OOP Java P1 Wk2

OOP Java Part 1 Week 2

The Week 2 covers loops and basics of methods. Earlier in my other previous posts, I mentioned that we could use either Netbeans, or Eclipse. I realized that in MOOC, it’s required to use NetBeans in order to unlock new exercises such as week 2 and other weeks exercises.

I copied and pasted all exercises from Eclipse to NetBeans. Some of my codes are not accepted, most of them are rejected because of simple String, and or whitespaces differences. I have to follow specifics, like typing the exact same String as they want me to, for example: same number of whitespaces in the String. I do enjoy using Eclipse, because sometime I have to do more programming where components of methods aren’t included and I make them myself. Netbeans have components and methods for us to use quickly . That kind of takes the fun away. Just because Netbeans have components ready, doesn’t mean it’s easier. It is a lot harder, because it wants us to program in a certain way. So have a ton of patience!

Week 2 begins!

This week covers basic mathematic programming. I am going to summarize and cover highlights where I wouldn’t have to repeat similar exercises. Some exercises will be similar and yet different.

EXERCISE #33

This exercise teaches how to do sum of all numbers between two inputted numbers. It is a simple programming.

public class TheSumBetweenTwoNumbers {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        System.out.print("First: ");
        int num1 = Integer.parseInt(reader.nextLine());

        System.out.print("Last: ");
        int num2 = Integer.parseInt(reader.nextLine());

        int sum = 0;

        while (num1 <= num2) {
            sum += num1;
            num1++;

            System.out.println(num1);
        }

        System.out.println("The sum is " + (sum));  
    }
}

Scanner does what it does… parsing input. Integer.parseInt() parses integer. First input is given a number num1, and then second input a second number num2. I want to add up all the numbers between num1 and num2, so I create a variable sum and set it to equal to 0.

The next step is setting up a while() loop. Its condition is true as long num1 is less or equals to num2. Starting with sum and add up to the num1. Notice how sum completely ignores num2. Next step is num1 increments by one every time a loop rolled. Finally, incremented num1 is printed.

The loop rolls on… until num1 equals to num2. num1 isn’t the sum of anything another than an increment by 1. However sum is adding up in every loop. Once the loop ended, sum is printed and is the desired result.

EXERCISE #34

Exercise #34 is the same idea as Exercise #33. It wants the result of factorial on a number. Factorial number symbol looks like this n!, and we cannot use do n!. The ! is somewhat an operator and cannot be used as in variable unless it is a string. I always could use n as a short for number. Take a look at this code below:

public class Factorial {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);

        System.out.print("Type a number: ");
        int n = Integer.parseInt(reader.nextLine());

        int count = 1;
        int product = 1;

        while (count <= n) {
                product *= count;
                count++;
        } System.out.println("Factorial is " + product);
    }
}

I am going to use product instead of sum. Isn’t that what product means after of all? Using while() again, with a condition being true as long count is less or equals to n. Setting count and product as 1. It’s what makes this little different from add and subtract. If I multiply any number to 0, it will always remains 0. I don’t want that. product ignores n, and multiplies count while becoming the new product. Same process as in Exercise #33, count increments by 1 until it equals to the given number n. Finally when the loop ends, product is printed as a result outside the loop.

EXERCISE #35

Almost same formula as two other exercises above. Just more tasks and more results expected. The tasks listed:

• program the power of n
• find the result
• sum up each results

I got to use this method Math.pow(number, power) and it returns double type where will have a decimal result. One thing great about MOOC, is that it taught me this new trick. Showing it to you below!

System.out.print("Type a number: ");
    int power = Integer.parseInt(reader.nextLine());
    int sum = 0;
    while(power >= 0) {
        int result = (int)Math.pow(2, power);
        System.out.println(result);
        sum += result;
        power--;
    }


    System.out.println("The result is " + sum);
}

Using (int)Math.pow(number, power) will automatically convert the double result into an int. What’s happening here is I created two variables power and sum. power will receive an input from a user. sum is set as 0. Using same while() loop as previous two exercises. The condition is slightly different, it’s the opposite. It’s true as long the power is greater than 0. While it’s true, the first step inside the loop is producing a result and then have it printed to get a better idea of what’s going on during the loops. sum adds up result. The power is decremented by 1 until the loop ends. sum is then printed as a final.

EXERCISE #36

I am going to close this post with Exercise #36. Will continue with week 2 on the next post. This Exercise #36 is fun and expects much more results. What this exercise wants…

• to be able to put in many inputs as user want
• end the while() loop with a trigger using -1 or any negative number
• send out a farewell string
• print the sum of all inputs except for a negative number
• count how many times the inputs were entered
• find the average out of all inputs
• find how many odd numbers were entered
• find how many even numbers were entered

Woo! I held my breath as long as I could, to list all these! Let’s do the first half then second half afterward. For a user to enter inputs as many as he/she wants, we will need to use a while() loop. It wants us to end the loop by entering a negative number. Easily done with break;! Send out a String? Not a problem! sum… hmm we have done all that in the previous exercises in this post.

First half:

public static void main(String[] args) {

    Scanner reader = new Scanner(System.in);

    System.out.print("Type numbers: ");
    int sum = 0;
    int num;

    while (true) {
        num = Integer.parseInt(reader.nextLine());
        if (num < 0) {
            System.out.println("Thank you and see you later!");
            num = 0;
            break;
        } else {
            sum += num;
        }
     }
    System.out.println("The sum is " + sum);
}

This is the first half, and let’s concise on what’s happening in that code.

• Two variables are entered
  • sum equals to 0, “intialized” to add up all inputs
  • num is the input that user enters
• while() loop with a condition saying always true
  • num inputs parser is put in there so user can put as many as inputs he/she want
• if conditional statment when num is less than 0
  • String printed
  • negative number num is set to 0 so it wouldn’t affect the sum
  • break; is triggered, ending the always true
• else conditional statment when when inputs are not and have not been negative
  • inputs are being added up to sum
• sum is printed when the loop ended

I hope that’s clear, and now second half coming up!

public static void main(String[] args) {
   Scanner reader = new Scanner(System.in);

    System.out.print("Type numbers: ");
    int i = 0;
    int sum = 0;
    int num;
    double average = 0;
    int oddNum = 0;
    int evenNum = 0;

    while (true) {
        num = Integer.parseInt(reader.nextLine());
        if (num < 0) {
            System.out.println("Thank you and see you later!");
            num = 0;
            break;
        } else {
            i++;
            sum += num;
            if (num % 2 != 0) {
                    oddNum++;
            } else {
                    evenNum++;
            }
        }
     }

    average = ((1.0*sum)/i);

    System.out.println("The sum is " + sum);
    System.out.println("How many numbers entered: " + i);
    System.out.println("Average: " + sum + " / " + i + " = " + average);
    System.out.println("Odd numbers: " + oddNum);
    System.out.println("Even numbers: " + evenNum);
}

Few variables are added. i to count how many times inputs are entered. Variable double average is added to calculate the average, using double to get the result in decimal. Two other variables count inputs that are either even numbers or odd numbers. All are “initialized” by setting to 0

Add the process inside else conditional statment, because it’s where loop continues rolling. i++ to increment by 1 to count the inputs. Add another branches of conditional statments inside the else, saying if input is not a even number, oddNum++ increments. Else evenNum++ increments.

Outside the loop when it’s done break;, average set to a mathematic equation saying ((1.0 * sum) / i). Now I think this is important to point out, just because average is assigned to a double type doesn’t mean it will give a correct answer. It will just result with a decimal. So I added 1.0, and then average will have the exact quotient.

Concluding… print all the results!