矩阵对角元素的平方

你给出了一个奇数维的整数矩阵。求两边对角线元素的平方。

例子：

Input  :  1 2 3
          4 5 6
          7 8 9
Output :  Diagonal one: 1 25 81
          Diagonal two: 9 25 49

Input  :  2 5 7  
          3 7 2
          5 6 9
Output :  Diagonal one : 4 49 81
          Diagonal two : 49 49 25

方法1：首先我们找到矩阵的对角元素，然后打印该元素的平方。

C++

// Simple CPP program to print squares of
// diagonal elements.
#include 
using namespace std;
  
#define MAX 100
  
// function of diagonal square
void diagonalsquare(int mat[][MAX], int row, 
                                  int column)
{
    // This loop is for finding square of first 
    // diagonal elements
    cout << "Diagonal one : ";
    for (int i = 0; i < row; i++)
    {
        for (int j = 0; j < column; j++)
  
            // if this condition will become true 
            // then we will get diagonal element
            if (i == j)
  
                // printing square of diagonal element
                cout << mat[i][j] * mat[i][j] << " ";
    }
  
    // This loop is for finding square of second 
    // side of diagonal elements
    cout << " \n\nDiagonal two : ";
    for (int i = 0; i < row; i++)
    {
        for (int j = 0; j < column; j++)
  
            // if this condition will become true
            // then we will get second side diagonal 
            // element
            if (i + j == column - 1)
  
                // printing square of diagonal element
                cout << mat[i][j] * mat[i][j] << "  ";
    }
}
  
// Driver code
int main()
{
    int mat[][MAX] = { { 2, 5, 7 },
                       { 3, 7, 2 }, 
                       { 5, 6, 9 } };
    diagonalsquare(mat, 3, 3);
    return 0;
}

Java

// Simple JAva program to print squares of
// diagonal elements.
import java.io.*;
  
class GFG
{
    static int MAX =100;
      
    // function of diagonal square
    static void diagonalsquare(int mat[][], int row, 
                                         int column)
    {
        // This loop is for finding square of first 
        // diagonal elements
        System.out.print( "Diagonal one : ");
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < column; j++)
      
                // if this condition will become true 
                // then we will get diagonal element
                if (i == j)
      
                    // printing square of diagonal element
                    System.out.print ( mat[i][j] * mat[i][j] +" ");
        }
        System.out.println();
      
        // This loop is for finding square of second 
        // side of diagonal elements
        System.out.print("Diagonal two : ");
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < column; j++)
      
                // if this condition will become true
                // then we will get second side diagonal 
                // element
                if (i + j == column - 1)
      
                    // printing square of diagonal element
                    System.out.print(mat[i][j] * mat[i][j] +" ");
        }
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int mat[][] = { { 2, 5, 7 },
                        { 3, 7, 2 }, 
                        { 5, 6, 9 } };
        diagonalsquare(mat, 3, 3);
          
    }
}
  
// This code is contributed by vt_m.

Python3

# Simple Python program 
# to print squares
# of diagonal elements.
  
# function of diagonal square
def diagonalsquare(mat, row, column) :
  
    # This loop is for finding square
    # of first diagonal elements
    print ("Diagonal one : ", end = "")
    for i in range(0, row) :
        for j in range(0, column) :
  
            # if this condition will 
            # become true then we will
            # get diagonal element
            if (i == j) : 
                # printing square of 
                # diagonal element
                print ("{} ".format(mat[i][j] * 
                                    mat[i][j]), end = "")
  
    # This loop is for finding 
    # square of second side 
    # of diagonal elements
    print (" \n\nDiagonal two : ", end = "")
    for i in range(0, row) :
        for j in range(0, column) :
  
            # if this condition will become
            # true then we will get second
            # side diagonal element
            if (i + j == column - 1) :
  
                # printing square of diagonal
                # element
                print ("{} ".format(mat[i][j] * 
                                    mat[i][j]), end = "")
  
  
# Driver code
mat = [[ 2, 5, 7 ],
        [ 3, 7, 2 ], 
        [ 5, 6, 9 ]]
diagonalsquare(mat, 3, 3)
  
# This code is contributed by 
# Manish Shaw(manishshaw1)

C#

// Simple C# program to print squares of
// diagonal elements.
using System;
  
class GFG
{
    //static int MAX =100;
      
    // function of diagonal square
    static void diagonalsquare(int [,]mat, int row, 
                                        int column)
    {
          
        // This loop is for finding
        // square of first 
        // diagonal elements
        Console.Write( "Diagonal one : ");
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < column; j++)
      
                // if this condition will become true 
                // then we will get diagonal element
                if (i == j)
      
                    // printing square of diagonal element
                    Console.Write ( mat[i,j] * mat[i,j] +" ");
        }
        Console.WriteLine();
      
        // This loop is for finding 
        // square of second side of
        // diagonal elements
        Console.Write("Diagonal two : ");
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < column; j++)
      
                // if this condition will become true
                // then we will get second side diagonal 
                // element
                if (i + j == column - 1)
      
                    // printing square of diagonal element
                    Console.Write(mat[i,j] * mat[i,j] +" ");
        }
    }
      
    // Driver code
    public static void Main () 
    {
        int [,]mat = {{ 2, 5, 7 },
                      { 3, 7, 2 }, 
                      { 5, 6, 9 }};
        diagonalsquare(mat, 3, 3);
          
    }
}
  
// This code is contributed by anuj_67.

