2023年国家基地“楚慧杯”网络安全实践能力竞赛初赛-Crypto+Misc WP ...

Misc

ez_zip

题目
4096个压缩包套娃
我的解答：
写个脚本直接解压即可：

1. import zipfile
3. name = '附件路径\\题目附件.zip'
4. for i in range(4097):
5.     f = zipfile.ZipFile(name , 'r')
6.     f.extractall(pwd=name[:-4].encode())
7.     name = f.filelist[0].filename
8.     print(name[:-4],end="")
9.     f.close()

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得到

1. +-+++-++ +-+++++- +-+-++-- +-++++-- +-+-+-++ +-+++--+ +----+-- ++--+++- ++--++++ +--+++-- ++--+-+- ++---+++ ++--++-+ ++--+-+- ++---+++ +--+++-- +--+++-- +--++--+ ++--+++- +--++-+- ++--+--- +--+++-- ++--+--+ ++--++-- ++--+++- +--++-+- ++--+-+- ++---++- ++--+++- ++--+++- +--++-+- +--++-++ ++--+--+ +--++++- +--+++-- +--+++-- ++--+-++ +--++-+- +--++-++ +-----+-

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一眼丁真01，将+改成0，-改成1

1. x='''01000100 01000001 01010011 01000011 01010100 01000110 01111011 00110001 00110000 01100011 00110101 00111000 00110010 00110101 00111000 01100011 01100011 01100110 00110001 01100101 00110111 01100011 00110110 00110011 00110001 01100101 00110101 00111001 00110001 00110001 01100101 01100100 00110110 01100001 01100011 01100011 00110100 01100101 01100100 01111101'''
2. s=x.split(' ')
3. print(''.join(chr(int(c,2)) for c in s))
5. 'DASCTF{10c58258ccf1e7c631e5911ed6acc4ed}'

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easy取证

题目

我的解答：
一个取证问题，简单来分析一下：
本人是在windows下使用，参考：https://blog.csdn.net/m0_68012373/article/details/127419463
我们先查看镜像信息

1. volatility.exe -f mem.raw imageinfo

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然后我们可以利用插件grep查找一下常见的信息，例如：zip,txt,docx,png,jpg等
测试之后发现桌面有个docx文档

1. volatility.exe -f mem.raw --profile Win7SP1x64 filescan | grep .docx

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我们需要提取出来

1. volatility.exe -f mem.raw --profile=Win7SP1x64 dumpfiles -Q 0x000000003dceaf20 -D ./

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得到

修改后缀为docx打开，复制内容到txt里面

一眼丁真snow隐写，但需要找到密码，我们使用mimikatz（mimikatz插件可以获取系统明文密码，网上有安装教程）获取密码
得到

1. H7Qmw_X+WB6BXDXa

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然后直接解码即可

1. SNOW.EXE -C -p "H7Qmw_X+WB6BXDXa" White.txt

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Crypto

so-large-e

题目

公钥

1. -----BEGIN PUBLIC KEY-----
2. MIIBIDANBgkqhkiG9w0BAQEFAAOCAQ0AMIIBCAKBgQCl7ZhtXDOIFdSnnejtOn2W
3. OdcqyzrvKMVFTIqSyPV3Tkj5m9ETc/rlvLJLcQvI0V6tr+u5Tq+zqWBQzsvRsvKt
4. +ap0JW8up0qD1nGIvcJVdsWAjdse7AH/N3+xg8NrH3nO0OIWzMpkGH+4TVsOBu8M
5. nhnR9SxTkDp+gUtHtPR/awKBgQChjp2PFfpCV4hrVyPJrKP2gHbW+o7mBNiUd3Av
6. Ucbkr7rA6Sj3tU33yGKIoXbmZC4rWNcuqrsoCPvIFl/YHYPKVDOl2PlLaDVi/Q5E
7. ymGkUfPXoZsScxvkZtkOt9XY4wWEeDMtxF/swnV0nhAUSJEHamFL3i0PAWf9uBdd
8. VheH4Q==
9. -----END PUBLIC KEY-----

