Dado N, comprueba si es un número narcisista o no.
Nota: Número narcisista es un número que es la suma de sus propios dígitos, cada uno elevado a la potencia del número de dígitos
Ejemplos:
Entrada: 153
Salida: sí
Explicación: 1^3+5^3+3^3=153Entrada: 1634
Salida: sí
Explicación: 1^4+6^4+3^4+4^4=1634
El enfoque será contar la cantidad de dígitos y luego extraer cada dígito y luego, al usar la función pow, podemos obtener la potencia de ese dígito y luego resumirlo al final y compararlo con el número original para verificar si es un narcisista. Número o no.
A continuación se muestra la implementación de la idea anterior.
C++
// CPP program for checking of
// Narcissistic number
#include <bits/stdc++.h>
using namespace std;
// function to count digits
int countDigit(int n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
// Returns true if n is Narcissistic number
bool check(int n)
{
// count the number of digits
int l = countDigit(n);
int dup = n;
int sum = 0;
// calculates the sum of digits
// raised to power
while (dup)
{
sum += pow(dup % 10, l);
dup /= 10;
}
return (n == sum);
}
// Driver code
int main()
{
int n = 1634;
if (check(n))
cout << "yes";
else
cout << "no";
return 0;
}
Java
// Java program for checking
// of Narcissistic number
import java.io.*;
import static java.lang.Math.*;
class narcissistic
{
// function to count digits
int countDigit(int n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
// Returns true if n is Narcissistic number
boolean check(int n)
{
// count the number of digits
int l = countDigit(n);
int dup = n;
int sum = 0;
// calculates the sum of
//digits raised to power
while(dup > 0)
{
sum += pow(dup % 10, l);
dup /= 10;
}
return (n == sum);
}
// Driver code
public static void main(String args[])
{
narcissistic obj = new narcissistic();
int n = 1634;
if (obj.check(n))
System.out.println("yes");
else
System.out.println("no");
}
}
//This code is contributed by Anshika Goyal.
Python3
# Python 3 program for checking of # Narcissistic number # function to count digits def countDigit(n) : if (n == 0) : return 0 return (1 + countDigit(n // 10)) # Returns true if n is Narcissistic number def check(n) : # Count the number of digits l = countDigit(n) dup = n; sm = 0 # Calculates the sum of digits # raised to power while (dup) : sm = sm + pow(dup % 10, l) dup = dup // 10 return (n == sm) # Driver code n = 1634 if (check(n)) : print( "yes") else : print( "no") # This code is contributed by Nikita Tiwari.
C#
// C# program for checking
// of Narcissistic number
using System;
class narcissistic
{
// function to count digits
int countDigit(int n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
// Returns true if n is Narcissistic number
bool check(int n)
{
// count the number of digits
int l = countDigit(n);
int dup = n;
int sum = 0;
// calculates the sum of
//digits raised to power
while(dup > 0)
{
sum += (int)Math.Pow(dup % 10, l);
dup /= 10;
}
return (n == sum);
}
// Driver code
public static void Main()
{
narcissistic obj = new narcissistic();
int n = 1634;
if (obj.check(n))
Console.WriteLine("yes");
else
Console.WriteLine("no");
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program for checking of
// Narcissistic number
// Function to count digits
function countDigit($n)
{
if ($n == 0)
return 0;
return (1 + countDigit($n / 10));
}
// Returns true if n is
// Narcissistic number
function check( $n)
{
// count the number of digits
$l = countDigit($n);
$dup = $n;
$sum = 0;
// calculates the sum of digits
// raised to power
while ($dup)
{
$sum += pow($dup % 10, $l);
$dup = (int)$dup / 10;
}
return ($n == $sum);
}
// Driver Code
$n = 1634;
if (check(!$n))
echo "yes";
else
echo "no";
// This code is contributed by akt_mit
?>
Javascript
<script>
// Javascript program for checking of
// Narcissistic number
// Function to count digits
function countDigit(n)
{
if (n == 0)
return 0;
return (1 + countDigit(n / 10));
}
// Returns true if n is
// Narcissistic number
function check( n)
{
// count the number of digits
let l = countDigit(n);
let dup = n;
let sum = 0;
// calculates the sum of digits
// raised to power
while (dup)
{
sum += Math.pow(dup % 10, l);
dup = parseINT(dup / 10);
}
return (n == sum);
}
// Driver Code
let n = 1634;
if (check(!n))
document.write("yes");
else
document.write("no");
// This code is contributed by _saurabh_jaiswal
</script>
Producción
yes
Complejidad de tiempo: O (logn)
Espacio Auxiliar: O(1)
Método 2: método simplificado usando una string
Tenemos que tomar la entrada como una string y atravesar la potencia de cálculo de la string de cada carácter con la longitud de la string. Nota: aquí la longitud de la string da el número de dígitos de ese número
A continuación se muestra la implementación del enfoque anterior.
