生成一个矩阵，其二次对角线的总和等于一个完美的平方

给定一个整数N ，任务是使用范围[1, N] 中的正整数生成一个维度为N x N的矩阵，使得第二对角线的和是一个完美的平方。

例子：

Input: N = 3
Output:
1 2 3
2 3 1
3 2 1
Explanation:
The sum of secondary diagonal = 3 + 3 + 3 = 9(= 3²).

Input: N = 7
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6
Explanation:
The sum of secondary diagonal = 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49(= 7²).

方法：由于生成的矩阵需要维度为N x N ，因此，为了使次对角线中元素的总和为一个完美的平方，其想法是在次对角线的每个索引处分配N 。因此，这条对角线上所有N 个元素的总和是N ² ，这是一个完美的平方。请按照以下步骤解决问题：

1. 初始化维度为N x N的矩阵mat[][] 。
2. 将矩阵的第一行初始化为 {1 2 3 … N}。
3. 现在的矩阵的剩余行，由1填充由矩阵中的前一行的布置的圆形左移的每一行。
4. 完成上述步骤后打印矩阵。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
void diagonalSumPerfectSquare(int arr[], int N)
{
     
    // Iterate for next N - 1 rows
    for(int i = 0; i < N; i++)
    {
         
        // Print the current row after
        // the left shift
        for(int j = 0; j < N; j++)
        {
            cout << (arr[(j + i) % 7]) << " ";
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
     
    // Given N
    int N = 7;
 
    int arr[N];
 
    // Fill the array with elements
    // ranging from 1 to N
    for(int i = 0; i < N; i++)
    {
        arr[i] = i + 1;
    }
 
    // Function Call
    diagonalSumPerfectSquare(arr, N);
}
 
// This code is contributed by gauravrajput1

Java

// Java program for the above approach
class GFG {
 
    // Function to print the matrix whose sum
    // of element in secondary diagonal is a
    // perfect square
    static void diagonalSumPerfectSquare(int[] arr,
                                         int N)
    {
 
        // Iterate for next N - 1 rows
        for (int i = 0; i < N; i++)
        {
 
            // Print the current row after
            // the left shift
            for (int j = 0; j < N; j++)
            {
                System.out.print(arr[(j + i) % 7] + " ");
            }
            System.out.println();
        }
    }
 
    // Driver Code
    public static void main(String[] srgs)
    {
 
        // Given N
        int N = 7;
 
        int[] arr = new int[N];
 
        // Fill the array with elements
        // ranging from 1 to N
        for (int i = 0; i < N; i++)
        {
            arr[i] = i + 1;
        }
 
        // Function Call
        diagonalSumPerfectSquare(arr, N);
    }
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to print the matrix whose sum
# of element in secondary diagonal is a
# perfect square
def diagonalSumPerfectSquare(arr, N):
   
    # Print the current row
    print(*arr, sep =" ")
     
    # Iterate for next N - 1 rows
    for i in range(N-1):
        
        # Perform left shift by 1
        arr = arr[i::] + arr[:i:]
         
        # Print the current row after
        # the left shift
        print(*arr, sep =" ")
 
# Driver Code
 
# Given N
N = 7
 
arr = []
 
# Fill the array with elements
# ranging from 1 to N
for i in range(1, N + 1):
    arr.append(i)
 
# Function Call
diagonalSumPerfectSquare(arr, N)

C#

// C# program for the
// above approach
using System;
class GFG {
 
    // Function to print the matrix whose sum
    // of element in secondary diagonal is a
    // perfect square
    static void diagonalSumPerfectSquare(int[] arr,
                                         int N)
    {
        // Iterate for next N - 1 rows
        for (int i = 0; i < N; i++)
        {
            // Print the current row after
            // the left shift
            for (int j = 0; j < N; j++)
            {
                Console.Write(arr[(j + i) % 7] + " ");
            }
            Console.WriteLine();
        }
    }
 
    // Driver Code
    public static void Main(String[] srgs)
    {
        // Given N
        int N = 7;
 
        int[] arr = new int[N];
 
        // Fill the array with elements
        // ranging from 1 to N
        for (int i = 0; i < N; i++) {
            arr[i] = i + 1;
        }
 
        // Function Call
        diagonalSumPerfectSquare(arr, N);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program to implement
// the above approach
   
   // Function to prlet the matrix whose sum
    // of element in secondary diagonal is a
    // perfect square
    function diagonalSumPerfectSquare( arr,N)
    {
  
        // Iterate for next N - 1 rows
        for (let i = 0; i < N; i++)
        {
  
            // Prlet the current row after
            // the left shift
            for (let j = 0; j < N; j++)
            {
                document.write(arr[(j + i) % 7] + " ");
            }
            document.write("<br/>");
        }
    }
 
 
// Driver Code
 
        // Given N
        let N = 7;
  
        let arr = new Array(N).fill(0);
  
        // Fill the array with elements
        // ranging from 1 to N
        for (let i = 0; i < N; i++)
        {
            arr[i] = i + 1;
        }
  
        // Function Call
        diagonalSumPerfectSquare(arr, N);
  
 // This code is contributed by avijitmondal1998.

输出

1 2 3 4 5 6 7 
2 3 4 5 6 7 1 
3 4 5 6 7 1 2 
4 5 6 7 1 2 3 
5 6 7 1 2 3 4 
6 7 1 2 3 4 5 
7 1 2 3 4 5 6 

时间复杂度： O(N ² )
辅助空间： O(N)

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