Answer:
The loop for the given scenario is shown.
while( num >= 0 )
{
if( num%2 == 0 )
sum_even = sum_even + num;
else
sum_odd = sum_odd + num;
cout << "Enter a positive number." << endl;
cin >> num;
}
Explanation:
The program uses three variables to hold the user input, sum of even numbers and sum of odd numbers.
int sum_even, sum_odd, num;
The sum variables are initialized.
sum_even = 0;
sum_odd = 0;
Only positive input from the user needs to be considered as per the question, hence, while loop is used.
This loop will execute only if the user inputs valid number. Upon entering the negative number which is less than 0, the loop terminates.
Following this, the sum of even numbers is displayed followed by the sum of negative numbers.
cout << sum_even << " " << sum_odd << endl;
The c++ program for the given problem statement is as follows.
#include <iostream>
using namespace std;
int main()
{
// variables declared for sum and input
int sum_even, sum_odd, num;
// sum initialized to 0
sum_even = 0;
sum_odd = 0;
cout << "This program calculates the sum of even and odd numbers which are positive. This program will end on invalid input." << endl;
cout << "Enter a positive number." << endl;
cin >> num;
// loop will terminate if the user inputs negative number
while( num >= 0 )
{
// for even numbers, input is added to sum_even else sum_odd
if( num%2 == 0 )
sum_even = sum_even + num;
else
sum_odd = sum_odd + num;
cout << "Enter a positive number." << endl;
cin >> num;
}
cout << sum_even << " " << sum_odd << endl;
return 0;
}
OUTPUT
This program calculates the sum of even and odd numbers which are positive. This program will end on invalid input.
Enter a positive number.
23
Enter a positive number.
34
Enter a positive number.
45
Enter a positive number.
67
Enter a positive number.
-69
34 135
