CS 1063 -- Review for Final Exam, Answers

CS 1063 Intro. to Programming
Topics and Review Questions for
Final Exam, Fall 2005
Answers

Date and Length of Exam: Tuesday, 13 December 2005, 1:30 pm - 4:15 pm.

Sample Questions: I will pick simple questions about many of the topics from the web pages and from class. Often it will involve giving you some code and asking you to explain, modify, or complete the code.

classes: Probably some question about writing a class to represent something, and writing a test class to create an instance of the given class. This is what we've been doing all semester long.

primitive types and operators:

Write Java code that will start with a certain number of seconds, and will convert to hours, minutes, and seconds, where "hours" and "minuters" are as large as possible:

int time = 74747;  // or some other value in seconds
int hours = ...;   // fill in for ...
int minutes = ...; // fill in for ...
int seconds = ...; // fill in for ... 
System.out.println(time + " seconds is\n   " +
   hours + " hours\n   " +
   minutes + " minutes\n   " +
   seconds + " seconds\n   ");

Two versions	Several runs
// Hours, minutes and seconds public class Time { public static void main(String[] args) { int time = 74747; // or some other value int hours = time/3600; int minutes = (time - hours3600)/60; int seconds = (time - hours3600 - minutes*60); System.out.println(time + " seconds is\n " + hours + " hours,\n " + minutes + " minutes,\n " + seconds + " seconds\n"); } } //--------------------------------------------------- // Hours, minutes and seconds public class Time { public static void main(String[] args) { int time = 74747; // or some other value int hours = time/3600; int minutes = time%3600/60; int seconds = (time%3600)%60; System.out.println(time + " seconds is\n " + hours + " hours,\n " + minutes + " minutes,\n " + seconds + " seconds\n"); } }	74747 seconds is 20 hours, 45 minutes, 47 seconds 33333 seconds is 9 hours, 15 minutes, 33 seconds 55555 seconds is 15 hours, 25 minutes, 55 seconds

Two versions

Several runs

// Hours, minutes and seconds
public class Time {

   public static void main(String[] args) {
      int time = 74747; // or some other value
      int hours = time/3600;
      int minutes = (time - hours*3600)/60;
      int seconds = (time - hours*3600 - minutes*60);
      System.out.println(time + " seconds is\n   " +
         hours + " hours,\n   " +
         minutes + " minutes,\n   " +
         seconds + " seconds\n");
   }
}
//---------------------------------------------------
// Hours, minutes and seconds
public class Time {

   public static void main(String[] args) {
      int time = 74747; // or some other value
      int hours = time/3600;
      int minutes = time%3600/60;
      int seconds = (time%3600)%60;
      System.out.println(time + " seconds is\n   " +
         hours + " hours,\n   " +
         minutes + " minutes,\n   " +
         seconds + " seconds\n");
   }
}

74747 seconds is
   20 hours,
   45 minutes,
   47 seconds

33333 seconds is
   9 hours,
   15 minutes,
   33 seconds

55555 seconds is
   15 hours,
   25 minutes,
   55 seconds

if-else:

Write a short if statement (no else) that will check if an int is inside the range from 2 to 10 (inclusive) and will print "In range" if this is so. See both under 3 below.

Write another short if statement (no else) that will check if an int is not inside the range from 2 to 10 (inclusive) and will print "Out of range" if this is so.

Is size between 2 and 10?
// Range.java: is size between 2 and 10 (inclusive)? import java.util.Scanner; public class Range { public static void main(String[] args) { Scanner in = new Scanner(System.in); while(true) { System.out.print("Type an int for size ---> "); int size = in.nextInt(); if (size == 0) break; // 0 terminates loop // Question 2 above if ( 2 <= size && size <= 10) System.out.println(size + " in range"); // Question 3 above (3 separate answers -- study them all!) if (!(2 <= size && size <= 10)) System.out.println(size + " out of range (version 1)"); if (!(2 <= size) \|\| !(size <= 10)) System.out.println(size + " out of range (version 2)"); if ( 2 > size \|\| size > 10) System.out.println(size + " out of range (version 3)"); } } }
A run (input in bold)
1 1 out of range (version 1) 1 out of range (version 2) 1 out of range (version 3) 2 2 in range 6 6 in range 10 10 in range 11 11 out of range (version 1) 11 out of range (version 2) 11 out of range (version 3) 0

Is size between 2 and 10?

