给定长度为N的数组arr [] ，任务是找到该数组所有子集的子集的总和。
例子：
Input: arr[] = {1, 1}
Output: 6
All possible subsets:
a) {} : 0
All the possible subsets of this subset
will be {}, Sum = 0
b) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
c) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
d) {1, 1} : 4
All the possible subsets of this subset
will be {}, {1}, {1} and {1, 1}, Sum = 0 + 1 + 1 + 2 = 4
Thus, ans = 0 + 1 + 1 + 4 = 6
Input: arr[] = {1, 4, 2, 12}
Output: 513
方法：在本文中，将讨论O(N * 2 N )时间复杂度用于解决给定问题的方法。
首先，生成数组的所有可能子集。总共将有2 N个子集。然后，对于每个子集，找到其所有子集的总和。
因为，可以观察到，在长度为L的数组中，每个元素在子集的总和中将恰好是2 (L – 1)倍。因此，每个元素的贡献将为其值的2 (L – 1)倍。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to sum of all subsets of a
// given array
void subsetSum(vector& c, int& ans)
{
int L = c.size();
int mul = (int)pow(2, L - 1);
for (int i = 0; i < c.size(); i++)
ans += c[i] * mul;
}
// Function to generate the subsets
void subsetGen(int* arr, int i, int n,
int& ans, vector& c)
{
// Base-case
if (i == n) {
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c, ans);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, ans, c);
c.push_back(arr[i]);
subsetGen(arr, i + 1, n, ans, c);
c.pop_back();
}
// Driver code
int main()
{
int arr[] = { 1, 1 };
int n = sizeof(arr) / sizeof(int);
// To store the final ans
int ans = 0;
vector c;
subsetGen(arr, 0, n, ans, c);
cout << ans;
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// To store the final ans
static int ans;
// Function to sum of all subsets of a
// given array
static void subsetSum(Vector c)
{
int L = c.size();
int mul = (int)Math.pow(2, L - 1);
for (int i = 0; i < c.size(); i++)
ans += c.get(i) * mul;
}
// Function to generate the subsets
static void subsetGen(int []arr, int i,
int n, Vector c)
{
// Base-case
if (i == n)
{
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, c);
c.add(arr[i]);
subsetGen(arr, i + 1, n, c);
c.remove(0);
}
// Driver code
public static void main(String []args)
{
int arr[] = { 1, 1 };
int n = arr.length;
Vector c = new Vector();
subsetGen(arr, 0, n, c);
System.out.println(ans);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation of the approach
# store the answer
c = []
ans = 0
# Function to sum of all subsets of a
# given array
def subsetSum():
global ans
L = len(c)
mul = pow(2, L - 1)
i = 0
while ( i < len(c)):
ans += c[i] * mul
i += 1
# Function to generate the subsets
def subsetGen(arr, i, n):
# Base-case
if (i == n) :
# Finding the sum of all the subsets
# of the generated subset
subsetSum()
return
# Recursively accepting and rejecting
# the current number
subsetGen(arr, i + 1, n)
c.append(arr[i])
subsetGen(arr, i + 1, n)
c.pop()
# Driver code
if __name__ == "__main__" :
arr = [ 1, 1 ]
n = len(arr)
subsetGen(arr, 0, n)
print (ans)
# This code is contributed by Arnab KunduC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// To store the final ans
static int ans;
// Function to sum of all subsets of a
// given array
static void subsetSum(List c)
{
int L = c.Count;
int mul = (int)Math.Pow(2, L - 1);
for (int i = 0; i < c.Count; i++)
ans += c[i] * mul;
}
// Function to generate the subsets
static void subsetGen(int []arr, int i,
int n, List c)
{
// Base-case
if (i == n)
{
// Finding the sum of all the subsets
// of the generated subset
subsetSum(c);
return;
}
// Recursively accepting and rejecting
// the current number
subsetGen(arr, i + 1, n, c);
c.Add(arr[i]);
subsetGen(arr, i + 1, n, c);
c.RemoveAt(0);
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 1 };
int n = arr.Length;
List c = new List();
subsetGen(arr, 0, n, c);
Console.WriteLine(ans);
}
}
// This code is contributed by Rajput-Ji 输出：
6