C语言实现DES加密解密

不念不忘少年蓝@ 2022-03-31 16:11 261阅读 0赞

### 一、实现原理 ###

先将明文转为16进制。然后每个16进制数取二进制的前四位。取16个数字。每16个16进制数加密一次。如果不够16个，就补0；

再将密文转换为2进制，每个字母取二进制的前4位。存完后一个64位。再将这个64位二进制数字通过pc-1表映射，变成56位二进制数字。C0等于前28位，D0等于后28位。然后通过表格。变换出C1-C16,D1-D16；然后将C1-C16,D1-D16组合得到C1D1-C16D16；再将C1D1-C16D16。通过PC-2表映射得到只有48位的K1-K16。到此16轮子密钥全部生成完毕。

我们再将64位2进制明文通过IP表映射，得到新的64位2进制明文。我们取L0等于新明文的前32位，R0等于新明文的后32位。然后我们计算Ln=Rn-1，Rn=Ln-1+f(Rn-1,Kn)。Ln很好算，如何计算Rn呢，首先将Rn-1与Kn异或。Rn-1是32位的但是Kn却是48位的。所以将Rn-1通过E表映射。变成48位。最后我们有8个6位的二进制数字，也就是48位。只不过被分成了8组。我们分别对这8组二进制数字进行S1-S8表的处理。让其变成8组4位数据。最终变成32位数据。最后在与Ln-1进行异或。处理16轮。得到最终的R16L16，然后通过IP-1表映射得到最终的64位数据。最后将这64位数据变成16进制输出即可得到密文。

如何解密呢？很简单，前面的步骤全部一样，只需要在最后将16轮密钥倒置使用即可解密。

### 二、系统功能描述 ###

功能有加密，随机生成密钥，解密。将明文储存在文件中，将密钥储存在文件中，将密文储存在文件中。

### 三、实现代码 ###

#include <bits/stdc++.h>
    using namespace std;
    string k;
    struct node
    {
        int c[80];
        int d[80];
        int cd[80];
        int k_n[80];
        int l[80];
        int r[80];
        node()
        {
            memset(c,0,sizeof(c));
            memset(d,0,sizeof(d));
            memset(k_n,0,sizeof(k_n));
            memset(l,0,sizeof(l));
            memset(r,0,sizeof(r));
        }
    };
    
    char mikey[18]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    
    //IP初始置换表
    int pc_ip[80]= {0,58,50,42,34,26,18,10,2,
                    60,52,44,36,28,20,12,4,
                    62,54,46,38,30,22,14,6,
                    64,56,48,40,32,24,16,8,
                    57,49,41,33,25,17,9,1,
                    59,51,43,35,27,19,11,3,
                    61,53,45,37,29,21,13,5,
                    63,55,47,39,31,23,15,7};
    
    //置换选择PC-1
    int pc_1[60]= {0,57,49,41,33,25,17,
                   9,1,58,50,42,34,26,
                   18,10,2,59,51,43,35,
                   27,19,11,3,60,52,44,
                   36,63,55,47,39,31,23,
                   15,7,62,54,46,38,30,
                   22,14,6,61,53,45,37,
                   29,21,13,5,28,20,12,4};
    
    //置换选择PC-2
    int pc_2[60]= {0,14,17,11,24,1,5,
                   3,28,15,6,21,10,
                   23,19,12,4,26,8,
                   16,7,27,20,13,2,
                   41,52,31,37,47,55,
                   30,40,51,45,33,48,
                   44,49,39,56,34,53,
                   46,42,50,36,29,32};
    
    //E盒扩展变换
    int pc_e[80]= {0,32,1,2,3,4,5,
                   4,5,6,7,8,9,
                   8,9,10,11,12,13,
                   12,13,14,15,16,17,
                   16,17,18,19,20,21,
                   20,21,22,23,24,25,
                   24,25,26,27,28,29,
                   28,29,30,31,32,1};
    
    //P盒置换表
    int pc_p[80]= {0,16,7,20,21,
                   29,12,28,17,
                   1,15,23,26,
                   5,18,31,10,
                   2,8,24,14,
                   32,27,3,9,
                   19,13,30,6,
                   22,11,4,25};
    
    //逆初始置换表
    int pc_ip_1[80]= {0,40,8,48,16,56,24,64,32,
                      39,7,47,15,55,23,63,31,
                      38,6,46,14,54,22,62,30,
                      37,5,45,13,53,21,61,29,
                      36,4,44,12,52,20,60,28,
                      35,3,43,11,51,19,59,27,
                      34,2,42,10,50,18,58,26,
                      33,1,41,9,49,17,57,25};
    
