Misc
phishing email
Challenge
某公司安全工程师告警捕获一起邮件钓鱼样本,请帮助排查分析
A security engineer from a certain company has detected a phishing email sample. Please assist in investigation and analysis.
百度网盘:https://pan.baidu.com/s/1L1jBvqcpEI03uBlYsm0KAQ?pwd=ywr7
Google Dirve: https://drive.google.com/file/d/1U9RRjLNT6YRbHcXUesbPXL2a591bSX4m/view?usp=drive_link
Solution
从邮件附件得到以下 XML 代码:
<?xml version="1.0" encoding="UTF-8"?><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 800 600" width="800" height="600"> <!-- Fake invoice design to look legitimate --> <rect width="800" height="600" fill="#f8f9fa"/> <rect x="50" y="50" width="700" height="500" fill="white" stroke="#dee2e6" stroke-width="2"/> <!-- Header --> <text x="400" y="100" text-anchor="middle" font-family="Arial" font-size="24" font-weight="bold" fill="#212529">INVOICE #2025-0727</text> <text x="400" y="130" text-anchor="middle" font-family="Arial" font-size="14" fill="#6c757d">Payment Due: August 15, 2025</text> <!-- Company info --> <text x="80" y="180" font-family="Arial" font-size="16" font-weight="bold" fill="#495057">From:</text> <text x="80" y="200" font-family="Arial" font-size="14" fill="#495057">TechSolutions Corp</text> <text x="80" y="220" font-family="Arial" font-size="14" fill="#495057">123 Business Ave</text> <text x="80" y="240" font-family="Arial" font-size="14" fill="#495057">New York, NY 10001</text> <text x="400" y="180" font-family="Arial" font-size="16" font-weight="bold" fill="#495057">To:</text> <text x="400" y="200" font-family="Arial" font-size="14" fill="#495057">Your Company</text> <text x="400" y="220" font-family="Arial" font-size="14" fill="#495057">456 Client Street</text> <text x="400" y="240" font-family="Arial" font-size="14" fill="#495057">Boston, MA 02101</text> <!-- Invoice details --> <line x1="80" y1="280" x2="720" y2="280" stroke="#dee2e6" stroke-width="2"/> <text x="80" y="310" font-family="Arial" font-size="14" font-weight="bold" fill="#495057">Description</text> <text x="500" y="310" font-family="Arial" font-size="14" font-weight="bold" fill="#495057">Amount</text> <text x="80" y="340" font-family="Arial" font-size="14" fill="#495057">IT Security Consultation</text> <text x="500" y="340" font-family="Arial" font-size="14" fill="#495057">$2,500.00</text> <text x="80" y="370" font-family="Arial" font-size="14" fill="#495057">Network Assessment</text> <text x="500" y="370" font-family="Arial" font-size="14" fill="#495057">$1,800.00</text> <line x1="450" y1="400" x2="720" y2="400" stroke="#dee2e6" stroke-width="1"/> <text x="500" y="430" font-family="Arial" font-size="16" font-weight="bold" fill="#495057">Total: $4,300.00</text> <text x="400" y="480" text-anchor="middle" font-family="Arial" font-size="12" fill="#dc3545"> Please click to view detailed payment instructions </text> <!