输出：

Diagonal one : 4 49 81  

Diagonal two : 49  49  25

时间复杂度O(n * n )

方法二：
一个有效的解决方案也与天真的方法相同，但在这种情况下，我们只需要一个循环来找到对角线元素，然后我们打印该元素的平方。

C++

// Efficient CPP program to print squares of
// diagonal elements.
#include 
using namespace std;
  
#define MAX 100
  
// function of diagonal square
void diagonalsquare(int mat[][MAX], int row,
                                 int column)
{
    // This loop is for finding of square of 
    // the first side of diagonal elements
    cout << " \nDiagonal one : ";
    for (int i = 0; i < row; i++)
    {
  
        // printing direct square of diagonal 
        // element there is no need to check 
        // condition
        cout << mat[i][i] * mat[i][i] << " ";
    }
  
    // This loop is for finding square of the 
    // second side of diagonal elements
    cout << " \n\nDiagonal two : ";
    for (int i = 0; i < row; i++)
    {
        // printing direct square of diagonal 
        // element in the second side
        cout << mat[i][row - i - 1] * mat[i][row - i - 1] 
            << " ";
    }
}
  
// Driver code
int main()
{
    int mat[][MAX] = { { 2, 5, 7 },
                    { 3, 7, 2 }, 
                    { 5, 6, 9 } };
    diagonalsquare(mat, 3, 3);
    return 0;
}

Java

// Efficient JAVA program to print squares of
// diagonal elements.
import java.io.*;
  
class GFG 
{
    static int MAX =100;
      
    // function of diagonal square
    static void diagonalsquare(int mat[][], int row, int column)
    {
        // This loop is for finding of square of 
        // the first side of diagonal elements
        System.out.print (" Diagonal one : ");
        for (int i = 0; i < row; i++)
        {
      
            // printing direct square of diagonal 
            // element there is no need to check 
            // condition
            System.out.print( mat[i][i] * mat[i][i] +" ");
        }
        System.out.println();
      
        // This loop is for finding square of the 
        // second side of diagonal elements
        System.out.print( " Diagonal two : ");
        for (int i = 0; i < row; i++)
        {
            // printing direct square of diagonal 
            // element in the second side
            System.out.print( mat[i][row - i - 1] *
                              mat[i][row - i - 1] + " ");
        }
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int mat[][] = { { 2, 5, 7 },
                        { 3, 7, 2 }, 
                        { 5, 6, 9 } };
        diagonalsquare(mat, 3, 3);
      
    }
} 
  
// This code is contributed by vt_m.

Python3

# Efficient Python program 
# to print squares of 
# diagonal elements.
  
# function of diagonal square
def diagonalsquare(mat, row, 
                   column) :
      
    # This loop is for finding
    # of square of the first 
    # side of diagonal elements
    print ("Diagonal one : ", 
                    end = "")
    for i in range(0, row) : 
  
        # printing direct square 
        # of diagonal element 
        # there is no need to 
        # check condition
        print (mat[i][i] *
               mat[i][i], end = " ")
      
  
    # This loop is for finding 
    # square of the second side
    # of diagonal elements
    print ("\n\nDiagonal two : ",
                        end = "")
      
    for i in range(0, row) :     
          
        # printing direct square 
        # of diagonal element in
        # the second side
        print (mat[i][row - i - 1] *
               mat[i][row - i - 1] , 
                          end = " ") 
  
# Driver code
mat = [[2, 5, 7 ],
       [3, 7, 2 ], 
       [5, 6, 9 ]]
diagonalsquare(mat, 3, 3)
      
# This code is contributed by 
# Manish Shaw(manishshaw1)

C＃

// Efficient C# program to print 
// squares of diagonal elements.
using System;
  
class GFG {
      
    static int MAX =100;
      
    // function of diagonal square
    static void diagonalsquare(int [,] mat, 
                               int row, 
                               int column)
    {
          
        // This loop is for finding of 
        // square of the first side of 
        // diagonal elements
        Console.Write ("Diagonal one : ");
          
        for (int i = 0; i < row; i++)
        {
      
            // printing direct square of diagonal 
            // element there is no need to check 
            // condition
            Console.Write(mat[i, i] * 
                          mat[i, i] +" ");
        }
        Console.WriteLine();
      
        // This loop is for finding square 
        // of the second side of diagonal
        // elements
        Console.Write("Diagonal two : ");
        for (int i = 0; i < row; i++)
        {
            // printing direct square of diagonal 
            // element in the second side
            Console.Write(mat[i, row - i - 1] *
                          mat[i, row - i - 1] + " ");
        }
    }
      
    // Driver code
    public static void Main () 
    {
        int [,] mat = new int[,]{{ 2, 5, 7 },
                                 { 3, 7, 2 }, 
                                 { 5, 6, 9 }};
        diagonalsquare(mat, 3, 3);
      
    }
} 
  
// This code is contributed by KRV.

输出：

Diagonal one : 4 49 81  

Diagonal two : 49 49 25 

时间复杂度O(n)