打印给定集合的所有子集的总和

给定一个整数数组，打印其中的所有子集的总和。输出总和可以以任何顺序打印。

例子 ：

Input : arr[] = {2, 3}
Output: 0 2 3 5

Input : arr[] = {2, 4, 5}
Output : 0 2 4 5 6 7 9 11

方法1(递归)
我们可以递归地解决这个问题。总共有2 ⁿ个子集。对于每个元素，我们考虑两个选择，我们将其包含在子集中，而没有将其包含在子集中。以下是基于此思想的递归解决方案。

C++

// C++ program to print sums of all possible
// subsets.
#include
using namespace std;
 
// Prints sums of all subsets of arr[l..r]
void subsetSums(int arr[], int l, int r,
                int sum=0)
{
    // Print current subset
    if (l > r)
    {
        cout << sum << " ";
        return;
    }
 
    // Subset including arr[l]
    subsetSums(arr, l+1, r, sum+arr[l]);
 
    // Subset excluding arr[l]
    subsetSums(arr, l+1, r, sum);
}
 
// Driver code
int main()
{
    int arr[] = {5, 4, 3};
    int n = sizeof(arr)/sizeof(arr[0]);
 
    subsetSums(arr, 0, n-1);
    return 0;
}

Java

// Java program to print sums
// of all possible subsets.
import java .io.*;
 
class GFG
{
     
    // Prints sums of all
    // subsets of arr[l..r]
    static void subsetSums(int []arr, int l,
                            int r, int sum )
    {
         
        // Print current subset
        if (l > r)
        {
            System.out.print(sum + " ");
            return;
        }
     
        // Subset including arr[l]
        subsetSums(arr, l + 1, r,
                   sum + arr[l]);
     
        // Subset excluding arr[l]
        subsetSums(arr, l + 1, r, sum);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int []arr = {5, 4, 3};
        int n = arr.length;
     
        subsetSums(arr, 0, n - 1, 0);
    }
}
 
// This code is contributed by anuj_67

Python3

# Python3 program to print sums of
# all possible subsets.
 
# Prints sums of all subsets of arr[l..r]
def subsetSums(arr, l, r, sum = 0):
     
    # Print current subset
    if l > r:
        print (sum, end = " ")
        return
 
    # Subset including arr[l]
    subsetSums(arr, l + 1, r, sum + arr[l])
 
    # Subset excluding arr[l]
    subsetSums(arr, l + 1, r, sum)
 
# Driver code
arr = [5, 4, 3]
n = len(arr)
subsetSums(arr, 0, n - 1)
 
# This code is contributed by Shreyanshi Arun.

C#

// C# program to print sums of all possible
// subsets.
using System;
 
class GFG {
     
    // Prints sums of all subsets of
    // arr[l..r]
    static void subsetSums(int []arr, int l,
                            int r, int sum )
    {
         
        // Print current subset
        if (l > r)
        {
            Console.Write(sum + " ");
            return;
        }
     
        // Subset including arr[l]
        subsetSums(arr, l+1, r, sum + arr[l]);
     
        // Subset excluding arr[l]
        subsetSums(arr, l+1, r, sum);
    }
     
    // Driver code
    public static void Main ()
    {
        int []arr = {5, 4, 3};
        int n = arr.Length;
     
        subsetSums(arr, 0, n-1,0);
    }
}
 
// This code is contributed by anuj_67

PHP

 $r)
    {
        echo $sum , " ";
        return;
    }
 
    // Subset including arr[l]
    subsetSums($arr, $l + 1, $r,
               $sum + $arr[$l]);
 
    // Subset excluding arr[l]
    subsetSums($arr, $l + 1, $r, $sum);
}
 
// Driver code
$arr = array(5, 4, 3);
$n = count($arr);
 
subsetSums($arr, 0, $n - 1);
 
// This code is contributed by anuj_67.
?>

Javascript

// Javascript program to program to print
// sums of all possible subsets.
 
// Prints sums of all
// subsets of arr[l..r]
function subsetSums(arr, l, r, sum)
{
     
    // Print current subset
    if (l > r)
    {
        document.write(sum + " ");
        return;
    }
   
    // Subset including arr[l]
    subsetSums(arr, l + 1, r,
               sum + arr[l]);
                
    // Subset excluding arr[l]
    subsetSums(arr, l + 1, r, sum);
}
     
// Driver code
let arr = [5, 4, 3];
let n = arr.length;
 
subsetSums(arr, 0, n - 1, 0);
 
// This code is contributed by code_hunt
     

---

输出 ：

12 9 8 5 7 4 3 0

方法2(迭代)
如上所述，总共有2 ⁿ个子集。这个想法是生成从0到2 ⁿ – 1的循环。对于每个数字，请选择与当前数字的二进制表示形式中的1s对应的所有数组元素。

C++

// Iterative C++ program to print sums of all
// possible subsets.
#include
using namespace std;
 
// Prints sums of all subsets of array
void subsetSums(int arr[], int n)
{
    // There are totoal 2^n subsets
    long long total = 1<

Java

// Iterative Java program to print sums of all
// possible subsets.
import java.util.*;
 
class GFG{
 
// Prints sums of all subsets of array
static void subsetSums(int arr[], int n)
{
     
    // There are totoal 2^n subsets
    int total = 1 << n;
 
    // Consider all numbers from 0 to 2^n - 1
    for(int i = 0; i < total; i++)
    {
       int sum = 0;
        
       // Consider binary reprsentation of
       // current i to decide which elements
       // to pick.
       for(int j = 0; j < n; j++)
          if ((i & (1 << j)) != 0)
              sum += arr[j];
        
       // Print sum of picked elements.
       System.out.print(sum + " ");
    }
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = new int[]{ 5, 4, 3 };
    int n = arr.length;
 
    subsetSums(arr, n);
}
}
 
// This code is contributed by spp____

Java脚本

 
    // Iterative Javascript program to print sums of all
    // possible subsets.
     
    // Prints sums of all subsets of array
    function subsetSums(arr, n)
    {
 
        // There are totoal 2^n subsets
        let total = 1 << n;
 
        // Consider all numbers from 0 to 2^n - 1
        for(let i = 0; i < total; i++)
        {
           let sum = 0;
 
           // Consider binary reprsentation of
           // current i to decide which elements
           // to pick.
           for(let j = 0; j < n; j++)
              if ((i & (1 << j)) != 0)
                  sum += arr[j];
 
           // Print sum of picked elements.
           document.write(sum + " ");
        }
    }
     
    let arr = [ 5, 4, 3 ];
    let n = arr.length;
  
    subsetSums(arr, n);
     

输出 ：

0 5 4 9 3 8 7 12 

感谢cfh在评论中建议上述迭代解决方案。
注意：我们实际上并没有创建子集来找到它们的总和，而是只是使用了递归来找到给定集合的非连续子集的总和。