Encuentra la suma de los dígitos de un número en lugares pares e impares

Dado un número N , la tarea es encontrar la suma de los dígitos de un número en lugares pares e impares.

Ejemplos: 

Entrada: N = 54873 
Salida: 
Suma impar = 16 
Suma par = 11

Entrada: N = 457892 
Salida: 
Suma impar = 20 
Suma par = 15  

Acercarse:  

• Primero, calcula el reverso del número dado.
• Al número inverso aplicamos el operador de módulo y extraemos su último dígito, que en realidad es el primer dígito de un número, por lo que es un dígito impar.
• El siguiente dígito será un dígito de posición par, y podemos tomar la suma en turnos alternos.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the reverse of a number
int reverse(int n)
{
    int rev = 0;
    while (n != 0) {
        rev = (rev * 10) + (n % 10);
        n /= 10;
    }
    return rev;
}
 
// Function to find the sum of the odd
// and even positioned digits in a number
void getSum(int n)
{
    n = reverse(n);
    int sumOdd = 0, sumEven = 0, c = 1;
 
    while (n != 0) {
 
        // If c is even number then it means
        // digit extracted is at even place
        if (c % 2 == 0)
            sumEven += n % 10;
        else
            sumOdd += n % 10;
        n /= 10;
        c++;
    }
 
    cout << "Sum odd = " << sumOdd << "\n";
    cout << "Sum even = " << sumEven;
}
 
// Driver code
int main()
{
    int n = 457892;
    getSum(n);
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return the reverse of a number
    static int reverse(int n)
    {
        int rev = 0;
        while (n != 0) {
            rev = (rev * 10) + (n % 10);
            n /= 10;
        }
        return rev;
    }
 
    // Function to find the sum of the odd
    // and even positioned digits in a number
    static void getSum(int n)
    {
        n = reverse(n);
        int sumOdd = 0, sumEven = 0, c = 1;
 
        while (n != 0) {
 
            // If c is even number then it means
            // digit extracted is at even place
            if (c % 2 == 0)
                sumEven += n % 10;
            else
                sumOdd += n % 10;
            n /= 10;
            c++;
        }
 
        System.out.println("Sum odd = " + sumOdd);
        System.out.println("Sum even = " + sumEven);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 457892;
        getSum(n);
    }
}
 
// This code is contributed by
// Surendra_Gangwar

# Python3 implementation of the approach
 
# Function to return the
# reverse of a number
def reverse(n):
    rev = 0
    while (n != 0):
        rev = (rev * 10) + (n % 10)
        n //= 10
    return rev
 
# Function to find the sum of the odd
# and even positioned digits in a number
def getSum(n):
 
    n = reverse(n)
    sumOdd = 0
    sumEven = 0
    c = 1
 
    while (n != 0):
 
        # If c is even number then it means
        # digit extracted is at even place
        if (c % 2 == 0):
            sumEven += n % 10
        else:
            sumOdd += n % 10
        n //= 10
        c += 1
 
    print("Sum odd =", sumOdd)
    print("Sum even =", sumEven)
 
# Driver code
n = 457892
getSum(n)
 
# This code is contributed
# by mohit kumar

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the reverse of a number
    static int reverse(int n)
    {
        int rev = 0;
        while (n != 0) {
            rev = (rev * 10) + (n % 10);
            n /= 10;
        }
        return rev;
    }
 
    // Function to find the sum of the odd
    // and even positioned digits in a number
    static void getSum(int n)
    {
        n = reverse(n);
        int sumOdd = 0, sumEven = 0, c = 1;
 
        while (n != 0) {
 
            // If c is even number then it means
            // digit extracted is at even place
            if (c % 2 == 0)
                sumEven += n % 10;
            else
                sumOdd += n % 10;
            n /= 10;
            c++;
        }
 
        Console.WriteLine("Sum odd = " + sumOdd);
        Console.WriteLine("Sum even = " + sumEven);
    }
 
    // Driver code
    public static void Main()
    {
        int n = 457892;
        getSum(n);
    }
}
 
// This code is contributed by
// Akanksha Rai

<?php
// PHP implementation of the above approach
 
// Function to return the reverse of a number
function reverse($n)
{
    $rev = 0;
    while ($n != 0)
    {
        $rev = ($rev * 10) + ($n % 10);
        $n = floor($n / 10);
    }
    return $rev;
}
 
// Function to find the sum of the odd
// and even positioned digits in a number
function getSum($n)
{
    $n = reverse($n);
    $sumOdd = 0; $sumEven = 0; $c = 1;
 
    while ($n != 0)
    {
 
        // If c is even number then it means
        // digit extracted is at even place
        if ($c % 2 == 0)
            $sumEven += $n % 10;
        else
            $sumOdd += $n % 10;
             
        $n = floor($n / 10);
        $c++;
    }
 
    echo "Sum odd = ", $sumOdd, "\n";
    echo "Sum even = ", $sumEven;
}
 
// Driver code
$n = 457892;
getSum($n);
 