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1. from Crypto.Util.number import *
2. from Crypto.PublicKey import RSA
3. from flag import flag
4. import random
6. m = bytes_to_long(flag)
8. p = getPrime(512)
9. q = getPrime(512)
10. n = p*q
11. e = random.getrandbits(1024)
12. assert size(e)==1024
13. phi = (p-1)*(q-1)
14. assert GCD(e,phi)==1
15. d = inverse(e,phi)
16. assert size(d)==269
18. pub = (n, e)
19. PublicKey = RSA.construct(pub)
20. with open('pub.pem', 'wb') as f :
21.     f.write(PublicKey.exportKey())
23. c = pow(m,e,n)
24. print('c =',c)
26. print(long_to_bytes(pow(c,d,n)))
29. #c = 6838759631922176040297411386959306230064807618456930982742841698524622016849807235726065272136043603027166249075560058232683230155346614429566511309977857815138004298815137913729662337535371277019856193898546849896085411001528569293727010020290576888205244471943227253000727727343731590226737192613447347860

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我的解答：
我们先分解公钥得到n,e

1. from gmpy2 import *
2. from Crypto.Util.number import *
3. from Crypto.PublicKey import RSA
6. # 公钥提取
7. with open("pub.pem","r",encoding="utf-8") as file:
8.     text=file.read()
9. key=RSA.import_key(text)
10. e=key.e
11. n=key.n
12. print(e)
13. print(n)
15. 113449247876071397911206070019495939088171696712182747502133063172021565345788627261740950665891922659340020397229619329204520999096535909867327960323598168596664323692312516466648588320607291284630435682282630745947689431909998401389566081966753438869725583665294310689820290368901166811028660086977458571233
16. 116518679305515263290840706715579691213922169271634579327519562902613543582623449606741546472920401997930041388553141909069487589461948798111698856100819163407893673249162209631978914843896272256274862501461321020961958367098759183487116417487922645782638510876609728886007680825340200888068103951956139343723

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发现这个e很大，我们根据题目代码可知它跟n一个数量级。而d很小，想到维纳攻击或低解密指数攻击。但尝试无果有报错。问题就在于他的私钥太小d < N^0.292并不满足维纳或低解密d的要求。
因此我们需要爆出真正的满足题目条件的d，利用LLL-attacks解出d
详细解析可参考：https://github.com/Gao-Chuan/RSA-and-LLL-attacks/blob/master/boneh_durfee.sage

