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[原创]看雪CTF_2018 团队赛 第五题 write up(吃我十六字阴阳风水秘术)
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2018-12-9 23:32 3223
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环境配置
系统 : Windows 7、nexus 5
程序 : CrackMe.KwaiChing.apk
要求 : 输入口令
使用工具 : 称骨算命、Android Studio、jadx、apktool
搜集线索
在测试机中安装该apk,发现程序需求一个key,如果以123
这样的随机数作为输入的话,会输出failed!!!
。
我们使用apktool反编译这个程序
apktool d CrackMe.KwaiChing.apk
然后在\res\values\strings.xml
位置查看字符串信息:
<string name="me">" s \u0009\u0009u \u0009\u0009\u0009c \u0009\u0009\u0009\u0009c \u0009\u0009\u0009\u0009\u0009e \u0009\u0009\u0009\u0009\u0009\u0009s \u0009\u0009\u0009\u0009\u0009\u0009\u0009s \u0009\u0009\u0009\u0009\u0009\u0009\u0009\u0009! \u0009\u0009\u0009\u0009\u0009\u0009\u0009\u0009\u0009! "</string> <string name="notMe">f \u0009a \u0009\u0009i \u0009\u0009\u0009l \u0009\u0009\u0009\u0009e \u0009\u0009\u0009\u0009\u0009d \u0009\u0009\u0009\u0009\u0009\u0009! \u0009\u0009\u0009\u0009\u0009\u0009\u0009! \u0009\u0009\u0009\u0009\u0009\u0009\u0009\u0009!</string> <string name="ok">驗證</string> <string name="search_menu_title">Search</string> <string name="status_bar_notification_info_overflow">999+</string> <string name="success00" /> <string name="success34">" 財穀有餘主得內助富貴之命 此命福氣果如何,僧道門中衣祿多; 離祖出家方為妙,朝晚拜佛念彌陀。 此命推來為人性躁; 與人做事反為不美; 離祖成家; 三番四次自成自立安享福; 直自三十六至四十六; 財不謀而自至; 福不求而自得; 有貴人助; 家庭安寧; 妻宮若要無刑; 猴、豬、羊、蛇不可配; 龍、虎、馬、牛方得安; 雖有二子; 終生帶暗方可. 兄弟六親如冰碳; 在家不得安然; 初限駁雜多端; 勞碌奔波不能聚錢; 常有憂愁. 壽元七十八歲; 死於三月中."</string>
看来存在有一个提示正确和一个提示错误的字符串,还有就是success34
的迷之字符串。我搜索了一下,居然发现了存有相关信息的网页。
看来是一个玄学题目(笑
开始分析
我们运行jadx:
D:\software\jadx-0.8.0(1)\lib>java -jar jadx-gui-0.8.0.jar
定位到p007cn.kwaiching.crackme包下的CrackMe类,发现这就是算法的主流程:
protected void onCreate(Bundle bundle) { super.onCreate(bundle); setContentView((int) R.layout.activity_fate_me); b(); this.n = (TextView) findViewById(R.id.fate); ((Button) findViewById(R.id.ok)).setOnClickListener(new OnClickListener() { public void onClick(View view) { try { CrackMe.this.a(); } catch (Exception unused) { CrackMe.this.n.setText(CrackMe.this.getString(R.string.notMe)); } } }); }
来看看a的流程:
private void a() { try { c(); if (this.j == 0 || this.i == 0 || this.h == 0) { this.n.setText(getString(R.string.notMe)); return; } d(); a(((e() + f()) + g()) + h()); } catch (Exception unused) { this.n.setText(getString(R.string.notMe)); } }
我们先来查看h函数的部分代码:
/* JADX WARNING: Removed duplicated region for block: B:17:0x0050 A:{Catch:{ Exception -> 0x005c }} */ private int h() { /* r6 = this; r0 = 2131165227; // 0x7f07002b float:1.7944665E38 double:1.0529355243E-314; r1 = 2131427370; // 0x7f0b002a float:1.8476354E38 double:1.05306504E-314; r2 = 0; r0 = r6.findViewById(r0); Catch:{ Exception -> 0x005c }
这样的代码就是jadx的反编译出了点问题,我们需要设置一下选项,让代码的可读性变高。
代码修饰
这里,在jadx的菜单中点击文件
->设置
->启用反混淆
,接着选择显示不一致的代码
。这时,可以发现代码漂亮了不少。
导入as查看
在jadx中查看,不如导入到android stdio(下文中简称为as)这样专业的ide中查看,这样能修改变量名,快速查看引用位置等等。
这里,在jadx的菜单中点击文件
->另存为 gradle项目
,选中一个任意一个空的文件夹,然后保存就可以。
流程分析
在as中,发现主流程是:
private void m162a() { try { m166c(); if (this.year == 0 || this.month == 0 || this.day == 0) { this.f125n.setText(getString(C0252R.string.notMe)); return; } m167d(); m163a(((m168e() + m169f()) + m170g()) + m171h()); } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); } }
m166c函数的流程是:
