show all subset of a sub

[mmag]

A program is a set of all subsets of this part of the show. Users can enter a number., For example, if n = 2 the output looks like this:
{}
{1}
{2}
{1,2}
or for n=3 we have:
{}
{1}
{2}
{3}
{1,2}
{1,3}
{2,3}
{1,2,3}
meantime this program should be solved with tow way.
recursive function and Non-recursive.Can also be used in solving the problem of bitwise operations.

[Nepomuk]

OK, so that's the task. What have you done so far? We won't do your homework for you but we're happy to help you solve it yourself.

[mmag]

my algorithm that is for per subset exists one number. these numbers are form 0 until (2^n)-1.allow me explain it with a example.
n=3
{ } 000 0
{1} 100 4
{2} 010 2
{3} 001 1
{1,2} 110 6
{1,3} 101 5
{2,3} 011 3
{1,2,3} 111 7
The third column show some number(fn) that i use in code.now we know per number of subset has to state(sn).0 or 1. this means it exists or not exists.now i should match fn and sn. and suggested way is use of bitwise operators.(operator &).and now i don't know what do and thing with else thing.

#include <iostream>
using namespace std;
#include <conio.h>
#include <string.h>
int pow(int a,int b)
{
    int p=1;
    for(int i=0;i<b;i++)
        p=p*a;
    return p;
}
struct ar
{
    int  binary;
    int content;
};
int main()
 
{
    int testcase=0;
    int n;
    ar *a;
    a=new ar [2000];
    cin>>testcase;
    int i=0;
    while(i<testcase)
    {
        cin>>n;
        for(int j=0,k=1;j<n;j++,k++)
        {
            a[j].content=k;
            a[j].binary=2;
        }
 
 
    for(int p=0;p<pow(2,n)-1;p++)
    {
        cout<<'{';
    for(int m=0;m<n;m++)
    {
 
        int b;
        b=a[m].binary&p;
        if(b==1)
        cout<<a[i].content<<' ';
 
    }
      cout<<'}';
    }
 
    cout<<endl;
    i++;
    }
    return 0;
}