Crypto
CRYPTO01
2023 年江蘇省領航杯 bd 原題：
題目：
from Crypto.Util.number import *
from secret import flag
p = getPrime(512)
q = getPrime(512)
n = p * q
d = getPrime(299)
e = inverse(d,(p-1)*(q-1))
m = bytes_to_long(flag)
c = pow(m,e,n)
hint1 = p >> (512-70)
hint2 = q >> (512-70)
print(f"n = {n}")
print(f"e = {e}")
print(f"c = {c}")
print(f"hint1 = {hint1}")
print(f"hint2 = {hint2}")
n = 95987463597889741532025162535631829592517704738860431905943824498597890101136796870879646153634795544527837591685182170270252555997933421564167468816667980089869165228796395618775798781717091178143300536302805947806332962230499807469654672313206953750808878098101882367253566423367338396717229488061237787619
e = 12761568528114005244342182138319275328501544878744699681091257281459893043102455333575012392492554249378138377894473691530001400901111860569849400611145049669197371290112261284014768486937294128410391332641648640840309135718123516725634190495261570498188176102281061084366456374907755950527213961595506707585
c = 35154471719082941146017277238175991504655570882040897713927696748547265178059291385527810510087300325678561502592029686486144851647754163110759942860051150957340793828236482243293747881251009434747066136505663782280758979333019882382608002308773317613847702048959887017010151916544703971113441862351668075919
hint1 = 1175980694459189065778
hint2 = 632846170973644915854
思路：
利用上高位的boneh_durfee攻擊
參考論文：
[367.pdf (iacr.org)](https://eprint.iacr.org/2023/367.pdf)
他們實驗時的指令碼：Codes of the manuscript—Practical Attacks on Small Private Exponent RSA—New Records and New Insi - Pastebin.com
exp:
先恢復 d
import time
time.clock = time.time
debug = True
strict = False
helpful_only = True
dimension_min = 7 # 如果晶格達到該尺寸，則停止移除
# 顯示有用向量的統計資料
def helpful_vectors(BB, modulus):
nothelpful = 0
for ii in range(BB.dimensions()[0]):
if BB[ii,ii] >= modulus:
nothelpful += 1
# 顯示帶有 0 和 X 的矩陣
def matrix_overview(BB, bound):
for ii in range(BB.dimensions()[0]):
a = ('%02d ' % ii)
for jj in range(BB.dimensions()[1]):
a += '0' if BB[ii,jj] == 0 else 'X'
if BB.dimensions()[0] < 60:
a += ' '
if BB[ii, ii] >= bound:
a += '~'
#print (a)
# 嘗試刪除無用的向量
# 從當前 = n-1（最後一個向量）開始
def remove_unhelpful(BB, monomials, bound, current):
# 我們從當前 = n-1（最後一個向量）開始
if current == -1 or BB.dimensions()[0] <= dimension_min:
return BB
# 開始從後面檢查
for ii in range(current, -1, -1):
# 如果它沒有用
if BB[ii, ii] >= bound:
affected_vectors = 0
affected_vector_index = 0
# 讓我們檢查它是否影響其他向量
for jj in range(ii + 1, BB.dimensions()[0]):
# 如果另一個向量受到影響：
# 我們增加計數
if BB[jj, ii] != 0:
affected_vectors += 1
affected_vector_index = jj
# 等級：0
# 如果沒有其他載體最終受到影響
# 我們刪除它
if affected_vectors == 0:
#print ("* removing unhelpful vector", ii)
BB = BB.delete_columns([ii])
BB = BB.delete_rows([ii])
monomials.pop(ii)
BB = remove_unhelpful(BB, monomials, bound, ii-1)
return BB
# 等級：1
#如果只有一個受到影響，我們會檢查
# 如果它正在影響別的向量
elif affected_vectors == 1:
affected_deeper = True
for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
# 如果它影響哪怕一個向量
# 我們放棄這個
if BB[kk, affected_vector_index] != 0:
affected_deeper = False
# 如果沒有其他向量受到影響，則將其刪除，並且
# 這個有用的向量不夠有用
#與我們無用的相比
if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