    //S盒
    int s_box[8][4][16]={
    	14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7,
    	0,15,7,4,14,2,13,1,10,6,12,11,9,5,3,8,
    	4,1,14,8,13,6,2,11,15,12,9,7,3,10,5,0,
    	15,12,8,2,4,9,1,7,5,11,3,14,10,0,6,13,
    
    	15,1,8,14,6,11,3,4,9,7,2,13,12,0,5,10,
    	3,13,4,7,15,2,8,14,12,0,1,10,6,9,11,5,
    	0,14,7,11,10,4,13,1,5,8,12,6,9,3,2,15,
    	13,8,10,1,3,15,4,2,11,6,7,12,10,5,14,9,
    
    	10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8,
    	13,7,0,9,3,4,6,10,2,8,5,14,12,11,15,1,
    	13,6,4,9,8,15,3,0,11,1,2,12,5,10,14,7,
    	1,10,13,0,6,9,8,7,4,15,14,3,11,5,2,12,
    
    	7,13,14,3,0,6,9,10,1,2,8,5,11,12,4,15,
    	13,8,11,5,6,15,0,3,4,7,2,12,1,10,14,9,
    	10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4,
    	3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14,
    
    	2,12,4,1,7,10,11,6,8,5,3,15,13,0,14,9,
    	14,11,2,12,4,7,13,1,5,0,15,10,3,9,8,6,
    	4,2,1,11,10,13,7,8,15,9,12,5,6,3,0,14,
    	11,8,12,7,1,14,2,13,6,15,0,9,10,4,5,3,
    
    	12,1,10,15,9,2,6,8,0,13,3,4,14,7,5,11,
    	10,15,4,2,7,12,9,5,6,1,13,14,0,11,3,8,
    	9,14,15,5,2,8,12,3,7,0,4,10,1,13,11,6,
    	4,3,2,12,9,5,15,10,11,14,1,7,6,0,8,13,
    
    	4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1,
    	13,0,11,7,4,9,1,10,14,3,5,12,2,15,8,6,
    	1,4,11,13,12,3,7,14,10,15,6,8,0,5,9,2,
    	6,11,13,8,1,4,10,7,9,5,0,15,14,2,3,12,
    
    	13,2,8,4,6,15,11,1,10,9,3,14,5,0,12,7,
    	1,15,13,8,10,3,7,4,12,5,6,11,0,14,9,2,
    	7,11,4,1,9,12,14,2,0,6,10,13,15,3,5,8,
    	2,1,14,7,4,10,8,13,15,12,9,0,3,5,6,11};
    
    ///F函数实现E盒扩展
    int f(int r[80],int kn[80])
    {
        int x=0;
        int e[80]= {0};
        int h=0,l=0,idx=0;
        //E盒扩展得到e(Ri);
        for(int i=1; i<=48; i++)
            e[i]=r[pc_e[i]];
        //异或运算得到k^(Ri);
        for(int i=1; i<=48; i++)
            r[i]=e[i]^kn[i];
        //S盒压缩将48位分为8组，每组6位，得到S(k^(Ri));
        for(int i=1; i<=48; i+=6)
        {
            h=r[i]*2+r[i+5]*1;
            l=r[i+1]*8+r[i+2]*4+r[i+3]*2+r[i+4]*1;
            e[++idx]=(s_box[x][h][l]>>3)&1;
            e[++idx]=(s_box[x][h][l]>>2)&1;
            e[++idx]=(s_box[x][h][l]>>1)&1;
            e[++idx]=s_box[x][h][l]&1;
            x++;
        }
        //P盒置换
        for(int i=1; i<=32; i++)
            r[i]=e[pc_p[i]];
    }
    
    int jiami(string m,int choice)
    {
    /*
    转二进制
    功能：将16个16进制转换成16个二进制并存到two数组
          k秘钥转换成二进制并存到two_k数组
    */
        int two[66]= {0},num,idx=0,two_k[66]= {0},two_kup[66]= {0};
        for(int i=0; i<16; i++) //将16个16进制转换成16个二进制:变成64位
        {
            if(m[i]>='0'&& m[i]<='9')
                num=m[i]-'0';
            else num=m[i]-'A'+10;
            two[++idx]=(num>>3)&1;
            two[++idx]=(num>>2)&1;
            two[++idx]=(num>>1)&1;
            two[++idx]=num&1;
        }
        idx=0;
        for(int i=0; i<16; i++) //k秘钥转2进制
        {
            if(k[i]>='0'&&k[i]<='9')
                num=k[i]-'0';
            else num=k[i]-'A'+10;
            two_k[++idx]=(num>>3)&1;
            two_k[++idx]=(num>>2)&1;
            two_k[++idx]=(num>>1)&1;
            two_k[++idx]=num&1;
        }
    
