Acwing 92. recursive implementation of exponential enumeration

from 1∼n this n Randomly select any number of integers , Output all possible options .

Input format
Enter an integer n.

Output format
Output one solution per line .

The numbers in the same row must be in ascending order , Two adjacent numbers are used exactly 1 Space between .

For a scheme without any number of options , Output blank lines .

This question has a custom calibrator （SPJ）, All walks of life （ Different options ） In any order .

Data range
1 ≤ n ≤ 15
sample input ：

3

sample output ：

// Blank line 
3
2
2 3
1
1 3
1 2
1 2 3

Ideas

We can divide each node into selected and unselected , Just walk for the last time .

#include <iostream>

using namespace std;

const int N = 20;

int n;

bool vis[N]; // Decide whether to choose or not 

void dfs(int u) // The first level is to filter the number 
{
    if (u > n) // You can't have an equal sign , If there is an equal sign, there will be less recursion , That is, the last layer cannot be recursive 
    {
        for (int i = 1; i <= n; i ++) // from 1 To n choice 
            if (vis[i]) cout << i << ' ';
        cout << '\n';
        return;
    }
    else
    {
        vis[u] = true; // Choose this number 
        dfs(u + 1);

        vis[u] = false; // Don't choose this number 
        dfs(u + 1);
    }
}

int main()
{
    cin >> n;
    dfs(1); // from 1 Start choosing , To n end , So we can't go from 0 Start ;
    return 0;
}