python实现SM4加解密 源码分析

 第一步：密钥拓展。

 第二步：加密

 第三步：解密（机密的过程与加密完全相同，只不过加密的时候密钥是从rk0开始的，解密是从rk31开始的）

# -*- coding: utf-8 -*-
# @Author  : 一根排骨
# @file: sm4.py
# @time: 2019/4/25 11.30
# @Software: VSCode

class sm:

    # 固定参数
    CK = [0x00070F15,0x1c232a31,0x383f464d,0x545b6269,
          0x70777e85,0x8c939aa1,0xa8afb6bd,0xc4cbd2d9,
          0xe0e7eef5,0xfc030a11,0x181f262d,0x343b4249,
          0x50575e65,0x6c737a81,0x888f969d,0xa4abb2b9,
          0xc0c7ced5,0xdce3eaf1,0xf8ff060d,0x141b2229,
          0x30373e45,0x4c535a61,0x686f767d,0x848b9299,
          0xa0a7aeb5,0xbcc3cad1,0xd8dfe6ed,0xf4fb0209,
          0x10171e25,0x2c333a41,0x484f565d,0x646b7279]
    # 常数 
    FK = [0xa3b1bac6,0x56aa3350,0x677d9197,0xb27022dc]
    # S盒
    SboxTable = [
    0xd6, 0x90, 0xe9, 0xfe, 0xcc, 0xe1, 0x3d, 0xb7, 0x16, 0xb6, 0x14, 0xc2, 0x28, 0xfb, 0x2c, 0x05,
    0x2b, 0x67, 0x9a, 0x76, 0x2a, 0xbe, 0x04, 0xc3, 0xaa, 0x44, 0x13, 0x26, 0x49, 0x86, 0x06, 0x99,
    0x9c, 0x42, 0x50, 0xf4, 0x91, 0xef, 0x98, 0x7a, 0x33, 0x54, 0x0b, 0x43, 0xed, 0xcf, 0xac, 0x62,
    0xe4, 0xb3, 0x1c, 0xa9, 0xc9, 0x08, 0xe8, 0x95, 0x80, 0xdf, 0x94, 0xfa, 0x75, 0x8f, 0x3f, 0xa6,
    0x47, 0x07, 0xa7, 0xfc, 0xf3, 0x73, 0x17, 0xba, 0x83, 0x59, 0x3c, 0x19, 0