Test 1 Solution

1. (10%) The following program fails to compile. Why does it fail to compile?

   public class Broken {
   
     public static void main(String[] args) {
       for (int i = 0; i < 10; i++){
         int result = 0;
         result = result + 1;
       }
       System.out.println(result);
     }
   
   }
   

   Answer: The variable result is used outside its scope, by the println statement.
2. (10%) The following program also fails to compile. Why does it fail to compile?

   import java.util.Scanner;
   
   public class BrokenA {
     static final double RATE = 0.5;
   
     public double nextYear(double x) {
       return x * RATE;
     }
   
     private static void main(String[] args) {
       int x = 0;
       String name = "Steve";
       Scanner keyboard = new Scanner(System.in);
       String value = keyboard.next();
       BrokenA  ba = new BrokenA();  
       double r = ba.nextYear(value);
       System.out.println(r);
     }
   }
   
   

   Answer:The method nextYear takes a double as an arugment but we are giving it a String.
3. (30%) What does the following program print out:

   public class RPS {
   
     public static void main(String[] args) {
       String p1 = "PPPSSSRRRPPP"; //notice, only 3 letters used
       String p2 = "RSPPSRPSRPSP";
       for (int i=0; i < p1.length(); i++){
         char c1 = p1.charAt(i);
         char c2 = p2.charAt(i);
         System.out.print (c1 + " " + c2 + " ");
         if (c1 == c2) {
           System.out.println("=");
         }
         else switch (c1) {
         case 'R':
           System.out.println((c2 == 'S') ? 1 : 2);
           break;
         case 'P':
           System.out.println((c2 == 'S') ? 2 : 1);
           break;
         case 'S':
           System.out.println((c2 == 'P') ? 1 : 2);
           break;
         };
       }
     }
   }
   

   Answer:

   P R 1
   P S 2
   P P =
   S P 1
   S S =
   S R 2
   R P 2
   R S 1
   R R =
   P P =
   P S 2
   P P =
   

   Its the game of rock-paper-scissors (RPS).

4. (50%) Write a program that:
   1. Asks the user for an integer. Assume he does enter an integer. If he enters 0 then the program stops.
   2. Otherwise, if it is a negative integer then the program converts to its positive counterpart (for example, -5 turns into 5).
   3. Otherwise, determine if there are two positive integers X and Y such that the number the user typed is equal to 3*X + 7*Y. If there are, print them, if not then print NONE FOUND. You will do this by brute-force: examine all realistic combinations of possible values of X and Y. (HINT: both of these will always be positive and smaller than...)
   Below is a sample interaction with the program. The users' inputs are in boldface:

   Enter number:3
   3= 3*1 + 7*0
   Enter number:-3
   3= 3*1 + 7*0
   Enter number:5
   NOT FOUND
   Enter number:7
   7= 3*0 + 7*1
   Enter number:11
   NOT FOUND
   Enter number:12
   12= 3*4 + 7*0
   Enter number:21
   21= 3*0 + 7*3
   Enter number:22
   22= 3*5 + 7*1
   Enter number:123456
   123456= 3*6 + 7*17634
   

   Answer:

   import java.util.Scanner;
   
   public class Legal {
   
     public static void main(String[] args) {
       Scanner keyboard = new Scanner(System.in);
       while (true) {
         System.out.print("Enter number:");
         int number = keyboard.nextInt();
         if (number == 0) break;
         if (number < 0) number = number * -1;
         int x =0;
         int y =0;
         boolean found = false; //a flag
         for (x = 0; x < number; x++){ 
           for (y=0; y < number; y++){ //number is a bit high here, what should we use?
             if (number == 3*x + 7*y) {
               found = true;
               break;}
           }
           if (found) break;
         }
         if (found) {
           System.out.println(number + "= 3*" + x + " + 7*" + y);
         }
         else {
           System.out.println("NOT FOUND");
         };
       }
     }
   }