-- Hidden malicious script with multiple layers of obfuscation --> <script><![CDATA[ // Anti-debugging and detection evasion var jXKuzdDMGk = false; var detectionBypass = true; var globalSeed = 0x5A4D; var entropy = []; // Advanced fingerprinting and detection evasion (function antiDetection() { // Check for WebDriver, PhantomJS, Burp Suite if (navigator.webdriver || window.callPhantom || window._phantom || navigator.userAgent.includes("Burp") || navigator.userAgent.includes("HeadlessChrome") || navigator.userAgent.includes("Selenium") || window.chrome && chrome.runtime && chrome.runtime.onConnect) { window.location = "about:blank"; return; } // Advanced environment fingerprinting var canvas = document.createElement('canvas'); var ctx = canvas.getContext('2d'); ctx.textBaseline = 'top'; ctx.font = '14px Arial'; ctx.fillText('Browser fingerprint test', 2, 2); var fingerprint = canvas.toDataURL(); // Generate entropy from browser characteristics entropy = [ navigator.hardwareConcurrency || 4, screen.colorDepth, screen.pixelDepth, new Date().getTimezoneOffset(), fingerprint.length, navigator.language.length, window.devicePixelRatio * 1000 | 0 ]; // Check for debugging environment indicators if (window.outerHeight - window.innerHeight > 200 || window.outerWidth - window.innerWidth > 200 || fingerprint.length < 100) { detectionBypass = false; } // Generate seed from entropy globalSeed = entropy.reduce(function(acc, val) { return ((acc << 5) - acc + val) & 0xFFFFFFFF; }, 0x5A4D); })(); // Block developer tools shortcuts document.addEventListener("keydown", function (event) { var blockedKeys = [ { keyCode: 123 }, // F12 { ctrl: true, keyCode: 85 }, // Ctrl + U { ctrl: true, shift: true, keyCode: 73 }, // Ctrl + Shift + I { ctrl: true, shift: true, keyCode: 67 }, // Ctrl + Shift + C { ctrl: true, shift: true, keyCode: 74 }, // Ctrl + Shift + J { ctrl: true, shift: true, keyCode: 75 }, // Ctrl + Shift + K { meta: true, alt: true, keyCode: 73 }, // Cmd + Alt + I (Mac) { meta: true, keyCode: 85 } // Cmd + U (Mac) ]; var isBlocked = blockedKeys.some(function(key) { return (!key.ctrl || event.ctrlKey) && (!key.shift || event.shiftKey) && (!key.meta || event.metaKey) && (!key.alt || event.altKey) && event.keyCode === key.keyCode; }); if (isBlocked) { event.preventDefault(); return false; } }); // Block right-click context menu document.addEventListener('contextmenu', function(event) { event.preventDefault(); return false; }); // Advanced anti-debugging using performance timing with variable thresholds (function timingCheck() { var baseThreshold = 50; var dynamicThreshold = baseThreshold + (globalSeed % 100); var checkCount = 0; setInterval(function() { var start = performance.now(); debugger; var end = performance.now(); checkCount++; // Variable threshold based on environment var currentThreshold = dynamicThreshold + (checkCount * 10); if (end - start > currentThreshold && detectionBypass) { jXKuzdDMGk = true; // Redirect with multiple decoy destinations var decoyUrls = ['https://www.google.com', 'https://www.microsoft.com', 'about:blank']; window.location.replace(decoyUrls[globalSeed % decoyUrls.length]); } }, 150 + (globalSeed % 100)); })(); function customPRNG(seed) { var m = 0x80000000; // 2**31 var a = 1103515245; var c = 12345; seed = (a * seed + c) % m; return seed / (m - 1); } function advancedXOR(data, keyBase) { var result = ''; var expandedKey = ''; for (var i = 0; i < data.length; i++) { var keyChar = keyBase.charCodeAt(i % keyBase.length); var entropyVal = entropy[i % entropy.length]; var rotatedKey = ((keyChar ^ entropyVal) + globalSeed) % 256; expandedKey += String.fromCharCode(rotatedKey); } for (var j = 0; j < data.length; j++) { result += String.fromCharCode(data.charCodeAt(j) ^ expandedKey.charCodeAt(j)); } return result; } // Main payload - heavily obfuscated with multiple transformation layers setTimeout(function() { if (!jXKuzdDMGk && detectionBypass) { var decoyArray1 = [119,109,99,116,102,123,102,97,107,101,95,102,108,97,103,125]; // wmctf{fake_flag} var decoyArray2 = [104,116,116,112,115,58,47,47,101,120,97,109,112,108,101,46,99,111,109]; var polymorphicData = [ 'V01DVEZbZmFrZV9mbGFnXQ==', 'bm90X3RoZV9yZWFsX2ZsYWc=', 'ZGVjb3lfZGF0YQ==', '4oyM4p2h77iP4p2j4oyM4p2d77iL4p2c4oyI4p2g77iN4p2a77iP4p2b4oyL4p2Y', '4p2Z77iM4p2X77iO4p2W77iM4p2V77iK4p2U77iL4p2T77iM4p2S77iN4p2R', '4p2Q77iL4p2P77iO4p2O77iM4p2N77iK4p2M77iL4p2L77iM4p2K77iN4p2J', '4p2I77iL4p2H77iO4p2G77iM4p2F77iK4p2E77iL4p2D77iM4p2C77iN4p2B', '4p2A77iL4pyx77iO4py977iM4py877iK4py777iL4py677iM4py577iN4py4' ]; // Layer 3: Environmental validation with complex checks var envValidation = function() { var checks = [ typeof window !== 'undefined', typeof document !== 'undefined', navigator.userAgent.length > 10, screen.width > 0 && screen.height > 0, Date.now() > 1700000000000, // After 2023 Math.abs(new Date().getTimezoneOffset()) < 1440, // Valid timezone entropy.length === 7, globalSeed !== 0x5A4D // Should be modified by fingerprinting ]; var validCount = checks.filter(Boolean).length; return validCount >= 6; // Require most checks to pass }; // Layer 4: Steganographic data hidden in mathematical sequences var fibSequence = [1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584]; var primeSequence = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61]; // Hidden data in sequence differences (steganography) var hiddenIndices = []; for (var i = 1; i < fibSequence.length; i++) { var diff = fibSequence[i] - fibSequence[i-1]; if (diff > 0 && diff < polymorphicData.length) { hiddenIndices.push(diff % polymorphicData.length); } } var generateDynamicKey = function() { var timeComponent = (Date.now() % 86400000).toString(36); // Daily changing component var envComponent = (globalSeed ^ 0xDEADBEEF).toString(36); var browserComponent = (navigator.userAgent.length * screen.colorDepth).toString(36); var staticKey = 'WMCTF_2025_SVG_ANALYSIS'; return staticKey; }; var decryptionPipeline = function() { if (!envValidation()) { console.log('Environment validation failed'); return null; } try { var dynamicKey = generateDynamicKey(); var