Un número piramidal cuadrado representa la suma de los cuadrados de los primeros números naturales . Los primeros números piramidales cuadrados son 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, …
Geométricamente, estos números representan el número de esferas que se apilarán para formar una pirámide con base cuadrada. Consulte esta imagen de Wiki para obtener más claridad.
Dado un número s (1 <= s <= 1000000000). Si s es la suma de los cuadrados de los primeros n números naturales, imprima n, de lo contrario imprima -1.
Ejemplos:
Input : 14 Output : 3 Explanation : 1*1 + 2*2 + 3*3 = 14 Input : 26 Output : -1
Una solución simple es ejecutar todos los números a partir de 1, calcular la suma actual. Si la suma actual es igual a la suma dada, devolvemos verdadero, de lo contrario, falso.
C++
// C++ program to check if a
// given number is sum of
// squares of natural numbers.
#include <iostream>
using namespace std;
// Function to find if the
// given number is sum of
// the squares of first n
// natural numbers
int findS(int s)
{
int sum = 0;
// Start adding squares of
// the numbers from 1
for (int n = 1; sum < s; n++)
{
sum += n * n;
// If sum becomes equal to s
// return n
if (sum == s)
return n;
}
return -1;
}
// Drivers code
int main()
{
int s = 13;
int n = findS(s);
n == -1 ? cout << "-1" : cout << n;
return 0;
}
C
// C program to check if a
// given number is sum of
// squares of natural numbers.
#include <stdio.h>
// Function to find if the
// given number is sum of
// the squares of first n
// natural numbers
int findS(int s)
{
int sum = 0;
// Start adding squares of
// the numbers from 1
for (int n = 1; sum < s; n++)
{
sum += n * n;
// If sum becomes equal to s
// return n
if (sum == s)
return n;
}
return -1;
}
// Drivers code
int main()
{
int s = 13;
int n = findS(s);
n == -1 ? printf("-1") : printf("%d",n);
return 0;
}
// This code is contributed by kothavvsaakash.
Java
// Java program to check if a
// given number is sum of
// squares of natural numbers.
class GFG
{
// Function to find if the
// given number is sum of
// the squares of first
// n natural numbers
static int findS(int s)
{
int sum = 0;
// Start adding squares of
// the numbers from 1
for (int n = 1; sum < s; n++)
{
sum += n * n;
// If sum becomes equal to s
// return n
if (sum == s)
return n;
}
return -1;
}
// Drivers code
public static void main(String[] args)
{
int s = 13;
int n = findS(s);
if (n == -1)
System.out.println("-1");
else
System.out.println(n);
}
}
Python3
# Python3 program to find if
# the given number is sum of
# the squares of first
# n natural numbers
# Function to find if the given
# number is sum of the squares
# of first n natural numbers
def findS (s):
_sum = 0
n = 1
# Start adding squares of
# the numbers from 1
while(_sum < s):
_sum += n * n
n+= 1
n-= 1
# If sum becomes equal to s
# return n
if _sum == s:
return n
return -1
# Driver code
s = 13
n = findS (s)
if n == -1:
print("-1")
else:
print(n)
C#
// C# program to check if a given
// number is sum of squares of
// natural numbers.
using System;
class GFG
{
// Function to find if the given
// number is sum of the squares
// of first n natural numbers
static int findS(int s)
{
int sum = 0;
// Start adding squares of
// the numbers from 1
for (int n = 1; sum < s; n++)
{
sum += n * n;
// If sum becomes equal
// to s return n
if (sum == s)
return n;
}
return -1;
}
// Drivers code
public static void Main()
{
int s = 13;
int n = findS(s);
if(n == -1)
Console.Write("-1") ;
else
Console.Write(n);
}
}
// This code is contribute by
// Smitha Dinesh Semwal
PHP
<?php
// PHP program to check if a
// given number is sum of
// squares of natural numbers.
// Function to find if the given number
// is sum of the squares of first n
// natural numbers
function findS($s)
{
$sum = 0;
// Start adding squares of
// the numbers from 1
for ($n = 1; $sum < $s; $n++)
{
$sum += $n * $n;
// If sum becomes equal to s
// return n
if ($sum == $s)
return $n;
}
return -1;
}
// Drivers code
$s = 13;
$n = findS($s);
if($n == -1)
echo("-1");
else
echo($n);
// This code is contributed by Ajit.
?>
Javascript
<script>
// JavaScript to check if a
// given number is sum of
// squares of natural numbers.
// Function to find if the
// given number is sum of
// the squares of first
// n natural numbers
function findS(s)
{
let sum = 0;
// Start adding squares of
// the numbers from 1
for (let n = 1; sum < s; n++)
{
sum += n * n;
// If sum becomes equal to s
// return n
if (sum == s)
return n;
}
return -1;
}
// Driver code
let s = 13;
let n = findS(s);
if (n == -1)
document.write("-1");
else
document.write(n);
// This code is contributed by souravghosh0416.
</script>
PRODUCCIÓN :
-1
Una solución alternativa es utilizar el método de Newton Raphson .
Sabemos que la suma de los cuadrados de los primeros n números naturales es n * (n + 1) * (2*n + 1) / 6 .
Podemos escribir las soluciones como
k * (k + 1) * (2*k + 1) / 6 = s
k * (k + 1) * (2*k + 1) – 6s = 0
Podemos encontrar las raíces de las cúbicas anteriores ecuación usando el método de Newton Raphson, luego verifique si la raíz es un número entero o no.
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA