I don't understand your question, like what do you mean?
The type of optimization that is being applied to this programming code is: D. Loop unrolling.
<h3>What is loop unrolling?</h3>
Loop unrolling is also referred to as loop unwinding and it can be defined as loop transformation technique that is typically designed and developed to optimize the number of instructions that are executed by a software program at a given period of time, especially at the expense of its binary size (space-time tradeoff).
This ultimately implies that, loop unrolling can be used by programmers to optimize the execution time of a software program, in order to reduce the number of branches and loop maintenance instructions.
Read more on loop unrolling here: brainly.com/question/26098908
Answer:
Explanation:
The following code is written in Java and like requested prompts the user for a number to continue or a letter to exit. Then loops and keeps adding all of the numbers to the sum variable and adding 1 to the count for each number entered. Finally, it prints the sum and the average using printf and the variables.
import java.util.Scanner;
class Examine1 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int sum = 0;
int count = 0;
System.out.println("Enter a number to continue or a letter to exit.");
while(in.hasNextInt()) {
System.out.println("Enter a number:");
sum += in.nextInt();
count += 1;
}
System.out.printf("The sum is %s.%n", sum);
System.out.printf("The average is %s.", (sum / count));
}
}
Here you go. The program will count entered zeros. Null is not happening here...
int main()
{
int NNeg = 0;
int NPos = 0;
int NZero = 0;
std::cout << "Enter 6 numbers:\n";
for (int i = 0; i < 6; i++) {
double number;
std::cin >> number;
if (number < 0.0) NNeg++;
else if (number > 0.0) NPos++;
else NZero++;
}
std::cout << "You entered:\n";
std::cout << NNeg << " negative numbers\n";
std::cout << NZero << " times zero\n";
std::cout << NPos << " positive numbers\n";
}
Answer:
Communication path basically define the path in which the information and messages can be exchange by using the efficient communication path.
There are simple formula for calculating the total number of communication channel that is :
Where, n is the number of stack holder.
Now, the maximum number of communication paths for a team of twenty people can be calculated as:
n=20
=\frac{20(20-1)}{2} = 190
Therefore, 190 is the total number of communication path.