矩阵对角元素的正方形 · GeeksForGeeks 中文系列教程

# 矩阵对角元素的正方形 > 原文： [https://www.geeksforgeeks.org/squares-of-matrix-diagonal-elements/](https://www.geeksforgeeks.org/squares-of-matrix-diagonal-elements/) 您给定了一个具有奇数维的整数矩阵。在两侧找到对角线元素的平方。 **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，找到矩阵的对角元素，然后打印该元素的平方。 ## 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 ``` **时间复杂度** O（n * n） **方法 2**：一种有效的解决方案也与朴素的方法相同，但是在这种方法中，我们只用一个循环找到对角线元素，然后打印该元素的平方。 ## 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; } ```