/* renamed from: c */ private void m166c() { try { String obj = ((EditText) findViewById(C0252R.id.code)).getText().toString(); this.year = 0; this.month = 0; this.day = 0; this.year = Integer.parseInt(obj.length() > 4 ? obj.substring(0, 4) : obj); if (this.year > 0 && this.year < 189) { this.year = 0; } if (this.year <= 1983 || this.year >= 2007) { this.year = 0; } this.month = Integer.parseInt(obj.length() > 6 ? obj.substring(4, 6) : obj); if (this.month < 1 || this.month > 12) { this.month = 0; } if (obj.length() > 8) { obj = obj.substring(6, 8); } this.day = Integer.parseInt(obj); if (this.day < 1 || this.day > 31) { this.day = 0; } } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); } }
稍微分析一下,可知程序把输入限定在:
- 年份:
1983
后,2007
前。 - 月份:
1
-12
- 日期:
1
-31
再来看看其他的函数:
private void m167d() { try { if (this.year == 1989 || this.year == 2004) { this.day = 31; } if (this.month == 1 || this.month == 4 || this.month == 5 || this.month == 7 || this.month == 10 || this.month == 11 || this.month == 12) { this.year = 1999; } if (this.year <= 1994 && (this.month == 2 || this.month == 6 || this.month == 8)) { this.month = 3; } if (this.year >= 1996 && (this.month == 2 || this.month == 6 || this.month == 8)) { this.month = 9; } if (this.year == 1995 && (this.day > this.month + 2 || this.month == this.day)) { this.month = 6; } this.my_year = this.year; this.my_month = this.month; this.my_day = this.day; } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); } }
这里其实是将可能的输入范围减少了,比如第一个判断,2004和1989年的所有可选日期都是31号。以此类推。
四个小函数
private int m168e() { try { return this.f115d[(this.my_year - 1900) % 60]; } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); return 0; } } /* renamed from: f */ private int m169f() { try { return this.f114c[this.my_month - 1]; } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); return 0; } } /* renamed from: g */ private int m170g() { try { return this.f113b[this.my_day - 1]; } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); return 0; } } /* JADX WARNING: Removed duplicated region for block: B:17:0x0050 A:{Catch:{ Exception -> 0x005c }} */ /* Code decompiled incorrectly, please refer to instructions dump. */ /* renamed from: h */ private int m171h() { try { Object obj; String other_str = ((EditText) findViewById(C0252R.id.code)).getText().toString(); other_str = other_str.substring(8, other_str.length()); int MONTH = this.my_month; int i2 = 0; while (i2 < this.f124m.length) { if (!other_str.equals(this.f124m[i2])) { i2++; } else if (MONTH == 2 && other_str.equals(this.f124m[6])) { return 63; } else { this.f122k = this.f112a[i2]; obj = 1; if (obj == null) { this.f125n.setText(getString(C0252R.string.notMe)); } return this.f122k; } } obj = null; if (obj == null) { } return this.f122k; } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); return 0; } } }
最本质的操作都是从固定数组中取值,然后返回相关值。以下是相关数组:
/* renamed from: a */ int[] f112a = new int[]{16, 6, 7, 10, 9, 16, 10, 8, 8, 9, 6, 6}; /* renamed from: b */ int[] f113b = new int[]{5, 10, 8, 15, 16, 15, 8, 16, 8, 16, 9, 17, 8, 17, 10, 8, 9, 18, 5, 15, 10, 9, 8, 9, 15, 18, 7, 8, 16, 6}; /* renamed from: c */ int[] f114c = new int[]{6, 7, 18, 9, 5, 16, 9, 15, 18, 8, 9, 5}; /* renamed from: d */ int[] f115d = new int[]{7, 7, 9, 12, 8, 7, 13, 5, 14, 5, 9, 17, 5, 7, 12, 8, 8, 6, 19, 6, 8, 16, 10, 6, 12, 9, 6, 7, 12, 5, 9, 8, 7, 8, 15, 9, 16, 8, 8, 19, 12, 6, 8, 7, 5, 15, 6, 16, 15, 7, 9, 12, 10, 7, 15, 6, 5, 14, 14, 9}; String[] f124m = new String[]{"23to01", "01to03", "03to05", "05to07", "07to09", "09to11", "11to13", "13to15", "15to17", "17to19", "19to21", "21to23"};
判断函数
这个判断函数,emmm...
