从给定数组中找到原始数组，其中第 i 个元素是前 i 个元素的平均值

给定一个数组arr[] 长度为N，任务是找到原始数组，使得给定数组 ( arr[i] ) 中的每个第i 个元素都是原始数组的前i个元素的平均值。

例子：

Input: arr = {4 3 3 3} 
Output: 4 2 3 3
Explanation: (4) / 1 = 1, (4+2) / 2 = 3, (4+2+3) / 3 = 3, (4+2+3+3) / 4 = 3

Input: arr = {2 6 8 10} 
Output: 2 10 12 16
Explanation: (2) / 1 = 2, (2+10) / 2 = 6, (2+10+12) / 3 = 8, (2+10+12+16) / 4 = 10

方法：给定的问题可以使用数学方法来解决。请按照以下步骤操作：

• 将变量sum初始化为数组arr的第一个元素
• 迭代数组arr从第 2 个索引到末尾，并在每次迭代时：
  • 将当前元素arr[i]与当前索引 + 1 (i + 1)相乘并从中减去sum的值
  • 将生成的当前元素添加到变量sum
• 修改后返回结果数组，因为它将是原始数组

C++

// C++ implementation for the above approach
 
#include 
using namespace std;
 
// Function to find the original
// array from the modified array
void findOriginal(int arr[], int N)
{
 
    // Initialize the variable sum
    // with the first element of array
    int sum = arr[0];
 
    for (int i = 1; i < N; i++) {
 
        // Calculate original element
        // from average of first i elements
        arr[i] = (i + 1) * arr[i] - sum;
 
        // Add current element to sum
        sum += arr[i];
    }
 
    // Print the array
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver function
int main()
{
 
    int arr[] = { 2, 6, 8, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Call the function
    findOriginal(arr, N);
 
    return 0;
}

Java

// Java implementation for the above approach
class GFG
{
   
  // Function to find the original
  // array from the modified array
  static void findOriginal(int arr[], int N)
  {
 
      // Initialize the variable sum
      // with the first element of array
      int sum = arr[0];
 
      for (int i = 1; i < N; i++) {
 
          // Calculate original element
          // from average of first i elements
          arr[i] = (i + 1) * arr[i] - sum;
 
          // Add current element to sum
          sum += arr[i];
      }
 
      // Print the array
      for (int i = 0; i < N; i++) {
          System.out.print(arr[i] + " ");
      }
  }
 
  // Driver function
  public static void main(String [] args)
  {
 
      int [] arr = new int [] { 2, 6, 8, 10 };
      int N = arr.length;
 
      // Call the function
      findOriginal(arr, N);
  }   
}
 
// This code is contributed by ihritik

Python3

# Python implementation for the above approach
 
# Function to find the original
# array from the modified array
def findOriginal(arr, N):
 
    # Initialize the variable sum
    # with the first element of array
    sum = arr[0]
 
    for i in range(1, N):
 
        # Calculate original element
        # from average of first i elements
        arr[i] = (i + 1) * arr[i] - sum
 
        # Add current element to sum
        sum = sum + arr[i]
     
 
    # Print the array
    for i in range (0, N):
        print(arr[i], end=" ")
  
 
# Driver function
 
arr= [ 2, 6, 8, 10 ]
N = len(arr)
 
# Call the function
findOriginal(arr, N)
 
 
# This code is contributed by ihritik

C#

// C# program for above approach
using System;
 
class GFG {
 
    // Function to find the original
    // array from the modified array
    static void findOriginal(int[] arr, int N)
    {
 
        // Initialize the variable sum
        // with the first element of array
        int sum = arr[0];
 
        for (int i = 1; i < N; i++) {
 
            // Calculate original element
            // from average of first i elements
            arr[i] = (i + 1) * arr[i] - sum;
 
            // Add current element to sum
            sum += arr[i];
        }
 
        // Print the array
        for (int i = 0; i < N; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int N = 4;
        int[] arr = { 2, 6, 8, 10 };
        findOriginal(arr, N);
    }
}
// This code is contributed by dwivediyash

Javascript

// JavaScript implementation for the above approach
 
// Function to find the original
// array from the modified array
function findOriginal(arr, N)
{
 
    // Initialize the variable sum
    // with the first element of array
    var sum = arr[0];
 
    for (var i = 1; i < N; i++) {
 
        // Calculate original element
        // from average of first i elements
        arr[i] = (i + 1) * arr[i] - sum;
 
        // Add current element to sum
        sum += arr[i];
    }
 
    // Print the array
    for (var i = 0; i < N; i++) {
           document.write(arr[i] + " ");
    }
}
 
// Driver code
var arr = [ 2, 6, 8, 10 ];
var N = arr.length;
 
// Call the function
findOriginal(arr, N);
 
// This code is contributed by AnkThon
 

输出

2 10 12 16 

时间复杂度： O(N)
辅助空间： O(1)