标签：process append adb activity 安卓 col 技术 decrypt bad
国外比赛，做着玩
官方wp：https://github.com/tghack/tg20hack
RE1:
Game of Keys
题目提供了两个文件，一个PYC文件，一个wordlist.txt
在线pyc反编译：http://tools.bugscaner.com/decompyle/
import base64 from itertools import cycle class myGame: def __init__(self, xdim=4, ydim=4): self.x = xdim self.y = ydim self.matrix = [] for i in range(self.x): row = [] for j in range(self.y): row.append(0) self.matrix.append(row) def make_keys(self, *args, **kwargs): words = [] with open(‘wordlist.txt‘) as (f): for line in f: words.append(line.strip()) for i in range(self.x): for j in range(self.y): self.matrix[j][i] = words[i + j] keyArray = [] keyArray.append(self.matrix[args[0]][args[1]]) keyArray.append(self.matrix[args[2]][args[3]]) key = ‘‘ for k in keyArray: key = key.strip() + str(k).strip() print(key) return key def checkdata(self, key): f = base64.b64decode(‘NSYDUhoVWQ8SQVcOAAYRFQkORA4FQVMDQQ5fQhUEWUYMDl4MHA==‘) data = f.decode(‘ascii‘) c = ‘‘.join((chr(ord(c) ^ ord(k)) for c, k in zip(data, cycle(key)))) print(‘%s ^ %s = %s‘ % (data, key, c)) if __name__ == ‘__main__‘: mgame = myGame(25, 25) x = input(‘input a number: ‘) y = input(‘input a number: ‘) x1 = input(‘input a number: ‘) y1 = input(‘input a number: ‘) data = mgame.make_keys(int(x), int(y), int(x1), int(y1)) mgame.checkdata(data)
大意就是输入数字，生成key，再生成flag字符串，爆破即可：
import base64 from itertools import cycle words = [] class myGame: def __init__(self, xdim=4, ydim=4): self.x = xdim self.y = ydim self.matrix = [] for i in range(self.x): row = [] for j in range(self.y): row.append(0) self.matrix.append(row) def make_keys(self, *args, **kwargs): global words for i in range(self.x): for j in range(self.y): self.matrix[j][i] = words[i + j] keyArray = [] keyArray.append(self.matrix[args[0]][args[1]]) keyArray.append(self.matrix[args[2]][args[3]]) key = ‘‘ for k in keyArray: key = key.strip() + str(k).strip() #print(key) return key def checkdata(self, key,f): data = f.decode(‘ascii‘) c = ‘‘.join((chr(ord(c) ^ ord(k)) for c, k in zip(data, cycle(key)))) #print(‘%s ^ %s = %s‘ % (data, key, c)) if(c[0:4]==‘TG20‘): print c return c def makeWord(): words = [] with open(‘wordlist.txt‘) as (f): for line in f: words.append(line.strip()) return words if __name__ == ‘__main__‘: mgame = myGame(25, 25) f = base64.b64decode(‘NSYDUhoVWQ8SQVcOAAYRFQkORA4FQVMDQQ5fQhUEWUYMDl4MHA==‘) fp = open(‘flag.txt‘,‘w‘) words = makeWord() for i in range(0,24): for j in range(0,24): for k in range(0,24): for l in range(0,24): data = mgame.make_keys(int(i), int(j), int(k), int(l)) tmp = mgame.checkdata(data,f) if(tmp[0:4]==‘TG20‘): fp.write(tmp+‘\n‘) fp.close()
最后拿到flag:
这里之前我还以为是the 交了之后没交上，直接用这个交，就对了。
RE2：Bad intentions
描述：Someone found this very old game lying around. Apparently there is an extremely funny joke in there somewhere.
Hint：Random title?
是一个安卓小游戏，界面：
JEB反编译，根据题目，找到一个JokeActivity
用ADB启动，上Frida拿flag：
adb shell am start -W -n no.tghack.gaiainvaders/.JokeActivity
frida脚本：
import frida, sys source = """ Java.perform(function() { var clazz = Java.use(‘no.tghack.gaiainvaders.JokeActivity‘); clazz.decrypt.implementation = function() { var flag = clazz.decrypt.apply(this,arguments); send(flag); return flag; } }); """ def on_message(message, data): if message[‘type‘] == ‘send‘: print("[*] {0}".format(message[‘payload‘])) else: print(message) process = frida.get_usb_device().attach(‘no.tghack.gaiainvaders‘) script = process.create_script(source) script.on(‘message‘, on_message) script.load() sys.stdin.read()
RE3:Chuck Norris
描述：
Ha-ha! Did you like the joke from the previous challenge? Here, have another one:
Chuck Norris once won a game of Space Invaders without shooting.
