构造一个矩阵，其总和等于对角线元素的总和

给定一个整数N ，任务是使用正整数和负整数构造一个大小为N ²且不包括0的矩阵，以使矩阵的总和等于矩阵对角线的总和。
例子：

Input: N = 2 
Output: 
1 -2 
2 4 
Explanation: 
Diagonal sum = (1 + 4) = 5 
Matrix sum = (1 – 2 + 2 + 4) = 5 
Input: N = 5 
Output: 
1 2 3 5 10 
3 1 4 -9 1 
-19 6 1 5 -8 
4 -7 2 1 12 
-17 1 1 1 1 
Explanation: 
Diagonal sum = (1 + 1 + 1 + 1 + 1) = 5 
Matrix sum = 5 
 

方法：
解决此问题的方法是遍历矩阵的所有索引，并在N个对角线位置打印一个正元素(例如y )，并在其余N个位置平均分配一个单值正负整数(例如x和-x ) ² – N个职位。
下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include  
using namespace std;
  
// Function to construct matrix with
// diagonal sum equal to matrix sum
void constructmatrix(int N)
{
    bool check = true;
  
    for (int i = 0; i < N; i++) {
  
        for (int j = 0; j < N; j++) {
  
            // If diagonal position
            if (i == j) {
                cout << 1 << " ";
            }
  
            else if (check) {
  
                // Positve element
                cout << 2 << " ";
                check = false;
            }
            else {
  
                // Negative element
                cout << -2 << " ";
                check = true;
            }
        }
  
        cout << endl;
    }
}
  
// Driver Code
int main()
{
    int N = 5;
  
    constructmatrix(5);
  
    return 0;
}

Java

// Java program to implement
// the above approach
public class Main {
  
    // Function to construct matrix with
    // diagonal sum equal to matrix sum
    public static void constructmatrix(int N)
    {
        boolean check = true;
  
        for (int i = 0; i < N; i++) {
  
            for (int j = 0; j < N; j++) {
  
                // If diagonal position
                if (i == j) {
                    System.out.print("1 ");
                }
                else if (check) {
  
                    // Positve element
                    System.out.print("2 ");
                    check = false;
                }
                else {
                    // Negative element
                    System.out.print("-2 ");
                    check = true;
                }
            }
  
            System.out.println();
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int N = 5;
  
        constructmatrix(5);
    }
}

Python3

# Python3 program to implement 
# the above approach 
  
# Function to construct matrix with 
# diagonal sum equal to matrix sum 
def constructmatrix(N): 
      
    check = bool(True) 
  
    for i in range(N):
        for j in range(N):
  
            # If diagonal position 
            if (i == j): 
                print(1, end = " ")
  
            elif (check): 
  
                # Positve element 
                print(2, end = " ")
                check = bool(False) 
                  
            else:
  
                # Negative element 
                print(-2, end = " ")
                check = bool(True) 
  
        print()
  
# Driver code
N = 5
constructmatrix(5)
  
# This code is contributed by divyeshrabadiya07

C#

// C# program to implement
// the above approach
using System;
  
class GFG{
      
// Function to construct matrix with
// diagonal sum equal to matrix sum
public static void constructmatrix(int N)
{
    bool check = true;
  
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
              
            // If diagonal position
            if (i == j)
            {
                Console.Write("1 ");
            }
            else if (check)
            {
                  
                // Positve element
                Console.Write("2 ");
                check = false;
            }
            else
            {
                  
                // Negative element
                Console.Write("-2 ");
                check = true;
            }
        }
        Console.WriteLine();
    }
}
  
// Driver Code
static public void Main ()
{
    int N = 5;
  
    constructmatrix(N);
}
}
  
// This code is contributed by piyush3010

输出：

1 2 -2 2 -2 
2 1 -2 2 -2 
2 -2 1 2 -2 
2 -2 2 1 -2 
2 -2 2 -2 1

时间复杂度： O(N ² )
辅助空间： O(1)