// This code is contributed by Ryuga
?>

<script>
 
//JavaScript implementation of the approach
 
    // Function to return the
    // reverse of a number
    function reverse(n) {
    let rev = 0;
    while (n != 0) {
        rev = (rev * 10) + (n % 10);
        n = Math.floor(n / 10);
    }
    return rev;
}
   
    // Function to find the sum of the odd
    // and even positioned digits in a number
    function getSum(n) {
        n = reverse(n);
        let sumOdd = 0, sumEven = 0, c = 1;
 
        while (n != 0) {
 
        // If c is even number then it means
        // digit extracted is at even place
        if (c % 2 == 0)
            sumEven += n % 10;
        else
            sumOdd += n % 10;
        n = Math.floor(n / 10);
        c++;
    }
  
    document.write("Sum odd = " + sumOdd);
    document.write("<br>");
    document.write("Sum even = " + sumEven);
}
      let n = 457892;
      // function call
      getSum(n);
 
// This code is contributed by Surbhi Tyagi
 
</script>

Sum odd = 20
Sum even = 15

Complejidad temporal: O(n)
Espacio auxiliar: O(1)

Otro enfoque: el problema se puede resolver sin invertir el número. Podemos extraer todos los dígitos del número uno por uno desde el final. Si el número original era impar, el último dígito debe estar en una posición impar, de lo contrario, estará en una posición par. Después de procesar un dígito, podemos invertir el estado de par a impar y viceversa.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of the odd
// and even positioned digits in a number
void getSum(int n)
{
 
    // If n is odd then the last digit
    // will be odd positioned
    bool isOdd = (n % 2 == 1) ? true : false;
 
    // To store the respective sums
    int sumOdd = 0, sumEven = 0;
 
    // While there are digits left process
    while (n != 0) {
 
        // If current digit is odd positioned
        if (isOdd)
            sumOdd += n % 10;
 
        // Even positioned digit
        else
            sumEven += n % 10;
 
        // Invert state
        isOdd = !isOdd;
 
        // Remove last digit
        n /= 10;
    }
 
    cout << "Sum odd = " << sumOdd << "\n";
    cout << "Sum even = " << sumEven;
}
 
// Driver code
int main()
{
    int n = 457892;
    getSum(n);
 
    return 0;
}

// Java implementation of the above approach
class GFG{
     
// Function to find the sum of the odd
// and even positioned digits in a number
static void getSum(int n)
{
     
    // If n is odd then the last digit
    // will be odd positioned
    boolean isOdd = (n % 2 == 1) ? true : false;
 
    // To store the respective sums
    int sumOdd = 0, sumEven = 0;
 
    // While there are digits left process
    while (n != 0)
    {
         
        // If current digit is odd positioned
        if (isOdd)
            sumOdd += n % 10;
 
        // Even positioned digit
        else
            sumEven += n % 10;
 
        // Invert state
        isOdd = !isOdd;
 
        // Remove last digit
        n /= 10;
    }
    System.out.println("Sum odd = " + sumOdd);
    System.out.println("Sum even = " + sumEven);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 457892;
    getSum(n);
}
}
 
// This code is contributed by jrishabh99

# Python3 implementation of the approach
 
# Function to find the sum of the odd
# and even positioned digits in a number
def getSum(n):
 
    # If n is odd then the last digit
    # will be odd positioned
    if (n % 2 == 1) :
        isOdd = True
    else:
        isOdd = False
 
    # To store the respective sums
    sumOdd = 0
    sumEven = 0
 
    # While there are digits left process
    while (n != 0) :
 
        # If current digit is odd positioned
        if (isOdd):
            sumOdd += n % 10
 
        # Even positioned digit
        else:
            sumEven += n % 10
 
        # Invert state
        isOdd = not isOdd
 
        # Remove last digit
        n //= 10
     
    print( "Sum odd = " , sumOdd )
    print("Sum even = " ,sumEven)
 
# Driver code
if __name__ =="__main__":
    n = 457892
    getSum(n)
 
# This code is contributed by chitranayal

// C# implementation of the above approach
using System;
 
class GFG{
     
// Function to find the sum of the odd
// and even positioned digits in a number
static void getSum(int n)
{
     
    // If n is odd then the last digit
    // will be odd positioned
    bool isOdd = (n % 2 == 1) ? true : false;
     
    // To store the respective sums
    int sumOdd = 0, sumEven = 0;
     
    // While there are digits left process
    while (n != 0)
    {
         
        // If current digit is odd positioned
        if (isOdd)
            sumOdd += n % 10;
  
        // Even positioned digit
        else
            sumEven += n % 10;
  
        // Invert state
        isOdd = !isOdd;
  
        // Remove last digit
        n /= 10;
    }
    Console.WriteLine("Sum odd = " + sumOdd);
    Console.Write("Sum even = " + sumEven);
}
 