这道题目的文件和第二题是一样的
JEB中查看到关键点：
一样的，HOOK：
frida脚本：
import frida, sys source = """ Java.perform(function() { var clazz = Java.use(‘no.tghack.gaiainvaders.Invader$Companion‘); clazz.getNumberOfInvaders.implementation = function() { return 0; } var clazzz = Java.use(‘no.tghack.gaiainvaders.EncryptionKt‘); clazzz.decrypt.implementation = function() { var flag = clazzz.decrypt.apply(this,arguments); send(flag); return flag; } }); """ def on_message(message, data): if message[‘type‘] == ‘send‘: print("[*] {0}".format(message[‘payload‘])) else: print(message) process = frida.get_usb_device().attach(‘no.tghack.gaiainvaders‘) script = process.create_script(source) script.on(‘message‘, on_message) script.load() sys.stdin.read()
MISC1：s2s messaging
描述:
Intercepted ship to ship communication. Can you find the message?
Hint:MQTT
流量分析题，提示了MQTT协议
wireshark过滤MQTT：
追踪TCP流：
提取json格式的data，base64解密后发现png头：
python解码：
查看图片获得flag:
MISC:.-.-.- .-.-.- .-.-.-
题目描述：
.---$‘4c 53 41 74 4c 53 34 67 4c 69 34 74 4c 53 30 67 4c 53 30 74 4c 53 30 67 65 79 34 74 4c 53 41 75 49 43 38 67‘-------------------------| /-$‘76 49 43 34 67 4c 69 34 75 49 43 30 74 4c 53 41 76 49 43 34 75 49 43 30 75 49 43 38 67 4c 53 30 74 49 43 34 75 4c 53 41 75 4c 53‘--\ | | | _________ | || | ______ | || | | || | ____ ____ | || | __ | || | ______ | || | ___ ____ | | | | | _ _ | | || | .‘ ___ | | || | | || | |_ || _| | || | / \ | || | .‘ ___ | | || | |_ ||_ _| | | | | |_/ | | \_| | || | / .‘ \_| | || | ______ | || | | |__| | | || | / /\ \ | || | / .‘ \_| | || | | |_/ / | | | | | | | || | | | ____ | || | |______| | || | | __ | | || | / ____ \ | || | | | | || | | __‘. | | | | _| |_ | || | \ `.___] _| | || | | || | _| | | |_ | || | _/ / \ \_ | || | \ `.___.‘\ | || | _| | \ \_ | | | | |_____| | || | `._____.‘ | || | | || | |____||____| | || ||____| |____|| || | `._____.‘ | || | |____||____| | | | | | || | | || | | || | | || | | || | | || | | | | | &-‘d3 d3 14 94 93 24 96 c4 76 43 35 c4 57 43 34 94 57 43 34 94 57 43 35 c4 57 14 97 c4 76 43‘$---------------------------------------/ | \-‘14 96 c4 76 03 35 c4 76 03 35 c4 47 14 96 c4 57 43 34 94 67 14 96 c4 57 03 34 94 57 14 96 c4 57 03 34 94 57 14 96 c4 76 43 35 c4‘$----/
.---$和
&-‘可以想到，这是asciidot
在线跑一下：
https://asciidots.herokuapp.com/index.html
HEX2ASCII后拿到：LSAtLS4gLi4tLS0gLS0tLS0gey4tLSAuIC8gLS4gLiAuIC0uLiAuIC0uLiAvIC4uLiAtLS0gLS0gLiAvIC4gLi4uIC0tLSAvIC4uIC0uIC8gLS0tIC4uLSAuLS4gLyAuLS4uIC4uIC4uLS4gLiB9IA==
base64解码：- --. ..--- ----- {.-- . / -. . . -.. . -.. / ... --- -- . / . ... --- / .. -. / --- ..- .-. / .-.. .. ..-. . }
摩丝解密：
TG20WENEEDEDSOMEESOINOURLIFE配合base64解码后的{}
得到：TG20{WE NEED SOME ESO IN OUR LIFE}
MISC:poke
拿到题目是一个很长的未知图腾：
解出是：remember to examine the
binwalk：
分离后：
拿到个这样的图片和另一张：
想到了pikalang：
https://www.dcode.fr/pikalang-language
这个图里可以分离到压缩包：
有密码，密码在另一张图的LSB里，使用zsteg:
拿到密码，解开：
这就是那张图的文本。
解码后：
结合之前位置图腾解出来的：remember to examine the
答案就是 TG20{remember to examine the foo bar dog closely}
misc:cubiksrube
这道题目是看了nepnep的nop的题解：
题目描述：
_sa
l{l
2ii
sG_ atu bpn o_u
Rre oh_ mir _re
yng o‘g Tc} dil
y0k
etb
co_
Oh, and take this! It might help you.