// Range.java: is size between 2 and 10 (inclusive)?
import java.util.Scanner;

public class Range {

   public static void main(String[] args) {
      Scanner in = new Scanner(System.in);
      while(true) {
         System.out.print("Type an int for size ---> ");
         int size = in.nextInt();

         if (size == 0) break; // 0 terminates loop

         // Question 2 above
         if (  2 <= size  && size <= 10)
            System.out.println(size + " in range");

         // Question 3 above (3 separate answers -- study them all!)
         if (!(2 <= size  && size <= 10))
            System.out.println(size + " out of range (version 1)");

         if (!(2 <= size) || !(size <= 10))
            System.out.println(size + " out of range (version 2)");

         if (  2 > size   ||   size > 10)
            System.out.println(size + " out of range (version 3)");
      }
   }
}

A run (input in bold)

1
1 out of range (version 1)
1 out of range (version 2)
1 out of range (version 3)
2
2 in range
6
6 in range
10
10 in range
11
11 out of range (version 1)
11 out of range (version 2)
11 out of range (version 3)
0

loops:

Translate the following while loop to a for loop that does the same thing:

int i = 1;
int sum = 0;
while (i < 10) {
   sum = sum + i*i;
   i++;
}
System.out.println("sum: " + sum);

See both under 5 below.

Translate the following for loop to a while loop that does the same thing:

for (int k = 10; k >= 1; k--) {
   System.out.print(k);
   if (k != 1) System.out.print(", ");
}
System.out.println();

Original Loops	Altered Loops
public class Loops1 { public static void main(String[] args) { // change to for int i = 1; int sum = 0; while (i < 10) { sum = sum + i*i; i++; } System.out.println("sum: " + sum); // change to while for (int k = 10; k >= 1; k--) { System.out.print(k); if (k != 1) System.out.print(", "); } System.out.println(); } }	public class Loops2 { public static void main(String[] args) { // the for loop int sum = 0; for (int i = 1; i < 10; i++) { sum = sum + i*i; } System.out.println("sum: " + sum); // the while loop int k = 10; while (k >= 1) { System.out.print(k); if (k != 1) System.out.print(", "); k--; } System.out.println(); } }
Common Output
sum: 285 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Original Loops

Altered Loops

public class Loops1 {
   public static void main(String[] args) {
      // change to for
      int i = 1;
      int sum = 0;
      while (i < 10) {
         sum = sum + i*i;
         i++;
      }
      System.out.println("sum: " + sum);

      // change to while

      for (int k = 10; k >= 1; k--) {
         System.out.print(k);
         if (k != 1) System.out.print(", ");

      }
      System.out.println();
   }
}

public class Loops2 {
   public static void main(String[] args) {
      // the for loop

      int sum = 0;
      for (int i = 1; i < 10; i++) {
         sum = sum + i*i;

      }
      System.out.println("sum: " + sum);

      // the while loop
      int k = 10;
      while (k >= 1) {
         System.out.print(k);
         if (k != 1) System.out.print(", ");
         k--;
      }
      System.out.println();
   }
}

Common Output

sum: 285
10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Redo the example that we worked on in class: use a for loop to calculate 100000 terms of the following sum. Compare the result with Math.PI*Math.PI/6.0.

1 + 1/2² + 1/3² + 1/4² + 1/5² + ...

File: Pi.java	File: PiTest.java
public class Pi { private int numTerms; public Pi(int terms) { numTerms = terms; } public double calcPi() { double sum = 0.0; for (int i = 1; i <= numTerms; i++) { sum = sum + 1.0/i/i; } return sum; } } // Note: we could have written // sum = sum + 1.0/((double)ii); // but NOT // sum = sum + 1.0/(ii); (WHY?)	import java.util.Scanner; public class PiTest { public static void main(String[] args) { Scanner in = new Scanner(System.in); System.out.print("Type an int for # terms ---> "); int terms = in.nextInt(); Pi pi = new Pi(terms); double piApprox = pi.calcPi(); System.out.println("Sum of series (approx): " + piApprox + ", with " + terms + " terms"); System.out.println("Sum (exact value): " + Math.PIMath.PI/6.0); System.out.println("Pi from series (approx): " + Math.sqrt(piApprox6.0) + ", with " + terms + " terms"); System.out.println("Pi (exact value): " + Math.PI); } }
Output of 3 Runs
Type an int for # terms ---> 10000 Sum of series (approx): 1.6448340718480652, with 10000 terms Sum (exact value): 1.6449340668482264 Pi from series (approx): 3.1414971639472147, with 10000 terms Pi (exact value): 3.141592653589793 Type an int for # terms ---> 100000 Sum of series (approx): 1.6449240668982423, with 100000 terms Sum (exact value): 1.6449340668482264 Pi from series (approx): 3.141583104326456, with 100000 terms Pi (exact value): 3.141592653589793 Type an int for # terms ---> 1000000 Sum of series (approx): 1.64493306684877, with 1000000 terms Sum (exact value): 1.6449340668482264 Pi from series (approx): 3.1415916986605086, with 1000000 terms Pi (exact value): 3.141592653589793

File: Pi.java

File: PiTest.java

public class Pi {
   private int numTerms;

   public Pi(int terms) {
      numTerms = terms;
   }

   public double calcPi() {
      double sum = 0.0;
      for (int i = 1; i <= numTerms; i++) {
         sum = sum + 1.0/i/i;
      }
      return sum;
   }
}
// Note: we could have written
//     sum = sum + 1.0/((double)i*i);
//  but NOT
//     sum = sum + 1.0/(i*i);   (WHY?)

