中心对称矩阵

给定大小为nxn的矩阵。任务是检查给定矩阵是否为中心对称矩阵。

在数学中，中心对称矩阵是关于其中心对称的矩阵。

例子：

Input : n = 3,
        m[][] = { { 1, 3, 5 },
                  { 6, 8, 6 },
                  { 5, 3, 1 } };
Output : Yes

Input : n = 3, 
        m[][] = { { 1, 3, 5 },
                  { 6, 8, 6 },
                  { 5, 3, 0 } };
Output : No

想法是检查每个元素m [i] [j]是否等于m [n – i – 1] [n – j – 1]。如果任何元素不满足上述条件，则打印“否”，否则打印“是”。

下面是上述方法的实现：

C++

// CPP Program to check whether given
// matrix is centrosymmetric or not.
#include 
using namespace std;
#define N 3
  
bool checkCentrosymmetricted(int n, int m[N][N])
{
    int mid_row;
  
    // Finding the middle row of the matrix
    if (n & 1)
        mid_row = n / 2 + 1;
    else
        mid_row = n / 2;
  
    // for each row upto middle row.
    for (int i = 0; i < mid_row; i++) {
  
        // If each element and its corresponding
        // element is not equal then return false.
        for (int j = 0; j < n; j++) {
            if (m[i][j] != m[n - i - 1][n - j - 1])
                return false;
        }
    }
  
    return true;
}
  
// Driven Program
int main()
{
    int n = 3;
    int m[N][N] = { { 1, 3, 5 },
                    { 6, 8, 6 },
                    { 5, 3, 1 } };
  
    (checkCentrosymmetricted(n, m) ? 
              (cout << "Yes") : (cout << "No"));
  
    return 0;
}

Java

// Java Program to check whether given
// matrix is centrosymmetric or not.
import java.io.*;
  
class GFG {
          
    static int N = 3;
      
    static boolean checkCentrosymmetricted(
                           int n, int m[][])
    {
        int mid_row;
      
        // Finding the middle row of the
        // matrix
        if ((n & 1)>0)
            mid_row = n / 2 + 1;
        else
            mid_row = n / 2;
      
        // for each row upto middle row.
        for (int i = 0; i < mid_row; i++)
        {
      
            // If each element and its 
            // corresponding element is
            // not equal then return false.
            for (int j = 0; j < n; j++)
            {
                if (m[i][j] !=
                  m[n - i - 1][n - j - 1])
                    return false;
            }
        }
      
        return true;
    }
      
    // Driven Program
    public static void main (String[] args)
    {
        int n = 3;
        int m[][] = { { 1, 3, 5 },
                      { 6, 8, 6 },
                      { 5, 3, 1 } };
  
    if(checkCentrosymmetricted(n, m))
        System.out.println( "Yes");
    else
        System.out.println( "No");
    }
}
  
// This code is contributed by anuj_67.

Python3

# Python3 Program to check whether given
# matrix is centrosymmetric or not.
  
def checkCentrosymmetricted( n, m):
  
    mid_row = 0;
  
    # Finding the middle row 
    # of the matrix
    if ((n & 1) > 0):
        mid_row = n / 2 + 1;
    else:
        mid_row = n / 2;
  
    # for each row upto middle row.
    for i in range(int(mid_row)): 
  
        # If each element and 
        # its corresponding
        # element is not equal 
        # then return false.
        for j in range(n):
            if (m[i][j] != m[n - i - 1][n - j - 1]):
                return False;
  
    return True;
      
# Driver Code
n = 3;
m = [[1, 3, 5 ],
     [ 6, 8, 6 ],
     [ 5, 3, 1 ]];
  
if(checkCentrosymmetricted(n, m)):
    print("Yes");
else:
    print("No");
          
# This code is contributed by mits

C#

// C# Program to check whether given
// matrix is centrosymmetric or not.
using System;
  
class GFG {
          
    ///static int N = 3;
      
    static bool checkCentrosymmetricted(
                        int n, int [,]m)
    {
        int mid_row;
      
        // Finding the middle row of the
        // matrix
        if ((n & 1)>0)
            mid_row = n / 2 + 1;
        else
            mid_row = n / 2;
      
        // for each row upto middle row.
        for (int i = 0; i < mid_row; i++)
        {
      
            // If each element and its 
            // corresponding element is
            // not equal then return false.
            for (int j = 0; j < n; j++)
            {
                if (m[i,j] !=
                m[n - i - 1,n - j - 1])
                    return false;
            }
        }
      
        return true;
    }
      
    // Driven Program
    public static void Main()
    {
        int n = 3;
        int [,]m = { { 1, 3, 5 },
                    { 6, 8, 6 },
                    { 5, 3, 1 } };
  
        if(checkCentrosymmetricted(n, m))
            Console.WriteLine( "Yes");
        else
            Console.WriteLine( "No");
    }
}
  
// This code is contributed by anuj_67.

输出

Yes

复杂度： O(N ² )