Dado un valor n, necesitamos imprimir el siguiente patrón en consecuencia, usando solo espacio extra constante.
Ejemplos:
Input : n = 1
Output : x
Input : n = 2
Output :
x
x x
x
Input: n = 5
Output:
x
x
o x
o x
x o x o x
x o
x o
x
x
Input: n = 6
Output:
x
x
o x
o x
x o x
x o x x o x
x o x
x o
x o
x
x
Input : n = 7;
Output :
x
x
o x
o x
x o x
x o x
x o x o x o x
x o x
x o x
x o
x o
x
x
Input : n = 8;
Output :
x
x
o x
o x
x o x
x o x
o x o x
x o x o o x o x
x o x o
x o x
x o x
x o
x o
x
x
Podemos dividir este problema en 3 partes:
1) Imprime la mitad superior con n-1 líneas para n impar o n-2 líneas para n par.
2) Imprime líneas intermedias, 1 línea para n impar o 3 líneas para n par.
3) Imprime la mitad inferior, con n-1 líneas para n impar o n-2 líneas para n par.
Para patrones tan complejos, puede ser más fácil si podemos usar la indexación basada en 1
y funciones separadas para imprimir caracteres que comiencen con x o o.
C++
// Author:: Satish Srinivas
#include <iostream>
using namespace std;
// print alternate x o beginning with x
void printx(int n)
{
for (int i = 1; i <= n; i++) {
if (i % 2 != 0)
cout << "x ";
else
cout << "o ";
}
return;
}
// print alternate x o beginning with o
void printo(int n)
{
for (int i = 1; i <= n; i++) {
if (i % 2 != 0)
cout << "o ";
else
cout << "x ";
}
return;
}
// print the pattern for n
void printPattern(int n)
{
// upper half
// n-1 lines for odd, n-2 lines for even
int x = n;
if (n % 2 == 0)
x = x - 1;
// number of spaces to leave in each line
int p = n - 1;
// number of characters in each line
int s = 1;
// prints double lines in each iteration
for (int i = 1; i <= (x - 1) / 2; i++) {
for (int j = 1; j <= p; j++) {
cout << " ";
}
if (i % 2 != 0)
printx(s);
else
printo(s);
cout << endl;
p++;
for (int j = 1; j <= p; j++)
cout << " ";
if (i % 2 != 0)
printx(s);
else
printo(s);
cout << endl;
p--;
s++;
}
// extra upper middle for even
if (n % 2 == 0) {
for (int i = 1; i <= p; i++)
cout << " ";
if (n % 4 != 0)
printx(n / 2);
else
printo(n / 2);
cout << endl;
}
// middle line
if (n % 2 != 0)
printx(n);
else {
if (n % 4 != 0) {
printx(n / 2);
printx(n / 2);
}
else {
printx(n / 2);
printo(n / 2);
}
}
cout << endl;
// extra lower middle for even
if (n % 2 == 0) {
cout << " ";
printx(n / 2);
cout << endl;
}
// lower half
p = 1;
if (n % 2 == 0) {
x--;
p = 2;
}
int q = x / 2;
// one line for each iteration
for (int i = 1; i <= x; i++) {
for (int j = 1; j <= p; j++)
cout << " ";
printx(q);
if (i % 2 == 0)
q--;
cout << endl;
p++;
}
cout << endl;
}
// Driver code
int main()
{
int n = 7;
printPattern(n);
n = 8;
printPattern(n);
return 0;
}
Java
// java program to Print symmetric
// double triangle pattern
class GFG
{
// print alternate x o beginning with x
static void printx(int n)
{
for (int i = 1; i <= n; i++) {
if (i % 2 != 0)
System.out.print("x ");
else
System.out.print("o ");
}
return;
}
// print alternate x o beginning with o
static void printo(int n)
{
for (int i = 1; i <= n; i++) {
if (i % 2 != 0)
System.out.print("o ");
else
System.out.print("x ");
}
return;
}
// print the pattern for n
static void printPattern(int n)
{
// upper half n-1 lines for
// odd, n-2 lines for even
int x = n;
if (n % 2 == 0)
x = x - 1;
// number of spaces to leave in each line
int p = n - 1;
// number of characters in each line
int s = 1;
// prints double lines in each iteration
for (int i = 1; i <= (x - 1) / 2; i++) {
for (int j = 1; j <= p; j++) {
System.out.print(" ");
}
if (i % 2 != 0)
printx(s);
else
printo(s);
System.out.println();
p++;
for (int j = 1; j <= p; j++)
System.out.print(" ");
if (i % 2 != 0)
printx(s);
else
printo(s);
System.out.println();
p--;
s++;
}
// extra upper middle for even
if (n % 2 == 0) {
for (int i = 1; i <= p; i++)
System.out.print(" ");
if (n % 4 != 0)
printx(n / 2);
else
printo(n / 2);
System.out.println();
}
// middle line
if (n % 2 != 0)
printx(n);
else {
if (n % 4 != 0) {
printx(n / 2);
printx(n / 2);
}
else {
printx(n / 2);
printo(n / 2);
}
}
System.out.println();
// extra lower middle for even
if (n % 2 == 0) {
System.out.print(" ");
printx(n / 2);
System.out.println();
}
// lower half
p = 1;
if (n % 2 == 0) {
x--;
p = 2;
}