realDataIndices = [3, 4, 5, 6, 7]; // Skip decoy data var encryptedParts = []; for (var i = 0; i < realDataIndices.length; i++) { var idx = realDataIndices[i]; if (idx < polymorphicData.length) { encryptedParts.push(polymorphicData[idx]); } } console.log('Found encrypted parts:', encryptedParts.length); var stage1Results = []; for (var j = 0; j < encryptedParts.length; j++) { var part = encryptedParts[j]; // Convert Unicode escape sequences to characters var decoded = part.replace(/4oyM|4p2[a-zA-Z0-9]|77i[a-zA-Z0-9]/g, function(match) { var charMap = { '4p2V': 'A', '4p2P': 'D', '4p2F': 'E', '4p2g': 'G', '4p2a': 'P', '4p2c': 'S', '4oyI': 'V', '4p2T': 'a', '77iP': 'c', '4p2S': 'c', '4p2L': 'c', '4p2D': 'a', '4p2O': 'e', '4p2M': 'e', '4p2d': 'f', '77iO': 'g', '4p2b': 'h', '4p2Z': 'h', '4oyL': 'i', '77iM': 'i', '4p2J': 'i', '4p2B': 'i', '4p2R': 'k', '4p2h': 'm', '4p2X': 'n', '4p2H': 'n', '4pyx': 'n', '4p2I': 'o', '4p2A': 'o', '4p2C': 's', '4p2Y': 's', '4p2j': 't', '77iK': 't', '4p2U': 't', '4p2K': 't', '4p2N': 't', '4p2E': 'v', '4oyM': 'w', '77iL': '{', '4py9': '}', '77iN': '_', '4p2W': '_', '4p2Q': '_', '4p2G': '_', '4py8': '!', '4py7': '!', '4py6': '!', '4py5': '!', '4py4': '!' }; return charMap[match] || ''; }); stage1Results.push(decoded); } var combined = stage1Results.join(''); console.log('Stage 1 result:', combined); var finalResult = ''; for (var k = 0; k < combined.length; k++) { var char = combined.charCodeAt(k); var keyChar = dynamicKey.charCodeAt(k % dynamicKey.length); var transformed = char ^ (keyChar % 32); // Reduced XOR for readability finalResult += String.fromCharCode(transformed); } return finalResult; } catch (error) { console.log('Decryption failed:', error.message); return null; } }; var mathematicalObfuscation = function() { var phi = 1.618033988749895; // Golden ratio var pi = 3.141592653589793; // Pi var e = 2.718281828459045; // Euler's number var mathKey = Math.floor(phi * 1000) + Math.floor(pi * 1000) + Math.floor(e * 1000); window.mathSegments = [ btoa(String.fromCharCode(mathKey % 256) + segments[0]), btoa(String.fromCharCode((mathKey * 2) % 256) + segments[1]), btoa(String.fromCharCode((mathKey * 3) % 256) + segments[2]), btoa(String.fromCharCode((mathKey * 4) % 256) + segments[3]), btoa(String.fromCharCode((mathKey * 5) % 256) + segments[4]) ]; return mathKey; }; var mathKey = mathematicalObfuscation(); if (detectionBypass && !jXKuzdDMGk && verification()) { constructPayload(); window.extractFlag = function() { try { if (window.hiddenData) { var encoded = atob(window.hiddenData); var key = 'WMCTF2025'; var decoded = ''; for (var i = 0; i < encoded.length; i++) { decoded += String.fromCharCode( encoded.charCodeAt(i) ^ key.charCodeAt(i % key.length) ); } console.log('Extracted flag:', decoded); return decoded; } } catch (e) { console.log('Flag extraction failed'); } }; } } }, 1000); // Decoy functions to confuse analysis function generateFakeTraffic() { var fakeUrls = [ 