C++14
// CPP program for checking of
// Narcissistic number
#include <bits/stdc++.h>
#include <string.h>
using namespace std;
string getResult(string st)
{
int sum = 0;
int length = st.length();
// Traversing through the string
for (int i = 0; i < length; i++)
{
// Since ascii value of numbers
// starts from 48 so we subtract it from sum
sum = sum + pow(st[i] - '0', length);
}
// Converting string to integer
int number = stoi(st);
// Comparing number and sum
if (number == sum)
return "yes";
else
return "no";
}
// Driver Code
int main()
{
string st = "153";
cout << getResult(st);
return 0;
}
Java
// Java program for checking of
// Narcissistic number
class GFG
{
static String getResult(String st)
{
int sum = 0;
int length = st.length();
// Traversing through the string
for (int i = 0; i < length; i++)
{
// Since ascii value of numbers
// starts from 48 so we subtract it from sum
sum = sum + (int)Math.pow(st.charAt(i) - '0', length);
}
// Converting string to integer
int number = Integer.parseInt(st);
// Comparing number and sum
if (number == sum)
return "yes";
else
return "no";
}
// Driver Code
public static void main(String []args)
{
String st = "153";
System.out.print(getResult(st));
}
}
// This code is contributed by rutvik_56.
Python3
# Python program for checking of # Narcissistic number def getResult(st): sum = 0 length = len(st) # Traversing through the string for i in st: # Converting character to int sum = sum + int(i) ** length # Converting string to integer number = int(st) # Comparing number and sum if (number == sum): return "true" else: return "false" # Driver Code # taking input as string st = "153" print(getResult(st))
C#
// C# program for checking of
// Narcissistic number
using System;
class GFG
{
static string getResult(string st)
{
int sum = 0;
int length = st.Length;
// Traversing through the string
for (int i = 0; i < length; i++)
{
// Since ascii value of numbers
// starts from 48 so we subtract it from sum
sum = sum + (int)Math.Pow(st[i] - '0', length);
}
// Converting string to integer
int number = int.Parse(st);
// Comparing number and sum
if (number == sum)
return "yes";
else
return "no";
}
// Driver Code
public static void Main(string []args)
{
string st = "153";
Console.Write(getResult(st));
}
}
// This code is contributed by pratham76.
Javascript
<script>
// Javascript program for checking of
// Narcissistic number
function getResult(st)
{
let sum = 0;
let length = st.length;
// Traversing through the string
for (let i = 0; i < length; i++)
{
// Since ascii value of numbers
// starts from 48 so we subtract it from sum
sum = sum + Math.pow(st[i] - '0', length);
}
// Converting string to integer
let number = parseInt(st, 10);
// Comparing number and sum
if (number == sum)
return "yes";
else
return "no";
}
let st = "153";
document.write(getResult(st));
</script>
Producción
yes
Complejidad de tiempo: O (nlogn)
Espacio Auxiliar: O(1)
Referencias: http://mathandmultimedia.com/2012/01/16/narcissistic-numbers/