int q = x / 2;
// one line for each iteration
for (int i = 1; i <= x; i++) {
for (int j = 1; j <= p; j++)
System.out.print(" ");
printx(q);
if (i % 2 == 0)
q--;
System.out.println();
p++;
}
System.out.println();
}
// Driver code
public static void main (String[] args)
{
int n = 7;
printPattern(n);
n = 8;
printPattern(n);
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to Print symmetric
# double triangle pattern
# Print alternate x o beginning with x
def printx(n):
for i in range(1, n + 1):
if (i % 2 != 0):
print("x ", end = "")
else:
print("o ", end = "")
return
# Print alternate x o beginning with o
def printo(n):
for i in range(1, n + 1):
if (i % 2 != 0):
print("o ", end = "")
else:
print("x ", end = "")
return
# Print the pattern for n
def printPattern(n):
# upper half
# n-1 lines for odd,
# n-2 lines for even
x = n
if (n % 2 == 0):
x = x - 1
# number of spaces to leave
# in each line
p = n - 1
# number of characters in each line
s = 1
# prints double lines in each iteration
for i in range(1, (x - 1) // 2 + 1):
for j in range(1, p + 1):
print(" ", end = "")
if (i % 2 != 0):
printx(s)
else:
printo(s)
print()
p += 1
for j in range(1, p + 1):
print(" ", end = "")
if (i % 2 != 0):
printx(s)
else:
printo(s)
print()
p -= 1
s += 1
# extra upper middle for even
if (n % 2 == 0):
for i in range(1, p + 1):
print(" ", end = "")
if (n % 4 != 0):
printx(n // 2)
else:
printo(n // 2)
print()
# middle line
if (n % 2 != 0):
printx(n)
else:
if (n % 4 != 0):
printx(n // 2)
printx(n // 2)
else:
printx(n // 2)
printo(n // 2)
print()
# extra lower middle for even
if (n % 2 == 0):
print(" ", end = "")
printx(n // 2)
print()
# lower half
p = 1
if (n % 2 == 0):
x-=1
p = 2
q = x // 2
# one line for each iteration
for i in range(1, x + 1):
for j in range(1, p + 1):
print(" ", end = "")
printx(q)
if (i % 2 == 0):
q -= 1
print()
p += 1
print()
# Driver code
n = 7
printPattern(n)
n = 8
printPattern(n)
# This code is contributed by mohit kumar
Javascript
<script>
// JavaScript program to Print symmetric
// double triangle pattern
// print alternate x o beginning with x
function printx(n) {
for (var i = 1; i <= n; i++) {
if (i % 2 !== 0) document.write("x ");
else document.write("o ");
}
return;
}
// print alternate x o beginning with o
function printo(n) {
for (var i = 1; i <= n; i++) {
if (i % 2 !== 0) document.write("o ");
else document.write("x ");
}
return;
}
// print the pattern for n
function printPattern(n) {
// upper half n-1 lines for
// odd, n-2 lines for even
var x = n;
if (n % 2 === 0) x = x - 1;
// number of spaces to leave in each line
var p = n - 1;
// number of characters in each line
var s = 1;
// prints double lines in each iteration
for (var i = 1; i <= (x - 1) / 2; i++) {
for (var j = 1; j <= p; j++) {
document.write(" ");
}
if (i % 2 !== 0) printx(s);
else printo(s);
document.write("<br>");
p++;
for (var j = 1; j <= p; j++) document.write(" ");
if (i % 2 != 0) printx(s);
else printo(s);
document.write("<br>");
p--;
s++;
}
// extra upper middle for even
if (n % 2 === 0) {
for (var i = 1; i <= p; i++) document.write(" ");
if (n % 4 !== 0) printx(n / 2);
else printo(n / 2);
document.write("<br>");
}
// middle line
if (n % 2 !== 0) printx(n);
else {
if (n % 4 !== 0) {
printx(n / 2);
printx(n / 2);
} else {
printx(n / 2);
printo(n / 2);
}
}
document.write("<br>");
// extra lower middle for even
if (n % 2 === 0) {
document.write(" ");
printx(n / 2);
document.write("<br>");
}
// lower half
p = 1;
if (n % 2 === 0) {
x--;
p = 2;
}
var q = x / 2;
// one line for each iteration
for (var i = 1; i <= x; i++) {
for (var j = 1; j <= p; j++) document.write(" ");
printx(q);
if (i % 2 === 0) q--;
document.write("<br>");
p++;
}
document.write("<br>");
}
// Driver code
var n = 7;
printPattern(n);
n = 8;
printPattern(n);
</script>
Producción:
x
x
o x
o x
x o x
x o x
x o x o x o x
x o x
x o x
x o
x o
x
x
x
x
o x
o x
x o x
x o x
o x o x
x o x o o x o x
x o x o
x o x
x o x
x o
x o
x
x
La complejidad de tiempo del programa anterior es 
.
Este artículo es una contribución de Satish Srinivas . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a contribuir@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA