Answer:
#include <iostream>
using namespace std;
int main()
{
int input = 0;
int count = 0;
int sum = 0;
int sumNegative = 0;
while (true) {
cout << "Enter a number: ";
cin >> input;
if (input == 0) break;
count++;
sum += input;
if (input < 0) {
sumNegative += input;
}
}
cout << "Count of the numbers: " << count << endl;
cout << "Sum of all the numbers: " << sum << endl;
cout << "Sum of the negative numbers: " << sumNegative << endl;
}
Explanation:
Your requirements regarding the sum and the negative numbers was a bit vague so I just did something you can probably adjust easily to your liking.
Hit the return key after you type in a cell
Answer:
a=4 , b=1
Explanation:
I'm not a computer science major at all but I think I can help you with this code.
Our program wants us to add 2 to a get new a value while also subtracting 1 from b value to obtain new b value. We we want to for for as long b is not 0 and a/b is nonnegative.
One round we get:
New a=0+2=2
New b=3-1=2
Let's see if we can go another round:
New a=2+2=4
New b=2-1=1
We can't go another round because b would be negative while a is positive which would make a/b negative. So our loop stops at this 2nd round.
a=4 , b=1
Other notes:
2nd choice makes no sense because a is always going to increase because of the addition on a and b was going to decrease because of the subtraction on it.
Third choice makes no sense because a/b doesn't even exist.
Fourth choice a/b is negative not nonnegative.