    
    ///密钥加密///
    /*
    求16个子密钥
    功能：由密钥生成16轮加密所用的子密钥
    */
        ///表4-5用pc_1得到k+(将密钥用pc_1置换)
        for(int i=1; i<=56; i++)
            two_kup[i]=two_k[pc_1[i]];
        //左右分组
        node c_and_d[20];
        for(int i=1; i<=28; i++) //得到c[0];
            c_and_d[0].c[i]=two_kup[i];
        for(int i=1,j=29; j<=56; i++,j++) //得到d[0];
            c_and_d[0].d[i]=two_kup[j];
    
        ///16轮生成每轮子密钥
        for(int i=1; i<=16; i++) //得到c[1]-c[15],d[1]-d[15],CnDn;
        {
            ///表4-6的把移一位和移两位的分开
    
            //如果为第1、2、9、16轮，则d循环左移1位
            if(i==1||i==2||i==9||i==16)
            {
                for(int j=1; j<=27; j++)
                    c_and_d[i].c[j]=c_and_d[i-1].c[j+1];
                c_and_d[i].c[28]=c_and_d[i-1].c[1];
                for(int j=1; j<=27; j++)
                    c_and_d[i].d[j]=c_and_d[i-1].d[j+1];
                c_and_d[i].d[28]=c_and_d[i-1].d[1];
            }
            //如果是其他轮次，则d循环左移2位
            else
            {
                for(int j=1; j<=26; j++)
                    c_and_d[i].c[j]=c_and_d[i-1].c[j+2];
                c_and_d[i].c[27]=c_and_d[i-1].c[1];
                c_and_d[i].c[28]=c_and_d[i-1].c[2];
                for(int j=1; j<=26; j++)
                    c_and_d[i].d[j]=c_and_d[i-1].d[j+2];
                c_and_d[i].d[27]=c_and_d[i-1].d[1];
                c_and_d[i].d[28]=c_and_d[i-1].d[2];
            }
            for(int j=1; j<=28; j++)
                c_and_d[i].cd[j]=c_and_d[i].c[j];
            for(int j=29,t=1; t<=28; j++,t++)
                c_and_d[i].cd[j]=c_and_d[i].d[t];
        }
    
        ///PC_2置换后打印16轮子密钥
        node k_16[20];
        for(int i=1; i<=16; i++) //得到k1-kn;
            for(int j=1; j<=48; j++)
                k_16[i].k_n[j]=c_and_d[i].cd[pc_2[j]];
    
    
    ///明文加密 71页图4-8明文加密过程///
    /*
    DES算法主函数
    功能：实现DES算法的16轮加密
    */
        int ip[80]= {0};
        //初始变换ip
        for(int i=1; i<=64; i++)
            ip[i]=two[pc_ip[i]];
    
        //初始化得到 l[0],r[0]
        node l_r[20];
        for(int i=1; i<=32; i++) //得到l;
            l_r[0].l[i]=ip[i];
        for(int i=1,j=33; j<=64; i++,j++) //得到r;
            l_r[0].r[i]=ip[j];
    
        ///choice=0时是加密操作，choice=1时是解密操作
        //进行16轮运算
        if(choice==0)
        {
            for(int i=1; i<=16; i++) //计算L1-L16，R1-R16;
            {
                for(int j=1; j<=32; j++)
                    //左右的左等于左右的右
                    l_r[i].l[j]=l_r[i-1].r[j];
                ///F函数包含E盒扩展、异或、S盒压缩、P盒置换
                f(l_r[i-1].r,k_16[i].k_n);
                ///左右合在一起，两者进行最终按位异或得到r[i]
                for(int j=1; j<=32; j++)
                    l_r[i].r[j]=l_r[i-1].l[j]^l_r[i-1].r[j];
            }
        }
        else
        {
            for(int i=1; i<=16; i++) //计算L1-L16，R1-R16;
            {
                for(int j=1; j<=32; j++)
                    l_r[i].l[j]=l_r[i-1].r[j];
                f(l_r[i-1].r,k_16[16-i+1].k_n);
                //^异或运算符
                for(int j=1; j<=32; j++)
                    l_r[i].r[j]=l_r[i-1].l[j]^l_r[i-1].r[j];
            }
        }
    
