Answer:
Check the explanation
Explanation:
10
5
3
21
2
-6
the output will be somthing like this:
2 21
You can then suppose that the list of integers will contain at least 2 values.
Answer:
If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.
Explanation:
Lets take an example of two strings abc and cba which are reverse of each other.
string1 = abc
string2 = cba
Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.
Push abc each character on a stack in the following order.
c
b
a
Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.
Insert cba each character on a stack in the following order.
a b c
First c is added to queue then b and then a.
Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.
First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match
c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.
One million is the answer.
You should follow a set of directions
Answer:
The code solution is written in Java.
- import java.util.Scanner;
-
- public class TestScore {
-
- public static void main(String[] args) {
-
- Scanner input = new Scanner(System.in);
-
- System.out.print("Please enter first score: ");
- double firstScore = input.nextDouble();
- System.out.println("Grade: " + determineGrade(firstScore));
-
- System.out.print("Please enter second score: ");
- double secondScore = input.nextDouble();
- System.out.println("Grade: " + determineGrade(secondScore));
-
- System.out.print("Please enter third score: ");
- double thirdScore = input.nextDouble();
- System.out.println("Grade: " + determineGrade(thirdScore));
-
- System.out.print("Please enter fourth score: ");
- double fourthScore = input.nextDouble();
- System.out.println("Grade: " + determineGrade(fourthScore));
-
- System.out.print("Please enter fifth score: ");
- double fifthScore = input.nextDouble();
- System.out.println("Grade: " + determineGrade(fifthScore));
-
- System.out.println("Average score: " + calcAverage(firstScore, secondScore, thirdScore, fourthScore, fifthScore));
-
- }
-
- public static double calcAverage(double score1, double score2, double score3, double score4, double score5){
- double average = (score1 + score2 + score3 + score4 + score5) / 5;
- return average;
- }
-
- public static String determineGrade(double score){
- if(score >= 90){
- return "A";
- }
- else if(score >= 80 ){
- return "B";
- }
- else if(score >=70){
- return "C";
- }
- else if(score >=60){
- return "D";
- }
- else{
- return "F";
- }
- }
- }
Explanation:
Firstly, create the method, <em>calcAverage()</em>, that takes five test scores. Within the method, calculate the average and return it as output. (Line 33 - 36)
Next, create another method, <em>determineGrade()</em>, which takes only one score and return the grade based on the range of the score. (Line 38 -54)
Once the two required methods are created, we are ready to prompt use for input five test scores using Java Scanner class. To use get user input, create a Scanner object (Line 7). Next, use getDouble() method to get an input score and assign it to variables firstScore, secondScore, thirdScore, fourthScore & fifthScore, respectively. Once a score input by user, call determineGrade() method by passing the input score as argument and immediately print out the return grade. (Line 9 - 27)
At last, call calcAverage() method by passing the first test score variables as argument and print out the returned average value. (Line 29).
