsigned __int8)temp[v13] )
{
if ( ++v13 == 15 )
return 1;
}
return 0;看见这个代码开始我是蒙逼的，，找不到代表输入的临时变量是哪个，，
只能大胆的猜测，v15既input为输入，str为yInS567!bcNOUV8vwCDefXYZadoPQRGx13ghTpqrsHklm2EFtuJKLzMijAB094W，取dest的值作为str的下标赋值给temp，然后与input进行比较，脚本如下。
s="yInS567!bcNOUV8vwCDefXYZadoPQRGx13ghTpqrsHklm2EFtuJKLzMijAB094W"
dest=[0x39,0x20,7,0xA,0x20,0x29,0x13,2,0x3A,0xC,0x11,0x31,0x3B,0xB,7]
s=list(s)
flag=""
for i in range(len(dest)):
flag+=s[dest[i]]
print flag
没想到真的是这样，，getflag。
0x3 人民的名义-抓捕赵德汉1
由于是jar文件，所以直接用jd-gui软件打开。
将输入的字符串载入checkPassword函数验证。
无用代码比较多，差点就被迷惑了~~
将输入的字符串进行md5加密，然后与fa3733c647dca53a66cf8df953c2d539进行比较。
由于以前做的md5加密的题比较多，这个题还挺简单的。
将fa3733c647dca53a66cf8df953c2d539解密md5即可。
0x4 人民的名义-抓捕赵德汉2
程序载入jd-hui后乱码比较多，这个题难度比上一个题提升了不少。
public static void main(String[] args)
{
JFrame frame = new JFrame("Key check");
JButton button = new JButton("Click to activate");
button.addActionListener(new ActionListener()
{
public void actionPerformed(ActionEvent ae)
{
String str = JOptionPane.showInputDialog(null, "Enter the product key: ",
"xxxx-xxxx-xxxx-xxxx", 1);
if (????????????.???(str)) {
JOptionPane.showMessageDialog(null, "Well done that was the correct key",
"Key check", 1);
} else {
JOptionPane.showMessageDialog(null, " Sorry that was the incorrect key \nRemember it is a crime to use software without paying for it",
"Key check", 1);
}
}
});
JPanel panel = new JPanel();
panel.add(button);
frame.add(panel);
frame.setSize(300, 100);
frame.setDefaultCloseOperation(3);
frame.setVisible(true);
}
}这里需要注意的一点就是，输入是以xxxx-xxxx-xxxx-xxxx格式进行输入的！
将验证函数修改一下如下，
public class iiiiiiiiiiii
{
static String key1 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
static String _ii = "ZYXWVUTSRQPONMLKJIHGFEDCBA";
public static boolean iii(String str)
{
if ((str != null) && (str.length() == 19))
{
key1 = System.arraycopy(_ii, 0, key1, 5, 5);
boolean keyGuessWrong = true;
int i = 0;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++) {
if (str.charAt(i + j) != key1.charAt(Start.operate(i + j, key1))) {
keyGuessWrong = false;
}
}
i += 5;
}
return keyGuessWrong;
}
return false;
}将i+j,和key1放到operate函数进行变换，然后与输入进行比较。
其中key1为start.main（）得到的字符串。
public static String main(String... args)
{
String x = "";
char[] arrayOfChar;
int j = (arrayOfChar = "v??��??????v?��??????����??".toCharArray()).length;
for (int i = 0; i < j; i++)
{
int $ = arrayOfChar[i];
x = x + (char)(($ >> 1) + 15);
}
return x;
}但是这有个乱码，，，很是难为了一番。最后经Simp1er表哥提示换种别工具可以得到这一个字符串，如下图。
最后，按照惯例附上脚本
key1="JsnatterrtJuaththovacke"
li=[0x76,0xC3,0x88,0xC2,0xBE,0xC2,0xA4,0xC3,0x8A,0xC3,0x8A,0xC2,0xAC,0xC3,0x86,
0xC3,0x86,0xC3,0x8A,0x76,0xC3,0x8C,0xC2]
key=""
for i in li:
key+=chr((i>>1)+15)
s=list(key1)
def digui(a):
if a>2:
return digui(a-1)+digui(a-2)
else:
return 1
flag=""
for i in range(4):
a=i*5
for j in range(4):
flag+=s[digui(a+j)%len(s)]
if i!=3:
flag+="-"
print flag附上题目及工具：http://pan.baidu.com/s/1eRI2HSA
密码：xpxn
911
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