长度为 K 的所有子序列的总和

给定一个数组 arr[]和一个整数 K ，任务是从给定的数组中找到所有 K 个长度子序列的总和。

例子：

Input: arr[] = {2, 3, 4}, K = 2 
Output: 18 
Explanation: 
There are 3 possible subsequences of length 2 which are {2, 3}, {2, 4} and {3, 4} 
The sum of all 2 length subsequences is 5 + 6 + 7 = 18

Input: arr[] = {7, 8, 9, 2}, K = 2 
Output: 78 
Explanation: 
There are 6 subsequences of length 2 which are {7, 8}, {7, 9}, {7, 2}, {8, 9}, {8, 2} and {9, 2}. 
The sum of all 2 length sub sequences is 15 + 16 + 9 + 17 + 10 + 11 = 78 
 

方法：
为了解决上面提到的问题，我们必须考虑所有K长度的子序列，即“n 选择 k”，即$k * nCk$

• 所有 K 个长度的子序列中的总元素数为$k * nCk$ ，每个元素出现的可能性都是一样的。
• 所以每个元素出现$((k * nCk) / n )$时间和它的贡献$arr[i] * ( (k*nCk)/n )$结果中。
• 因此，所有 K 个长度的子序列之和为$sum(array) * ( (k * nCk) / n )$

下面是上述方法的实现：

C++

// C++ implementation to find sum
// of all subsequences of length K
  
#include 
using namespace std;
  
int fact(int n);
  
// Function to find nCr
int nCr(int n, int r)
{
    return fact(n)
           / (fact(r)
              * fact(n - r));
}
  
// Function that returns
// factorial of n
int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
  
// Function for finding sum
// of all K length subsequences
int sumSubsequences(
    int arr[], int n, int k)
{
  
    int sum = 0;
  
    // Calculate the sum of array
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
    int kLengthSubSequence;
  
    // Calculate nCk
    kLengthSubSequence = nCr(n, k);
  
    int ans
        = sum
          * ((k * kLengthSubSequence)
             / n);
  
    // Return the final result
    return ans;
}
  
// Driver code
int main()
{
  
    int arr[] = { 7, 8, 9, 2 };
  
    int K = 2;
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << sumSubsequences(arr, n, K);
    return 0;
}

Java

// Java implementation to find sum
// of all subsequences of length K
class GFG{
  
// Function to find nCr
static int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
  
// Function that returns
// factorial of n
static int fact(int n)
{
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
  
// Function for finding sum
// of all K length subsequences
static int sumSubsequences(int arr[], 
                           int n, int k)
{
    int sum = 0;
  
    // Calculate the sum of array
    for (int i = 0; i < n; i++) 
    {
        sum += arr[i];
    }
    int kLengthSubSequence;
  
    // Calculate nCk
    kLengthSubSequence = nCr(n, k);
  
    int ans = sum * ((k * kLengthSubSequence) / n);
  
    // Return the final result
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 7, 8, 9, 2 };
  
    int K = 2;
  
    int n = arr.length;
  
    System.out.print(sumSubsequences(arr, n, K));
}
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 implementation to find sum 
# of all subsequences of length K
  
# Function to find nCr 
def nCr(n, r):
      
    return fact(n) / (fact(r) * 
                      fact(n - r))
  
# Function that returns 
# factorial of n
def fact(n):
      
    res = 1
    for i in range(2, n + 1):
        res = res * i 
    return res
      
# Function for finding sum 
# of all K length subsequences
def sumSubsequences(arr, n, k):
      
    sum = 0
      
    # Calculate the sum of array 
    for i in range(0, n):
        sum = sum + arr[i]
      
    # Calculate nCk     
    kLengthSubSequence = nCr(n, k)
    ans = sum * ((k * kLengthSubSequence) / n);
      
    # Return the final result 
    return ans
  
# Driver Code 
arr = [ 7, 8, 9, 2 ]
k = 2
n = len(arr)
  
print(sumSubsequences(arr, n, k))
  
# This code is contributed by skylags

C#

// C# implementation to find sum
// of all subsequences of length K
using System;
  
class GFG{
      
// Function to find nCr
static int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
      
// Function that returns
// factorial of n
static int fact(int n)
{
    int res = 1;
      
    for(int i = 2; i <= n; i++)
       res = res * i;
    return res;
}
      
// Function for finding sum
// of all K length subsequences
static int sumSubsequences(int[] arr, 
                           int n, int k)
{
    int sum = 0;
      
    // Calculate the sum of array
    for(int i = 0; i < n; i++) 
    {
       sum += arr[i];
    }
      
    int kLengthSubSequence;
      
    // Calculate nCk
    kLengthSubSequence = nCr(n, k);
    int ans = sum * ((k * kLengthSubSequence) / n);
      
    // Return the final result
    return ans;
}
  
// Driver code
static void Main() 
{
    int[] arr = { 7, 8, 9, 2 };
    int K = 2;
    int n = arr.Length;
          
    Console.Write(sumSubsequences(arr, n, K));
}
}
  
// This code is contributed by divyeshrabadiya07

Javascript

  
// Javascript implementation to find sum
// of all subsequences of length K
  
// Function to find nCr
function nCr(n, r)
{
    return fact(n) / (fact(r) * 
           fact(n - r));
}
  
// Function that returns
// factorial of n
function fact(n)
{
    var res = 1;
    for(var i = 2; i <= n; i++)
        res = res * i;
          
    return res;
}
  
// Function for finding sum
// of all K length subsequences
function sumSubsequences(arr, n, k)
{
    var sum = 0;
  
    // Calculate the sum of array
    for(var i = 0; i < n; i++)
    {
        sum += arr[i];
    }
    var kLengthSubSequence;
  
    // Calculate nCk
    kLengthSubSequence = nCr(n, k);
  
    var ans = sum * ((k * 
              kLengthSubSequence) / n);
  
    // Return the final result
    return ans;
}
  
// Driver code
var arr = [ 7, 8, 9, 2 ];
var K = 2;
var n = arr.length;
  
document.write(sumSubsequences(arr, n, K));
  
// This code is contributed by noob2000
  

输出：

78

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