打印给定蒙版的所有子蒙版

给定一个整数N ，任务是打印由N的二进制表示形式中存在的置位位形成的该组的所有子集。

例子：

Input: N = 5
Output: 5 4 1 
Explanation:
Binary representation of N is “101”, Therefore all the required subsets are {“101”, “100”, “001”, “000”}.

Input: N = 25
Output: 25 24 17 16 9 8 1 
Explanation:
Binary representation of N is “1101”. Therefore, all the required subsets are {“11001”, “11000”, “10001”, “10000”, “01001”, “01000”, “0001”, “0000”}.

天真的方法：最简单的方法是遍历[0，1 <<((N中的设置位的数量) ]范围内的每个掩码，并检查是否除N中的位之外未设置其他位。然后打印。
时间复杂度： O(2 ^{(N中的设置位数)} )
辅助空间： O(1)

高效方法：上述方法可以仅通过遍历它们的掩模n中的子集中的submasks进行优化。

• Suppose S is the current submask which is the subset of mask N. Then, it can be observed that by assigning S = (S – 1) & N, the next submask of N can be obtained which is less than S.
• In S – 1, it flips all the bits present on the right of the rightmost set bit including rightmost set bit of S.
• Therefore, after performing Bitwise & with N, a submask of N is obtained.
• Therefore, S = (S – 1) & N gives the next submask of N which is less than S.

请按照以下步骤解决问题：

• 初始化变量，例如S =N。
• 在S> 0时进行迭代，并在每次迭代中打印S的值。
• 分配S =(S – 1)＆N。

下面是上述方法的实现：

C++

// C++ Program for above approach
#include 
using namespace std;
 
// Function to print the submasks of N
void SubMasks(int N)
{
    for (int S = N; S; S = (S - 1) & N) {
        cout << S << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 25;
    SubMasks(N);
 
    return 0;
}

Java

// Java Program for above approach
import java.util.*;
class GFG
{
   
// Function to print the submasks of N
static void SubMasks(int N)
{
    for (int S = N; S > 0; S = (S - 1) & N)
    {
        System.out.print(S + " ");
    }
}
 
// Driver Code
public static void main(String args[])
{
    int N = 25;
    SubMasks(N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Python3

# Python3 program for the above approach
 
# Function to print the submasks of N
def SubMasks(N) :
    S = N
    while S > 0:
        print(S,end=' ')
        S = (S - 1) & N
 
# Driven Code
if __name__ == '__main__':
    N = 25
    SubMasks(N)
     
    # This code is contributed by bgangwar59.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to print the submasks of N
static void SubMasks(int N)
{
    for (int S = N; S > 0; S = (S - 1) & N)
    {
        Console.Write(S + " ");
    }
}
 
// Driver Code
static public void Main()
{
    int N = 25;
    SubMasks(N);
}
}
 
// This code is contributed by Code_hunt.

Javascript

 
// JavaScript program of the above approach
 
// Function to print the submasks of N
function SubMasks(N)
{
    for (let S = N; S > 0; S = (S - 1) & N)
    {
        document.write(S + " ");
    }
}
 
    // Driver Code
     
    let N = 25;
    SubMasks(N);
 

输出：

25 24 17 16 9 8 1

时间复杂度： O(2 ^{(N中的设置位数)} )
辅助空间： O(1)