由前N个自然数组成的集合的所有子集的乘积

给定数字N ，任务是从由前N个自然数形成的集合的所有可能子集中找到所有元素的乘积。
例子：

Input: N = 2 
Output: 4 
Possible subsets are {{1}, {2}, {1, 2}}. 
Product of elements in subsets = {1} * {2} * {1 * 2} = 4
Input: N = 3 
Output: 1296 
Possible subsets are {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} 
Product of elements in subsets = 1 * 2 * 3 * (1 * 2) * (1 * 3) * (2 * 3) * (1 * 2 * 3) = 1296 
 

天真的方法：一个简单的解决方案是生成前N个自然数的所有子集。然后，对于每个子集，计算其乘积，最后返回每个子集的整体乘积。
高效方法：

• 可以看出，原始数组的每个元素在所有子集中出现2 ^{(N – 1)}次。
• 因此，最终答案中任何元素arr _i的贡献将是

i * 2(N – 1)

• 因此，所有子集的多维数据集之和为

12N-1 * 22N-1 * 32N-1......N2N-1

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
int product(int N)
{
    int ans = 1;
    int val = pow(2, N - 1);
 
    for (int i = 1; i <= N; i++) {
        ans *= pow(i, val);
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int N = 2;
 
    cout << product(N);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to find the product of all elements
    // in all subsets in natural numbers from 1 to N
    static int product(int N)
    {
        int ans = 1;
        int val = (int)Math.pow(2, N - 1);
     
        for (int i = 1; i <= N; i++) {
            ans *= (int)Math.pow(i, val);
        }
     
        return ans;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int N = 2;
     
        System.out.println(product(N));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to find the product of all elements
# in all subsets in natural numbers from 1 to N
def product(N) :
    ans = 1;
    val = 2 **(N - 1);
 
    for i in range(1, N + 1) :
        ans *= (i**val);
     
    return ans;
 
 
# Driver Code
if __name__ == "__main__" :
 
    N = 2;
 
    print(product(N));
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to find the product of all elements
    // in all subsets in natural numbers from 1 to N
    static int product(int N)
    {
        int ans = 1;
        int val = (int)Math.Pow(2, N - 1);
     
        for (int i = 1; i <= N; i++) {
            ans *= (int)Math.Pow(i, val);
        }
     
        return ans;
    }
     
    // Driver Code
    public static void Main (string[] args)
    {
        int N = 2;
     
        Console.WriteLine(product(N));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

// javascript implementation of the approach
 
// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
function product( N)
{
    let ans = 1;
    let val = Math.pow(2, N - 1);
    for (let i = 1; i <= N; i++)
    {
        ans *= Math.pow(i, val);
    }
    return ans;
}
 
// Driver Code
 
    let N = 2;
    document.write(product(N));
 
// This code is contributed by todaysgaurav
 

输出：

4