Dado un número n, comprobar si es el número de Stella Octangula o no. Un número de la forma 
donde n es un número entero (0, 1, 2, 3, 4, …) se llama Stella Octangula. Los primeros números de Stella Octangula son 0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, …
Los números de Stella octangula que son cuadrados perfectos son 1 y 9653449 .
Dado un número x, comprueba si es Stella octangula.
Ejemplos:
Input: x = 51 Output: Yes For n = 3, the value of expression n(2n2 - 1) is 51 Input: n = 53 Output: No
Una solución simple es ejecutar un bucle a partir de n = 0. Para cada n, compruebe si n(2n 2 – 1) es igual a x. Ejecutamos el bucle mientras el valor de n(2n 2 – 1) es menor o igual que x.
Una solución eficiente es utilizar la búsqueda binaria ilimitada . Primero encontramos un valor de n tal que n(2n 2 – 1) es mayor que x usando duplicaciones repetidas. Luego aplicamos la búsqueda binaria .
C++
// Program to check if a number is Stella
// Octangula Number
#include <bits/stdc++.h>
using namespace std;
// Returns value of n*(2*n*n - 1)
int f(int n) {
return n*(2*n*n - 1);
}
// Finds if a value of f(n) is equal to x
// where n is in interval [low..high]
bool binarySearch(int low, int high, int x)
{
while (low <= high) {
long long mid = (low + high) / 2;
if (f(mid) < x)
low = mid + 1;
else if (f(mid) > x)
high = mid - 1;
else
return true;
}
return false;
}
// Returns true if x isStella Octangula Number.
// Else returns false.
bool isStellaOctangula(int x)
{
if (x == 0)
return true;
// Find 'high' for binary search by
// repeated doubling
int i = 1;
while (f(i) < x)
i = i*2;
// If condition is satisfied for a
// power of 2.
if (f(i) == x)
return true;
// Call binary search
return binarySearch(i/2, i, x);
}
// driver code
int main()
{
int n = 51;
if (isStellaOctangula(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Program to check if
// a number is Stella
// Octangula Number
import java.io.*;
class GFG
{
// Returns value of
// n*(2*n*n - 1)
static int f(int n)
{
return n * (2 * n *
n - 1);
}
// Finds if a value of
// f(n) is equal to x
// where n is in
// interval [low..high]
static boolean binarySearch(int low,
int high,
int x)
{
while (low <= high)
{
int mid = (low + high) / 2;
if (f(mid) < x)
low = mid + 1;
else if (f(mid) > x)
high = mid - 1;
else
return true;
}
return false;
}
// Returns true if x
// is Stella Octangula
// Number.Else returns
// false.
static boolean isStellaOctangula(int x)
{
if (x == 0)
return true;
// Find 'high' for
// binary search by
// repeated doubling
int i = 1;
while (f(i) < x)
i = i * 2;
// If condition is
// satisfied for a
// power of 2.
if (f(i) == x)
return true;
// Call binary search
return binarySearch(i / 2,
i, x);
}
// Driver code
public static void main (String[] args)
{
int n = 51;
if (isStellaOctangula(n))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed
// by anuj_67.
Python3
# Python3 program to check if a number
# is Stella octangula number
# Returns value of n*(2*n*n - 1)
def f(n):
return n * (2 * n * n - 1);
# Finds if a value of f(n) is equal to x
# where n is in interval [low..high]
def binarySearch(low, high, x):
while (low <= high):
mid = int((low + high) // 2);
if (f(mid) < x):
low = mid + 1;
elif (f(mid) > x):
high = mid - 1;
else:
return True;
return False;
# Returns true if x isStella octangula
# number. Else returns false.
def isStellaOctangula(x):
if (x == 0):
return True;
# Find 'high' for binary search
# by repeated doubling
i = 1;
while (f(i) < x):
i = i * 2;
# If condition is satisfied for a
# power of 2.
if (f(i) == x):
return True;
# Call binary search
return binarySearch(i / 2, i, x);
# Driver code
n = 51;
if (isStellaOctangula(n) == True):
print("Yes");
else:
print("No");
# This code is contributed by Code_Mech
C#
// Program to check if
// a number is Stella
// Octangula Number
using System;
class GFG
{
// Returns value of
// n*(2*n*n - 1)
static int f(int n)
{
return n * (2 * n *
n - 1);
}
// Finds if a value of
// f(n) is equal to x
// where n is in
// interval [low..high]
static bool binarySearch(int low,
int high,
int x)
{
while (low <= high)
{
int mid = (low + high) / 2;
if (f(mid) < x)
low = mid + 1;
else if (f(mid) > x)
high = mid - 1;
else
return true;
}
return false;
}
// Returns true if x
// is Stella Octangula
// Number.Else returns
// false.
static bool isStellaOctangula(int x)
{
if (x == 0)
return true;
// Find 'high' for
// binary search by
// repeated doubling
int i = 1;
while (f(i) < x)
i = i * 2;
// If condition is
// satisfied for a
// power of 2.
if (f(i) == x)
return true;
// Call binary search
return binarySearch(i / 2,
i, x);
}
// Driver code
public static void Main ()
{
int n = 51;
if (isStellaOctangula(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed
// by anuj_67.
Javascript
<script>
// JavaScript program to check if
// a number is Stella
// Octangula Number
// Returns value of
// n*(2*n*n - 1)
function f(n)
{
return n * (2 * n *
n - 1);
}
// Finds if a value of
// f(n) is equal to x
// where n is in
// interval [low..high]
function binarySearch(low, high, x)
{
while (low <= high)
{
let mid = (low + high) / 2;
if (f(mid) < x)
low = mid + 1;
else if (f(mid) > x)
high = mid - 1;
else
return true;
}
return false;
}
// Returns true if x
// is Stella Octangula
// Number.Else returns
// false.
function isStellaOctangula(x)
{
if (x == 0)
return true;
// Find 'high' for
// binary search by
// repeated doubling
let i = 1;
while (f(i) < x)
i = i * 2;
// If condition is
// satisfied for a
// power of 2.
if (f(i) == x)
return true;
// Call binary search
return binarySearch(i / 2,
i, x);
}
// Driver code
let n = 51;
if (isStellaOctangula(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by souravghosh0416.
</script>
Producción:
Yes
Publicación traducida automáticamente
Artículo escrito por jaideeppyne1997 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA