aes.cpp

此程序是实现AES算法的

#include<stdio.h>
#include<string.h>
#include<malloc.h>
#include<math.h>

int nb,nk,nr;
int w[400][4],plaintext[8][4],keytext[8][4];
int firstrcon[15]={1,2,4,6,8,32,64,128,27,54,108,216,171,79,158}; //the first byte of rcon
int tablebytesub[16][16]={
                           {99,124,119,123,242,107,111,193,48,1,103,43,254,215,171,118},
                           {202,180,201,125,250,89,71,240,173,212,162,175,156,164,114,192},
                           {183,253,147,38,54,63,247,204,52,165,229,241,113,216,49,21},
                           {4,199,35,195,24,150,5,154,7,19,128,226,235,39,178,117},
                           {9,131,44,26,27,110,90,160,82,59,218,179,41,227,47,132},
                           {83,209,0,237,32,252,177,91,106,203,190,57,74,76,88,207},
                           {208,239,170,251,67,77,51,133,69,249,2,127,80,60,159,168},
                           {81,163,68,143,146,157,56,245,188,182,218,33,16,255,243,210},
                           {205,12,19,236,95,151,68,23,196,167,126,61,100,93,25,115},
                           {96,129,79,220,34,42,144,136,70,238,184,20,222,94,11,219},
                           {224,50,58,10,73,6,36,92,194,211,172,98,145,149,228,121},
                           {231,200,55,109,141,213,78,169,108,86,244,234,101,122,174,8},
                           {186,120,37,46,28,166,180,198,232,221,116,31,75,189,139,138},
                           {225,248,152,17,105,217,142,148,155,30,135,233,206,85,40,223},
                           {140,161,137,13,191,230,66,104,65,153,45,15,176,84,187,22}
                           };
                         

 int bytesub(int k)
{
   int kk=k%16,kkk=k/16;
   return tablebytesub[kkk][kk];
}

int keyexpansion()
{
  int i,j,k;
  for(i=0;i<nk;i++)   // every row is a word && start from 0
    for(j=0;j<4;j++)    // 4 element 
      w[i][j]=keytext[i][j];
        
 if(nk<6)
 {    
  for(i=nk;i<nk*(nr+1);i+=nk)
  {
    for(j=0;j<4;j++)
    {                        
       w[i][j]=bytesub(w[i-1][(j+1)%4])^w[i-nk][j];
       if(j==0)
       w[i][j]=(w[i][j]^firstrcon[i/nk])%256;
    }
   for(j=1;j<nk&&i+j<nb*(nr+1);j++)
   {
      for(k=0;k<4;k++)
        w[i+j][k]=(w[i+j-nk][k]^w[i+j-1][k])%256;
   }
  }
 }
  
  else
 {
  //?????????????????????????????????????????????                                   
 }                            
  
 return 0;
} 
       
int shfitrow()    /////obey the order of 0 1  2  3 
{
   int i,j;
   int numofshift[4]={0,1,2,3};
   for(i=0;i<4;i++)
  {
   while(numofshift[i]--)
   {
     int t=plaintext[0][i];
     for(j=0;j<nb-1;j++)
       plaintext[j][i]=plaintext[j+1][i];
    plaintext[j][i]=t;
   }
  }
  return 0; 
} 
int inlog[256]={
                1,3,5,15,17,51,85,255,26,46,114,150,161,248,19,53,
                95,225,56,72,216,115,149,164,247,2,6,10,30,34,102,170,
                229,52,92,228,55,89,235,38,106,190,217,112,144,171,230,49,
                83,245,4,12,20,60,68,204,79,209,104,184,211,110,178,205,
                76,212,103,169,224,59,77,215,98,166,241,8,24,40,120,136,
                131,158,185,208,107,189,220,127,129,152,179,206,73,219,118,154,
                181,196,87,249,16,48,80,240,11,29,39,105,187,214,97,163,
                254,25,43,125,135,146,173,236,47,113,147,174,233,32,96,160,
                251,22,58,78,210,109,183,194,93,231,50,86,250,21,63,65,
                159,186,213,100,172,239,42,126,130,157,188,223,122,142,137,128,
                155,182,193,88,232,35,101,175,234,37,111,177,200,67,197,84,
                252,31,33,99,165,244,7,9,27,45,119,153,176,203,70,202,
                 69,207,74,222,121,139,134,145,168,227,62,66,198,81,243,14,
                 18,54,90,238,41,123,141,140,143,138,133,148,167,242,13,23,
                 57,75,221,124,132,151,162,253,28,36,108,180,199,82,246,1
                };
int logfun[256]={
                 0,0,25,1,50,2,26,198,75,199,27,104,51,238,223,3,
                 100,4,224,14,52,141,129,239,76,113,8,200,248,105,28,193,
                 125,194,29,181,249,185,39,106,77,228,166,114,154,201,9,120,
                 101,47,138,5,33,15,225,36,18,240,130,69,53,147,218,142,
                 150,143,219,189,54,208,206,148,19,92,210,241,64,70,131,56,
                 102,221,253,48,191,6,139,98,179,37,226,152,34,136,145,16,
                 126,110,72,195,163,182,30,66,58,107,40,84,250,133,61,186,
                 43,121,10,21,155,159,94,202,78,212,172,229,243,115,167,87,
                 175,88,168,80,244,234,214,116,79,174,233,213,231,230,173,232,
                 44,215,117,122,235,22,11,245,89,203,95,176,156,169,81,160,
                 127,12,246,111,23,196,73,236,216,67,31,45,164,118,123,183,
                 204,187,62,90,251,96,177,134,59,82,161,108,170,85,41,157,
                 151,178,135,144,97,190,220,252,188,149,207,205,55,63,91,209,
                 83,57,132,60,65,162,109,71,20,42,158,93,86,242,211,171,
                 68,17,146,217,35,32,46,137,180,124,184,38,119,153,227,165,
                 103,74,237,222,197,49,254,24,13,99,140,128,192,247,112,7
                 };
                 
