# 矩陣對角元素的正方形 > 原文： [https://www.geeksforgeeks.org/squares-of-matrix-diagonal-elements/](https://www.geeksforgeeks.org/squares-of-matrix-diagonal-elements/) 您給定了一個具有奇數維的整數矩陣。 在兩側找到對角線元素的平方。 **示例**： ``` 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，找到矩陣的對角元素，然后打印該元素的平方。 ## C++ ```cpp // Simple CPP program to print squares of // diagonal elements. #include <iostream> 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 ```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 ```py # 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# ```cs // 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\. ``` ## PHP ```php <?php // Simple PHP program to print squares // of diagonal elements. //$MAX = 100; // function of diagonal square function diagonalsquare($mat, $row,? ????????????????????????????$column) { ????// This loop is for finding square ????// of first diagonal elements ????echo "Diagonal one : "; ????for ( $i = 0; $i < $row; $i++) ????{ ????????for ( $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 ????????????????echo $mat[$i][$j] * $mat[$i][$j]? ??????????????????????????????????????????, " "; ????} ????// This loop is for finding square of second? ????// side of diagonal elements ????echo " \n\nDiagonal two : "; ????for ( $i = 0; $i < $row; $i++) ????{ ????????for ($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 ????????????????echo $mat[$i][$j] * $mat[$i][$j], ????????????????????????????????????????????" "; ????} } // Driver code ????$mat = array(array( 2, 5, 7 ), ?????????????????array( 3, 7, 2 ),? ?????????????????array( 5, 6, 9 ) ); ????diagonalsquare($mat, 3, 3); // This code is contributed by anuj_67\. ?> ``` Output: ``` Diagonal one : 4 49 81 Diagonal two : 49 49 25 ``` **時間復雜度** O（n * n） **方法 2**： 一種有效的解決方案也與樸素的方法相同，但是在這種方法中，我們只用一個循環找到對角線元素，然后打印該元素的平方。 ## C++ ``` // Efficient CPP program to print squares of // diagonal elements. #include <iostream> 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; } ```