        int R16L16[80]= {0}; //得到R16L16;
        //F函数过后左右合在一起
        for(int i=1; i<=32; i++)
            R16L16[i]=l_r[16].r[i];
        for(int i=33,j=1; j<=32; j++,i++)
            R16L16[i]=l_r[16].l[j];
    
        int ans[80]= {0}; //得到最终变换;
        //进行ip逆置换
        for(int i=1; i<=64; i++)
            ans[i]=R16L16[pc_ip_1[i]];
    
        ///choice=0时是加密操作，choice=1时是解密操作
        int tem_num;
        if(choice==0)
        {
            //字节转换成字符输出密文
            for(int i=1; i<=64; i+=4)
            {
                tem_num=ans[i]*8+ans[i+1]*4+ans[i+2]*2+ans[i+3]*1;
                if(tem_num>=10)
                    printf("%c",(tem_num-10)+'A');
                else printf("%c",tem_num+'0');
            }
        }
        else
        {
            int change[1000];
            int pos=0;
            for(int i=1; i<=64; i+=4)
            {
                tem_num=ans[i]*8+ans[i+1]*4+ans[i+2]*2+ans[i+3]*1;
                change[++pos]=tem_num;
            }
            int jieans=0;
            for(int i=1; i<=16; i+=2)
            {
                jieans+=change[i];
                jieans*=16;
                jieans+=change[i+1];
                //printf("%d %d\n",change[i],change[i+1]);
                printf("%c",jieans);
                jieans=0;
            }
        }
    }
    
    int main()
    {
        while(1)
        {
            cout<<"***********************"<<endl;
            cout<<"******加密请输入1*******"<<endl;
            cout<<"******解密请输入2*******"<<endl;
            cout<<"******退出请输入3*******"<<endl;
            cout<<"***********************"<<endl;
            int choice;
            cin>>choice;
            if(choice==1)
            {
                k="";
                getchar();
                srand((int)time(0));
                cout<<"******请输入明文:******"<<endl;
                char say[1000];
                gets(say);
                cout<<"*******正在产生随机密钥:******"<<endl;
                int num;
                for(int i=1;i<=16;i++)
                    num=rand()%16,k+=mikey[num];
                cout<<"生成成功,密钥为:"<<k<<endl;
                printf("密文为:");
                int len=strlen(say),idx=0;
                char tensix[2000],tem[1000],now=0;
                for(int i=0; i<len; i++) //文本转16进制
                {
                    while(say[i])
                    {
                        if(say[i]%16>=10) tem[++idx]=((say[i]%16)-10)+'A';
                        else tem[++idx]=(say[i]%16)+'0';
                        say[i]/=16;
                    }
                    //倒序存入
                    for(int j=idx; j>=1; j--)
                        tensix[++now]=tem[j];
                    idx=0;
                }
                while(now%16)//不够16位补0；
                {
                    ++now;
                    tensix[now]='0';
                }
                int sum=0;
                string m="";
                for(int i=1; i<=now; i++) //每64位加密一次；
                {
                    sum++;
                    m+=tensix[i];
                    //cout<<m<<endl;
                    if(sum==16)
                    {
                        sum=0;
                        jiami(m,0);
                        m="";
                    }
                }
                printf("\n");
            }
            else if(choice==2)
            {
                getchar();
                cout<<"******请输入密文:******"<<endl;
                char tensix[1000];
                gets(tensix);
                cout<<"******请输入秘钥:******"<<endl;
                cin>>k;
                printf("明文为:");
                int len=strlen(tensix);
                int sum=0;
                string m="";
                for(int i=0; i<len; i++) //每64位加密一次；
                {
                    sum++;
                    m+=tensix[i];
                    if(sum==16)
                    {
                        sum=0;
                        jiami(m,1);
                        m="";
                    }
                }
                printf("\n");
            }
            else
                return 0;
        }
        return 0;
    }

### 四、实现结果 ###

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2R5d182NjY2NjY_size_16_color_FFFFFF_t_70][]

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2R5d182NjY2NjY_size_16_color_FFFFFF_t_70]: /images/20220331/500b5b355bd04d3b952c1467a5e516ea.png