#include <iostream>
#include <string>
#include <memory.h>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cmath>
using namespace std;
//定义一些常变量
const int M = 26; //定义集合{a,b,...,z}的26个英文字母
//行和列均为5
const int ROW = 5;
const int COL = 5;
//定义5*5的加密矩阵
int K[ROW][COL];
//定义5*5的解密矩阵
int D[ROW][COL];
int P[ROW]; //明文单元
int C[ROW]; //密文单元
int F[ROW]; //密文解密后的单元
//三元组gcd(a,b) = ax + by = d
struct GCD
{
int x;
int y;
int d;
};
class Hill_Cipher
{
public:
//产生随机矩阵
void random_Matrix();
//求矩阵的行列式
int Det(int matrix[ROW][ROW], int row);
//求两个数的最大公约数
int gcd(int a, int b);
/*
*判断矩阵K是否在模26的情况下可逆
*因为矩阵在模26的情形下存在可逆矩阵的充分必要条件是
*gcd(det K,26) = 1
*/
bool Inverse(int matrix[ROW][ROW]);
//矩阵相乘
void multiphy(int matrix[ROW][ROW], int p[ROW], int row);
//求出伴随矩阵
void adjoint_matrix(int matrix[ROW][ROW], int row);
//将明文加密为密文
string encryption(string plaintext);
//将密文解密为明文(为了辨识清楚,我们统一以小写字母作为明文,大写字母作为密文)
string deciphering(string ciphertext);
//欧几里得算法求模的逆
GCD extended_Euclid(int a, int b);
//模逆运算
int inverse(int a, int m);
//由于C++不存在负数取模的内置函数,现在自己设定一个
//定义一个模M的值
int Mod(int a);
};
void Hill_Cipher::random_Matrix()
{
int i, j;
for (i = 0; i < ROW; i++)
{
for (j = 0; j < COL; j++)
{
K[i][j] = rand() % 26; //产生一个5*5模26的矩阵
}
}
}
//求矩阵的行列式
int Hill_Cipher::Det(int matrix[ROW][ROW], int row)
{
int i, j;
int cofa[ROW][ROW]; //用于存放余子阵
int l; //l为所递归的余子阵的行
int p = 0, q = 0;
int sum = 0;
//由于行和列相同(方阵),所以行列式的值一定存在,故不需要判断是否为方阵
//递归基
if (row == 1)
return matrix[0][0];
for (i = 0; i < row; i++)
{
for (l = 0; l < row - 1; l++)
{
if (l < i)
p = 0;
else
p = 1;
for (j = 0; j < row - 1; j++)
{
cofa[l][j] = matrix[l + p][j + 1];
}
}
//相当于(-1)^i
if (i % 2 == 0)
q = 1;
else
q = (-1);
sum = sum + matrix[i][0] * q * Det(cofa, row - 1);
}
return sum;
}
//求两个数的最大公约数
int Hill_Cipher::gcd(int a, int b)
{
int temp;
//交换两个数的大小,使得a为较大数
if (a < b)
{
temp = a;
a = b;
b = temp;
}
while (a % b)
{
temp = b;
b = a % b;
a = temp;
}
return b;
}
/*
*判断矩阵K是否在模26的情况下可逆
*因为矩阵在模26的情形下存在可逆矩阵的充分必要条件是
*gcd(det K,26) = 1
*/
bool Hill_Cipher::Inverse(int matrix[ROW][ROW])
{
if (gcd(Det(matrix, ROW), M) == 1)
return true;
else
return false;
}
void Hill_Cipher::multiphy(int matrix[ROW][ROW], int p[ROW], int row)
{
int i, j;
//先将密文单元清零
memset(C, 0, sizeof(C));
for (i = 0; i < ROW; i++)
{
for (j = 0; j < ROW; j++)
{
C[i] += P[j] * K[j][i];
}
}
}
//将明文加密为密文
string Hill_Cipher::encryption(string plaintext)
{
int i;
string ciphertext;
//将字符串转化为明文数组
for (i = 0; i < ROW; i++)
{
P[i] = plaintext[i] - 'a';
}
multiphy(K, P, ROW);
//将密文数组转化为密文
for (i = 0; i < ROW; i++)
//这里先将其模26,再翻译为对应的字母
{
C[i] = Mod(C[i]);
ciphertext += C[i] + 'A';
}
return ciphertext;
}
//求出伴随矩阵
void Hill_Cipher::adjoint_matrix(int matrix[ROW][ROW], int row)
{
int i, j, k, l;
int p, q;
p = q = 0;
int temp[ROW][ROW];
for (i = 0; i < ROW; i++)
{
for (j = 0; j < ROW; j++)
{
for (k = 0; k < ROW - 1; k++)
{
if (k < i)
p = 0;
else
p = 1;
for (l = 0; l < ROW - 1; l++)
{
if (l < j)
q = 0;
else
q = 1;
temp[k][l] = matrix[k + p][l + q];
}
}
D[j][i] = (int)pow(-1, (double)i + j) * Det(temp, ROW - 1);
D[j][i] = Mod(D[j][i]);