#print ("* removing unhelpful vectors", ii, "and", affected_vector_index)
BB = BB.delete_columns([affected_vector_index, ii])
BB = BB.delete_rows([affected_vector_index, ii])
monomials.pop(affected_vector_index)
monomials.pop(ii)
BB = remove_unhelpful(BB, monomials, bound, ii-1)
return BB
# nothing happened
return BB
"""
Returns:
* 0,0 if it fails
* -1，-1 如果 "strict=true"，並且行列式不受約束
* x0,y0 the solutions of `pol`
"""
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
"""
Boneh and Durfee revisited by Herrmann and May
在以下情況下找到解決方案：
* d < N^delta
* |x|< e^delta
* |y|< e^0.5
每當 delta < 1 - sqrt（2）/2 ~ 0.292
"""
# substitution (Herrman and May)
PR.<u, x, y> = PolynomialRing(ZZ) #多項式環
Q = PR.quotient(x*y + 1 - u) # u = xy + 1
polZ = Q(pol).lift()
UU = XX*YY + 1
# x-移位
gg = []
for kk in range(mm + 1):
for ii in range(mm - kk + 1):
xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
gg.append(xshift)
gg.sort()
# 單項式 x 移位列表
monomials = []
for polynomial in gg:
for monomial in polynomial.monomials(): #對於多項式中的單項式。單項式（）：
if monomial not in monomials: # 如果單項不在單項中
monomials.append(monomial)
monomials.sort()
# y-移位
for jj in range(1, tt + 1):
for kk in range(floor(mm/tt) * jj, mm + 1):
yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
yshift = Q(yshift).lift()
gg.append(yshift) # substitution
# 單項式 y 移位列表
for jj in range(1, tt + 1):
for kk in range(floor(mm/tt) * jj, mm + 1):
monomials.append(u^kk * y^jj)
# 構造格 B
nn = len(monomials)
BB = Matrix(ZZ, nn)
for ii in range(nn):
BB[ii, 0] = gg[ii](0, 0, 0)
for jj in range(1, ii + 1):
if monomials[jj] in gg[ii].monomials():
BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)
#約化格的原型
if helpful_only:
# #自動刪除
BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
# 重置維度
nn = BB.dimensions()[0]
if nn == 0:
print ("failure")
return 0,0
# 檢查向量是否有幫助
if debug:
helpful_vectors(BB, modulus^mm)
# 檢查行列式是否正確界定
det = BB.det()
bound = modulus^(mm*nn)
if det >= bound:
print ("We do not have det < bound. Solutions might not be found.")
print ("Try with highers m and t.")
if debug:
diff = (log(det) - log(bound)) / log(2)
print ("size det(L) - size e^(m*n) = ", floor(diff))
if strict:
return -1, -1
else:
print ("det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")
# display the lattice basis
if debug:
matrix_overview(BB, modulus^mm)
# LLL
if debug:
print ("optimizing basis of the lattice via LLL, this can take a long time")
#BB = BB.BKZ(block_size=25)
BB = BB.LLL()
if debug:
print ("LLL is done!")
# 替換向量 i 和 j ->多項式 1 和 2
if debug:
print ("在格中尋找線性無關向量")
found_polynomials = False
for pol1_idx in range(nn - 1):
for pol2_idx in range(pol1_idx + 1, nn):
# 對於i and j, 構造兩個多項式
PR.<w,z> = PolynomialRing(ZZ)
pol1 = pol2 = 0
for jj in range(nn):
pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)
# 結果
PR.<q> = PolynomialRing(ZZ)
rr = pol1.resultant(pol2)
if rr.is_zero() or rr.monomials() == [1]:
continue
else:
print ("found them, using vectors", pol1_idx, "and", pol2_idx)
found_polynomials = True
break
if found_polynomials:
break
if not found_polynomials:
print ("no independant vectors could be found. This should very rarely happen...")
