2016年大三上-网络安全-密码学-DES算法
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/* *Copyright (c) 2014,烟台大学计算机学院 *All rights reserved. *文件名称:Annpion.cpp *作者:王耀鹏 *完成日期:2016年10月27日 *版本号:v1.0 * *问题描述:密码学-DES算法 *输入描述:64位密文和64位密钥 *输出描述:加密后的密文 */ #include <iostream>#include <math.h>using namespace std;int IP[]= {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 };int IPE[]= {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 };int PC_1[]= {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 };int PC_2[]= {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 };int E[]= {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 };int P[]= {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 };int Shift[]= {1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1};void ByteToBits(int x,char *D) //字节转比特{ char tmp[5]; int len=0; while(x) { tmp[len++]=x%2+'0'; x/=2; } int num=0; if(len<4)num=4-len; int i; for(i=0; i<num; ++i)D[i]='0'; for(int j=len-1; i<4; ++i,j--)D[i]=tmp[j]; D[4]='\0';}int BitsToByte(char *D,int n) //比特转字节{ int x,temp=n-1; double s=0,m=0; while(temp>=0) { x=D[temp--]-'0'; s+=x*pow(2,m++); } return s;}void myPermuation(char *In,char *Out,int *P,int n) //通过置换表进行置换{ for(int i=0; i<n; ++i) { *(Out+i)=*(In+(*(P+i)-1)); }}void myXOR(char *In1,char *In2,int n,char *Out) //字符串In1,In2异或，结果保存在字符串Out中{ for(int i=0; i<n; ++i) { if(In1[i]!=In2[i]) Out[i]='1'; else Out[i]='0'; } Out[n]='\0';}void S_Box(char *In1,char *In2,char *Out) //可以将S盒数据放在函数体内{ int S[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 } }; int i,j,s=0; char a[8][6],Out1[50],Out2[50],Out3[50]; myPermuation(In1,Out1,E,48); myXOR(In2,Out1,48,Out2); for(i=0; i<8; ++i) for(j=0; j<6; ++j) a[i][j]=Out2[i*6+j]; for(i=0; i<8; ++i) { int x,y,temp; char b[4]; b[0]=a[i][0]; b[1]=a[i][5]; x=BitsToByte(b,2); for(j=0; j<4; ++j) b[j]=a[i][j+1]; y=BitsToByte(b,4); temp=S[i][x][y]; ByteToBits(temp,b); for(j=0; j<4; ++j) Out3[s++]=b[j]; } myPermuation(Out3,Out,P,32);}void myShift(char *In,char *Out,int n,int s) //字符串循环左移{ int temp,i; char a[100]; for(i=0; i<n; ++i) { a[i]=In[i]; a[i+n]=In[i]; } temp=s%n; for(i=0; i<n; ++i) Out[i]=a[i+temp]; Out[i]='\0';}void StringCopy(char *a,char *b,int n) //字符串复制{ int i; for(i=0; i<n; ++i) { a[i]=b[i]; }}int main(){ int i,j; char M_In[100]= {"0000000100100011010001010110011110001001101010111100110111101111"},M_Out[70],K_In[100]= {"0001001100110100010101110111100110011011101111001101111111110001"},K_Out[60]; myPermuation(M_In,M_Out,IP,64); myPermuation(K_In,K_Out,PC_1,56); //密文和密钥的初始置换 char L0[35],L1[35],R0[35],R1[35],C0[30],C1[30],D0[30],D1[30],Key_In[60],Key_Out[50]; for(i=0; i<32; ++i) { L0[i]=M_Out[i]; R0[i]=M_Out[i+32]; } for(i=0; i<28; ++i) { C0[i]=K_Out[i]; D0[i]=K_Out[28+i]; } //初始置换后的字符串左右半边的分离 for(i=0; i<16; ++i) // 进行16轮加密过程 { myShift(C0,C1,28,Shift[i]); myShift(D0,D1,28,Shift[i]); for(j=0; j<28; ++j) { Key_In[j]=C1[j]; Key_In[j+28]=D1[j]; } myPermuation(Key_In,Key_Out,PC_2,48); //获取子密钥 StringCopy(L1,R0,32); S_Box(R0,Key_Out,Key_In); myXOR(L0,Key_In,32,R1); //左右半边子密文通过S盒进行轮函数加密 StringCopy(C0,C1,28); //参数复用 StringCopy(D0,D1,28); StringCopy(L0,L1,32); StringCopy(R0,R1,32); } for(i=0; i<32; ++i) { M_In[i]=R0[i]; M_In[i+32]=L0[i]; } myPermuation(M_In,M_Out,IPE,64); //最后一轮轮函数加密后通过置换表得到最后的加密结果。 for(i=0;i<64;++i) { if(i%8==0) cout<<endl; cout<<M_Out[i]; } return 0;}运行结果：
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