Programa para Números de Armstrong

Dado un número x , determine si el número dado es el número de Armstrong o no. 

Un entero positivo de n dígitos se denomina número de Armstrong de orden n (el orden es el número de dígitos) si. 

abcd… = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n) + …. 

Ejemplo: 

Entrada : 153

Salida : Sí

153 es un número de Armstrong.

1*1*1 + 5*5*5 + 3*3*3 = 153

Entrada : 120

Salida : Sí

120 no es un número de Armstrong.

1*1*1 + 2*2*2 + 0*0*0 = 9

Entrada: 1253

Salida : Sí

1253 no es un número de Armstrong

1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723

Entrada: 1634

Salida : Sí

1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634

Enfoque: La idea es primero contar los dígitos de los números (o encontrar el orden). Sea n el número de dígitos. Para cada dígito r en el número de entrada x, calcule r ⁿ . Si la suma de todos estos valores es igual a n, devuelve verdadero, de lo contrario, falso.

C++

// C++ program to determine whether the number is // Armstrong number or not #include <bits/stdc++.h> using namespace std;   /* Function to calculate x raised to the power y */ int power(int x, unsigned int y) {     if (y == 0)         return 1;     if (y % 2 == 0)         return power(x, y / 2) * power(x, y / 2);     return x * power(x, y / 2) * power(x, y / 2); }   /* Function to calculate order of the number */ int order(int x) {     int n = 0;     while (x) {         n++;         x = x / 10;     }     return n; }   // Function to check whether the given number is // Armstrong number or not bool isArmstrong(int x) {     // Calling order function     int n = order(x);     int temp = x, sum = 0;     while (temp) {         int r = temp % 10;         sum += power(r, n);         temp = temp / 10;     }       // If satisfies Armstrong condition     return (sum == x); }   // Driver Program int main() {     int x = 153;     cout << boolalpha << isArmstrong(x) << endl;     x = 1253;     cout << boolalpha << isArmstrong(x) << endl;     return 0; }

C

// C program to find Armstrong number   #include <stdio.h>   /* Function to calculate x raised to the power y */ int power(int x, unsigned int y) {     if (y == 0)         return 1;     if (y % 2 == 0)         return power(x, y / 2) * power(x, y / 2);     return x * power(x, y / 2) * power(x, y / 2); }   /* Function to calculate order of the number */ int order(int x) {     int n = 0;     while (x) {         n++;         x = x / 10;     }     return n; }   // Function to check whether the given number is // Armstrong number or not int isArmstrong(int x) {     // Calling order function     int n = order(x);     int temp = x, sum = 0;     while (temp) {         int r = temp % 10;         sum += power(r, n);         temp = temp / 10;     }       // If satisfies Armstrong condition     if (sum == x)         return 1;     else         return 0; }   // Driver Program int main() {     int x = 153;     if (isArmstrong(x) == 1)         printf("True\n");     else         printf("False\n");       x = 1253;     if (isArmstrong(x) == 1)         printf("True\n");     else         printf("False\n");       return 0; }

Java

// Java program to determine whether the number is // Armstrong number or not public class Armstrong {     /* Function to calculate x raised to the        power y */     int power(int x, long y)     {         if (y == 0)             return 1;         if (y % 2 == 0)             return power(x, y / 2) * power(x, y / 2);         return x * power(x, y / 2) * power(x, y / 2);     }       /* Function to calculate order of the number */     int order(int x)     {         int n = 0;         while (x != 0) {             n++;             x = x / 10;         }         return n;     }       // Function to check whether the given number is     // Armstrong number or not     boolean isArmstrong(int x)     {         // Calling order function         int n = order(x);         int temp = x, sum = 0;         while (temp != 0) {             int r = temp % 10;             sum = sum + power(r, n);             temp = temp / 10;         }           // If satisfies Armstrong condition         return (sum == x);     }       // Driver Program     public static void main(String[] args)     {         Armstrong ob = new Armstrong();         int x = 153;         System.out.println(ob.isArmstrong(x));         x = 1253;         System.out.println(ob.isArmstrong(x));     } }