1. from __future__ import print_function
2. import time
4. ############################################
5. # Config
6. ##########################################
8. """
9. Setting debug to true will display more informations
10. about the lattice, the bounds, the vectors...
11. """
12. debug = True
14. """
15. Setting strict to true will stop the algorithm (and
16. return (-1, -1)) if we don't have a correct
17. upperbound on the determinant. Note that this
18. doesn't necesseraly mean that no solutions
19. will be found since the theoretical upperbound is
20. usualy far away from actual results. That is why
21. you should probably use `strict = False`
22. """
23. strict = False
25. """
26. This is experimental, but has provided remarkable results
27. so far. It tries to reduce the lattice as much as it can
28. while keeping its efficiency. I see no reason not to use
29. this option, but if things don't work, you should try
30. disabling it
31. """
32. helpful_only = True
33. dimension_min = 7 # stop removing if lattice reaches that dimension
35. ############################################
36. # Functions
37. ##########################################
39. # display stats on helpful vectors
40. def helpful_vectors(BB, modulus):
41.     nothelpful = 0
42.     for ii in range(BB.dimensions()[0]):
43.         if BB[ii,ii] >= modulus:
44.             nothelpful += 1
46.     print(nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")
48. # display matrix picture with 0 and X
49. def matrix_overview(BB, bound):
50.     for ii in range(BB.dimensions()[0]):
51.         a = ('%02d ' % ii)
52.         for jj in range(BB.dimensions()[1]):
53.             a += '0' if BB[ii,jj] == 0 else 'X'
54.             if BB.dimensions()[0] < 60:
55.                 a += ' '
56.         if BB[ii, ii] >= bound:
57.             a += '~'
58.         print(a)
60. # tries to remove unhelpful vectors
61. # we start at current = n-1 (last vector)
62. def remove_unhelpful(BB, monomials, bound, current):
63.     # end of our recursive function
64.     if current == -1 or BB.dimensions()[0] <= dimension_min:
65.         return BB
67.     # we start by checking from the end
68.     for ii in range(current, -1, -1):
69.         # if it is unhelpful:
70.         if BB[ii, ii] >= bound:
71.             affected_vectors = 0
72.             affected_vector_index = 0
73.             # let's check if it affects other vectors
74.             for jj in range(ii + 1, BB.dimensions()[0]):
75.                 # if another vector is affected:
76.                 # we increase the count
77.                 if BB[jj, ii] != 0:
78.                     affected_vectors += 1
79.                     affected_vector_index = jj
81.             # level:0
82.             # if no other vectors end up affected
83.             # we remove it
84.             if affected_vectors == 0:
85.                 print("* removing unhelpful vector", ii)
86.                 BB = BB.delete_columns([ii])
87.                 BB = BB.delete_rows([ii])
88.                 monomials.pop(ii)
89.                 BB = remove_unhelpful(BB, monomials, bound, ii-1)
90.                 return BB
92.             # level:1
93.             # if just one was affected we check
94.             # if it is affecting someone else
95.             elif affected_vectors == 1:
96.                 affected_deeper = True
97.                 for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
98.                     # if it is affecting even one vector
99.                     # we give up on this one
100.                     if BB[kk, affected_vector_index] != 0:
101.                         affected_deeper = False
102.                 # remove both it if no other vector was affected and
103.                 # this helpful vector is not helpful enough
104.                 # compared to our unhelpful one
105.                 if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
106.                     print("* removing unhelpful vectors", ii, "and", affected_vector_index)
107.                     BB = BB.delete_columns([affected_vector_index, ii])
108.                     BB = BB.delete_rows([affected_vector_index, ii])
109.                     monomials.pop(affected_vector_index)
110.                     monomials.pop(ii)
111.                     BB = remove_unhelpful(BB, monomials, bound, ii-1)
112.                     return BB
113.     # nothing happened
114.     return BB
116. """
117. Returns:
118. * 0,0   if it fails
119. * -1,-1 if `strict=true`, and determinant doesn't bound
120. * x0,y0 the solutions of `pol`
121. """
122. def boneh_durfee(pol, modulus, mm, tt, XX, YY):
123.     """
124.     Boneh and Durfee revisited by Herrmann and May
125.    
126.     finds a solution if:
127.     * d < N^delta
128.     * |x| < e^delta
129.     * |y| < e^0.5
130.     whenever delta < 1 - sqrt(2)/2 ~ 0.292
131.     """
133.     # substitution (Herrman and May)
134.     PR.<u, x, y> = PolynomialRing(ZZ)
135.     Q = PR.quotient(x*y + 1 - u) # u = xy + 1
136.     polZ = Q(pol).lift()
138.     UU = XX*YY + 1
140.     # x-shifts
141.     gg = []
142.     for kk in range(mm + 1):
143.         for ii in range(mm - kk + 1):
144.             xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
145.             gg.append(xshift)
146.     gg.sort()
148.     # x-shifts list of monomials
149.     monomials = []
150.     for polynomial in gg:
151.         for monomial in polynomial.monomials():
152.             if monomial not in monomials:
153.                 monomials.append(monomial)
154.     monomials.sort()
155.    
156.     # y-shifts (selected by Herrman and May)
157.     for jj in range(1, tt + 1):
158.         for kk in range(floor(mm/tt) * jj, mm + 1):
159.             yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
160.             yshift = Q(yshift).lift()
161.             gg.append(yshift) # substitution
162.    
163.     # y-shifts list of monomials
164.     for jj in range(1, tt + 1):
165.         for kk in range(floor(mm/tt) * jj, mm + 1):
