REVERSE-PRACTICE-BUUCTF-31
REVERSE-PRACTICE-BUUCTF-31
- [羊城杯 2020]login
- [羊城杯 2020]Bytecode
- [羊城杯 2020]babyre
- [ACTF新生賽2020]fungame
[羊城杯 2020]login
exe程序,運行后輸入,無殼,ida分析
沒找到主要邏輯,在字符串窗口看到一些“py”的字樣,應該是python打包成了exe
用pyinstxtractor.py將exe解包,得到了這些文件
login文件缺少magic number,用struct文件的magic number(頭部的12個字節)補充,保存,改后綴名為.pyc
用uncompyle6反編譯login.pyc,得到python源碼
z3解方程組,
from z3 import * a1=Int('a1') a2=Int('a2') a3=Int('a3') a4=Int('a4') a5=Int('a5') a6=Int('a6') a7=Int('a7') a8=Int('a8') a9=Int('a9') a10=Int('a10') a11=Int('a11') a12=Int('a12') a13=Int('a13') a14=Int('a14') s=Solver() s.add(a1 * 88 + a2 * 67 + a3 * 65 - a4 * 5 + a5 * 43 + a6 * 89 + a7 * 25 + a8 * 13 - a9 * 36 + a10 * 15 + a11 * 11 + a12 * 47 - a13 * 60 + a14 * 29 == 22748) s.add(a1 * 89 + a2 * 7 + a3 * 12 - a4 * 25 + a5 * 41 + a6 * 23 + a7 * 20 - a8 * 66 + a9 * 31 + a10 * 8 + a11 * 2 - a12 * 41 - a13 * 39 + a14 * 17 == 7258) s.add(a1 * 28 + a2 * 35 + a3 * 16 - a4 * 65 + a5 * 53 + a6 * 39 + a7 * 27 + a8 * 15 - a9 * 33 + a10 * 13 + a11 * 101 + a12 * 90 - a13 * 34 + a14 * 23 == 26190) s.add(a1 * 23 + a2 * 34 + a3 * 35 - a4 * 59 + a5 * 49 + a6 * 81 + a7 * 25 + (a8 *128) - a9 * 32 + a10 * 75 + a11 * 81 + a12 * 47 - a13 * 60 + a14 * 29 == 37136) s.add(a1 * 38 + a2 * 97 + a3 * 35 - a4 * 52 + a5 * 42 + a6 * 79 + a7 * 90 + a8 * 23 - a9 * 36 + a10 * 57 + a11 * 81 + a12 * 42 - a13 * 62 - a14 * 11 == 27915) s.add(a1 * 22 + a2 * 27 + a3 * 35 - a4 * 45 + a5 * 47 + a6 * 49 + a7 * 29 + a8 * 18 - a9 * 26 + a10 * 35 + a11 * 41 + a12 * 40 - a13 * 61 + a14 * 28 == 17298) s.add(a1 * 12 + a2 * 45 + a3 * 35 - a4 * 9 - a5 * 42 + a6 * 86 + a7 * 23 + a8 * 85 - a9 * 47 + a10 * 34 + a11 * 76 + a12 * 43 - a13 * 44 + a14 * 65 == 19875) s.add(a1 * 79 + a2 * 62 + a3 * 35 - a4 * 85 + a5 * 33 + a6 * 79 + a7 * 86 + a8 * 14 - a9 * 30 + a10 * 25 + a11 * 11 + a12 * 57 - a13 * 50 - a14 * 9 == 22784) s.add(a1 * 8 + a2 * 6 + a3 * 64 - a4 * 85 + a5 * 73 + a6 * 29 + a7 * 2 + a8 * 23 - a9 * 36 + a10 * 5 + a11 * 2 + a12 * 47 - a13 * 64 + a14 * 27 == 9710) s.add(a1 * 67 - a2 * 68 + a3 * 68 - a4 * 51 - a5 * 43 + a6 * 81 + a7 * 22 - a8 * 12 - a9 * 38 + a10 * 75 + a11 * 41 + a12 * 27 - a13 * 52 + a14 * 31 == 13376) s.add(a1 * 85 + a2 * 63 + a3 * 5 - a4 * 51 + a5 * 44 + a6 * 36 + a7 * 28 + a8 * 15 - a9 * 6 + a10 * 45 + a11 * 31 + a12 * 7 - a13 * 67 + a14 * 78 == 24065) s.add(a1 * 47 + a2 * 64 + a3 * 66 - a4 * 5 + a5 * 43 + a6 * 112 + a7 * 25 + a8 * 13 - a9 * 35 + a10 * 95 + a11 * 21 + a12 * 43 - a13 * 61 + a14 * 20 == 27687) s.add(a1 * 89 + a2 * 67 + a3 * 85 - a4 * 25 + a5 * 49 + a6 * 89 + a7 * 23 + a8 * 56 - a9 * 92 + a10 * 14 + a11 * 89 + a12 * 47 - a13 * 61 - a14 * 29 == 29250) s.add(a1 * 95 + a2 * 34 + a3 * 62 - a4 * 9 - a5 * 43 + a6 * 83 + a7 * 25 + a8 * 12 - a9 * 36 + a10 * 16 + a11 * 51 + a12 * 47 - a13 * 60 - a14 * 24 == 15317) if s.check():print(s.model()) # [a2 = 24,a13 = 88, a6 = 43,a9 = 52,a14 = 33,a5 = 104,a12 = 74,a7 = 28,a1 = 119, a10 = 108, a11 = 88, a8 = 91, a4 = 7, a3 = 10]寫將位置換回以及逆異或運算腳本即可得到flag
import hashlib data=[119,24,10,7,104,43,28,91,52,108,88,74,88,33] index=[2,1,0,3,4,5,6,7,9,8,10,11,12,13] flag=[0]*14 for i in range(len(flag)):flag[index[i]]=data[i] for i in range(len(flag)-2,-1,-1):flag[i]^=flag[i+1] flag_str=''.join(chr(i) for i in flag) print(flag_str) # U_G07_th3_k3y! h=hashlib.md5() h.update(flag_str.encode(encoding='utf-8')) print(h.hexdigest()) # 58964088b637e50d3a22b9510c1d1ef8[羊城杯 2020]Bytecode
txt文件給了python的字節碼,翻譯成源碼
#coding:utf-8 en=[3,37,72,9,6,132] output=[101,96,23,68,112,42,107,62,96,53,176,179,98,53,67,29,41,120,60,106,51,101,178,189,101,48] print('welcome to GWHT2020') flag=raw_input('please input your flag:') str=flag def func0(): # 驗證輸入的長度a = len(str)if a < 38:print('lenth wrong!') def func1(): # 驗證輸入的前5個字符if (((ord(str[0])*2020+ord(str[1]))*2020+ord(str[2]))*2020+ord(str[3]))*2020+ord(str[4])==1182843538814603:print('good!continue\xe2\x80\xa6\xe2\x80\xa6') def func2(): # 驗證輸入花括號{}內的前26個字符x=[]k=5for i in range(13):b=ord(str[k])c=ord(str[k+1])a11=c^en[i%6]a22=b^en[i%6]x.append(a11)x.append(a22)k+=2if x==output:print('good!continue\xe2\x80\xa6\xe2\x80\xa6') def func3(): # 驗證輸入花括號{}內的后6個字符l=len(str)a1=ord(str[l-7])a2=ord(str[l-6])a3 = ord(str[l - 5])a4 = ord(str[l - 4])a5 = ord(str[l - 3])a6 = ord(str[l - 2])if a1*3+a2*2+a3*5==1003 and a1*4+a2*7+a3*9==2013 and a1+a2*8+a3*2==1109 and a4*3+a5*2+a6*5==671 and a4*4+a5*7+a6*9==1252 and a4+a5*8+a6*2==644:print('congraduation!you get the right flag!') func0() func1() func2() func3()func1驗證輸入的前5個字符,寫爆破腳本,得到"GWHT{"
for i in range(32,127):for j in range(32,127):for k in range(32, 127):for m in range(32, 127):for n in range(32, 127):if (((i*2020+j)*2020+k)*2020+m)*2020+n==1182843538814603:print(chr(i)+chr(j)+chr(k)+chr(m)+chr(n))break #GWHT{func2驗證輸入花括號{}內的前26個字符,寫逆腳本,得到"cfa2b87b3f746a8f0ac5c5963f"
en=[3,37,72,9,6,132] output=[101,96,23,68,112,42,107,62,96,53,176,179,98,53,67,29,41,120,60,106,51,101,178,189,101,48] k=0 flag=[] for i in range(13):c1=output[k+1]^en[i%6]c2=output[k]^en[i%6]flag.append(c1)flag.append(c2)k+=2 print(''.join(chr(i) for i in flag)) # cfa2b87b3f746a8f0ac5c5963ffunc3驗證輸入花括號{}內的后6個字符,用z3解方程,轉成字符串,得到"aeff73"
from z3 import * a1=Int('a1') a2=Int('a2') a3=Int('a3') a4=Int('a4') a5=Int('a5') a6=Int('a6') s=Solver() s.add(a1*3+a2*2+a3*5==1003) s.add(a1*4+a2*7+a3*9==2013) s.add(a1+a2*8+a3*2==1109) s.add(a4*3+a5*2+a6*5==671) s.add(a4*4+a5*7+a6*9==1252) s.add(a4+a5*8+a6*2==644) if s.check():print(s.model()) # [a5 = 55, a2 = 101, a6 = 51, a3 = 102, a4 = 102, a1 = 97] data=[97,101,102,102,55,51] print(''.join(chr(i) for i in data)) # aeff73最后加上一個’}’,于是flag為"GWHT{cfa2b87b3f746a8f0ac5c5963faeff73}"
[羊城杯 2020]babyre
elf文件,無殼,ida分析
main函數,首先sub_402563函數進行一段SMC,獲取輸入,輸入長度限為16,對輸入進行DES加密,密鑰動態調試可得,比較DES加密過的輸入(密文)與已知的byte_6040C0,驗證成功后,將未經DES加密過的輸入傳入sub_40272D函數作為AES加密的密鑰
調試得到DES密鑰為b'\xAD\x52\xF2\x4C\xE3\x2C\x20\xD6',密文為b'\x0A\xF4\xEE\xC8\x42\x8A\x9B\xDB\xA2\x26\x6F\xEE\xEE\xE0\xD8\xA2',分別用ECB模式和CBC模式解DES,兩次解密結果的拼接即為第一次正確的輸入
進入sub_40272D函數,獲取輸入,第一次的輸入作為AES的密鑰,對輸入進行常規的AES.ECB加密,密文異或運算,然后還有個相鄰兩個元素參與的運算給byte_6040D0賦值,最后byte_6040D0與已知的res比較
寫逆運算腳本即可得到flag
兩次輸入,驗證成功,再md5一下,提交成功
[ACTF新生賽2020]fungame
exe程序,運行后輸入,無殼,ida分析
main函數,給v3和x填充0,x大小為36,只填充了24個0
sub_401340函數,獲取輸入,對輸入的前16個字符進行驗證
sub_4013BA函數,兩次copy
查找x的交叉引用,除了main和sub_4013BA函數,第三處在sub_40233D函數
再次輸入,對輸入進行常規的base64編碼,結果與已知的v0比較驗證
寫腳本,但是提交失敗
最后參考Mz1師傅的wp:re | [ACTF新生賽2020]fungame
總結
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