Dado el valor de la longitud, imprima el patrón X en un cuadro usando # y ” “
Ejemplos:
Input : 10
Output : ##########
## ##
# # # #
# # # #
# ## #
# ## #
# # # #
# # # #
## ##
##########
Input : 7
Output : #######
## ##
# # # #
# # #
# # # #
## ##
#######
A continuación se muestra la implementación para imprimir X en un patrón de caja rectangular:
C++
// CPP code to print the above
// specified pattern
#include <bits/stdc++.h>
using namespace std;
// Function to print pattern
void pattern(int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || i == n - 1 ||
j == 0 || j == n - 1 ||
i == j || i == n - 1 - j)
cout << "#";
else
cout << " ";
}
cout << endl;
}
}
// Driver program
int main()
{
int n = 9;
pattern(n);
return 0;
}
Java
// Java code to print the above
// specified pattern
class GFG {
// Function to print pattern
static void pattern(int n)
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (i == 0 || i == n - 1 ||
j == 0 || j == n - 1 ||
i == j || i == n - 1 - j)
System.out.print("#");
else
System.out.print(" ");
}
System.out.println();
}
}
// Driver Code
public static void main(String args[])
{
int n = 9;
pattern(n);;
}
}
// This code is contributed by Sam007
Python 3
# Python3 code to print the above
# specified pattern
# Function to print pattern
def pattern(n):
for i in range(0, n):
for j in range(0, n):
if (i == 0 or i == n - 1
or j == 0 or j == n - 1
or i == j or i == n - 1 - j):
print( "#", end="")
else:
print( " ",end="")
print("")
# Driver program
n = 9
pattern(n)
# This code is contributed by Smitha.
C#
// C# code to print the above
// specified pattern
using System;
public class GFG {
// Function to print pattern
static void pattern(int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || i == n - 1 ||
j == 0 || j == n - 1 ||
i == j || i == n - 1 - j)
Console.Write("#");
else
Console.Write(" ");
}
Console.WriteLine();
}
}
// Driver code
public static void Main()
{
int n = 9;
pattern(n);
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP code to print the above
// specified pattern
// Function to print pattern
function pattern($n)
{
for ($i = 0; $i < $n; $i++) {
for ($j = 0; $j < $n; $j++) {
if ($i == 0 || $i == $n - 1 ||
$j == 0 || $j == $n - 1 ||
$i == $j || $i == $n - 1 - $j)
echo "#";
else
echo " ";
}
echo "\n";
}
}
// Driver Code
$n = 9;
pattern($n);
// This code is contributed by nitin mittal
?>
Javascript
<script>
// JavaScript code to print the above
// specified pattern
// Function to print pattern
function pattern(n) {
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
if (
i === 0 ||
i === n - 1 ||
j === 0 ||
j === n - 1 ||
i === j ||
i === n - 1 - j
)
document.write("#");
else document.write(" ");
}
document.write("<br>");
}
}
// Driver code
var n = 9;
pattern(n);
</script>
Producción:
######### ## ## # # # # # # # # # # # # # # # # # # # ## ## #########