用前 N^2 个自然数为输入 N 构造一个矩阵 N x N

给定一个整数N ，任务是构造一个大小为N x N的矩阵M[][] ，其中的数字范围为[1, N^2] ，条件如下：

1. 矩阵 M 的元素应该是 1 到 N^2 之间的整数。
2. 矩阵 M 的所有元素都是成对不同的。
3. 对于包含行 r 到 r+a 和列 c 到 c+a(含)中的单元格的每个方形子矩阵，对于某些有效整数 r,c 和 a>=0： M(r,c)+M(r+a, c+a) 是偶数，M(r,c+a)+M(r+a,c) 是偶数。

例子：

Input: N = 2 
Output: 
1 2 
4 3 
Explanation: 
This matrix has 5 square submatrix and 4 of them ([1], [2], [3], [4]) have a=0 so they satisfy the conditions. 
The last square submatrix is the whole matrix M where r=c=a=1. We can see that M_{(1, 1)}+M_{(2, 2)}=1+3=4 and M_{(1, 2)}+M_{(2, 1)}=2+4=6 are both even.
Input: N = 4 
Output: 
1 2 3 4 
8 7 6 5 
9 10 11 12 
16 15 14 13 
 

方法：我们知道，当两个数的奇偶性相同时，它们的和是偶数。假设M _{(i, j)}的奇偶性是奇数，这意味着M _{(i+1, j+1)} , M _{(i+1, j-1)} , M _{(i-1, j+1 )}的奇偶性₎ , M _{(i-1, j-1)}必须是奇数。
下面是 N = 4 生成大小为 4×4 的矩阵的说明：

所以从上面的插图中，我们必须在棋盘图案中填充矩阵。我们可以通过两种方式填写：

• 所有黑色单元格都有一个奇数整数，而白色单元格有一个偶数整数。
• 所有黑色单元格都有一个偶数，而白色单元格有一个奇数。

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 desired matrix
void UniqueMatrix(int N)
{
    int element_value = 1;
    int i = 0;
 
    // element_value will start from 1
    // and go up to N ^ 2
 
    // i is row number and it starts
    // from 0 and go up to N-1
     
    // Iterate ove all [0, N]
    while(i < N)
    {
         
        // If is even
        if(i % 2 == 0)
        {
            for(int f = element_value;
                    f < element_value + N; f++)
            {
                 
                // If row number is even print
                // the row in forward order
                cout << f << " ";
            }
            element_value += N;
        }
        else
        {
            for(int k = element_value + N - 1;
                    k > element_value - 1; k--)
            {
                 
                // If row number is odd print
                // the row in reversed order
                cout << k << " ";
 
            }
            element_value += N;
        }
        cout << endl;
        i = i + 1;
    }
}
 
// Driver Code
int main()
{
     
    // Given matrix size
    int N = 4;
 
    // Function call
    UniqueMatrix(N);
}
 
// This code is contributed by chitranayal

Java

// Java program for the above approach
public class Gfg
{
   
    // Function to print the desired matrix
    public static void UniqueMatrix(int N)
    {
        int element_value = 1;
        int i = 0;
         
        // element_value will start from 1
        // and go up to N ^ 2
   
        // i is row number and it starts
        // from 0 and go up to N-1
       
        // Iterate ove all [0, N]
        while(i < N)
        {
            // If is even
            if(i % 2 == 0)
            {
                for(int f = element_value;
                    f < element_value + N; f++)
                {
                   
                    // If row number is even print
                    // the row in forward order
                    System.out.print(f+" ");
                }
                element_value += N;
            }
            else
            {
                for(int k = element_value + N - 1;
                    k > element_value - 1; k--)
                {
                   
                    // If row number is odd print
                    // the row in reversed order
                    System.out.print(k+" ");
                }
                element_value += N;
            }
            System.out.println();
            i = i + 1;
        }
    }
   
    // Driver Code
    public static void main(String []args)
    {
       
        // Given matrix size
        int N = 4;
       
        // Function call
        UniqueMatrix(N);
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program for the above approach
 
# Function to print the desired matrix
def UniqueMatrix(N):
 
    element_value = 1
    i = 0
 
    # element_value will start from 1
    # and go up to N ^ 2
 
    # i is row number and it starts
    # from 0 and go up to N-1
     
    # Iterate ove all [0, N]
    while(i < N):
         
        # If is even
        if(i % 2 == 0):
 
            for f in range(element_value, element_value + N, 1):
 
                # If row number is even print
                # the row in forward order
                print(f, end =' ')
            element_value += N
 
        else:
 
            for k in range(element_value + N-1, element_value-1, -1):
 
                # if row number is odd print
                # the row in reversed order
                print(k, end =' ')
 
            element_value += N
 
        print()
        i = i + 1
 
 
# Driver Code
 
# Given Matrix Size
N = 4
 
# Function Call
UniqueMatrix(N)

C#

// C# program for the above approach
using System;
class GFG
{
    static void UniqueMatrix(int N)
    {
        int element_value = 1;
        int i = 0;
          
        // element_value will start from 1
        // and go up to N ^ 2
    
        // i is row number and it starts
        // from 0 and go up to N-1
        
        // Iterate ove all [0, N]
        while(i < N)
        {
           
            // If is even
            if(i % 2 == 0)
            {
                for(int f = element_value;
                    f < element_value + N; f++)
                {
                   
                    // If row number is even print
                    // the row in forward order
                    Console.Write(f + " ");
                }
                element_value += N;
            }
            else
            {
                for(int k = element_value + N - 1;
                    k > element_value - 1; k--)
                {
                    Console.Write(k + " ");
                }
                element_value += N;
            }
            Console.WriteLine();
            i = i + 1;
        }
    }
   
  // Driver code
  static public void Main ()
  {
    // Given matrix size
    int N = 4;
 
    // Function call
    UniqueMatrix(N);
  }
}
 
// This code is contributed by rag2127

Javascript

// Java script  program for the above approach
 
    // Function to print the desired matrix
    function  UniqueMatrix( N)
    {
        let element_value = 1;
        let i = 0;
         
        // element_value will start from 1
        // and go up to N ^ 2
 
        // i is row number and it starts
        // from 0 and go up to N-1
     
        // Iterate ove all [0, N]
        while(i < N)
        {
            // If is even
            if(i % 2 == 0)
            {
                for(let f = element_value;
                    f < element_value + N; f++)
                {
                 
                    // If row number is even print
                    // the row in forward order
                    document.write(f+" ");
                }
                element_value += N;
            }
            else
            {
                for(let k = element_value + N - 1;
                    k > element_value - 1; k--)
                {
                 
                    // If row number is odd print
                    // the row in reversed order
                    document.write(k+" ");
                }
                element_value += N;
            }
            document.write("<br>");
            i = i + 1;
        }
    }
 
    // Driver Code
     
        // Given matrix size
        let N = 4;
     
        // Function call
        UniqueMatrix(N);
 
// contributed by sravan kumar

输出：

1 2 3 4 
8 7 6 5 
9 10 11 12 
16 15 14 13

时间复杂度： O(N^2)
辅助空间： O(N^2)

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