参考引用中未提及MD4加密算法的实现方式。不过，MD4是一种较早期的哈希算法，以下是一个Python实现MD4加密的示例代码： ```python import struct # 定义常量 A = 0x67452301 B = 0xefcdab89 C = 0x98badcfe D = 0x10325476 # 定义辅助函数 def F(x, y, z): return (x & y) | ((~x) & z) def G(x, y, z): return (x & y) | (x & z) | (y & z) def H(x, y, z): return x ^ y ^ z def rotate_left(x, n): return ((x << n) | (x >> (32 - n))) & 0xFFFFFFFF def md4(message): # 填充消息 original_length = len(message) * 8 message += b'\x80' while (len(message) * 8) % 512 != 448: message += b'\x00' message += struct.pack('<Q', original_length) # 初始化变量 a = A b = B c = C d = D # 处理消息块 for i in range(0, len(message), 64): block = message[i:i+64] X = [struct.unpack('<I', block[j:j+4])[0] for j in range(0, 64, 4)] aa = a bb = b cc = c dd = d # 第一轮 for j in range(16): k = j s = [3, 7, 11, 19][j % 4] a = rotate_left((a + F(b, c, d) + X[k]), s) a, b, c, d = d, a, b, c # 第二轮 for j in range(16): k = (j % 4) * 4 + (j // 4) s = [3, 5, 9, 13][j % 4] a = rotate_left((a + G(b, c, d) + X[k] + 0x5a827999), s) a, b, c, d = d, a, b, c # 第三轮 for j in range(16): k = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15][j] s = [3, 9, 11, 15][j % 4] a = rotate_left((a + H(b, c, d) + X[k] + 0x6ed9eba1), s) a, b, c, d = d, a, b, c a = (a + aa) & 0xFFFFFFFF b = (b + bb) & 0xFFFFFFFF c = (c + cc) & 0xFFFFFFFF d = (d + dd) & 0xFFFFFFFF # 输出结果 result = struct.pack('<IIII', a, b, c, d) return result.hex() # 示例使用 message = b"Hello, World!" md4_hash = md4(message) print("MD4 Hash:", md4_hash) ```