输出一个集合所有子集的元素和（Print sums of all subsets of a given set）

原文地址：http://www.geeksforgeeks.org/print-sums-subsets-given-set/

Given an array of integers, print sums of all subsets in it. Output sums can be printed in any order.

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

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

Method 1 (Recursive)

We can recursively solve this problem. There are total 2n subsets. For every element, we consider two choices, we include it in a subset and we don’t include it in a subset. Below is recursive solution based on this idea.

方法一（递归）
我们可以递归地解决这个问题，这个集合共有2n个子集。对于每一个元素，我们考虑两种选择，子集中包含这个元素或者不包含这个元素。下面是基于这个想法的递归方法。

// C++ program to print sums of all possible
// subsets.
#include<bits/stdc++.h>
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;
}

Output:

12 9 8 5 7 4 3 0

Method 2 (Iterative)

As discussed above, there are total 2n subsets. The idea is generate loop from 0 to 2n – 1. For every number, pick all array elements which correspond to 1s in binary representation of current number.
方法二（迭代）
正如前面所讨论的，这个集合有2n个子集。这个想法就是从0到2n – 1循环。对于每个数字，挑出所有的数组元素，这些元素表示出二进制的时候每一位全部是1.

// Iterative C++ program to print sums of all
// possible subsets.
#include<bits/stdc++.h>
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<<n;

// Consider all numbers from 0 to 2^n - 1
for (long long i=0; i<total; i++)
{
long long 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))
sum += arr[j];

// Print sum of picked elements.
cout << sum << " ";
}
}

// Driver code
int main()
{
int arr[] = {5, 4, 3};
int n = sizeof(arr)/sizeof(arr[0]);

subsetSums(arr, 0, n-1);
return 0;
}
Output:

12 9 8 5 7 4 3 0

Thanks to cfh for suggesting above iterative solution in a comment.

Note: We haven’t actually created sub-sets to find their sums rather we have just used recursion to find sum of non-contiguous sub-sets of the given set.

The above mentioned techniques can be used to perform various operations on sub-sets like multiplication, division, XOR, OR, etc, without actually creating and storing the sub-sets and thus making the program memory efficient.

注意：我们还没有真正创建子集来得到它们的值，而是我们只是用递归得到已知集合子集的非连续和。
以上所提到的技术可以用于各种子集的操作，例如：乘法、除法、XOR、OR等，没有真正创建并存储子集，因此可以使得代码使用内存更有效。