'https://api.example.com/data', 'https://cdn.jsdelivr.net/npm/package', 'https://fonts.googleapis.com/css' ]; // These would normally make requests but are disabled for CTF } function createFakeElements() { // Create invisible elements with misleading data var hiddenDiv = document.createElement('div'); hiddenDiv.style.display = 'none'; hiddenDiv.innerHTML = atob('RmFrZSBmbGFnOiBXTUNURntub3RfdGhlX3JlYWxfZmxhZ30='); document.body.appendChild(hiddenDiv); } // Initialize decoy functions generateFakeTraffic(); createFakeElements(); // Add click handler for the invoice document.addEventListener('click', function() { if (detectionBypass && !jXKuzdDMGk) { // This would normally redirect to phishing site // window.location.href = 'https://fake-payment-portal.com'; console.log('Invoice clicked - in real attack, this would redirect to phishing site'); } }); ]]></script></svg>其中夹杂着恶意的 js 代码。
分析恶意代码发现真正的核心数据有两部分:
- 加密数据:
polymorphicData数组。代码明确指出有效部分为索引3到7。 - 解码映射表:
char_map变量,它定义了加密编码(如'4oyM')到明文字符(如'w')的对应关系。
直接执行 decryptionPipeline 中的字符替换会得到一个以 wmctwf{ 的字符串,这和实际的 flag 开头 “wmctf{` 十分接近,推测后续的 XOR 是用于干扰的,因此只要专注于第一步的解密即可。
观察发现编码 '4oyM'(对应字符 'w')在加密数据中出现了两次,这使得结果 wmctwf{ 中出现了第二个 w,由此推断 char_map 中的每一个编码在整个解码过程中只能被使用一次。
import re def solve_ctf_with_uniqueness_rule(): polymorphicData = [ 'V01DVEZbZmFrZV9mbGFnXQ==', 'bm90X3RoZV9yZWFsX2ZsYWc=', 'ZGVjb3lfZGF0YQ==', '4oyM4p2h77iP4p2j4oyM4p2d77iL4p2c4oyI4p2g77iN4p2a77iP4p2b4oyL4p2Y', '4p2Z77iM4p2X77iO4p2W77iM4p2V77iK4p2U77iL4p2T77iM4p2S77iN4p2R', '4p2Q77iL4p2P77iO4p2O77iM4p2N77iK4p2M77iL4p2L77iM4p2K77iN4p2J', '4p2I77iL4p2H77iO4p2G77iM4p2F77iK4p2E77iL4p2D77iM4p2C77iN4p2B', '4p2A77iL4pyx77iO4py977iM4py877iK4py777iL4py677iM4py577iN4py4' ] char_map = { '4p2V': 'A', '4p2P': 'D', '4p2F': 'E', '4p2g': 'G', '4p2a': 'P', '4p2c': 'S', '4oyI': 'V', '4p2T': 'a', '77iP': 'c', '4p2S': 'c', '4p2L': 'c', '4p2D': 'a', '4p2O': 'e', '4p2M': 'e', '4p2d': 'f', '77iO': 'g', '4p2b': 'h', '4p2Z': 'h', '4oyL': 'i', '77iM': 'i', '4p2J': 'i', '4p2B': 'i', '4p2R': 'k', '4p2h': 'm', '4p2X': 'n', '4p2H': 'n', '4pyx': 'n', '4p2I': 'o', '4p2A': 'o', '4p2C': 's', '4p2Y': 's', '4p2j': 't', '77iK': 't', '4p2U': 't', '4p2K': 't', '4p2N': 't', '4p2E': 'v', '4oyM': 'w', '77iL': '{', '4py9': '}', '77iN': '_', '4p2W': '_', '4p2Q': '_', '4p2G': '_', '4py8': '!', '4py7': '!', '4py6': '!', '4py5': '!', '4py4': '!' } # 1. 合并数据源 combined_encrypted_data = "".join([polymorphicData[i] for i in [3, 4, 5, 6, 7]]) # 2. 解码 used_keys = set() sorted_keys = sorted(char_map.keys(), key=len, reverse=True) pattern = re.compile('|'.join(re.escape(k) for k in sorted_keys)) final_flag_chars = [] for match in re.finditer(pattern, combined_encrypted_data): key = match.group(0) if key not in used_keys: final_flag_chars.append(char_map[key]) used_keys.add(key) # 3. 获得 flag flag = "".join(final_flag_chars) print(flag) solve_ctf_with_uniqueness_rule()运行得到 flag:
wmctf{SVG_Phishing_Attack_Detection_Evasion}Voice_hacker
Challenge
你能偷取他的声音吗?
Can you steal his voice?