private void m163a(int i) { if (i > 34 || i < 34) { this.f125n.setText(getString(C0252R.string.notMe)); return; } try { this.f125n.setText(String.format("%s%s", new Object[]{getString(C0252R.string.me), this.f123l[i]})); ((Button) findViewById(C0252R.id.ok)).setEnabled(false); } catch (Exception unused) { this.f125n.setText(getString(C0252R.string.notMe)); } }
可知需求i值为34了。
逻辑推理
我们知道,作者需求我们输入生辰八字,如果生辰八字按某种特定方式取值相加的结果为34,那么夺旗成功。
计算一下遍历所有可能的计算次数:
2007-1983=24 24 * 12 * 31 * 12 = 107136
十万次的运算对于计算机来说不算什么,如果发现可能的结果太多,再添加m167d函数中的判断缩小取值范围,还有m171h函数中的这一句:
MONTH == 2 && other_str.equals(this.f124m[6])
其实表示,月份为2时,时辰索引值不能取6,否则失败。
编写代码
根据以上逻辑,我们做python代码如下:
hour_weight_list = [16, 6, 7, 10, 9, 16, 10, 8, 8, 9, 6, 6] day_weight_list = [5, 10, 8, 15, 16, 15, 8, 16, 8, 16, 9, 17, 8, 17, 10, 8, 9, 18, 5, 15, 10, 9, 8, 9, 15, 18, 7, 8, 16, 6] month_weight_list = [6, 7, 18, 9, 5, 16, 9, 15, 18, 8, 9, 5] # begin at 1960 year_weight_list = [7, 7, 9, 12, 8, 7, 13, 5, 14, 5, 9, 17, 5, 7, 12, 8, 8, 6, 19, 6, 8, 16, 10, 6, 12, 9, 6, 7, 12, 5, 9, 8, 7, 8, 15, 9, 16, 8, 8, 19, 12, 6, 8, 7, 5, 15, 6, 16, 15, 7, 9, 12, 10, 7, 15, 6, 5, 14, 14, 9] date_list = [] for year_index in range(1984,2007): for month_index in range(1,len(month_weight_list)+1): for day_index in range(1,len(day_weight_list)+1): for hour_index in range(0,len(hour_weight_list)): year_weight = year_weight_list[(year_index - 1900) % 60] month_weight = month_weight_list[month_index-1] day_weight = day_weight_list[day_index-1] hour_weight = hour_weight_list[hour_index] year = year_index month = month_index if year == 1989 or year == 2004: continue if month == 1 or month == 4 or month == 5 or month == 7 or month == 10 or month == 11 or month == 12: year = 1999 year_weight = year_weight_list[(year - 1900) % 60] if year <= 1994 and (month == 2 or month == 6 or month == 8): month = 3 month_weight = month_weight_list[month-1] if year >= 1996 and (month == 2 or month == 6 or month == 8): month = 9 month_weight = month_weight_list[month-1] if year == 1995 and (day_index > month + 2 or month == day_index): month = 6 month_weight = month_weight_list[month-1] if ( year_weight + month_weight + day_weight + hour_weight ) == 34: date = str(year_index) + '/' + str(month_index) + '/' + str(day_index) + '/' + str(hour_index) if date_list.count(date) == 0: date_list.append(date) print "find_times = " + str(len(date_list)) print date_list[0] print date_list[1]
夺旗成功
最终运算结果如下:
➜ play_ground python count_date.py find_times = 2 1995/2/3/3 1995/2/3/6
根据月份为2,索引值不能为6的推论,我们得到了唯一的flag1995020305to07
,输入flag,程序提示成功,并弹出一段迷之算命结果。
95年的小伙,和我同龄,还不错的。不过,果然还是少看些算命,好好学习,天天向上比较好哦。
参考
[培训]《安卓高级研修班(网课)》月薪三万计划,掌握调试、分析还原ollvm、vmp的方法,定制art虚拟机自动化脱壳的方法