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