从加密数组中查找原始数组（其他元素之和的数组）

从给定的大小为 n 的加密数组中查找原始数组。加密数组是通过将原始数组的每个元素替换为剩余数组元素的总和来获得的。
例子 ：

Input :  arr[] = {10, 14, 12, 13, 11}
Output : {5, 1, 3, 2, 4}
Original array {5, 1, 3, 2, 4}
Encrypted array is obtained as:
= {1+3+2+4, 5+3+2+4, 5+1+2+4, 5+1+3+4, 5+1+3+2}
= {10, 14, 12, 13, 11}
Each element of original array is replaced by the 
sum of the remaining array elements.  

Input : arr[] = {95, 107, 103, 88, 110, 87}
Output : {23, 11, 15, 30, 8, 31}

方法纯粹基于算术观察，如下所示：

Let n = 4, and
the original array be ori[] = {a, b, c, d}
encrypted array is given as:
arr[] = {b+c+d, a+c+d, a+b+d, a+b+c}

Elements of encrypted array are :
arr[0] = (b+c+d), arr[1] = (a+c+d), 
arr[2] = (a+b+d), arr[3] = (a+b+c)
add up all the elements
sum =  arr[0] + arr[1] + arr[2] + arr[3]
       = (b+c+d) + (a+c+d) + (a+b+d) + (a+b+c)
       = 3(a+b+c+d) 
Sum of elements of ori[] = sum / n-1
                        = sum/3 
                        = (a+b+c+d)
Thus, for a given encrypted array arr[] of size n, the sum of 
the elements of the original array ori[] can be calculated as:
sum =  (arr[0]+arr[1]+....+arr[n-1]) / (n-1)

Then, elements of ori[] are calculated as:
ori[0] = sum - arr[0]
ori[1] = sum - arr[1] 
        .
        .
ori[n-1] = sum - arr[n-1]                      

下面是上述步骤的实现。

C++

// C++ implementation to find original array
// from the encrypted array
#include 
using namespace std;
 
// Finds and prints the elements of the original
// array
void findAndPrintOriginalArray(int arr[], int n)
{
    // total sum of elements
    // of encrypted array
    int arr_sum = 0;
    for (int i=0; i

---

Java

import java.util.*;
 
class GFG {
 
    // Finds and prints the elements of the original
    // array
    static void findAndPrintOriginalArray(int arr[], int n)
    {
       
        // total sum of elements
        // of encrypted array
        int arr_sum = 0;
        for (int i = 0; i < n; i++) {
            arr_sum += arr[i];
        }
 
        // total sum of elements
        // of original array
        arr_sum = arr_sum / (n - 1);
 
        // calculating and displaying
        // elements of original array
        for (int i = 0; i < n; i++) {
            System.out.print(arr_sum - arr[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 10, 14, 12, 13, 11 };
        int n = arr.length;
        findAndPrintOriginalArray(arr, n);
    }
}
 
// This code is contributed by rj13to.

Python 3

# Python 3 implementation to find
# original array from the encrypted
# array
 
# Finds and prints the elements of
# the original array
def findAndPrintOriginalArray(arr, n):
 
    # total sum of elements
    # of encrypted array
    arr_sum = 0
    for i in range(0, n):
        arr_sum += arr[i]
 
    # total sum of elements
    # of original array
    arr_sum = int(arr_sum / (n - 1))
 
    # calculating and displaying
    # elements of original array
    for i in range(0, n):
        print((arr_sum - arr[i]),
                       end = " ")
 
# Driver program to test above
arr = [10, 14, 12, 13, 11]
n = len(arr)
findAndPrintOriginalArray(arr, n)
 
# This code is contributed By Smitha

C#

// C# program to find original
// array from the encrypted array
using System;
 
class GFG {
     
    // Finds and prints the elements
    // of the original array
    static void findAndPrintOriginalArray(int []arr,
                                          int n)
    {
         
        // total sum of elements
        // of encrypted array
        int arr_sum = 0;
        for (int i = 0; i < n; i++)
            arr_sum += arr[i];
 
        // total sum of elements
        // of original array
        arr_sum = arr_sum / (n - 1);
 
        // calculating and displaying
        // elements of original array
        for (int i = 0; i < n; i++)
        Console.Write(arr_sum - arr[i] + " ");
    }
 
    // Driver Code
    public static void Main (String[] args)
    {
        int []arr = {10, 14, 12, 13, 11};
        int n =arr.Length;
        findAndPrintOriginalArray(arr, n);
    }
}
 
// This code is contributed by parashar...

Javascript

    // Javascript program to find original
    // array from the encrypted array
     
    // Finds and prints the elements
    // of the original array
    function findAndPrintOriginalArray(arr, n)
    {
          
        // total sum of elements
        // of encrypted array
        let arr_sum = 0;
        for (let i = 0; i < n; i++)
            arr_sum += arr[i];
  
        // total sum of elements
        // of original array
        arr_sum = parseInt(arr_sum / (n - 1), 10);
  
        // calculating and displaying
        // elements of original array
        for (let i = 0; i < n; i++)
            document.write(arr_sum - arr[i] + " ");
    }
     
    let arr = [10, 14, 12, 13, 11];
    let n =arr.length;
    findAndPrintOriginalArray(arr, n);
     
    // This code is contributed by rameshtravel07.

输出 ：

5 1 3 2 4

时间复杂度： O(N)