Алгоритм шифрования WAKE - C++ - Ответ 3817686

#include <iostream>
#include <stdio.h>
 
using namespace std;
void cypher (long V[],long n,long k[],long r[],long t[]);
void genkey (long t[], long k[]);
 
int main()
{
    
    long V[7], n, k[4], r[6], t[256];
    int j = 1;
    n = -4;
    for (int i = 1; i<7; i++) 
    r[i] = rand()%10;
    for (int count = 0; count < 4; count ++)
    {
    k[count] = rand()%512;
}
    for (int count_t = 0; count_t < 256; count_t ++){
        t[count_t] = rand()%255;
        cout << "t= " << t[count_t] << "\n";
        }
    
    V[0] = 1;
    V[1] = 3;
    V[2] = 5;
    V[3] = 7;
    V[4] = 21;
    V[5] = 43;
    V[6] = 12;
    V[7] = 17;
    
    genkey (t, k);
    cypher (V,n,k,r,t);
    for(int i = 0; i <=7; i++)
    cout << V[i] << "\t";
    n = -4;
    cout << "\n\n";
    cypher (V,n,k,r,t);
    for(int i = 0; i <=7; i++)
    cout << V[i] << "\t";
    system ("PAUSE");
    
    return 0;
}
 
void cypher (long V[],long n,long k[],long r[],long t[]){
long r1,r2,r3,r4,r5,r6,d,*e,m=0x00ffffff ;
 
r3=k[0] ; r4=k[1] ; r5=k[2] ; r6=k[3] ;
if (n<0) d= -1 ; else d=1 ;
e=V+n ;
while (V-e) {
 r1 = *V ;
 r2 = r1^r6 ;
 *V = r2 ; // Change into r1 for decoding.
 V += d ;
 r3 = r3+r2 ;
 r3 = (r3>>8&m)^t[r3&255] ;
 r4 = r4+r3 ;
 r4 = (r4>>8&m)^t[r4&255] ;
 r5 = r5+r4 ;
 r5 = (r5>>8&m)^t[r5&255] ;
 r6 = r6+r5 ;
 r6 = (r6>>8&m)^t[r6&255] ; }
r[0] =r3 ; r[1]=r4 ; r[2]=r5 ; r[3]=r6 ; }
 
 
void genkey (long t[], long k[]){
long x, z, p ;
static long tt[10]= {
0x726a8f3b,                                 // table
0xe69a3b5c,
0xd3c71fe5,
0xab3c73d2,
0x4d3a8eb3,
0x0396d6e8,
0x3d4c2f7a,
0x9ee27cf3, } ;
for (p=0 ; p<4 ; p++) t[p]=k[p] ;          // copy k
for (p=4 ; p<256 ; p++) {
  x=t[p-4]+t[p-1] ;                        // fill t
  t[p]=x>>3 ^ tt[x&7] ; }
 
for (p=0 ; p<23 ;
 p++) t[p]+=t[p+89] ;           // mix first entries
x=t[33] ; z=t[59] | 0x01000001 ;
z=z&0xff7fffff ;
for (p=0 ; p<256 ; p++) {       //change top byte to
  x=(x&0xff7fffff)+z ;          // a permutation etc
  t[p]=t[p] & 0x00ffffff ^ x ; }
 
t[256]=t[0] ; x&=255 ;
for (p=0 ; p<256 ; p++) {     // further change perm.
  t[p]=t[x=(t[p^x]^x)&255] ;  // and other digits
  t[x]=t[p+1] ;  }     }