Sum of Even & Odd Digits Problem Statement

Create a program that accepts an integer N and outputs two sums: the sum of all even digits and the sum of all odd digits found in the integer.

Explanation:

Digits refer to the numbers themselves, not their positional values. For example, if N is "13245", the even digits are 2 & 4, and the odd digits are 1, 3 & 5.

Input:

Integer N

Output:

Sum_of_Even_Digits Sum_of_Odd_Digits

First print the sum of even digits, followed by the sum of odd digits, separated by a space.

Example:

Input:

13245

Output:

6 9

Explanation:

In the input 13245, the even digits are 2 and 4, summing to 6. The odd digits are 1, 3, and 5, summing to 9.

Constraints:

0

Question

Sum of Even & Odd Digits Problem Statement

Create a program that accepts an integer N and outputs two sums: the sum of all even digits and the sum of all odd digits found in the integer.

Explanation:

Digits refer to the numbers themselves, not their positional values. For example, if N is "13245", the even digits are 2 & 4, and the odd digits are 1, 3 & 5.

Input:

Integer N

Output:

Sum_of_Even_Digits Sum_of_Odd_Digits

First print the sum of even digits, followed by the sum of odd digits, separated by a space.

Example:

Input:

Output:

6 9

Explanation:

In the input 13245, the even digits are 2 and 4, summing to 6. The odd digits are 1, 3, and 5, summing to 9.

Constraints:

0 <= N <= 10^8

Accepted Answer

k=[1,3,2,4,5]
even = []
odd = []
for i in k:
    #print(i)
    if i % 2 == 0:
        even.append(i)
    else:
        odd.append(i)

print("odd: ", odd)
print("even: ", even)

Accepted Answer

import java.util.Scanner;

public class SumEvenAndOddDigits {

public static void main(String[] args) {

Scanner scanner = new Scanner(System.in);

System.out.print("Enter an integer N: ");

int N = scanner.nextInt();

int evenSum = 0;

int oddSum = 0;

while (N != 0) {

int digit = N % 10;

if (digit % 2 == 0) {

evenSum += digit;

} else {

oddSum += digit;

}

N /= 10;

}

System.out.println("Sum of even digits: " + evenSum);

System.out.println("Sum of odd digits: " + oddSum);

}

Accepted Answer

a = int(input())
n = []
for _ in range(a):
    m = int(input())
    n.append(m)
even = 0
odd = 0
for x in range(0, len(n)):
    if n[x] % 2 == 0:
        even = even + n[x]
    else:
        odd = odd + n[x]
print(even)
print(odd)

Make it easy like this

Accepted Answer

Namepace SumOfEvenAndOddDigits

{

class Program

{

static void Main(string[] args)

{

Console.Write("Enter an integer: ");

int number = int.Parse(Console.ReadLine());

int sumEven = 0;

int sumOdd = 0;

// Convert the number to a string to access individual digits

string numString = number.ToString();

// Loop through each character in the string (representing each digit)

foreach (char digitChar in numString)

{

int digit = int.Parse(digitChar.ToString());

if (digit % 2 == 0)

{

// Even digit

sumEven += digit;

}

else

{

// Odd digit

sumOdd += digit;

}

Console.WriteLine("Sum _of_ even digits: " + sumEven);

Console.WriteLine("Sum of _odd digits: " + sumOdd);

}

Accepted Answer

def sum_of_even_and_odd_digits(N):

even_sum = 0

odd_sum = 0

while N > 0:

digit = N % 10 # Extract the last digit

if digit % 2 == 0: # Check if the digit is even

even_sum += digit

else:

odd_sum += digit

N //= 10 # Remove the last digit from the number

return even_sum, odd_sum

# Main program

N = int(input("Enter an integer: "))

even_sum, odd_sum = sum_of_even_and_odd_digits(N)

print("Sum of even digits:", even_sum)

print("Sum of odd digits:", odd_sum)