R‘ F2 L2 B2 U‘ B U‘ L R‘
B D‘ U D L‘ L‘ L‘ L2 U F2 U‘ U D2 F‘ R‘ B‘ B D2 L2 B2 F‘ D‘ R‘ L2 L‘ U‘ U‘ R D L2 D R B‘ F‘ U R2 R F L U‘
这是三阶魔方六面公式，可以写暴力拧魔方脚本，脚本也是nop的，这里改了拧的过程，反着拧
就是U‘ 变成 U U U
U U U变成U‘
再从后往前拧：
u = [[‘_‘,‘s‘,‘a‘], [‘l‘,‘{‘,‘l‘], [‘2‘,"i",‘i‘]] f = [[‘a‘,‘t‘,‘u‘], [‘o‘,‘h‘,‘_‘], [‘o‘,"‘",‘g‘]] d = [[‘y‘,‘0‘,‘k‘], [‘e‘,‘t‘,‘b‘], [‘c‘,"o",‘_‘]] b = [[‘o‘,‘_‘,‘u‘], [‘_‘,‘r‘,‘e‘], [‘d‘,"i",‘l‘]] l = [[‘s‘,‘G‘,‘_‘], [‘R‘,‘r‘,‘e‘], [‘y‘,"n",‘g‘]] r = [[‘b‘,‘p‘,‘n‘], [‘m‘,‘i‘,‘r‘], [‘T‘,"c",‘}‘]] def roll(matrix): #顺时针转 matrix = [[row[col] for row in matrix] for col in reversed(range(len(matrix[0])))] matrix = [[row[col] for row in matrix] for col in reversed(range(len(matrix[0])))] matrix = [[row[col] for row in matrix] for col in reversed(range(len(matrix[0])))] return matrix def PrintOne(w): for i in range(3): for j in range(3): print(w[i][j],end=‘ ‘) print(‘‘) def PrintAll(): global u global f global b global d global l global r m = 12 n = 9 temp = [[‘ ‘] * m for i in range(n) ] for i in range(3): for j in range(3): temp[i][j+3] = u[i][j] for i in range(3): for j in range(3): temp[i+3][j] = l[i][j] for i in range(3): for j in range(3): temp[i+3][j+3] = f[i][j] for i in range(3): for j in range(3): temp[i+3][j+6] = r[i][j] for i in range(3): for j in range(3): temp[i+3][j+9] = b[i][j] for i in range(3): for j in range(3): temp[i+6][j+3] = d[i][j] for i in range(9): for j in range(12): print(temp[i][j],end=‘ ‘) print(‘‘) def roll4(s): global u global f global b global d global l global r if(s == ‘F‘): f = roll(f) x = u[2][0] y = u[2][1] z = u[2][2] u[2][0] = l[2][2] u[2][1] = l[1][2] u[2][2] = l[0][2] l[2][2] = d[0][2] l[1][2] = d[0][1] l[0][2] = d[0][0] d[0][2] = r[0][0] d[0][1] = r[1][0] d[0][0] = r[2][0] r[0][0] = x r[1][0] = y r[2][0] = z elif(s == ‘U‘): u = roll(u) x = b[0][0] y = b[0][1] z = b[0][2] b[0][0] = l[0][0] b[0][1] = l[0][1] b[0][2] = l[0][2] l[0][0] = f[0][0] l[0][1] = f[0][1] l[0][2] = f[0][2] f[0][0] = r[0][0] f[0][1] = r[0][1] f[0][2] = r[0][2] r[0][0] = x r[0][1] = y r[0][2] = z elif(s == ‘D‘): d = roll(d) x = f[2][0] y = f[2][1] z = f[2][2] f[2][0] = l[2][0] f[2][1] = l[2][1] f[2][2] = l[2][2] l[2][0] = b[2][0] l[2][1] = b[2][1] l[2][2] = b[2][2] b[2][0] = r[2][0] b[2][1] = r[2][1] b[2][2] = r[2][2] r[2][0] = x r[2][1] = y r[2][2] = z elif(s == ‘L‘): l = roll(l) x = u[0][0] y = u[1][0] z = u[2][0] u[0][0] = b[2][2] u[1][0] = b[1][2] u[2][0] = b[0][2] b[2][2] = d[0][0] b[1][2] = d[1][0] b[0][2] = d[2][0] d[2][0] = f[2][0] d[1][0] = f[1][0] d[0][0] = f[0][0] f[0][0] = x f[1][0] = y f[2][0] = z elif(s == ‘L‘): l = roll(l) x = u[0][0] y = u[1][0] z = u[2][0] u[0][0] = b[2][2] u[1][0] = b[1][2] u[2][0] = b[0][2] b[2][2] = d[0][0] b[1][2] = d[1][0] b[0][2] = d[2][0] d[2][0] = f[2][0] d[1][0] = f[1][0] d[0][0] = f[0][0] f[0][0] = x f[1][0] = y f[2][0] = z elif(s == ‘R‘): r = roll(r) x = u[0][2] y = u[1][2] z = u[2][2] u[0][2] = f[0][2] u[1][2] = f[1][2] u[2][2] = f[2][2] f[0][2] = d[0][2] f[1][2] = d[1][2] f[2][2] = d[2][2] d[0][2] = b[2][0] d[1][2] = b[1][0] d[2][2] = b[0][0] b[0][0] = z b[1][0] = y b[2][0] = x elif(s == ‘B‘): b = roll(b) x = u[0][0] y = u[0][1] z = u[0][2] u[0][0] = r[0][2] u[0][1] = r[1][2] u[0][2] = r[2][2] r[0][2] = d[2][2] r[1][2] = d[2][1] r[2][2] = d[2][0] d[2][0] = l[0][0] d[2][1] = l[1][0] d[2][2] = l[2][0] l[0][0] = z l[1][0] = y l[2][0] = x roll4(‘U‘) roll4(‘L‘) roll4(‘L‘) roll4(‘L‘) roll4(‘F‘) roll4(‘F‘) roll4(‘F‘) roll4(‘R‘) roll4(‘U‘) roll4(‘U‘) roll4(‘U‘) roll4(‘F‘) roll4(‘B‘) roll4(‘R‘) roll4(‘R‘) roll4(‘R‘) roll4(‘D‘) roll4(‘D‘) roll4(‘D‘) roll4(‘L‘) roll4(‘L‘) roll4(‘D‘) roll4(‘D‘) roll4(‘D‘) roll4(‘R‘) roll4(‘R‘) roll4(‘R‘) roll4(‘U‘) roll4(‘U‘) roll4(‘L‘) roll4(‘L‘) roll4(‘L‘) roll4(‘R‘) roll4(‘D‘) roll4(‘F‘) roll4(‘B‘) roll4(‘B‘) roll4(‘L‘) roll4(‘L‘) roll4(‘D‘) roll4(‘D‘) roll4(‘R‘) roll4(‘F‘) roll4(‘D‘) roll4(‘D‘) roll4(‘F‘) roll4(‘F‘) roll4(‘U‘) roll4(‘U‘) roll4(‘U‘) roll4(‘L‘) roll4(‘D‘) roll4(‘D‘) roll4(‘D‘) roll4(‘U‘) roll4(‘U‘) roll4(‘U‘) roll4(‘D‘) roll4(‘B‘) roll4(‘B‘) roll4(‘B‘) roll4(‘R‘) roll4(‘L‘) roll4(‘L‘) roll4(‘L‘) # roll4(‘U‘) roll4(‘B‘) roll4(‘B‘) roll4(‘B‘) roll4(‘U‘) roll4(‘B‘) roll4(‘B‘) roll4(‘L‘) roll4(‘L‘) roll4(‘F‘) roll4(‘F‘) roll4(‘R‘) PrintAll()
标签：process append adb activity 安卓 col 技术 decrypt bad
原文地址：https://www.cnblogs.com/basstorm/p/12665281.html