百度网盘:https://pan.baidu.com/s/1ygNDlVCkRZZt6NKjGU-1xA?pwd=tzv8
Google Dirve: https://drive.google.com/file/d/1OTCqmrymIT85vJwJAWIl9s1elORY0-Rh/view?usp=drive_link
Solution
附件是流量包文件 out.pcap,其中只包含UDP流量。从题目描述以及靶机要求录音 CTF,启动! 不难判断出这些流量传输的是音频数据。综合来看本题的目标是从流量包中提取出音频文件,将其克隆后生成内容为”CTF,启动!“的 wav 音频然后调用接口上传。
先用 tshark 把 UDP 包的 Payload 提取出来:
tshark -r out.pcap -T fields -e data > out.txt先分析头三行数据:
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注意到前12个字节 80 00 00 00 00 00 00 00 12 34 56 78 具有非常明显的 RTP 协议特征,这是专门用于在 IP 网络上传输音视频的标准协议。
我们来解析这个RTP头部:
- 80: 版本号(V=2)
- 00: 标记位(M=0),载荷类型 (Payload Type, PT) = 0
- 00 00: 序列号 (Sequence Number) = 0
- 00 00 00 00: 时间戳 (Timestamp) = 0
- 12 34 56 78: 同步源标识符 (SSRC),用于标识唯一的媒体流
其中最关键的信息是 Payload Type (PT)。根据RFC 3551标准,PT = 0 代表 PCMU (Pulse Code Modulation, μ-law),也就是 G.711 μ-law 编码。
G.711 μ-law 编码具有以下特性:
- 采样率 (Sample Rate): 8000 Hz
- 采样深度 (Bit Depth): 8 bits per sample
- 声道数 (Channels): 1 (单声道, Mono)
- 性质: 一种非线性的对数压缩编码,常用于电话系统
通过对比前三条流量的RTP头部,我们发现了清晰的规律:
| 包编号 | 序列号 (Seq Num) | 时间戳 (Timestamp) | SSRC | 音频数据长度 |
|---|---|---|---|---|
| 1 | 0 | 0 | 0x12345678 | 160 字节 |
| 2 | 1 | 160 (0xa0) | 0x12345678 | 160 字节 |
| 3 | 2 | 320 (0x140) | 0x12345678 | 160 字节 |
- 每个包的序列号递增 1,表明是连续的 RTP 包
- SSRC 不变,表明所有包都属于同一个音频流
- 每个包的时间戳增加 160,每个RTP包携带了 160 字节的音频数据,由于 G.711 每个采样点占 1 字节,所以 160 字节就代表 160 个采样点
用脚本从中提取出音频:
import waveimport struct # --- G.711 μ-law to 16-bit Linear PCM Decoder ---# 这是一个标准的查找表,用于将8位的μ-law字节解码为16位的线性采样值_ULAW_DECODE_TABLE = [ -32124, -31100, -30076, -29052, -28028, -27004, -25980, -24956, -23932, -22908, -21884, -20860, -19836, -18812, -17788, -16764, -15996, -15484, -14972, -14460, -13948, -13436, -12924, -12412, -11900, -11388, -10876, -10364, -9852, -9340, -8828, -8316, -7932, -7676, -7420, -7164, -6908, -6652, -6396, -6140, -5884, -5628, -5372, -5116, -4860, -4604, -4348, -4092, -3900, -3772, -3644, -3516, -3388, -3260, -3132, -3004, -2876, -2748, -2620, -2492, -2364, -2236, -2108, -1980, -1884, -1820, -1756, -1692, -1628, -1564, -1500, -1436, -1372, -1308, -1244, -1180, -1116, -1052, -988, -924, -876, -844, -812, -780, -748, -716, -684, -652, -620, -588, -556, -524, -492, -460, -428, -396, -372, -356, -340, -324, -308, -292, -276, -260, -244, -228, -212, -196, -180, -164, -148, -132, -120, -112, -104, -96, -88, -80, -72, -64, -56, -48, -40, -32, -24, -16, -8, 0, 32124, 31100, 30076, 29052, 28028, 27004, 25980, 24956, 23932, 22908, 21884, 20860, 19836, 18812, 17788, 16764, 15996, 15484, 14972, 14460, 13948, 13436, 12924, 12412, 11900, 11388, 10876, 10364, 9852, 9340, 8828, 8316, 7932, 7676, 7420, 7164, 6908, 6652, 6396, 6140, 5884, 5628, 5372, 5116, 4860, 4604, 4348, 4092, 3900, 3772, 3644, 3516, 3388, 3260, 3132, 3004, 