Nested Control Statements in java

For the next example, we once again formulate an algorithm by using pseudocode and topdown,

stepwise refinement, and write a corresponding Java program. We’ve seen that control

statements can be stacked on top of one another (in sequence). In this case study, we

examine the only other structured way control statements can be connected—namely, by

nesting one control statement within another.

Consider the following problem statement:

A college offers a course that prepares students for the state licensing exam for real

estate brokers. Last year, ten of the students who completed this course took the exam.

The college wants to know how well its students did on the exam. You’ve been asked to

write a program to summarize the results. You’ve been given a list of these 10 students.

Next to each name is written a 1 if the student passed the exam or a 2 if the student

failed.

Your program should analyze the results of the exam as follows:

1. Input each test result (i.e., a 1 or a 2). Display the message “Enter result” on the screen

each time the program requests another test result.

2. Count the number of test results of each type.

3. Display a summary of the test results, indicating the number of students who passed and

the number who failed.

4. If more than eight students passed the exam, print the message “Bonus to instructor!”

After reading the problem statement carefully, we make the following observations:

1. The program must process test results for 10 students. A counter-controlled loop

can be used, because the number of test results is known in advance.

2. Each test result has a numeric value—either a 1 or a 2. Each time it reads a test

result, the program must determine whether it’s a 1 or a 2. We test for a 1 in our

algorithm. If the number is not a 1, we assume that it’s a 2. (Exercise 4.24 considers

the consequences of this assumption.)

3. Two counters are used to keep track of the exam results—one to count the number

of students who passed the exam and one to count the number who failed.

4. After the program has processed all the results, it must decide whether more than

eight students passed the exam.

Let’s proceed with top-down, stepwise refinement. We begin with a pseudocode representation

of the top:

1  Initialize passes to zero

2  Initialize failures to zero

3  Initialize student counter to one

   

4  While student counter is less than or equal to 10

5  Prompt the user to enter the next exam result

6  Input the next exam result

7  If the student passed

8  Add one to passes

9  Else

10  Add one to failures

11  Add one to student counter

12 Print the number of passes

13 Print the number of failures

14 If more than eight students passed

15 Print “Bonus to instructor!”

the program in java is :-

import java.util.Scanner;

public class Apples {

public static void main(String[] args){

int counter = 0;

float sum = 0;

float studentGrade =0;

float average = 0;

int pass = 0;

int fails = 0;

while (counter <= 10){

System.out.println("enter student grade ");

Scanner input = new Scanner(System.in); // creating object from scanner class

studentGrade = input.nextInt();

sum += studentGrade;

counter ++;

if (studentGrade > 60)

pass ++;

else

fails ++;

}

average = sum / counter;

System.out.printf("the average is %f " , average);

if (pass > 8)

System.out.println(" \n bonus to the instructor");

else

System.out.println("lame");

}

}

thanks  <3