一、任意给定两个素数p和q，p！= q，记 N = p * q ，构造Zn*，问（编程解决）：
1、是否每个元素都有inverse？是否成群？ 2、这个集合有多少元素？

利用有关群的理论来解答 : )

# coding=utf-8
# author:tyty
# date  :2018/3/23

from __future__ import division
import random

p = 0 #prime 1
q = 0 #prime 2

for j in range(100000):
    while(1):
        i = random.randint(1, 30)#u can set the random int whatever u want
        for num in range(2, i):
            if (i % num == 0):
                break
        else:
            p = i
            break
    while(1):
        k = random.randint(1, 30)#u can set the random int whatever u want
        for m in range(2, k):
            if (k % m == 0):
                break
        else:
            q = k
            break
    if (q != p): break
print ("random prime p: ", p, ", q: ", q)

#mod N
N = p * q
print ("  N : ", N )

#judge if sealed
#!!! set the range[1, N-1]
for i in range(1, N):
    for j in range(1, N):
        modNum = (i * j) % N
        #print (modNum)
        if ( modNum >= N and modNum < 0):
            print ("It's not sealed. Not a Group!")
print ("It's sealed!")

#judge if can be combined arbitrarily
for i in range(1, N):
    for j in range(1, N):
        for k in range(1, N):
            if (((((i*j) % N) * k) % N ) != ((((j*k) % N) * i) % N )):
                print("It's can not be combined. Not a Group!")
print ("It can be combined arbitratily!")

#we can see e is the 1
print ("Obviously, the E is 1 in the mod n* multiply exp")

#judge whether everyone has a inverse or not
inversenum = [199 for i in range(N)]
for i in range(1, N):
    for j in range(1, N):
        modNum = (i * j) % N
        if (modNum == 1): #has inverse
            inversenum[i] = 1
ifInverse = 1

#print (len(inversenum))
for i in range(1, len(inversenum)):
    #print (inversenum[i])
    if (inversenum[i] != 1):#one is not a inverse!
        ifInverse = 0
        print ("Not everyone has a inverse..., It's not a Group!")
        break
if (ifInverse):
    print ("Everyone in this set has a inverse! Also it's a Group!")

二、写一个程序，实现AES的S-box的构造。
　　难，主要难点在于求有限域上的乘法逆元，参考《密码编码学与网络安全-原理与实践（第七版）》P101 5.6.3求乘法逆元章节，没有完全理解。但是还是想尝试使用扩充欧几里得算法求得每个s盒初始化元素在有限域上的乘法逆元，如下博客链接写的很棒，关于扩充欧几里得和怎样求乘法逆元https://blog.csdn.net/Stray_Lambs/article/details/52133141
　　没写出来嘿嘿，:)
　　于是就选择《密码编码学与网络安全-原理与实践（第七版）》P104 5.6.5 使用生成元求乘法逆元的，一开始写了个很渣的，后来参考到http://www.docin.com/p-929525675.html 上有个很棒的用C语言实现代码。
　　另外，github上https://github.com/Skycker/AES 写的步骤也很明确，只是Sbox那儿直接过程黑盒过去，贴表上去就完事，其他步骤棒棒！
　　下面是整体实现：

# coding=utf-8
# author:tyty
# date  :2018/3/23

Sbox  = [0 for i in range(256)]　#sbox array
l_t   = [0 for i in range(256)]  #log table
m_t   = [0 for i in range(256)]  #pow table - decrease dimension
mid_t = [0 for i in range(256)]  #inverse table

b     = [0xf1, 0xe3, 0xc7, 0x8f, 0x1f, 0x3e, 0x7c, 0xf8]  #b vector

#initialize
p = 1
'''
Step1. the inverse of multiply operation in GF(2^8)
'''
for i in range(256):
#2 mapping table
    l_t[i] = p
    m_t[p] = i
    if (p & 0x80): #whether the b7 is 1 or not
        p = p ^ (p << 1) ^ (0x11b) #generator p - 1
    else :
        p = p ^ (p << 1) ^ 0
    #print (p)

for i in range(256):
    if (i):
        mid_t[i] = l_t[255 - m_t[i]] #255-m_t[i] is the location of inverse in table
                                     #mid_t is the val of inverse
    else:
        mid_t[i] = 0 #{00} to {00}
    #print (mid_t[i])

'''
Step2. the operation of matrix
'''
for i in range(256):
    t = 0
    tab = 0
    for j in range(8):
        m = mid = (b[j] & mid_t[i])#press b vector
        #matrix multiplication
        for k in range(8):
            n = mid>>1
            if (m != (n << 1)):
                t+=1 #locate t
            mid = n
            m = mid
        if (t % 2 > 0):
            temp = 1
            for k in range(j):
                temp = temp << 1 # carry
            tab += temp          # sum 
        t = 0
    Sbox[i] = tab ^ 0x63 #plus 0x63

print("Here is the Sbox!")#print 
for i in range(0, 257):
    if (i == 0):
        continue
    else :
        print ("%x" % Sbox[i - 1], end = "")
    if (i % 16 == 0):
        print()

代码质量差，多多见谅 : )