                 
int mixcolumn()
{
  int i,j,k,sum;
  int ofnum[4]={2,3,1,1},inlog[4]={25,1,0,0};
  int a[4];
  for(i=0;i<nb;i++)
  {
    for(j=0;j<4;j++)
     a[j]=plaintext[i][j];
    for(j=0;j<4;j++)
    {
      for(sum=0,k=0;k<4;k++)
        sum^=inlog[(logfun[ofnum[(k+j)%4]]+logfun[a[k]])%255];
       plaintext[i][j]=sum;
     }
  }
  return 0;
}
                 
int main()
{
                         
   int i,j,k,sum;
   int *ch,*key;
   while(1)
   {
         /////////////////////////////////////////////////////////////
   printf("请输入nb和nk和nr的值\n"); 
   scanf("%d%d%d",&nb,&nk,&nr);
   ch=(int *)malloc(sizeof(int)*nb*32);
   key=(int *)malloc(sizeof(int)*nk*32);
   printf("请输入明文：\n"); 
   for(i=1;i<=nb*32;i++)
     scanf("%d",&ch[i]);
   printf("请输入密钥:\n");  
   for(i=1;i<=nk*32;i++)
     scanf("%d",&key[i]);
    
    for(k=1,i=0,j=0,sum=0;k<=nb*32;k++)
    {
     sum+=ch[k]*(int)pow(2,(8-k)%8);
     if(k%8==0)
     {
       plaintext[i][j]=sum;
       sum=0;
       j=(j+1)%4;
       if(k%32==0)
        i++;
     }
    }
    for(k=1,i=0,j=0,sum=0;k<=nk*32;k++)
    {
      sum+=key[k]*(int)pow(2,(8-k)%8);
      if(k%8==0)
      {
         keytext[i][j]=sum;
         sum=0;
         j=(j+1)%4;
         if(k%32==0)
           i++;
      }
    }
       /////////////////////////////////////////////////////////////////
    for(i=0;i<nb;i++)// 初始圈密钥相加 
      for(j=0;j<4;j++)
       plaintext[i][j]^=w[i][j];
    int start=nb;
    while(nr--)
    {
      for(i=0;i<nb;i++)     // the submit of  bytesub;
       for(j=0;j<4;j++)
        plaintext[i][j]=bytesub(plaintext[i][j]);
        
     shfitrow();  //行移位 
     
     if(!nr)
       mixcolumn();    // 列混淆 
     
     for(i=0;i<nb;i++)
      for(j=0;j<4;j++)
        plaintext[i][j]^=w[start++][j];
    }
       ////////////////////////////////////////////////////////////
    for(i=0;i<nb;i++)
      for(j=0;j<4;j++)
       {
          for(k=3;k>=0;k--)
          {
            ch[i*32+j*4+k]=plaintext[i][j]%2;
            plaintext[i][j]/=2;
          }     
       }
    printf("密文是\n");   
    for(i=0;i<nb*32;i++)
     printf("%d",ch[i]);
    printf("\n"); 
   
   }  
    return 0;
}