166.             monomials.append(u^kk * y^jj)
168.     # construct lattice B
169.     nn = len(monomials)
170.     BB = Matrix(ZZ, nn)
171.     for ii in range(nn):
172.         BB[ii, 0] = gg[ii](0, 0, 0)
173.         for jj in range(1, ii + 1):
174.             if monomials[jj] in gg[ii].monomials():
175.                 BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)
177.     # Prototype to reduce the lattice
178.     if helpful_only:
179.         # automatically remove
180.         BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
181.         # reset dimension
182.         nn = BB.dimensions()[0]
183.         if nn == 0:
184.             print("failure")
185.             return 0,0
187.     # check if vectors are helpful
188.     if debug:
189.         helpful_vectors(BB, modulus^mm)
190.    
191.     # check if determinant is correctly bounded
192.     det = BB.det()
193.     bound = modulus^(mm*nn)
194.     if det >= bound:
195.         print("We do not have det < bound. Solutions might not be found.")
196.         print("Try with highers m and t.")
197.         if debug:
198.             diff = (log(det) - log(bound)) / log(2)
199.             print("size det(L) - size e^(m*n) = ", floor(diff))
200.         if strict:
201.             return -1, -1
202.     else:
203.         print("det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")
205.     # display the lattice basis
206.     if debug:
207.         matrix_overview(BB, modulus^mm)
209.     # LLL
210.     if debug:
211.         print("optimizing basis of the lattice via LLL, this can take a long time")
213.     BB = BB.LLL()
215.     if debug:
216.         print("LLL is done!")
218.     # transform vector i & j -> polynomials 1 & 2
219.     if debug:
220.         print("looking for independent vectors in the lattice")
221.     found_polynomials = False
222.    
223.     for pol1_idx in range(nn - 1):
224.         for pol2_idx in range(pol1_idx + 1, nn):
225.             # for i and j, create the two polynomials
226.             PR.<w,z> = PolynomialRing(ZZ)
227.             pol1 = pol2 = 0
228.             for jj in range(nn):
229.                 pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
230.                 pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)
232.             # resultant
233.             PR.<q> = PolynomialRing(ZZ)
234.             rr = pol1.resultant(pol2)
236.             # are these good polynomials?
237.             if rr.is_zero() or rr.monomials() == [1]:
238.                 continue
239.             else:
240.                 print("found them, using vectors", pol1_idx, "and", pol2_idx)
241.                 found_polynomials = True
242.                 break
243.         if found_polynomials:
244.             break
246.     if not found_polynomials:
247.         print("no independant vectors could be found. This should very rarely happen...")
248.         return 0, 0
249.    
250.     rr = rr(q, q)
252.     # solutions
253.     soly = rr.roots()
255.     if len(soly) == 0:
256.         print("Your prediction (delta) is too small")
257.         return 0, 0
259.     soly = soly[0][0]
260.     ss = pol1(q, soly)
261.     solx = ss.roots()[0][0]
263.     #
264.     return solx, soly
266. def example():
267.     ############################################
268.     # How To Use This Script
269.     ##########################################
271.     #
272.     # The problem to solve (edit the following values)
273.     #
275.     # the modulus
276.     e = 113449247876071397911206070019495939088171696712182747502133063172021565345788627261740950665891922659340020397229619329204520999096535909867327960323598168596664323692312516466648588320607291284630435682282630745947689431909998401389566081966753438869725583665294310689820290368901166811028660086977458571233
277.     N = 116518679305515263290840706715579691213922169271634579327519562902613543582623449606741546472920401997930041388553141909069487589461948798111698856100819163407893673249162209631978914843896272256274862501461321020961958367098759183487116417487922645782638510876609728886007680825340200888068103951956139343723
279.     # the hypothesis on the private exponent (the theoretical maximum is 0.292)
280.     delta = .27 # this means that d < N^delta
282.     #
283.     # Lattice (tweak those values)
284.     #
286.     # you should tweak this (after a first run), (e.g. increment it until a solution is found)
287.     m = 6 # size of the lattice (bigger the better/slower)
289.     # you need to be a lattice master to tweak these
290.     t = int((1-2*delta) * m)  # optimization from Herrmann and May
291.     X = 2*floor(N^delta)  # this _might_ be too much
292.     Y = floor(N^(1/2))    # correct if p, q are ~ same size
294.     #
295.     # Don't touch anything below
296.     #
298.     # Problem put in equation
299.     P.<x,y> = PolynomialRing(ZZ)
300.     A = int((N+1)/2)
301.     pol = 1 + x * (A + y)
303.     #
304.     # Find the solutions!
305.     #
307.     # Checking bounds
308.     if debug:
309.         print("=== checking values ===")
310.         print("* delta:", delta)
311.         print("* delta < 0.292", delta < 0.292)
312.         print("* size of e:", int(log(e)/log(2)))
313.         print("* size of N:", int(log(N)/log(2)))
314.         print("* m:", m, ", t:", t)
316.     # boneh_durfee
317.     if debug:
318.         print("=== running algorithm ===")
319.         start_time = time.time()
321.     solx, soly = boneh_durfee(pol, e, m, t, X, Y)
323.     # found a solution?
324.     if solx > 0:
325.         print("=== solution found ===")
326.         if False:
327.             print("x:", solx)
328.             print("y:", soly)
330.         d = int(pol(solx, soly) / e)
331.         print("private key found:", d)
332.     else:
333.         print("=== no solution was found ===")
335.     if debug:
336.         print(("=== %s seconds ===" % (time.time() - start_time)))
338. if __name__ == "__main__":
339.     example()

