📜  所有子集的平均值之和

📅  最后修改于: 2021-04-29 10:11:14             🧑  作者: Mango

给定一个由N个整数元素组成的数组arr,任务是找到该数组所有子集的平均值之和。

例子 :

Input  : arr[] = [2, 3, 5]
Output : 23.33 
Explanation : Subsets with their average are, 
[2]        average = 2/1 = 2
[3]        average = 3/1 = 3
[5]        average = 5/1 = 5
[2, 3]        average = (2+3)/2 = 2.5
[2, 5]        average = (2+5)/2 = 3.5
[3, 5]        average = (3+5)/2 = 4
[2, 3, 5]    average = (2+3+5)/3 = 3.33

Sum of average of all subset is, 
2 + 3 + 5 + 2.5 + 3.5 + 4 + 3.33 = 23.33

一个幼稚的解决方案是遍历所有可能的子集,获取所有子集的平均值,然后将它们一个接一个地添加,但是这将花费指数时间,并且对于较大的数组将是不可行的。

我们可以举个例子,

arr = [a0, a1, a2, a3]
sum of average = 
a0/1 + a1/1 + a2/2 + a3/1 +
(a0+a1)/2 + (a0+a2)/2 + (a0+a3)/2 + (a1+a2)/2 +
 (a1+a3)/2 + (a2+a3)/2 + 
(a0+a1+a2)/3 + (a0+a2+a3)/3 + (a0+a1+a3)/3 + 
 (a1+a2+a3)/3 +
(a0+a1+a2+a3)/4

If S = (a0+a1+a2+a3), then above expression 
can be rearranged as below,
sum of average = (S)/1 + (3*S)/2 + (3*S)/3 + (S)/4

分子的系数可以解释如下,假设我们对K个元素的子集进行迭代,则分母为K,分子为r * S,其中“ r”表示在迭代时添加特定数组元素的次数。大小相同的子集。通过检查,我们可以看到r为nCr(N – 1,n – 1),因为在求和一个元素后,我们需要从(N – 1)个元素中选择(n – 1)个元素,因此每个元素都有一个当考虑相同大小的子集时,nCr(N – 1,n – 1)的频率,因为所有元素都参加相等的次数,这也将是S的频率,并且将成为最终表达式中的分子。
在下面的代码中,nCr是使用动态编程方法实现的,您可以在此处了解更多信息,

C++
// C++ program to get sum of average of all subsets
#include 
using namespace std;
  
// Returns value of Binomial Coefficient C(n, k)
int nCr(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;
  
    // Calculate value of Binomial Coefficient in bottom
    // up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
  
            // Calculate value using previously stored
            // values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
    return C[n][k];
}
  
// method returns sum of average of all subsets
double resultOfAllSubsets(int arr[], int N)
{
    double result = 0.0; // Initialize result
  
    // Find sum of elements
    int sum = 0;
    for (int i = 0; i < N; i++)
        sum += arr[i];
  
    // looping once for all subset of same size
    for (int n = 1; n <= N; n++)
  
        /* each element occurs nCr(N-1, n-1) times while
           considering subset of size n  */
        result += (double)(sum * (nCr(N - 1, n - 1))) / n;
  
    return result;
}
  
// Driver code to test above methods
int main()
{
    int arr[] = { 2, 3, 5, 7 };
    int N = sizeof(arr) / sizeof(int);
    cout << resultOfAllSubsets(arr, N) << endl;
    return 0;
}


Java
// java program to get sum of
// average of all subsets
import java.io.*;
  
class GFG {
  
    // Returns value of Binomial
    // Coefficient C(n, k)
    static int nCr(int n, int k)
    {
        int C[][] = new int[n + 1][k + 1];
        int i, j;
  
        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;
  
                // Calculate value using
                // previously stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
        return C[n][k];
    }
  
    // method returns sum of average of all subsets
    static double resultOfAllSubsets(int arr[], int N)
    {
        // Initialize result
        double result = 0.0;
  
        // Find sum of elements
        int sum = 0;
        for (int i = 0; i < N; i++)
            sum += arr[i];
  
        // looping once for all subset of same size
        for (int n = 1; n <= N; n++)
  
            /* each element occurs nCr(N-1, n-1) times while
            considering subset of size n */
            result += (double)(sum * (nCr(N - 1, n - 1))) / n;
  
        return result;
    }
  
    // Driver code to test above methods
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 5, 7 };
        int N = arr.length;
        System.out.println(resultOfAllSubsets(arr, N));
    }
}
  
// This code is contributed by vt_m


Python3
# Python3 program to get sum
# of average of all subsets
  
# Returns value of Binomial
# Coefficient C(n, k)
def nCr(n, k):
  
    C = [[0 for i in range(k + 1)]
            for j in range(n + 1)]
  
    # Calculate value of Binomial 
    # Coefficient in bottom up manner
    for i in range(n + 1):
      
        for j in range(min(i, k) + 1):
          
            # Base Cases
            if (j == 0 or j == i):
                C[i][j] = 1
  
            # Calculate value using 
            # previously stored values
            else:
                C[i][j] = C[i-1][j-1] + C[i-1][j]
      
    return C[n][k]
  
# Method returns sum of
# average of all subsets
def resultOfAllSubsets(arr, N):
  
    result = 0.0 # Initialize result
  
    # Find sum of elements
    sum = 0
    for i in range(N):
        sum += arr[i]
  
    # looping once for all subset of same size
    for n in range(1, N + 1):
  
        # each element occurs nCr(N-1, n-1) times while
        # considering subset of size n */
        result += (sum * (nCr(N - 1, n - 1))) / n
  
    return result
  
# Driver code 
arr = [2, 3, 5, 7]
N = len(arr)
print(resultOfAllSubsets(arr, N))
  
  
# This code is contributed by Anant Agarwal.


C#
// C# program to get sum of
// average of all subsets
using System;
  
class GFG {
      
    // Returns value of Binomial
    // Coefficient C(n, k)
    static int nCr(int n, int k)
    {
        int[, ] C = new int[n + 1, k + 1];
        int i, j;
  
        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.Min(i, k); j++) 
            {
                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;
  
                // Calculate value using
                // previously stored values
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
            }
        }
        return C[n, k];
    }
  
    // method returns sum of average 
    // of all subsets
    static double resultOfAllSubsets(int[] arr, int N)
    {
        // Initialize result
        double result = 0.0;
  
        // Find sum of elements
        int sum = 0;
        for (int i = 0; i < N; i++)
            sum += arr[i];
  
        // looping once for all subset 
        // of same size
        for (int n = 1; n <= N; n++)
  
            /* each element occurs nCr(N-1, n-1) times while
               considering subset of size n */
            result += (double)(sum * (nCr(N - 1, n - 1))) / n;
  
        return result;
    }
  
    // Driver code to test above methods
    public static void Main()
    {
        int[] arr = { 2, 3, 5, 7 };
        int N = arr.Length;
        Console.WriteLine(resultOfAllSubsets(arr, N));
    }
}
  
// This code is contributed by Sam007


PHP


输出 :

63.75