Python

# Python program to determine whether the number is # Armstrong number or not   # Function to calculate x raised to the power y     def power(x, y):     if y == 0:         return 1     if y % 2 == 0:         return power(x, y/2)*power(x, y/2)     return x*power(x, y/2)*power(x, y/2)   # Function to calculate order of the number     def order(x):       # variable to store of the number     n = 0     while (x != 0):         n = n+1         x = x/10     return n   # Function to check whether the given number is # Armstrong number or not     def isArmstrong(x):     n = order(x)     temp = x     sum1 = 0     while (temp != 0):         r = temp % 10         sum1 = sum1 + power(r, n)         temp = temp/10       # If condition satisfies     return (sum1 == x)     # Driver Program x = 153 print(isArmstrong(x)) x = 1253 print(isArmstrong(x))

Python3

# python3 >= 3.6 for typehint support # This example avoids the complexity of ordering # through type conversions & string manipulation     def isArmstrong(val: int) -> bool:     """val will be tested to see if its an Armstrong number.     Arguments:         val {int} -- positive integer only.     Returns:         bool -- true is /false isn't     """       # break the int into its respective digits     parts = [int(_) for _ in str(val)]       # begin test.     counter = 0     for _ in parts:         counter += _**3     return (counter == val)     # Check Armstrong number print(isArmstrong(153))   print(isArmstrong(1253))

C#

// C# program to determine whether the // number is Armstrong number or not using System;   public class GFG {       // Function to calculate x raised     // to the power y     int power(int x, long y)     {         if (y == 0)             return 1;         if (y % 2 == 0)             return power(x, y / 2) * power(x, y / 2);           return x * power(x, y / 2) * power(x, y / 2);     }       // Function to calculate     // order of the number     int order(int x)     {         int n = 0;         while (x != 0) {             n++;             x = x / 10;         }         return n;     }       // Function to check whether the     // given number is Armstrong number     // or not     bool isArmstrong(int x)     {           // Calling order function         int n = order(x);         int temp = x, sum = 0;         while (temp != 0) {             int r = temp % 10;             sum = sum + power(r, n);             temp = temp / 10;         }           // If satisfies Armstrong condition         return (sum == x);     }       // Driver Code     public static void Main()     {         GFG ob = new GFG();         int x = 153;         Console.WriteLine(ob.isArmstrong(x));         x = 1253;         Console.WriteLine(ob.isArmstrong(x));     } }   // This code is contributed by Nitin Mittal.

JavaScript

<script>       // JavaScript program to determine whether the     // number is Armstrong number or not           // Function to calculate x raised     // to the power y     function power(x, y)     {         if( y == 0)             return 1;         if (y % 2 == 0)             return power(x, parseInt(y / 2, 10)) *                    power(x, parseInt(y / 2, 10));                                return x * power(x, parseInt(y / 2, 10)) *                    power(x, parseInt(y / 2, 10));     }         // Function to calculate     // order of the number     function order(x)     {         let n = 0;         while (x != 0)         {             n++;             x = parseInt(x / 10, 10);         }         return n;     }         // Function to check whether the     // given number is Armstrong number     // or not     function isArmstrong(x)     {                     // Calling order function         let n = order(x);         let temp = x, sum = 0;         while (temp != 0)         {             let r = temp % 10;             sum = sum + power(r, n);             temp = parseInt(temp / 10, 10);         }             // If satisfies Armstrong condition         return (sum == x);     }           let x = 153;     if(isArmstrong(x))     {         document.write("True" + "</br>");     }     else{         document.write("False" + "</br>");     }     x = 1253;     if(isArmstrong(x))     {         document.write("True");     }     else{         document.write("False");     }   </script>

Producción

true
false

El enfoque anterior también se puede implementar de una manera más corta como:

C++

#include <iostream> using namespace std;   int main() {           int n = 153;         int temp = n;         int p = 0;           /*function to calculate           the sum of individual digits          */         while (n > 0) {               int rem = n % 10;             p = (p) + (rem * rem * rem);             n = n / 10;         }           /* condition to check whether            the value of P equals            to user input or not. */           if (temp == p) {             cout<<("Yes. It is Armstrong No.");         }         else {             cout<<("No. It is not an Armstrong No.");         }     return 0; }   // This code is contributed by sathiyamoorthics19

C

#include <stdio.h>   int main() {           int n = 153;         int temp = n;         int p = 0;           /*function to calculate           the sum of individual digits          */         while (n > 0) {               int rem = n % 10;             p = (p) + (rem * rem * rem);             n = n / 10;         }           /* condition to check whether            the value of P equals            to user input or not. */           if (temp == p) {             printf("Yes. It is Armstrong No.");         }         else {             printf("No. It is not an Armstrong No.");         }     return 0; }   // This code is contributed by sathiyamoorthics19

Java

// Java program to determine whether the number is // Armstrong number or not   public class ArmstrongNumber {     public static void main(String[] args)     {           int n = 153;         int temp = n;         int p = 0;           /*function to calculate           the sum of individual digits          */         while (n > 0) {               int rem = n % 10;             p = (p) + (rem * rem * rem);             n = n / 10;         }           /* condition to check whether            the value of P equals            to user input or not. */         if (temp == p) {             System.out.println("Yes. It is Armstrong No.");         }         else {             System.out.println(                 "No. It is not an Armstrong No.");         }     } }

Producción

Yes. It is Armstrong No.