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之后直接解即可

1. import gmpy2
2. from Crypto.Util.number import *
4. d=663822343397699728953336968317794118491145998032244266550694156830036498673227937
5. c = 6838759631922176040297411386959306230064807618456930982742841698524622016849807235726065272136043603027166249075560058232683230155346614429566511309977857815138004298815137913729662337535371277019856193898546849896085411001528569293727010020290576888205244471943227253000727727343731590226737192613447347860
6. n=116518679305515263290840706715579691213922169271634579327519562902613543582623449606741546472920401997930041388553141909069487589461948798111698856100819163407893673249162209631978914843896272256274862501461321020961958367098759183487116417487922645782638510876609728886007680825340200888068103951956139343723
7. m=pow(c,d,n)
8. print(long_to_bytes(m))
9. #DASCTF{6f4fadce-5378-d17f-3c2d-2e064db4af19}

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matrixequation

题目

1. from sage.all import *
2. import string
3. from myflag import finalflag, flag
5. A = matrix([[0 for i in range(11)] for i in range(11)])
6. for k in range(len(flag)):
7.         i, j = 5*k // 11, 5*k % 11
8.         A[i, j] = alphabet.index(flag[k])
9. from hashlib import md5
10. assert(finalflag == 'DASCTF{' + f'{md5(flag.encode()).hexdigest()}' + '}')
11. key = getKey()
12. R, leftmatrix, matrixU = key
13. tmpMatrix = leftmatrix * matrixU
14. X = A + R
15. Y = tmpMatrix * X
16. E = leftmatrix.inverse() * Y
18. f = open('output','w')
19. f.write(str(E)+'\n')
20. f.write(str(leftmatrix * matrixU * leftmatrix)+'\n')
21. f.write(str(f'{leftmatrix.inverse() * tmpMatrix**2 * leftmatrix}\n'))
22. f.write(str(f'{R.inverse() * tmpMatrix**8}\n'))
24. A = matrix([[0 for i in range(11)] for i in range(11)])
25. for k in range(len(flag)):
26.         i, j = 5*k // 11, 5*k % 11
27.         A[i, j] = alphabet.index(flag[k])
28. from hashlib import md5
29. assert(finalflag == 'DASCTF{' + f'{md5(flag.encode()).hexdigest()}' + '}')
30. key = getKey()
31. R, leftmatrix, matrixU = key
32. tmpMatrix = leftmatrix * matrixU
33. X = A + R
34. Y = tmpMatrix * X
35. E = leftmatrix.inverse() * Y
37. f = open('output','w')
38. f.write(str(E)+'\n')
39. f.write(str(leftmatrix * matrixU * leftmatrix)+'\n')
40. f.write(str(f'{leftmatrix.inverse() * tmpMatrix**2 * leftmatrix}\n'))
41. f.write(str(f'{R.inverse() * tmpMatrix**8}\n'))

