C语言实现DES加密解密
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
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<<"***********************"<>choice;
if(choice==1)
{
k="";
getchar();
srand((int)time(0));
cout<<"******请输入明文:******"<=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<>k;
printf("明文为:");
int len=strlen(tensix);
int sum=0;
string m="";
for(int i=0; i
四、实现结果