return 0, 0
rr = rr(q, q)
# solutions
soly = rr.roots()
if len(soly) == 0:
print ("Your prediction (delta) is too small")
return 0, 0
soly = soly[0][0]
ss = pol1(q, soly)
solx = ss.roots()[0][0]
return solx, soly
def example():
############################################
# 隨機生成資料
##########################################
#start_time =time.perf_counter
start =time.clock()
size=512
length_N = 2*size;
ss=0
s=70;
M=1 # the number of experiments
delta = 299/1024
# p = random_prime(2^512,2^511)
for i in range(M):
# p = random_prime(2^size,None,2^(size-1))
# q = random_prime(2^size,None,2^(size-1))
# if(p<q):
# temp=p
# p=q
# q=temp
N = 95987463597889741532025162535631829592517704738860431905943824498597890101136796870879646153634795544527837591685182170270252555997933421564167468816667980089869165228796395618775798781717091178143300536302805947806332962230499807469654672313206953750808878098101882367253566423367338396717229488061237787619
e = 12761568528114005244342182138319275328501544878744699681091257281459893043102455333575012392492554249378138377894473691530001400901111860569849400611145049669197371290112261284014768486937294128410391332641648640840309135718123516725634190495261570498188176102281061084366456374907755950527213961595506707585
c = 35154471719082941146017277238175991504655570882040897713927696748547265178059291385527810510087300325678561502592029686486144851647754163110759942860051150957340793828236482243293747881251009434747066136505663782280758979333019882382608002308773317613847702048959887017010151916544703971113441862351668075919
hint1 = 1175980694459189065778
hint2 = 632846170973644915854
# print ("p真實高",s,"位元：", int(p/2^(512-s)))
# print ("q真實高",s,"位元：", int(q/2^(512-s)))
# N = p*q;
# 解密指數d的指數( 最大0.292)
m = 7 # 格大小（越大越好/越慢）
t = round(((1-2*delta) * m)) # 來自 Herrmann 和 May 的最佳化
X = floor(N^delta) #
Y = floor(N^(1/2)/2^s) # 如果 p、 q 大小相同，則正確
for l in range(int(hint1),int(hint1)+1):
print('\n\n\n l=',l)
pM=l;
p0=pM*2^(size-s)+2^(size-s)-1;
q0=N/p0;
qM=int(q0/2^(size-s))
A = N + 1-pM*2^(size-s)-qM*2^(size-s);
#A = N+1
P.<x,y> = PolynomialRing(ZZ)
pol = 1 + x * (A + y) #構建的方程
# Checking bounds
#if debug:
#print ("=== 核對資料 ===")
#print ("* delta:", delta)
#print ("* delta < 0.292", delta < 0.292)
#print ("* size of e:", ceil(log(e)/log(2))) # e的bit數
# print ("* size of N:", len(bin(N))) # N的bit數
#print ("* size of N:", ceil(log(N)/log(2))) # N的bit數
#print ("* m:", m, ", t:", t)
# boneh_durfee
if debug:
##print ("=== running algorithm ===")
start_time = time.time()
solx, soly = boneh_durfee(pol, e, m, t, X, Y)
if solx > 0:
#print ("=== solution found ===")
if False:
print ("x:", solx)
print ("y:", soly)
d_sol = int(pol(solx, soly) / e)
ss=ss+1
print ("=== solution found ===")
print ("p的高位元為：",l)
print ("q的高位元為：",qM)
print ("d=",d_sol)
if debug:
print("=== %s seconds ===" % (time.time() - start_time))
#break
print("ss=",ss)
#end=time.process_time
end=time.clock()
print('Running time: %s Seconds'%(end-start))
if __name__ == "__main__":
example()
然後再正常解 rsa
from Crypto.Util.number import *
import gmpy2
n = 95987463597889741532025162535631829592517704738860431905943824498597890101136796870879646153634795544527837591685182170270252555997933421564167468816667980089869165228796395618775798781717091178143300536302805947806332962230499807469654672313206953750808878098101882367253566423367338396717229488061237787619
e = 12761568528114005244342182138319275328501544878744699681091257281459893043102455333575012392492554249378138377894473691530001400901111860569849400611145049669197371290112261284014768486937294128410391332641648640840309135718123516725634190495261570498188176102281061084366456374907755950527213961595506707585
c = 35154471719082941146017277238175991504655570882040897713927696748547265178059291385527810510087300325678561502592029686486144851647754163110759942860051150957340793828236482243293747881251009434747066136505663782280758979333019882382608002308773317613847702048959887017010151916544703971113441862351668075919
hint1 = 1175980694459189065778
hint2 = 632846170973644915854
d = 994872951830622609173239108988480436496396666299008546372509127121113351371824434704770953
m = pow(c,d,n)
print(long_to_bytes(m))
flag： wdflag{a14a7fa0-4e7b-4624-ad0c-36c31dd8012e}
CRYPTO02
題目：
# coding: utf-8
#!/usr/bin/env python2
import gmpy2
import random
import binascii
from hashlib import sha256
from sympy import nextprime
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from Crypto.Util.number import long_to_bytes
from FLAG import flag
#flag = 'wdflag{123}'
def victory_encrypt(plaintext, key):
key = key.upper()
key_length = len(key)
plaintext = plaintext.upper()
ciphertext = ''
for i, char in enumerate(plaintext):
if char.isalpha():
shift = ord(key[i % key_length]) - ord('A')
encrypted_char = chr((ord(char) - ord('A') + shift) % 26 + ord('A'))
ciphertext += encrypted_char
else:
ciphertext += char
return ciphertext
victory_key = "WANGDINGCUP"
victory_encrypted_flag = victory_encrypt(flag, victory_key)
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
a = 0
b = 7
xG = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
yG = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
G = (xG, yG)
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
h = 1
zero = (0,0)
dA = nextprime(random.randint(0, n))
if dA > n:
print("warning!!")
