通过执行三次给定操作将所有数组元素减少为零

给定一个大小为N的数组arr[] ，任务是通过恰好 3 次应用以下操作将每个数组元素转换为0 ：

• 选择一个子数组。
• 将子数组的每个元素增加其长度的整数倍。

最后，打印所涉及的子数组的第一个和最后一个索引，以及在每个步骤中添加到子数组的每个索引的元素。

例子：

Input: arr[] = {1, 3, 2, 4}
Output:
Operation 1: 1 1
Added elements: 1
Operation 2: 3 4
Added elements: 4 2 
Operation 3: 2 4
Added elements: -3 -6 -6
Explanation:
Step 1: Select subarray {arr[1]} and add -1 to the only element in the subarray. Hence, the array arr[] modifies to {0, 3, 2, 4}.
Step 2: Select subarray {arr[3], arr[4]} and add {4, 2} to the subarray elements respectively. Hence, the array arr[] modifies to {0, 3, 6, 6}.
Step 3: Select subarray {arr[2], arr[4]} and add {-3, -6, -6} to the subarray elements respectively. Hence, the array arr[] modifies to {0, 0, 0, 0}

Input: arr[] = { 5 }
Output:
Operation 1 : 1 1
Added elements: -5
Operation 2 : 1 1
Added elements: 5
Operation 3 : 1 1
Added elements: -5

方法：根据以下观察可以解决给定的问题：

Operation 1: Select subarray {arr[1], .., arr[N]}. Set A[i] = A[i] – N * A[i] = (N – 1) * (-A[i]). 
After operation 1, each element is a multiple of (N – 1).
Operation 2: Select subarray {arr[1], .. arr[N – 1]}. Add / Subtract a multiple of (N – 1) to all values of the subarray till they reduce to 0.
Operation 3: Select subarray {arr[N]}, of size 1. Add / Subtract a multiple of 1 to make A[N] = 0.

请按照以下步骤解决问题：

• 如果N等于1 ，则分三步分别打印-arr[0], +arr[0], -arr[0] 。
• 否则，执行以下操作：
  • 打印1 N 。从数组的每个元素中减去N * arr[i]打印减去的元素。
  • 打印1 N – 1 。小号ubtract（N – 1）* ARR [I]从子阵列的每个元素，并打印所述经减去值。
  • 最后，打印N 。小号ubtract ARR [I – 1]从元件。打印arr[i – 1] 。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to reduce all
// array elements to zero
void ConvertArray(int arr[], int N)
{
    // If size of array is 1
    if (N == 1) {
 
        // First operation
        cout << "Operation 1 : " << 1
             << " " << 1 << endl;
        cout << "Added elements: "
             << -1 * arr[0] << endl;
        cout << endl;
 
        // 2nd Operation
        cout << "Operation 2 : "
             << 1 << " " << 1 << endl;
        cout << "Added elements: "
             << 1 * arr[0] << endl;
        cout << endl;
 
        // 3rd Operation
        cout << "Operation 3 : "
             << 1 << " " << 1 << endl;
        cout << "Added elements: "
             << -1 * arr[0] << endl;
    }
 
    // Otherwise
    else {
 
        // 1st Operation
        cout << "Operation 1 : "
             << 1 << " " << N << endl;
        cout << "Added elements: ";
        for (int i = 0; i < N; i++) {
            cout << -1 * arr[i] * N << " ";
        }
        cout << endl;
        cout << endl;
 
        // 2nd Operation
        cout << "Operation 2 : "
             << 1 << " " << N - 1 << endl;
        cout << "Added elements: ";
        for (int i = 0; i < N - 1; i++) {
            cout << arr[i] * (N - 1) << " ";
        }
        cout << endl;
        cout << endl;
 
        // 3rd Operation
        cout << "Operation 3 : " << N
             << " " << N << endl;
        cout << "Added elements: ";
        cout << arr[N - 1] * (N - 1) << endl;
    }
}
 
// Driver Code
int main()
{
    // Input
    int arr[] = { 1, 3, 2, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to make all
    // array elements equal to 0
    ConvertArray(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
  // Function to reduce all
  // array elements to zero
  static void ConvertArray(int arr[], int N)
  {
 
    // If size of array is 1
    if (N == 1) {
 
      // First operation
      System.out.println("Operation 1 : " + 1
                         + " " + 1 );
      System.out.println("Added elements: "
                         + -1 * arr[0] );
      System.out.println();
 
      // 2nd Operation
      System.out.println("Operation 2 : "
                         + 1 + " " + 1 );
      System.out.println("Added elements: "
                         + 1 * arr[0] );
      System.out.println();
 
      // 3rd Operation
      System.out.println("Operation 3 : "
                         + 1 + " " + 1 );
      System.out.println("Added elements: "
                         + -1 * arr[0] );
    }
 
    // Otherwise
    else {
 
      // 1st Operation
      System.out.println("Operation 1 : "
                         + 1 + " " + N );
      System.out.print("Added elements: ");
      for (int i = 0; i < N; i++) {
        System.out.print(-1 * arr[i] * N + " ");
      }
      System.out.println();
      System.out.println();
 
      // 2nd Operation
      System.out.println("Operation 2 : "
                         + 1 + " " + (N - 1) );
      System.out.print("Added elements: ");
      for (int i = 0; i < N - 1; i++) {
        System.out.print(arr[i] * (N - 1) + " ");
      }
      System.out.println();
      System.out.println();
 
      // 3rd Operation
      System.out.println("Operation 3 : " + N
                         + " " + N );
      System.out.print("Added elements: ");
      System.out.println(arr[N - 1] * (N - 1) );
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    // Input
    int arr[] = { 1, 3, 2, 4 };
    int N = arr.length;
 
    // Function call to make all
    // array elements equal to 0
    ConvertArray(arr, N);
  }
}
 
// This code is contributed by souravghosh0416.

