import random import hashlib from typing import Any from Crypto.PublicKey import RSA from Crypto.Cipher import PKCS1_v1_5, AES # 随机产生4个字符组成的字符串 def four_random_chart() -> str: return hex(int(65536 * (1 + random.random())))[2:][1:] # PKCS#1 v1.5 填充 + RSA 加密 def parse_jsbn_bigint(n_obj:dict[Any, int]) -> int: DB = 28 DV = 1 << DB t = n_obj['t'] result = 0 for i in range(t): result += n_obj[i] * (DV ** i) return result def encrypt_data(plaintext:str) -> str: cipher = PKCS1_v1_5.new(public_key) encrypted = cipher.encrypt(plaintext.encode('utf-8')) # 转换为十六进制，确保偶数长度 hex_result = encrypted.hex() if len(hex_result) % 2 == 1: hex_result = '0' + hex_result return hex_result def RSA_jiami_r(str_16:str) -> str: global public_key # 你的数据 n_data = { 0: 134982529, 1: 254232810, 2: 164556709, 3: 234907349, 4: 134685994, 5: 35463984, 6: 258277946, 7: 12518857, 8: 44638621, 9: 93783641, 10: 212253739, 11: 62792472, 12: 186688352, 13: 109500232, 14: 182488077, 15: 261196188, 16: 26354094, 17: 103248217, 18: 106891695, 19: 165771045, 20: 41530993, 21: 263704736, 22: 111785174, 23: 12753611, 24: 232116673, 25: 155524985, 26: 218291229, 27: 122452343, 28: 248250238, 29: 118739550, 30: 251169095, 31: 129059733, 32: 149835464, 33: 5498868, 34: 71719731, 35: 154456417, 36: 49635, 't': 37, 's': 0 } e = 65537 n = parse_jsbn_bigint(n_data) # 构造公钥 public_key = RSA.construct((n, e)) encrypted = encrypt_data(str_16) return encrypted # AES加密 # 加密模式: AES-CBC # 密钥长度: 128位 # IV: 固定为 "0000000000000000" def parse_string_to_wordarray(text:str) -> list[int]: """将字符串转换为 WordArray 格式""" length = len(text) words = [] for i in range(length): # 计算在 words 数组中的索引 word_index = i >> 2 # 相当于 i // 4 # 确保 words 数组足够长 while len(words) <= word_index: words.append(0) # 获取字符的 ASCII 码 char_code = ord(text[i]) & 0xFF # 计算位移量 shift = 24 - (i % 4) * 8 # 将字符添加到对应的 word 中 words[word_index] |= char_code << shift return words def AES_O(plaintext:str, str_16:str) -> list[int]: # 密钥 key_words = parse_string_to_wordarray(str_16) key = b''.join(w.to_bytes(4, 'big') for w in key_words) # IV iv = b'0000' * 4 # "0000000000000000" # 填充(PKCS7) pad_len = 16 - len(plaintext) % 16 plaintext_padded = plaintext.encode() + bytes([pad_len] * pad_len) # 加密 cipher = AES.new(key, AES.MODE_CBC, iv) ciphertext = cipher.encrypt(plaintext_padded) # 结果是字节数组 return list(ciphertext) # 自定义base64编码 def geetest_base64_encode(data:list[int]) -> dict[str, Any]: """极验自定义Base64编码""" # 配置 charset = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789()' pad_char = '.' # 位掩码 (这些是打乱的) masks = [7274496, 9483264, 19220, 235] mask_bits = 24 # 总位数 def extract_bits(value, mask): """根据掩码提取位""" result = 0 # 从高位到低位遍历 for i in range(mask_bits - 1, -1, -1): # 如果掩码的第i位是1 if (mask >> i) & 1: # 提取value的第i位,添加到结果 result = (result << 1) | ((value >> i) & 1) return result encoded = "" padding = "" length = len(data) # 每3字节一组 i = 0 while i < length: if i + 2 < length: # 完整3字节 block = (data[i] << 16) | (data[i + 1] << 8) | data[i + 2] # 使用4个掩码提取 encoded += charset[extract_bits(block, masks[0])] encoded += charset[extract_bits(block, masks[1])] encoded += charset[extract_bits(block, masks[2])] encoded += charset[extract_bits(block, masks[3])] i += 3 else: # 处理剩余 remainder = length - i if remainder == 2: block = (data[i] << 16) | (data[i + 1] << 8) encoded += charset[extract_bits(block, masks[0])] encoded += charset[extract_bits(block, masks[1])] encoded += charset[extract_bits(block, masks[2])] padding = pad_char elif remainder == 1: block = data[i] << 16 encoded += charset[extract_bits(block, masks[0])] encoded += charset[extract_bits(block, masks[1])] padding = pad_char + pad_char break return { "res": encoded, "end": padding } def encrypt_string(e:str, t:list[int], n:str) -> str: """ JS加密函数的Python实现 参数: e: 原始字符串 t: 加密参数数组 n: 十六进制字符串 """ if not t or not n: return e o = 0 # 偏移量 i = e # 结果字符串 s = t[0] # 12 a = t[2] # 98 _ = t[4] # 43 # 每次读取2个字符（十六进制） while o < len(n): r = n[o:o + 2] # 取2个字符 if len(r) < 2: break o += 2 # 解析十六进制 c = int(r, 16) # 转换为字符 l = chr(c) # 计算插入位置: (s * c^2 + a * c + _) % len(e) u = (s * c * c + a * c + _) % len(e) # 在位置u插入字符 i = i[:u] + l + i[u:] return i def simple_md5(message:str) -> str: """ 简化版MD5实现，结构更清晰 """ # 使用内置hashlib验证结果 def verify_result(msg): return hashlib.md5(msg.encode()).hexdigest() # 轮移位常量 shifts = [ 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21 ] # K常数（与JavaScript版本中的常数对应） K = [ 0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee, 0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501, 0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be, 0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821, 0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa, 0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8, 0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed, 0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a, 0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c, 0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70, 0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05, 0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665, 0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039, 0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1, 0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1, 0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391 ] # 实际实现... return verify_result(message)