def addition(t1, t2):
if t1 == zero:
return t2
if t2 == zero:
return t2
(m1, n1) = t1
(m2, n2) = t2
if m1 == m2:
if n1 == 0 or n1 != n2:
return zero
else:
k = (3 * m1 * m1 + a) % p * gmpy2.invert(2 * n1 , p) % p
else:
k = (n2 - n1 + p) % p * gmpy2.invert((m2 - m1 + p) % p, p) % p
m3 = (k * k % p - m1 - m2 + p * 2) % p
n3 = (k * (m1 - m3) % p - n1 + p) % p
return (int(m3),int(n3))
def multiplication(x, k):
ans = zero
t = 1
while(t <= k):
if (k &t )>0:
ans = addition(ans, x)
x = addition(x, x)
t <<= 1
return ans
def getrs(z, k):
(xp, yp) = P
r = xp
s = (z + r * dA % n) % n * gmpy2.invert(k, n) % n
return r,s
z1 = random.randint(0, p)
z2 = random.randint(0, p)
k = random.randint(0, n)
P = multiplication(G, k)
hA = multiplication(G, dA)
r1, s1 = getrs(z1, k)
r2, s2 = getrs(z2, k)
print("r1 = {}".format(r1))
print("r2 = {}".format(r2))
print("s1 = {}".format(s1))
print("s2 = {}".format(s2))
print("z1 = {}".format(z1))
print("z2 = {}".format(z2))
key = sha256(long_to_bytes(dA)).digest()
cipher = AES.new(key, AES.MODE_CBC)
iv = cipher.iv
encrypted_flag = cipher.encrypt(pad(victory_encrypted_flag.encode(), AES.block_size))
encrypted_flag_hex = binascii.hexlify(iv + encrypted_flag).decode('utf-8')
print("Encrypted flag (AES in CBC mode, hex):", encrypted_flag_hex)
# output
# r1 = 11455446275324978121918764201975007376236576456980676251003353868423934779741
# r2 = 11455446275324978121918764201975007376236576456980676251003353868423934779741
# s1 = 40939314972385973234538799073526528418921185412931089406906904671430814294251
# s2 = 20521265832237322311837491925934175279870547563516618597235100170259508682349
# z1 = 86373733089658748931377346977504497606857910789811212034370600969224209571891
# z2 = 114906325375159287808320541183977807561041047925384600130919891200501605749979
# ('Encrypted flag (AES in CBC mode, hex):', u'57608ba208813e738e7a354399e77272017548f9abf0da7a179a1136cda57579720b68a3ed46d85f5997c35af18f42175f43e856a0f64d964e7ab1e8a672b689')
思路：
**直接拷問 ai**
先求出，dA
k = ((z1 - z2) * inverse(s1 - s2, n)) % n
# print(k)
dA = ((s1 * k - z1) * inverse(r1, n)) % n
# print(dA)
接下來，我們使用 AES 金鑰和 CBC 模式解密 encrypted_flag：
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
import binascii
# 解密 AES 加密的 flag
encrypted_flag_hex = '57608ba208813e738e7a354399e77272017548f9abf0da7a179a1136cda57579720b68a3ed46d85f5997c35af18f42175f43e856a0f64d964e7ab1e8a672b689'
iv_and_encrypted_flag = binascii.unhexlify(encrypted_flag_hex)
iv = iv_and_encrypted_flag[:16]
encrypted_flag = iv_and_encrypted_flag[16:]
cipher = AES.new(key, AES.MODE_CBC, iv=iv)
decrypted_flag = unpad(cipher.decrypt(encrypted_flag), AES.block_size)
最後一步是解密勝利密碼。由於勝利密碼是一個簡單的多表替換密碼，我們只需要反轉加密過程即可解密：
def victory_decrypt(ciphertext, key):
key = key.upper()
key_length = len(key)
plaintext = ''
for i, char in enumerate(ciphertext):
if char.isalpha():
shift = ord(key[i % key_length]) - ord('A')
decrypted_char = chr((ord(char) - ord('A') - shift) % 26 + ord('A'))
plaintext += decrypted_char
else:
plaintext += char
return plaintext
# 解密勝利密碼
victory_key = "WANGDINGCUP"