import java.util.Scanner;

public class PiTest {
   public static void main(String[] args) {
      Scanner in = new Scanner(System.in);
      System.out.print("Type an int for # terms ---> ");
      int terms = in.nextInt();

      Pi pi = new Pi(terms);
      double piApprox = pi.calcPi();
      System.out.println("Sum of series (approx):  " +
         piApprox + ", with " + terms + " terms");
      System.out.println("Sum (exact value):       " +
         Math.PI*Math.PI/6.0);
      System.out.println("Pi from series (approx): " +
         Math.sqrt(piApprox*6.0) + ", with " +
         terms + " terms");
      System.out.println("Pi (exact value):        " +
         Math.PI);
   }
}

Output of 3 Runs

Type an int for # terms ---> 10000
Sum of series (approx):  1.6448340718480652, with 10000 terms
Sum (exact value):       1.6449340668482264
Pi from series (approx): 3.1414971639472147, with 10000 terms
Pi (exact value):        3.141592653589793

Type an int for # terms ---> 100000
Sum of series (approx):  1.6449240668982423, with 100000 terms
Sum (exact value):       1.6449340668482264
Pi from series (approx): 3.141583104326456, with 100000 terms
Pi (exact value):        3.141592653589793

Type an int for # terms ---> 1000000
Sum of series (approx):  1.64493306684877, with 1000000 terms
Sum (exact value):       1.6449340668482264
Pi from series (approx): 3.1415916986605086, with 1000000 terms
Pi (exact value):        3.141592653589793

Write a code segment in Java that will start with an amount of money (a double amount), and a number of years (an int years), and an interest rate (a double rate). Use a loop to get the amount of money at the end of this time, applying the interest each year. Compare this amount with the amount you would have if you applied the interest 4 times a year. (You apply 0.25*rate a number times equal to 4*years.) This problem worked out to be too elaborate. Here is a simple loop that computes compound interest inside the main method:

First Data	Second Data
public class Comp { public static void main(String[] args) { double amount = 1000; int years = 4; double rate = 10.0; for(int i = 1; i <= years; i++) { amount = amount + amount*rate/100.0; System.out.print("End of year " + i); System.out.println(", amount: " + amount); } System.out.println(); } }	public class Comp { public static void main(String[] args) { double amount = 1000; int years = 6; double rate = 30.0; for(int i = 1; i <= years; i++) { amount = amount + amount*rate/100.0; System.out.print("End of year " + i); System.out.println(", amount: " + amount); } System.out.println(); } }
First run	Second run
End of year 1, amount: 1100.0 End of year 2, amount: 1210.0 End of year 3, amount: 1331.0 End of year 4, amount: 1464.1	End of year 1, amount: 1300.0 End of year 2, amount: 1690.0 End of year 3, amount: 2197.0 End of year 4, amount: 2856.1 End of year 5, amount: 3712.93 End of year 6, amount: 4826.809

First Data

Second Data

public class Comp {
   public static void main(String[] args) {
      double amount = 1000;
      int years = 4;
      double rate = 10.0;

      for(int i = 1; i <= years; i++) {
         amount = amount + amount*rate/100.0;
         System.out.print("End of year " + i);
         System.out.println(", amount: " + 
            amount);
      }
      System.out.println();
   }
}

public class Comp {
   public static void main(String[] args) {
      double amount = 1000;
      int years = 6;
      double rate = 30.0;

      for(int i = 1; i <= years; i++) {
         amount = amount + amount*rate/100.0;
         System.out.print("End of year " + i);
         System.out.println(", amount: " + 
            amount);
      }
      System.out.println();
   }
}

First run

Second run

End of year 1, amount: 1100.0
End of year 2, amount: 1210.0
End of year 3, amount: 1331.0
End of year 4, amount: 1464.1

End of year 1, amount: 1300.0
End of year 2, amount: 1690.0
End of year 3, amount: 2197.0
End of year 4, amount: 2856.1
End of year 5, amount: 3712.93
End of year 6, amount: 4826.809

For a solution to the full problem, see Compound interest.

Here are some more sample questions: Practice with loops. Answers.
Just focus on the "First Starting Point" and the "Second Starting Point": this amounts to 6 exercises for "while" loops and 6 for "for" loops.

Everything (classes, if-else, loops, operators):
1. Activity 3: the TeacherSalary Class. (taken from the Week 15 lecture).
2. Two old Project 3 assignments along with solutions: Review Projects. (This is the Week 16 lecture.)

CS 1063 Intro. to Programming Topics and Review Questions forFinal Exam, Fall 2005 Answers

CS 1063 Intro. to Programming
Topics and Review Questions for
Final Exam, Fall 2005
Answers