Sum of even & odd
Write a program to input an integer N and print the sum of all its even digits and sum of all its odd digits separately.

Digits mean numbers, not the places! That is, if the given integer is "13245", even digits are 2 & 4 and odd digits are 1, 3 & 5.

Input format :

Integer N

Output format :

Sum_of_Even_Digits Sum_of_Odd_Digits (Print first even sum and then odd sum separated by space)

Constraints

0

Question

Sum of even & odd

Write a program to input an integer N and print the sum of all its even digits and sum of all its odd digits separately.

Digits mean numbers, not the places! That is, if the given integer is "13245", even digits are 2 & 4 and odd digits are 1, 3 & 5.

Input format :

 Integer N

Output format :

Sum_of_Even_Digits Sum_of_Odd_Digits
(Print first even sum and then odd sum separated by space)

Constraints

0 <= N <= 10^8

Accepted Answer

Space Complexity: O(1)Explanation: Time Complexity: O(1)Explanation:

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)

Write a program to input an integer N and print the sum of all its even digits and sum of all its odd digits separately.

Digits mean numbers, not the places! That is, if the given integer is "13245", even digits are 2 & 4 and odd digits are 1, 3 & 5.

Input format :

Output format :

Constraints

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