// Driver code   
static public void Main ()
{
    int n = 457892;
     
    getSum(n);
}
}
 
// This code is contributed by offbeat

<script>
 
// Javascript implementation of the approach
 
// Function to find the sum of the odd
// and even positioned digits in a number
function getSum(n)
{
 
    // If n is odd then the last digit
    // will be odd positioned
    let isOdd = (n % 2 == 1) ? true : false;
 
    // To store the respective sums
    let sumOdd = 0, sumEven = 0;
 
    // While there are digits left process
    while (n != 0) {
 
        // If current digit is odd positioned
        if (isOdd)
            sumOdd += n % 10;
 
        // Even positioned digit
        else
            sumEven += n % 10;
 
        // Invert state
        isOdd = !isOdd;
 
        // Remove last digit
        n = Math.floor(n/10);
    }
 
    document.write("Sum odd = " + sumOdd + "<br>");
    document.write("Sum even = " + sumEven);
}
 
// Driver code
  
    let n = 457892;
    getSum(n);
 
 
// This code is contributed by Mayank Tyagi
 
</script>

Sum odd = 20
Sum even = 15

Complejidad temporal: O(n)
Espacio auxiliar: O(1)

Método #3: Usando el método string():

1. Convierta el entero en string. Recorra la string y almacene la suma de todos los índices pares en una variable y la suma de todos los índices impares en otra variable.

A continuación se muestra la implementación:

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of the odd
// and even positioned digits in a number
void getSum(int n)
{
     
    // To store the respective sums
    int sumOdd = 0, sumEven = 0;
 
    // Converting integer to string
    string num = to_string(n);
 
    // Traversing the string
    for(int i = 0; i < num.size(); i++)
    {
        if (i % 2 == 0)
            sumOdd = sumOdd + (int(num[i]) - 48);
        else
            sumEven = sumEven + (int(num[i]) - 48);
    }
    cout << "Sum odd = " << sumOdd << "\n";
    cout << "Sum even = " << sumEven << "\n";
}
 
// Driver code
int main()
{
    int n = 457892;
    getSum(n);
     
    return 0;
}
 
// This code is contributed by souravmahato348

// Java implementation of the approach
  
import java.util.*;
  
class GFG{
  
static void getSum(int n)
{
    // To store the respective sum
    int sumOdd = 0;
    int sumEven = 0;
  
    // Converting integer to String
    String num = String.valueOf(n);
  
    // Traversing the String
    for(int i = 0; i < num.length(); i++)
        if (i % 2 == 0)
            sumOdd = sumOdd + (num.charAt(i) - '0');
        else
            sumEven = sumEven + (num.charAt(i) - '0');
  
    System.out.println("Sum odd = " + sumOdd);
    System.out.println("Sum even = " + sumEven);
}
  
// Driver code
public static void main(String[] args)
{
    int n = 457892;
    getSum(n);
}
}
 
// Code contributed by swarnalii

# Python3 implementation of the approach
 
# Function to find the sum of the odd
# and even positioned digits in a number
def getSum(n):
 
    # To store the respective sums
    sumOdd = 0
    sumEven = 0
     
    # Converting integer to string
    num = str(n)
     
    # Traversing the string
    for i in range(len(num)):
        if(i % 2 == 0):
            sumOdd = sumOdd+int(num[i])
        else:
            sumEven = sumEven+int(num[i])
 
    print("Sum odd = ", sumOdd)
    print("Sum even = ", sumEven)
 
 
# Driver code
if __name__ == "__main__":
    n = 457892
    getSum(n)
 
# This code is contributed by vikkycirus

// C# implementation of the approach
using System;
 
class GFG{
 
static void getSum(int n)
{
     
    // To store the respective sum
    int sumOdd = 0;
    int sumEven = 0;
 
    // Converting integer to String
    String num = n.ToString();
 
    // Traversing the String
    for(int i = 0; i < num.Length; i++)
        if (i % 2 == 0)
            sumOdd = sumOdd + (num[i] - '0');
        else
            sumEven = sumEven + (num[i] - '0');
 
    Console.WriteLine("Sum odd = " + sumOdd);
    Console.WriteLine("Sum even = " + sumEven);
}
 
// Driver code
public static void Main()
{
    int n = 457892;
    getSum(n);
}
}
 
// This code is contributed by subhammahato348

<script>
// Javascript implementation of the approach
 
function getSum(n)
{
    // To store the respective sum
    let sumOdd = 0;
    let sumEven = 0;
   
    // Converting integer to String
    let num = (n).toString();
   
    // Traversing the String
    for(let i = 0; i < num.length; i++)
        if (i % 2 == 0)
            sumOdd = sumOdd + (num[i] - '0');
        else
            sumEven = sumEven + (num[i] - '0');
   
    document.write("Sum odd = " + sumOdd+"<br>");
    document.write("Sum even = " + sumEven+"<br>");
}
 
// Driver code
let n = 457892;
getSum(n);
 
// This code is contributed by unknown2108
</script>

Sum odd = 20
Sum even = 15

Complejidad temporal: O(n)
Espacio auxiliar: O(n)