Encuentra ^el número n de Armstrong 

Entrada : 9

Salida : 9

Entrada : 10

Salida : 153

C++

// C++ Program to find // Nth Armstrong Number #include <bits/stdc++.h> #include <math.h> using namespace std;   // Function to find Nth Armstrong Number int NthArmstrong(int n) {     int count = 0;       // upper limit from integer     for (int i = 1; i <= INT_MAX; i++) {         int num = i, rem, digit = 0, sum = 0;           // Copy the value for num in num         num = i;           // Find total digits in num         digit = (int)log10(num) + 1;           // Calculate sum of power of digits         while (num > 0) {             rem = num % 10;             sum = sum + pow(rem, digit);             num = num / 10;         }         // Check for Armstrong number         if (i == sum)             count++;         if (count == n)             return i;     } }   // Driver Function int main() {     int n = 12;     cout << NthArmstrong(n);     return 0; }   // This Code is Contributed by 'jaingyayak'

Java

// Java Program to find // Nth Armstrong Number import java.lang.Math;   class GFG {       // Function to find Nth Armstrong Number     static int NthArmstrong(int n)     {         int count = 0;           // upper limit from integer         for (int i = 1; i <= Integer.MAX_VALUE; i++) {             int num = i, rem, digit = 0, sum = 0;               // Copy the value for num in num             num = i;               // Find total digits in num             digit = (int)Math.log10(num) + 1;               // Calculate sum of power of digits             while (num > 0) {                 rem = num % 10;                 sum = sum + (int)Math.pow(rem, digit);                 num = num / 10;             }               // Check for Armstrong number             if (i == sum)                 count++;             if (count == n)                 return i;         }         return n;     }       // Driver Code     public static void main(String[] args)     {         int n = 12;         System.out.println(NthArmstrong(n));     } }   // This code is contributed by Code_Mech.

Python3

# Python3 Program to find Nth Armstrong Number import math import sys   # Function to find Nth Armstrong Number     def NthArmstrong(n):       count = 0       # upper limit from integer     for i in range(1, sys.maxsize):           num = i         rem = 0         digit = 0         sum = 0           # Copy the value for num in num         num = i           # Find total digits in num         digit = int(math.log10(num) + 1)           # Calculate sum of power of digits         while(num > 0):             rem = num % 10             sum = sum + pow(rem, digit)             num = num // 10           # Check for Armstrong number         if(i == sum):             count += 1         if(count == n):             return i     # Driver Code n = 12 print(NthArmstrong(n))   # This code is contributed by chandan_jnu

C#

// C# Program to find // Nth Armstrong Number using System;   class GFG {       // Function to find Nth Armstrong Number     static int NthArmstrong(int n)     {         int count = 0;           // upper limit from integer         for (int i = 1; i <= int.MaxValue; i++) {             int num = i, rem, digit = 0, sum = 0;               // Copy the value for num in num             num = i;               // Find total digits in num             digit = (int)Math.Log10(num) + 1;               // Calculate sum of power of digits             while (num > 0) {                 rem = num % 10;                 sum = sum + (int)Math.Pow(rem, digit);                 num = num / 10;             }               // Check for Armstrong number             if (i == sum)                 count++;             if (count == n)                 return i;         }         return n;     }       // Driver Code     public static void Main()     {         int n = 12;         Console.WriteLine(NthArmstrong(n));     } }   // This code is contributed by Code_Mech.

