DES加密算法，c语言实现

```cpp
//des算法...真的有点难，debug哭了 
//参考博客https://www.cnblogs.com/songwenlong/p/5944139.html 
#include<bits/stdc++.h>
using namespace std;
char str[9];//输入的明文，** 
int express[65];//输入初始明文的二进制 
int l0[49],r0[49];//初始 
int l1[49],r1[49];//置换后 
int key[65];//初始**的二进制
int mov[20]={1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1};//左移位数表 
int c0[30],d0[30];//两块**
int c1[30],d1[30];//循环左移后
int key1[65];//选择置换生成的48位子** 
int er0[35];//e扩展置换后的r0 
int ans[50];//存放异或的结果 
int skey[50];//s盒替代后的结果
int pkey[50];//p盒替代后的结果 
int cipher[65];//密文 
int ip[65]={ 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
}; //ip置换表 
int nip[65]={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
}; //逆ip置换表 
int pc1[60]={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
};//置换选择表1 
int pc2[60]={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
};//置换选择表2 
int e[50]={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
};//E扩展置换表 
int s[8][4][16]={//s盒替代表 
    14,4,13,1,2,15,11,8,3,10,6,12,5,9,0,7,//1
    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,//2
    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,
    13,8,10,1,3,15,4,2,11,6,7,12,0,5,14,9,
    
    10,0,9,14,6,3,15,5,1,13,12,7,11,4,2,8,//3
    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,//4
    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,//5
    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,//6
    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,//7
    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,//8
    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
}; 
int p[32]={16,7,20,21,29,12,28,17,//p盒替代 
			 1,15,23,26,5,18,31,10,
    		 2,8,24,14,32,27,3,9,
			 19,13,30,6,22,11,4,25
};

void binarystring(char c,int k)//明文转换为二进制 
{
  int i;
  for(i=0;i<8;i++)
  { if (c & 0x80) express[k++]=1; else express[k++]=0;
    c <<= 1;
  }
}
void binarystring0(char c,int k)//**转换为二进制 
{
  int i;
  for(i=0;i<8;i++)
  { if (c & 0x80) key[k++]=1; else key[k++]=0;
    c <<= 1;
  }
}

void IPreplacement(){//ip置换加分组
	for(int i=0;i<32;i++){
		l0[i]=express[ip[i]];
	}
	for(int i=0;i<32;i++){
		r0[i]=express[ip[i+32]];
	}
}
void getkey(){//得到实际初始** 
	for(int i=0;i<28;i++){
		c0[i]=key[pc1[i]];
	} 
	for(int i=0;i<28;i++){
		d0[i]=key[pc1[i+28]];
	} 	
}
//Ln = R(n - 1)；
//Rn = L(n - 1)⊕f(Rn-1,kn-1)？？？ 
int f(int k){//**置换(Kn的生成，n=0~16)；扩展置换；S-盒代替；P-盒置换

	for(int i=0;i<28;i++){
		c1[i]=c0[(i+mov[k])%28];
		d1[i]=d0[(i+mov[k])%28];
	}
	for(int i=0;i<28;i++){
		c0[i]=c1[i];
		d0[i]=d1[i];
	}
	
	int t[60];
	for(int i=1;i<=28;i++){
		t[i]=c1[i-1];
	}
	for(int i=29;i<=56;i++){
		t[i]=d1[i-29];
	}

	for(int i=0;i<48;i++){//**置换 
		key1[i]=t[pc2[i]];
	}

	for(int i=0;i<48;i++){//e扩展置换 	
		er0[i]=r0[e[i]-1];
	}
	
	for(int i=0;i<48;i++){
		ans[i]=er0[i]^key1[i];
	} 
	
	int num=0;
	for(int i=0;i<8;i++){//s盒替代 
		int x=i*6;
		int hang=ans[x]*2+ans[x+5];
		int lie=ans[x+1]*8+ans[x+2]*4+ans[x+3]*2+ans[x+4];
		int oo=s[i][hang][lie];
		skey[num+3]=oo%2;
		oo=oo/2;
		skey[num+2]=oo%2;
		oo=oo/2;
		skey[num+1]=oo%2;
		oo=oo/2;
		skey[num]=oo%2;
		num=num+4;
	}

	
	for(int i=0;i<32;i++){//p盒替代
		pkey[i]=skey[p[i]-1];
	}
	for(int i=0;i<32;i++){
		r1[i]=pkey[i]^l0[i];
	} 
	for(int i=0;i<32;i++){
		l1[i]=r0[i];
	} 
	for(int i=0;i<32;i++){
		r0[i]=r1[i];
		l0[i]=l1[i];
	}
}

int fin(){//逆ip置换输出 
	int t[66];
	for(int i=0;i<32;i++){
		t[i]=r0[i];
		t[i+32]=l0[i];
	}
	for(int i=0;i<64;i++){
		cipher[i]=t[nip[i]-1];
	} 
    printf("加密结果为："); 
	for(int i=0;i<64;i+=4){//转十六进制输出 
		int t=cipher[i+3]+cipher[i+2]*2+cipher[i+1]*4+cipher[i]*8; 
		printf("%X",t); 
	}
//	for(int i=0;i<64;i+=8){//ascll码转字符输出 
//		int t=cipher[i+7]+cipher[i+6]*2+cipher[i+5]*4+cipher[i+4]*8+cipher[i+3]*16+cipher[i+2]*32+cipher[i+1]*64+cipher[i]*128;
//		//printf("%d",t);
//		cout<<(char)t;
//	}
}

int main(){
	printf("请输入64位明文：");
	scanf("%s",str);
	for(int i=0;i<8;i++){
		binarystring(str[i],i*8+1);
	}
	printf("请输入64位**：");
	scanf("%s",str);
	for(int i=0;i<8;i++){
		binarystring0(str[i],i*8+1);
	}
	IPreplacement();
	getkey();
    for(int i=0;i<16;i++){
    	f(i);
	}
	fin();
}