}
}
}
//将密文解密为明文(为了辨识清楚,我们统一以小写字母作为明文,大写字母作为密文)
string Hill_Cipher::deciphering(string ciphertext)
{
//求出矩阵的逆
string text;
int determinant = Det(K, ROW);
int inver = inverse(determinant, 26);
adjoint_matrix(K, ROW); //伴随矩阵
cout << "行列式的值: " << determinant << endl;
int i, j;
memset(F, 0, sizeof(F));
for (i = 0; i < ROW; i++)
{
for (j = 0; j < ROW; j++)
{
F[i] += C[j] * D[j][i];
}
F[i] *= inver;
F[i] = Mod(F[i]); //算到的结果要模去26
}
for (i = 0; i < ROW; i++)
text += F[i] + 'a';
return text;
}
GCD Hill_Cipher::extended_Euclid(int a, int b)
{
GCD aa, bb;
if (b == 0)
{
aa.x = 1;
aa.y = 0;
aa.d = a;
return aa;
}
else
{
bb = extended_Euclid(b, a % b);
aa.x = bb.y;
aa.y = bb.x - (a / b) * bb.y;
aa.d = bb.d;
}
return aa;
}
int Hill_Cipher::inverse(int a, int m)
{
GCD aa;
aa = extended_Euclid(a, m);
return aa.x;
}
int Hill_Cipher::Mod(int a)
{
return a >= 0 ? a % M : (M + a % M);
}
int main()
{
int i, j;
Hill_Cipher hh;
cout << "使用希尔密码进行消息的加解密:" << endl;
//srand()函数产生一个以当前时间开始的随机种子.以保证每次产生的随机数矩阵都不相同
srand((unsigned)time(0));
hh.random_Matrix();
while (!hh.Inverse(K))
{
hh.random_Matrix();
}
cout << "随机产生5*5的矩阵:" << endl;
for (i = 0; i < ROW; i++)
{
for (j = 0; j < COL; j++)
{
printf("%2d ", K[i][j]);
}
cout << endl;
}
cout << "该矩阵模26可逆,因此可以作为密钥." << endl;
cout << endl;
//利用所选密钥，对给定的5元明文信息进行加解密
string plaintext, ciphertext;
cout << "请输入5元明文信息:" << endl;
cin >> plaintext;
ciphertext = hh.encryption(plaintext);
cout << endl;
cout << "该明文通过希尔密码法加密过后,输出的密文消息为:" << endl;
cout << ciphertext << endl;
cout << endl;
cout << "***输入0:退出 ***" << endl;
cout << "***输入1:查看明文空间对***" << endl;
cout << "***输入2:查看密文空间对***" << endl;
cout << "***输入3:查看密钥 ***" << endl;
cout << "***输入4:将消息解密 ***" << endl;
cout << "***输入5:查看菜单 ***" << endl;
char c;
while (cin >> c)
{
if (c == '0')
{
cout << endl;
cout << "退出" << endl;
break;
}
else if (c == '1')
{
cout << "明文空间:" << endl;
for (i = 0; i < ROW; i++)
cout << P[i] << " ";
cout << endl;
cout << endl;
}
else if (c == '2')
{
cout << "密文空间:" << endl;
for (i = 0; i < ROW; i++)
cout << C[i] << " ";
cout << endl;
cout << endl;
}
else if (c == '3')
{
cout << "密钥:" << endl;
for (i = 0; i < ROW; i++)
{
for (j = 0; j < ROW; j++)
{
printf("%2d ", K[i][j]);
}
cout << endl;
}
cout << endl;
}
else if (c == '4')
{
hh.adjoint_matrix(K, ROW);
string ss;
ss = hh.deciphering(ciphertext);
cout << "该密文解密过后,显示的原来的明文消息:" << endl;
cout << ss << endl;
cout << endl;
}
else
{
cout << "***输入0:退出 ***" << endl;
cout << "***输入1:查看明文空间对***" << endl;
cout << "***输入2:查看密文空间对***" << endl;
cout << "***输入3:查看密钥 ***" << endl;
cout << "***输入4:将消息解密 ***" << endl;
cout << "***输入5:查看菜单 ***" << endl;
}
}
return 0;
}
转载于：https://blog.csdn.net/cbacq/article/details/78337051?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522165017424616780269816148%2522%252C%2522scm%2522%253A%252220140713.130102334…%2522%257D&request_id=165017424616780269816148&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2alltop_positive~default-1-78337051.142v9control,157v4control&utm_term=%E5%B8%8C%E5%B0%94%E5%AF%86%E7%A0%81c%2B%2B&spm=1018.2226.3001.4187