Knapsack 0-1 C++ binary & WE

[DANILIN]

Knapsack 0-1 C++ binary & WE

Classic Knapsack problem is solved in many ways

My newest program synthesizes all ciphers from 0 & 1
adding an extra register and 0 remain on left in cipher

Feature of algorithm:
converts whole into a string
and converts string into a whole
what will help in other programs

Number of comparisons decreases from N! to 2^N
for example N=10 & N!=3628800 >> 2^N=1024

Random values origin are automatically assigned
quantity and quality and integral of value is obtained
and in general: integral of quantity and quality

#include <iostream> // KNAPSACK 0-1 DANILIN
using namespace std; int main()
{ setlocale (LC_ALL, "RUS");
  srand(time(NULL)); // rextester.com/VCBSQ91995
 
{ int n=7; int G=5; int a=2;  
  int dec, i, h, k, max, m; 
  for (i=0; i<n; i++) a=2*a; string e[a]; // 2^n 
  int L[n], C[n], j[n], q[a], d[a];  
 
cout << "#  Amo Price" << endl << endl; 
for (i=0; i<n; i++)
{ L[i]=1+(rand() % 3); C[i]=10+(rand() % 9); j[i]=0;
  cout << i+1 << "   " << L[i] << "   " << C[i] << endl; 
} 
for (i=0; i<a; i++) { q[i]=0; d[i]=0;}
cout << endl; 
 
cout << "Mx Amo Price Chifer" << endl << endl; 
for (h = a-1; h>(a-1)/2; h--)
  { dec=h; while (dec > 0)
      { string s(""); s += '0'+dec%2;   
        e[h] = s + e[h]; dec/=2; 
      }
if (e[h] == "") {e[h] = "0";}
e[h]= e[h].substr(1, e[h].size()-1);
 
for (k=0; k<n; k++)
{ j[k] = atoi((char*)(e[h].substr(k,1)).c_str()); 
  q[h]=q[h]+L[k]*j[k]*C[k];
  d[h]=d[h]+L[k]*j[k];
}
 
if (d[h] <= G)
cout << G << "  " << d[h] << "  " << q[h] << "  " << e[h] << endl; 
} cout << endl;
 
max=0; m=1;
for (i=0; i<a; i++)
{ if (d[i]<=G && q[i]>max){ max=q[i]; m=i;}
}
cout << "Mx Price Cipher" << endl << endl; 
cout << d[m] << "  " << q[m] << "  " << e[m] << endl << endl;}
system("pause");
}