Python3

# Python 3 program of the above approach
 
# Function to reduce all
# array elements to zero
def ConvertArray(arr, N):
   
    # If size of array is 1
    if (N == 1):
       
        # First operation
        print("Operation 1 :",1,1)
        print("Added elements:",-1 * arr[0])
        print("\n",end = "")
 
        # 2nd Operation
        print("Operation 2 :",1,1)
        print("Added elements:",1 * arr[0])
        print("\n",end = "")
 
        # 3rd Operation
        print("Operation 3 :",1,1)
        print("Added elements:",-1 * arr[0])
        print("\n",end = "")
 
    # Otherwise
    else:
       
        # 1st Operation
        print("Operation 1 :",1,N)
        print("Added elements:",end = " ")
        for i in range(N):
            print(-1 * arr[i] * N,end = " ")
        print("\n")
 
        # 2nd Operation
        print("Operation 2 :",1,N - 1)
        print("Added elements:",end = " ")
        for i in range(N - 1):
            print(arr[i] * (N - 1),end = " ")
        print("\n")
 
        # 3rd Operation
        print("Operation 3 : ",N,N)
        print("Added elements:",end = " ")
        print(arr[N - 1] * (N - 1))
 
# Driver Code
if __name__ == '__main__':
   
    # Input
    arr =  [1, 3, 2, 4]
    N =  len(arr)
     
    # Function call to make all
    # array elements equal to 0
    ConvertArray(arr, N)
     
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
class GFG{
 
  // Function to reduce all
  // array elements to zero
  static void ConvertArray(int[] arr, int N)
  {
 
    // If size of array is 1
    if (N == 1) {
 
      // First operation
      Console.WriteLine("Operation 1 : " + 1
                         + " " + 1 );
      Console.WriteLine("Added elements: "
                         + -1 * arr[0] );
      Console.WriteLine();
 
      // 2nd Operation
      Console.WriteLine("Operation 2 : "
                         + 1 + " " + 1 );
      Console.WriteLine("Added elements: "
                         + 1 * arr[0] );
      Console.WriteLine();
 
      // 3rd Operation
      Console.WriteLine("Operation 3 : "
                         + 1 + " " + 1 );
      Console.WriteLine("Added elements: "
                         + -1 * arr[0] );
    }
 
    // Otherwise
    else {
 
      // 1st Operation
      Console.WriteLine("Operation 1 : "
                         + 1 + " " + N );
      Console.Write("Added elements: ");
      for (int i = 0; i < N; i++) {
        Console.Write(-1 * arr[i] * N + " ");
      }
      Console.WriteLine();
      Console.WriteLine();
 
      // 2nd Operation
      Console.WriteLine("Operation 2 : "
                         + 1 + " " + (N - 1) );
      Console.Write("Added elements: ");
      for (int i = 0; i < N - 1; i++) {
        Console.Write(arr[i] * (N - 1) + " ");
      }
      Console.WriteLine();
      Console.WriteLine();
 
      // 3rd Operation
      Console.WriteLine("Operation 3 : " + N
                         + " " + N );
      Console.Write("Added elements: ");
      Console.WriteLine(arr[N - 1] * (N - 1) );
    }
  }
 
// Driver Code
public static void Main(string[] args)
{
   
    // Input
    int[] arr = { 1, 3, 2, 4 };
    int N = arr.Length;
 
    // Function call to make all
    // array elements equal to 0
    ConvertArray(arr, N);
}
}
 
// This code is contributed by code_hunt.

Javascript

// javascript program for the above approach
 
    // Function to reduce all
    // array elements to zero
    function ConvertArray(arr, N)
    {
 
        // If size of array is 1
        if (N == 1)
        {
 
            // First operation
            document.write("Operation 1 : " + 1 + " " + 1+ "<br/>");
            document.write("Added elements: " + -1 * arr[0]+"<br/>");
            document.write("<br/>");
 
            // 2nd Operation
            document.write("Operation 2 : " + 1 + " " + 1+"<br/>");
            document.write("Added elements: " + 1 * arr[0]+"<br/>");
            document.write("<br/>");
 
            // 3rd Operation
            document.write("Operation 3 : " + 1 + " " + 1+"<br/>");
            document.write("Added elements: " + -1 * arr[0]+"<br/>");
        }
 
        // Otherwise
        else
        {
 
            // 1st Operation
            document.write("Operation 1 : " + 1 + " " + N+"<br/>");
            document.write("Added elements: ");
            for (i = 0; i < N; i++) {
                document.write(-1 * arr[i] * N + " ");
            }
            document.write("<br/>");
            document.write("<br/>");
 
            // 2nd Operation
            document.write("Operation 2 : " + 1 + " " + (N - 1)+"<br/>");
            document.write("Added elements: ");
            for (i = 0; i < N - 1; i++) {
                document.write(arr[i] * (N - 1) + " ");
            }
            document.write("<br/>");
            document.write("<br/>");
 
            // 3rd Operation
            document.write("Operation 3 : " + N + " " + N+"<br/>");
            document.write("Added elements: ");
            document.write(arr[N - 1] * (N - 1)+"<br/>");
        }
    }
 
    // Driver code
     
        // Input
        var arr = [ 1, 3, 2, 4 ];
        var N = arr.length;
 
        // Function call to make all
        // array elements equal to 0
        ConvertArray(arr, N);
 
// This code is contributed by todaysgaurav.

输出：

Operation 1 : 1 4
Added elements: -4 -12 -8 -16 

Operation 2 : 1 3
Added elements: 3 9 6 

Operation 3 : 4 4
Added elements: 12

时间复杂度： O(N)
辅助空间： O(1)

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