Merkle-Hellman背包密码算法

版权声明：转载请注明出处。 https://blog.csdn.net/u014427196/article/details/49434947

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <time.h>

using namespace std;

long long gcd(long long a, long long b)
{
    if (b == 0) return a;
    else return gcd(b, a%b);
}

long long ext_gcd(long long a, long long b, long long &x, long long &y)
{
    if (a == 0 && b == 0) return -1;
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    long long d = ext_gcd(b, a%b, y, x);
    y -= a / b*x;
    return d;
}

long long mod_reverse(long long a, long long n)
{
    long long x, y;
    long long d = ext_gcd(a, n, x, y);
    if (d == 1)
        return (x%n + n) % n;
    else
        return -1;
}

struct Hellman
{
    int n;
    int a[100010];
    int b[100010];
    int c[100010];
    void solve()
    {
        cout << "*******************************" << endl;
        printf("本数据假定递增序列（1,3,5,10）\nM=20，W=7\n明文为13的加密解秘过程\n");
        cout << "*******************************" << endl<<endl<<endl;

        //scanf("%d", &n);//输入长度n
        int sum = 0;
        int pos = 1;
        n = 4;

        //随机生成超递增序列
        /*
        while (1)
        {
            srand(time(NULL));
            int tmp = rand() % 1000;
            if (tmp > sum)
            {
                a[pos++] = tmp;//私钥
                sum += tmp;
            }
            if (pos == n + 1) break;
        }
        */

        sum = 19;
        a[1] = 1;
        a[2] = 3;
        a[3] = 5;
        a[4] = 10;

        //随机生成M W
        int M = double(rand() / RAND_MAX)* (sum + 1001 - sum + 1) + sum + 1;
        int W = M - 1;
        /*
        while (gcd(M, W) != 1)
        {
            srand(time(0));
            int aa = 1;
            int bb = M - 1;
            W = rand() / RAND_MAX * (bb - aa) + aa;
        }
        */

        M = 20;
        W = 7;

        //公钥
        for (int i = 1;i <= n;i++)
        {
            b[i] = W * a[i] % M;
        }

        ///////取明文
        cout << "请输入明文：" << endl;
        int clear;
        scanf("%d", &clear);
        int len = 1;
        int num[100010];
        int tmp = clear;
        cout << "加密：" << endl;
        cout << "start" << endl;
        cout << ".............." << endl;
        cout << "end" << endl;
        //获得比特流
        while (tmp)
        {
            if (tmp & 1)
                num[len++] = 1;
            else
                num[len++] = 0;
            tmp >>= 1;
        }


        //加密c
        int c = 0;
        for (int i = 1;i < len;i++)
            c += num[len - i] * b[i];

        cout << "加密得到的数字c为：" << endl;
            cout << c << endl;

        //int d = mod_reverse(W, M) * c % M;
        int d = 3;
        c = c*d%M;
        cout << "加密得到的转换的c为：" << endl;
        cout << c << endl;
        int ans[10010];
        memset(ans, 0, sizeof(ans));

        for (int i = n;i >= 1;i--)
        {
            if (c >= a[i])
            {
                ans[i] = 1;
                c = c - a[i];
            }
        }
        cout << "解密："<< endl;
        int tt = 0;
        for (int i = 1;i <= n;i++)
        {
            printf("%d", ans[i]);
            if (ans[i]) tt += pow(2,n-i);
        }
        cout << "----" << tt << endl;
    }
}hellman;

int main()
{
    hellman.solve();
    return 0;
}