Question

Write an application that asks a user to type an even number or the sentinel value 999 to stop. When the user types an even number, display the message "Good job!" and then ask for another input. When the user types an odd number, display an error message, "x is not an even number", and then ask for another input. When the user types the sentinel value 999, end the program

Answer 1

Answer:

int value;

while(1){

scanf("%d\n", &value);

if(value == 999)

break;

if(value%2 == 0)

printf("Good job");

else if(value%2 == 1)

printf("%d is not an even number", value);

}

Explanation:

I am going to write a C code for this.

The while(1) runs until the break command. So

int value;

while(1){

scanf("%d\n", &value);

if(value == 999)

break;

if(value%2 == 0)

printf("Good job");

else if(value%2 == 1)

printf("%d is not an even number", value);

}

Answer 2

Answer:

 do{

    std::cout << "input an even number: ";

    std::cin >> n;

    even=n%2;

    if (even==0)

    std::cout << "Good job!\n";

    else

    std::cout << n <<" is not an even number\n";

    }while(n!=999);

Explanation:

#include <iostream>

#include <string>

int main()

{ int n=0,even;

 do{

    std::cout << "input an even number: ";  //print on screen to input the number

    std::cin >> n;  //put that value on n variable

    even=n%2;

    if (even==0)  //if even is zero the number is even

    std::cout << "Good job!\n";

    else  //if even is not zero the number is odd

    std::cout << n <<" is not an even number\n";

    }while(n!=999);  // we will repeat the proccess while n is diffrente from 999

}