decrypted_flag = victory_decrypt(decrypted_flag.decode(), victory_key)
print("Decrypted flag:", decrypted_flag)
完整exp：
from Crypto.Util.number import *
import gmpy2
from hashlib import sha256
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
a = 0
b = 7
xG = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
yG = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
G = (xG, yG)
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
h = 1
zero = (0,0)
r1 = 11455446275324978121918764201975007376236576456980676251003353868423934779741
r2 = 11455446275324978121918764201975007376236576456980676251003353868423934779741
s1 = 40939314972385973234538799073526528418921185412931089406906904671430814294251
s2 = 20521265832237322311837491925934175279870547563516618597235100170259508682349
z1 = 86373733089658748931377346977504497606857910789811212034370600969224209571891
z2 = 114906325375159287808320541183977807561041047925384600130919891200501605749979
c = '57608ba208813e738e7a354399e77272017548f9abf0da7a179a1136cda57579720b68a3ed46d85f5997c35af18f42175f43e856a0f64d964e7ab1e8a672b689'
k = ((z1 - z2) * inverse(s1 - s2, n)) % n
# print(k)
dA = ((s1 * k - z1) * inverse(r1, n)) % n
# print(dA)
key = sha256(long_to_bytes(dA)).digest()
print(key)
# 解密aes
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
import binascii
# 解密 AES 加密的 flag
encrypted_flag_hex = '57608ba208813e738e7a354399e77272017548f9abf0da7a179a1136cda57579720b68a3ed46d85f5997c35af18f42175f43e856a0f64d964e7ab1e8a672b689'
iv_and_encrypted_flag = binascii.unhexlify(encrypted_flag_hex)
iv = iv_and_encrypted_flag[:16]
encrypted_flag = iv_and_encrypted_flag[16:]
cipher = AES.new(key, AES.MODE_CBC, iv=iv)
decrypted_flag = unpad(cipher.decrypt(encrypted_flag), AES.block_size)
def victory_decrypt(ciphertext, key):
key = key.upper()
key_length = len(key)
plaintext = ''
for i, char in enumerate(ciphertext):
if char.isalpha():
shift = ord(key[i % key_length]) - ord('A')
decrypted_char = chr((ord(char) - ord('A') - shift) % 26 + ord('A'))
plaintext += decrypted_char
else:
plaintext += char
return plaintext
# 解密勝利密碼
victory_key = "WANGDINGCUP"
decrypted_flag = victory_decrypt(decrypted_flag.decode(), victory_key)
# 轉小寫
print("Decrypted flag:", decrypted_flag.lower())
flag： wdflag{8ed62e3b409bb214e6fdb0a78f63569c}
Misc
MISC01
描述：
某單位網路遭到非法的攻擊，安全人員對流量調查取證之後儲存了關鍵證據，發現人員的定位資訊存在洩露，請對其進行分析。flag為使用者位置資訊進行32位md5雜湊值
提交的flag格式：wdflag
學習：Diameter協議摘要 - StevensFollower - 部落格園
Diameter協議-CSDN部落格
思路：
分析流量包發現是一個 diameter 協議分析，先過濾一下 diameter，
然後往下看 AVP 其中有條訊息表示的是地理位置，
可以發現 E-UTRAN-Cell-Global-Identit 就是的 val，就表示洩露的資訊，可以複製出來丟給 chat 分析一下
最後 md5 加密一下就行
flag： wdflag{0d1e868390ea06522d2c9f04c7ade736}
MISC03
題目：
近日某公司伺服器遭到惡意攻擊，隨後公司立即對流量監測系統中遭受攻擊時段的流量進行了取證，但是公司某一網路安全實習生進行分析溯源後並未找到攻擊者的攻擊IP，於是公司決定將這個任務重新交給更具經驗的你來進行，接手工作後，你立即對其進行了深入調查！
提交的flag格式：wdflag
分析流量包，過濾 http 流，檢視 200 的返回包，可以發現，這裡 hacker.php 惡意操作了
ip 為 39.168.5.60
flag： wdflag{39.168.5.60}
MISC04
IrisCTF-2024-Challenges/peanoscramble at main · IrisSec/IrisCTF-2024-Challenges
IrisCTF2024 | 1r0ny
（皮亞諾）曲線 【分形幾何】05.Peano（皮亞諾）曲線 - 知乎
推測應該是按著曲線的軌跡將畫素還原到原來的位置，編寫指令碼
exp：
from PIL import Image
