Question

find the sum of +ve even numbers and -ve odd numbers from the users input if input is 0 stop the process

Answer 1

1. #include <stdio.h>

3. void main()

4. {

5.     int i, num, odd_sum = 0, even_sum = 0;

7.     printf("Enter the value of num\n");

8.     scanf("%d", &num);

9.     for (i = 1; i <= num; i++)

10.     {

11.         if (i % 2 == 0)

12.             even_sum = even_sum + i;

13.         else

14.             odd_sum = odd_sum + i;

15.     }

16.     printf("Sum of all odd numbers  = %d\n", odd_sum);

17.     printf("Sum of all even numbers = %d\n", even_sum);

18. }

Answer 2

#include<bits/stdc++.h>
using namespace std ;
int main(){
    int n , i , a=0 , b=0 ;
    cin>> n ;
    if( n != 0 ){
        for( i=0 ; i<=n ; i++ ){
            if( i%2==0 ){
                a = a + i ;
             }else if( i%2==1 ){
              b = b + i ;
            }
        }
        cout<< a << endl << b ;
    }else
        cout<< endl ;
}

Answer 3

#include<iostream>
int main()
{
    int eve = 0;
    int odd = 0;
    int num = 1;
    std::cout << "Enter the Number: ";
    std::cin >> num;
    if (num % 2 == 0)
        eve += num;
    else
        odd -= num;
    while (num != 0)
    {
        std::cout << "\nEnter the number: ";
        std::cin >> num;
        if (num % 2 == 0)
            eve += num;
        else
            odd -= num;
    }
    if (num == 0)
        return 0;
    std::cout << "\nSUM OF +ve EVEN: " << eve;
    std::cout << "\nSUM OF -ve ODD: " << odd;
}