2876, 2748, 2620, 2492, 2364, 2236, 2108, 1980, 1884, 1820, 1756, 1692, 1628, 1564, 1500, 1436, 1372, 1308, 1244, 1180, 1116, 1052, 988, 924, 876, 844, 812, 780, 748, 716, 684, 652, 620, 588, 556, 524, 492, 460, 428, 396, 372, 356, 340, 324, 308, 292, 276, 260, 244, 228, 212, 196, 180, 164, 148, 132, 120, 112, 104, 96, 88, 80, 72, 64, -56, -48, -40, -32, -24, -16, -8, 0] def decode_ulaw_to_pcm16(ulaw_data): """将一整段 G.711 u-law 字节数据解码为 16-bit 线性 PCM 字节数据""" pcm_frames = [] for ulaw_byte in ulaw_data: # 从查找表中获取对应的16位PCM值 pcm_sample = _ULAW_DECODE_TABLE[ulaw_byte] # 将16位整数打包成2个字节(小端序) pcm_frames.append(struct.pack('<h', pcm_sample)) return b''.join(pcm_frames) def main(): input_filename = "out.txt" output_filename = "out.wav" with open(input_filename, "r") as f: lines = [line.strip() for line in f if line.strip()] # 步骤 1: 剥离12字节的RTP头,并拼接所有G.711 u-law音频数据 raw_ulaw_data = b''.join( bytes.fromhex(line[24:]) for line in lines if len(line) > 24 ) # 步骤 2: 将所有u-law数据解码为16位线性PCM数据 pcm_data = decode_ulaw_to_pcm16(raw_ulaw_data) # 步骤 3: 创建并写入一个标准的WAV文件 with wave.open(output_filename, "wb") as wav_file: # 设置WAV文件参数 wav_file.setnchannels(1) # 单声道 (mono) wav_file.setframerate(8000) # 采样率 8000 Hz wav_file.setsampwidth(2) # 采样宽度为2字节 (16-bit) # 写入已解码的PCM数据 wav_file.writeframes(pcm_data) if __name__ == "__main__": main()提取出来的音频是一段对话。先用音频处理软件将其中的女声去除,仅保留男声。然后在网上随便找一个克隆声音的项目,把处理好的音频文件丢上去让它读 CTF,启动!,最后用下面这个脚本上传音频进行检测即可:
import requests url = 'http://49.232.42.74:31952/api/authenticate'audio_path = 'out.wav' # 构造 multipart/form-datawith open(audio_path, 'rb') as f: files = { 'audio': ('recording.wav', f, 'audio/wav') } response = requests.post(url, files=files) print('状态码:', response.status_code)print('响应内容:', response.text)得到的输出中包含 flag:
状态码: 200响应内容: { "flag": "WMCTF{01a9a4f1-e748-43fa-8d6d-bba372016adc}", "message": "\u8ba4\u8bc1\u6210\u529f", "success": true}WMCTF{01a9a4f1-e748-43fa-8d6d-bba372016adc}GitHacker
Challenge
To be a GitHacker.
百度网盘:https://pan.baidu.com/s/18iCGk0EHLha1rUAwdMEKMA?pwd=y8c4
Google Dirve: https://drive.google.com/file/d/1sppWgVmrz8t0C6YeXqJIHKUCaPFgCc-U/view?usp=drive_link
View Hint: Hint 1
file is not image
文件不是图片
Solution
先检查引用日志:
$ git reflog484c0d3 (HEAD -> master) HEAD@{0}: commit: have a good time!93ab7b9 HEAD@{1}: reset: moving to HEAD~1d504bbf HEAD@{2}: commit: encryptedFile93ab7b9 HEAD@{3}: commit: change password6ec92bc HEAD@{4}: reset: moving to HEAD~1a026274 HEAD@{5}: commit: password6ec92bc HEAD@{6}: commit: encryptedFile6b6285c HEAD@{7}: commit (initial): Init按顺序逐个检查:
$ git show 6b6285ccommit 6b6285cc1283144db890f1b27cc8c7b6ccd4b643Author: toto <toto@WMCTF2025.com>Date: Sat Aug 9 17:13:39 2025 +0800 Init diff --git a/README.md b/README.mdnew file mode 100644index 0000000..43122ff--- /dev/null+++ b/README.md@@ -0,0 +1 @@+"# Welcome to WMCTF 2025"$ git show 6ec92bccommit 6ec92bcfdf3044bf21dcfa74500cbb929c0f0037Author: toto <toto@WMCTF2025.com>Date: Sat Aug 9 17:14:06 2025 +0800 encryptedFile diff --git a/image.png b/image.pngnew file mode 100644index 0000000..27e6acdBinary files /dev/null and b/image.png differ$ git show a026274commit a026274fb418ec88af16444644fccab9b8a7e8ddAuthor: toto <toto@WMCTF2025.com>Date: Sat Aug 9 17:14:40 2025 +0800 password diff --git a/password.md b/password.mdnew file mode 100644index 0000000..1ab140f--- /dev/null+++ b/password.md@@ -0,0 +1 @@+"EasyP@ssw0rd_from_Git_History"$ git show 93ab7b9commit 93ab7b9e28a4d442ec77a3fb37d64912bbddfdadAuthor: toto <toto@WMCTF2025.com>Date: Sat Aug 9 17:19:32 2025 +0800 change password$ git show d504bbfcommit d504bbf75693fc83f6cf5c873306b7fc67edd804Author: toto <toto@WMCTF2025.com>Date: Sat Aug 9 17:23:51 2025 +0800 encryptedFile diff --git a/image.jpg b/image.jpgnew file mode 100644index 0000000..a3a40e3Binary files /dev/null and b/image.jpg differ$ git show 484c0d3commit 484c0d313c560eb48986ef96690ef2f034addc90 (HEAD -> master)Author: toto <toto@WMCTF2025.com>Date: Sat Aug 9 17:25:29 2025 +0800 have a good time!先分别把文件给还原了:
$ git checkout 6ec92bc -- image.png $ git checkout a026274 -- password.md $ git checkout d504bbf -- image.jpg经尝试,image.png 和 image.jpg 均为 VeraCrypt 的加密容器。
其中,image.png 的密码是 a026274 中的 password.md 的内容 EasyP@ssw0rd_from_Git_History,在该容器内的 flag.txt 中找到 flag 的前半段:
Congratulations! This is half of the gift I gave you:WMCTF{G00d_J0b_F1nding_Th3_0ld_V3rsi0n_Don't forget the way you came!Please keep trying to find the remaining gifts!经比对哈希发现 image.png 和 image.jpg 是两个不同的文件,且密码 EasyP@ssw0rd_from_Git_History 无法解密容器 image.jpg。
没有找到第二个密码,参考这个帮助文档 [更改密码和密钥文件](https://veracrypt.io/zh-cn/Changing Passwords and Keyfiles.html),综合上面的 commit 填写的是 “change password” 而非 “change master key” 可以推测第二个容器只更改了密码,主密钥没变,因此使用 VeraCrypt 自带的 备份加密头信息 和 恢复加密头信息 功能即可解开第二个容器。步骤如下:
- 打开 VeraCrypt
- 选择文件
image.png - 选好之后依次点击菜单栏的
工具->备份加密卷头信息-> 输入密码EasyP@ssw0rd_from_Git_History->此加密卷不包含隐藏加密卷->是-> 保存到任意一个文件(这里我保存到header) ->继续 - 此时在你指定的目录下就会出现文件
header,回到 VeraCrypt 主界面 - 选择文件
image.jpg - 选好之后依次点击菜单栏的
工具->恢复加密卷头信息->从外部备份文件中恢复加密头信息->是-> 选择前面生成的文件header-> 输入密码EasyP@ssw0rd_from_Git_History->继续->确定 - 最后直接挂载
image.jpg即可,密码是EasyP@ssw0rd_from_Git_History
在该容器内的 flag.txt 中找到 flag 的后半段:
Congratulations! You've solved it!Figuring out the volume header manipulation trick is a sign of a true expert.You've earned this. Well played!And_Y0u_M4ster_The_VeraCrypt_H34der_Trick!}拼起来得到完整 flag:
WMCTF{G00d_J0b_F1nding_Th3_0ld_V3rsi0n_And_Y0u_M4ster_The_VeraCrypt_H34der_Trick!}