from tqdm import tqdm
def peano(n):
if n == 0:
return [[0,0]]
else:
in_lst = peano(n - 1)
lst = in_lst.copy()
px,py = lst[-1]
lst.extend([px - i[0], py + 1 + i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px + i[0], py + 1 + i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px + 1 + i[0], py - i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px - i[0], py - 1 - i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px + i[0], py - 1 - i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px + 1 + i[0], py + i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px - i[0], py + 1 + i[1]] for i in in_lst)
px,py = lst[-1]
lst.extend([px + i[0], py + 1 + i[1]] for i in in_lst)
return lst
order = peano(6)
img = Image.open('1.png')
width, height = img.size
block_width = width # // 3
block_height = height # // 3
new_image = Image.new("RGB", (width, height))
for i, (x, y) in tqdm(enumerate(order)):
# 根據列表順序獲取新的座標
new_x, new_y = i % width, i // width
# 獲取原影像素
pixel = img.getpixel((x, height - 1 - y))
# 在新影像中放置畫素
new_image.putpixel((new_x, new_y), pixel)
new_image.save("rearranged_image.jpg")
flag： wdflag{793a67fd-58b2-4267-860a-b94ca4842761}
Reverse
REVERSE01
一個安卓逆向題，分析一下主程式
載入了一個 so 檔案
逆向一下 so 程式
在 sub_940 發現 sm4 加密，key 是 16 位元組
key 的後一半
key：A11223574689900Z
提取密文
85FCBC1D8A32907CB6CACAAAA9109957E38C80651EADF138C3FE849A62DA68B36EFF51FCD8B69FEB7AE0738619A4CA13
flag： wdflag{f28268d3-35b2-4d43-bdeb-a03ad186346a}
REVERSE02
思路：
查殼，然後使用 ida64 開啟，分析一下 main 函式，將 flag 分成了 4 段，解釋在註釋裡面
直接進行解密即可，base64 表，追蹤 base64_encode 這個函式，就能發現
exp：
# 第一段
s2 = [0x70, 0xC2, 0x6C, 0xCA]
s1 = [ord(c) for c in "nppl"]
s = s2 + s1
dest = [s[i] // 2 for i in range(8)]
flag1 = ''.join(chr(i) for i in dest)
print('flag1 = ', flag1)
# 第二段
# 定義 v11 和 v22
v11 = [ord('9'), ord('['), 23, 16, 127, 13, 71, 6]
v22 = [ord(c) for c in "XorrLord"]
# 初始化 dest 陣列
dest = [0] * 8
flag2 = []
for i in range(8):
for j in range(1, 128):
if j ^ v22[i] == v11[i]:
flag2.append(j)
break
# 計算 dest 的第8到第15個元素
# for j in range(8):
# dest[j] = v11[j - 8] ^ v22[j - 8]
# print("dest =", ''.join(chr(i) for i in dest))
flag2 = ''.join(chr(i) for i in flag2)
print('flag2 = ', flag2)
# 第三段
import base64
import string
str1 = "PVLlQVPhPFW"
string1 = "CDEFGHIJKLMNOPQRSTUVWXYZABabcdefghijklmnopqrstuvwxyz0123456789+/"
string2 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
# 轉換字串
translated_str = str1.translate(str.maketrans(string1,string2))
# 為 Base64 解碼新增填充
padding_needed = len(translated_str) % 4
if padding_needed:
translated_str += '=' * (4 - padding_needed)
# 解碼並列印結果
try:
flag3 = base64.b64decode(translated_str).decode()
print("flag3 = ", flag3)
except Exception as e:
print("解碼錯誤:", e)
# 第四段
from Crypto.Cipher import AES
key = b"AesMasterAesMast"
c = [0xA7, 0x20, 0x15, 0xFF, 0xCD, 0xF4, 0x50, 0xDE, 0x78, 0xBE, 0xA7, 0x94, 0x7F, 0xBF, 0xAC, 0xC5]
cipher = AES.new(key, AES.MODE_ECB)
flag4 = cipher.decrypt(bytes(c)).decode()
print("flag4 = ", flag4)
flag = flag1 + flag2 + flag3 + flag4
print("flag = ", "wdflag{" + flag )
print(len(flag))
flag： wdflag{8a6e7886a4eb3b5b52e93a4506d28a04}