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1. [53 19 40  7 22 46 69 11 31 48 57]
2. [66 47 13 39  1 34 26 15 52 18 55]
3. [47  1  9 14 58 34 40 27  9  1 36]
4. [36 19 38 30 39 30 21 27  1  4 13]
5. [25 10 52 45 69 63  3 64 13 51 44]
6. [48  2 64 19 49 51 39 29 22 35 17]
7. [69 48 27 17 15 42 60 42 15 43  7]
8. [39 37 56  5 49 57 51 49  3 53 25]
9. [50 39  9 30 19 63 20 12  8 55 35]
10. [24 26 58 56 43 70 66 27 32 70 59]
11. [ 9 34 18 31 65  3 48 39 39 40 53]
12. [62 17 50 41  4  7  5  4 20 15 45]
13. [ 8 59 38 18 42 31 60 58 35  1 46]
14. [25 20 45 31 69 45 59 34 46 63 69]
15. [ 6  2 65 44 53 59 11 52 37 48 40]
16. [33 52  4  9  6  7 34  4 59 59  6]
17. [65 64 53 22 49 22 58 50 48 66 25]
18. [ 0 43 42 68 16 35 20 69  1 14 18]
19. [23 61 44 20 30 38 10 68 52 39  4]
20. [ 1  8 31 31 11 54 41 70 49 40  1]
21. [58 29 28 60 23 13 67 33 14 15  2]
22. [23 25 70  1 13 57 52 21 54 64 26]
23. [10  5 15 49 12 53 46 23 44 40 67]
24. [41 21 62 22 40 69 59  0 28 42 52]
25. [ 9 66 44  3 48  2 53  8 46 52  4]
26. [ 8 50 15 56 16 31 47  5 70  6 43]
27. [59 34 48  5 55 50 61 39 38  7 60]
28. [46 46  9 36 12 49 48 30 64 30 45]
29. [62 33  2 19  3 15 69 25  0 51 69]
30. [27 10 48 26 11 32  1 40 29 59 16]
31. [18 60 33 64 15 41 22  9 11 60 22]
32. [32 52 15 27  1 63 55 54 70 17 52]
33. [22 27 33 38 68 36 59 17 64 18 65]
34. [43 68 27 18 44 32 47 21 46 44 14]
35. [12 52 44 61 26 34 53 36 18  3 61]
36. [10 59 14  2 42 63 31  9 53  4 55]
37. [63 48 59 11 54  9 54 50 68  5 28]
38. [66 12 58 68 52 50  5 39 19  6 70]
39. [42 23 24 54 54 64  3 16 20 67 28]
40. [60 68 63 63 34  7  0 36  3 22 68]
41. [10 23  0  9 64  0 52  1 24 52 21]
42. [65 52 42  9 43 39 15  3 36 28 28]
43. [21 32 35 69 49 55  0 23  4 32 42]
44. [61 52 49 46 50 34 70 35 39  1 16]

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我的解答：
线性代数问题，题目把flag转成了矩阵A，我们要做的就是求出A
已知题目信息（由于变量较长，这里用缩略单词代替）

R, S = random_matrix
S = L * U
E = L^(-1) * S * (A + R)
a = L * U * L
b = L^(-1) * S^2 * L
c = R^(-1) * S^8

求解过程

b * a^(-1) = L^(-1) * S  (S = L * U = U * L)
a * b^(-1) * E = A + R   (上式^(-1) * E)
a * b * a^(-1) = S^2     (define ss)
c * ss^(-4) = R^(-1)
a * b^(-1) * E - c * ss^(-4) = A  (solve)

exp:

1. import re
2. import string
4. alphabet = string.printable[:71]
5. p = len(alphabet)
7. with open('output', 'r') as f:
8.   data = f.read().split('\n')[:-1]
10. data = [re.findall(r'\d+', d) for d in data]
11. data = [[Integer(di) for di in d] for d in data]
13. n = 11
14. E = matrix(GF(p), data[:11])
15. LUL = matrix(GF(p), data[11: 22])
16. ULUL = matrix(GF(p), data[22: 33])
17. RiLU8 = matrix(GF(p), data[33: 44])
19. Ui = LUL * ULUL^(-1)
20. Ri = RiLU8 * Ui * (ULUL^(-1))^3 * LUL^(-1)
21. U = Ui^(-1)
22. assert Ri * (LUL * U)^4 == RiLU8
24. R = Ri^(-1)
25. AR = Ui * E
26. A = AR - R
27. print(A)
30. flag = ''
31. for k in range(24):
32.   i, j = 5*k // 11, 5*k % 11
33.   flag += alphabet[A[i, j]]
34. print(flag)
35.   
37. from hashlib import md5
38. flag = 'DASCTF{' + f'{md5(flag.encode()).hexdigest()}' + '}'
39. print(flag)

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