Write a program segment with a do-while loop that displays whether a user-entered integer is even or odd. The code should then ask the user if he or she wants to test another number. The loop should repeat as long as the user enters 'Y' or 'y' . Use a logical OR operator in the do-while loop test expression

Following is the program in C++ program

#include<iostream> // header file

using namespace std; // namespace  

int main() // main function  

{

int num1 = 0; // variable declaration  

char test;// variable declaration  

do

{

cout << " Enter the number: ";

cin >> num1; //Read the input by the user  

if (num1 % 2 == 0) // check the condition of even

cout << " number is even.\n";

else // check the condition of odd

cout << "Number is odd.\n";

cout << " Do you wanted the another test of number (y/n)? ";

cin >> test; //Read the input by user

} while (test == 'y' || test == 'Y'); //terating the loop

return 0;    

}

Output:

Enter the number:45

number is even.

Do you wanted the another test of number (y/n) Y

Enter the number:5

Number is odd.

Do you wanted the another test of number (y/n) n

Explanation:

Following are the description of program

answer; i believe that the correct answer is (speed)(c)