所有子数组的总和|套装1

给定一个大小为n的整数数组’arr []’，找到给定数组的所有子数组的和。
例子 ：

Input   : arr[] = {1, 2, 3}
Output  : 20
Explanation : {1} + {2} + {3} + {2 + 3} + 
              {1 + 2} + {1 + 2 + 3} = 20

Input  : arr[] = {1, 2, 3, 4}
Output : 50

方法1(简单解决方案)
一个简单的解决方案是生成所有子数组并计算它们的总和。
以下是上述想法的实现。

C++

// Simple C++ program to compute sum of
// subarray elements
#include
using namespace std;
 
// Computes sum all sub-array
long int SubArraySum(int arr[], int n)
{
    long int result = 0,temp=0;
 
    // Pick starting point
    for (int i=0; i 

---

Java

// Simple Java program to compute sum of
// subarray elements
class GFG {
     
    // Computes sum all sub-array
    public static long SubArraySum(int arr[], int n)
    {
        long result = 0,temp=0;
      
        // Pick starting point
        for (int i = 0; i < n; i ++)
        {
            // Pick ending point
            temp=0;
            for (int j = i; j < n; j ++)
            {
                // sum subarray between current
                // starting and ending points
                temp+=arr[j];
                result += temp ;
            }
        }
        return result ;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 3} ;
        int n = arr.length;
        System.out.println("Sum of SubArray : " +
                          SubArraySum(arr, n));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python 3

# Python3 program to compute
# sum of subarray elements
 
# Computes sum all sub-array
def SubArraySum(arr, n):
    temp,result = 0,0
 
    # Pick starting point
    for i in range(0, n):
 
        # Pick ending point
        temp=0;
        for j in range(i, n):
 
            # sum subarray between
            # current starting and
            # ending points
            temp+=arr[j]
            result += temp
    return result
 
# driver program
arr = [1, 2, 3]
n = len(arr)
print ("Sum of SubArray :"
       ,SubArraySum(arr, n))
 
# This code is contributed by
# TheGameOf256.

C#

// Simple C# program to compute sum of
// subarray elements
using System;
 
class GFG {
     
    // Computes sum all sub-array
    public static long SubArraySum(int []arr,
                                      int n)
    {
        long result = 0,temp=0;
     
        // Pick starting point
        for (int i = 0; i < n; i ++)
        {
            // Pick ending point
            temp=0;
            for (int j = i; j < n; j ++)
            {
                 
                // sum subarray between current
                // starting and ending points
                temp+=arr[j];
                result += temp ;
            }
        }
        return result ;
    }
     
    // Driver Code
    public static void Main()
    {
        int []arr = {1, 2, 3} ;
        int n = arr.Length;
        Console.Write("Sum of SubArray : " +
                        SubArraySum(arr, n));
    }
}
 
// This code is contributed by nitin mittal

Javascript

 
// Simple Javascript program to compute sum of
// subarray elements
 
// Computes sum all sub-array
function SubArraySum(arr, n)
{
    let result = 0,temp=0;
 
    // Pick starting point
    for (let i=0; i <n; i++)
    {
        // Pick ending point
        temp=0;
        for (let j=i; j<n; j++)
        {
            // sum subarray between current
            // starting and ending points
            temp+=arr[j];
            result += temp ;
        }
    }
    return result ;
}
 
// driver program to test above function
  
    let arr = [1, 2, 3] ;
    let n = arr.length;
    document.write("Sum of SubArray : "
    + SubArraySum(arr, n) + "<br>");
     
// This code is contributed by Mayank Tyagi
 

---

---

输出：

Sum of SubArray : 20

时间复杂度： O(n ² )
方法2(有效的解决方案)
如果我们仔细观察，就会观察到一种模式。举个例子

arr[] = [1, 2, 3], n = 3
All subarrays :  [1], [1, 2], [1, 2, 3], 
                 [2], [2, 3], [3]
here first element 'arr[0]' appears 3 times    
     second element 'arr[1]' appears 4 times  
     third element 'arr[2]' appears 3 times

Every element arr[i] appears in two types of subsets:
i)  In subarrays beginning with arr[i]. There are 
    (n-i) such subsets. For example [2] appears
    in [2] and [2, 3].
ii) In (n-i)*i subarrays where this element is not
    first element. For example [2] appears in 
    [1, 2] and [1, 2, 3].

Total of above (i) and (ii) = (n-i) + (n-i)*i 
                            = (n-i)(i+1)
                                  
For arr[] = {1, 2, 3}, sum of subarrays is:
  arr[0] * ( 0 + 1 ) * ( 3 - 0 ) + 
  arr[1] * ( 1 + 1 ) * ( 3 - 1 ) +
  arr[2] * ( 2 + 1 ) * ( 3 - 2 ) 

= 1*3 + 2*4 + 3*3 
= 20

以下是上述想法的实现。

C++

// Efficient C++ program to compute sum of
// subarray elements
#include
using namespace std;
 
//function compute sum all sub-array
long int SubArraySum( int arr[] , int n )
{
    long int result = 0;
 
    // computing sum of subarray using formula
    for (int i=0; i

Java

// Efficient Java program to compute sum of
// subarray elements
class GFG {
     
    //function compute sum all sub-array
    public static long SubArraySum( int arr[] , int n )
    {
        long result = 0;
      
        // computing sum of subarray using formula
        for (int i=0; i

的Python 3

#Python3 code to compute
# sum of subarray elements
 
#function compute sum
# all sub-array
def SubArraySum(arr, n ):
    result = 0
 
    # computing sum of subarray
    # using formula
    for i in range(0, n):
        result += (arr[i] * (i+1) * (n-i))
 
    # return all subarray sum
    return result
 
# driver program
arr = [1, 2, 3]
n = len(arr)
print ("Sum of SubArray : ",
      SubArraySum(arr, n))
 
# This code is contributed by
# TheGameOf256.

C＃

// Efficient C# program
// to compute sum of
// subarray elements
using System;
 
class GFG
{
         
    // function compute
    // sum all sub-array
    public static long SubArraySum(int []arr ,
                                   int n )
    {
        long result = 0;
     
        // computing sum of
        // subarray using formula
        for (int i = 0; i < n; i++)
            result += (arr[i] *
                      (i + 1) * (n - i));
     
        // return all subarray sum
        return result ;
    }
     
    // Driver code
    static public void Main ()
    {
        int []arr = {1, 2, 3} ;
        int n = arr.Length;
        Console.WriteLine("Sum of SubArray: " +
                          SubArraySum(arr, n));
    }
}
 
// This code is contributed by akt_mit

输出 ：

Sum of SubArray : 20