PHP

<?php // PHP Program to find // Nth Armstrong Number   // Function to find // Nth Armstrong Number function NthArmstrong($n) {     $count = 0;           // upper limit     // from integer     for($i = 1;         $i <= PHP_INT_MAX; $i++)     {         $num = $i;         $rem;         $digit = 0;         $sum = 0;                   // Copy the value         // for num in num         $num = $i;                   // Find total         // digits in num         $digit = (int) log10($num) + 1;                   // Calculate sum of         // power of digits         while($num > 0)         {             $rem = $num % 10;             $sum = $sum + pow($rem,                               $digit);             $num = (int)$num / 10;         }                   // Check for         // Armstrong number         if($i == $sum)             $count++;         if($count == $n)             return $i;     } }   // Driver Code $n = 12; echo NthArmstrong($n);   // This Code is Contributed // by akt_mit ?>

JavaScript

<script>   // Javascript program to find Nth Armstrong Number   // Function to find Nth Armstrong Number function NthArmstrong(n) {     let count = 0;       // Upper limit from integer     for(let i = 1; i <= Number.MAX_VALUE; i++)     {         let num = i, rem, digit = 0, sum = 0;           // Copy the value for num in num         num = i;           // Find total digits in num         digit = parseInt(Math.log10(num), 10) + 1;           // Calculate sum of power of digits         while(num > 0)         {             rem = num % 10;             sum = sum + Math.pow(rem, digit);             num = parseInt(num / 10, 10);         }           // Check for Armstrong number         if (i == sum)             count++;         if (count == n)             return i;     }     return n; }   // Driver code let n = 12;   document.write(NthArmstrong(n));   // This code is contributed by rameshtravel07     </script>

C

// C Program to find // Nth Armstrong Number #include <stdio.h> #include <math.h> #include<limits.h>   // Function to find Nth Armstrong Number int NthArmstrong(int n) {     int count = 0;       // upper limit from integer     for (int i = 1; i <= INT_MAX; i++) {         int num = i, rem, digit = 0, sum = 0;           // Copy the value for num in num         num = i;           // Find total digits in num         digit = (int)log10(num) + 1;           // Calculate sum of power of digits         while (num > 0) {             rem = num % 10;             sum = sum + pow(rem, digit);             num = num / 10;         }         // Check for Armstrong number         if (i == sum)             count++;         if (count == n)             return i;     } }   // Driver Function int main() {     int n = 12;     printf("%d", NthArmstrong(n));     return 0; }   // This Code is Contributed by //'sathiyamoorthics19'

Producción

371

Uso de strings numéricas:

Al considerar el número como una string, podemos acceder a cualquier dígito que queramos y realizar operaciones

Java

public class armstrongNumber {       public void isArmstrong(String n)     {           char[] s = n.toCharArray();         int size = s.length;         int sum = 0;           for (char num : s)         {             int temp = 1;               int i                 = Integer.parseInt(Character.toString(num));               for (int j = 0; j <= size - 1; j++)             {                 temp *= i;             }               sum += temp;         }           if (sum == Integer.parseInt(n))         {             System.out.println("True");         }           else         {             System.out.println("False");         }     }       public static void main(String[] args)     {           armstrongNumber am = new armstrongNumber();         am.isArmstrong("153");         am.isArmstrong("1253");     } }

Python3

def armstrong(n):     number = str(n)       n = len(number)     output = 0     for i in number:         output = output+int(i)**n       if output == int(number):         return(True)     else:         return(False)     print(armstrong(153)) print(armstrong(1253))

C#

using System; public class armstrongNumber {     public void isArmstrong(String n)     {         char[] s = n.ToCharArray();         int size = s.Length;         int sum = 0;         foreach(char num in s)         {             int temp = 1;             int i = Int32.Parse(char.ToString(num));             for (int j = 0; j <= size - 1; j++) {                 temp *= i;             }             sum += temp;         }         if (sum == Int32.Parse(n)) {             Console.WriteLine("True");         }         else {             Console.WriteLine("False");         }     }       public static void Main(String[] args)     {         armstrongNumber am = new armstrongNumber();         am.isArmstrong("153");         am.isArmstrong("1253");     } }   // This code is contributed by umadevi9616

JavaScript

<script> function armstrong(n){     let number = new String(n)       n = number.length     let output = 0     for(let i of number)       output = output + parseInt(i)**n       if (output == parseInt(number))         return("True" + "<br>")     else         return("False" + "<br>") }           document.write(armstrong(153)) document.write(armstrong(1253))   // This code is contributed by _saurabh_jaiswal. </script>

Producción

True
False

Complejidad temporal: O(n).
Espacio Auxiliar: O(1).

Referencias:  
http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/chap04/arms.html  
http://www.programiz.com/c-programming/examples/check-armstrong-number
 

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