This is my code
import java.util.*;
import java.io.*;
public class eCheck10A
{
public static void main(String[] args)
{
PrintStream out = System.out;
Scanner in = new Scanner(System.in);
out.print("Enter your integers");
out.println("Negative = sentinel");
List<Integer> aList = new ArrayList<Integer>();
for (int n1 = in.nextInt(); n1 > 0; n1 = in.nextInt())
{
if(n1 < 0)
{
break;
}
}
}
}
if i want to take all the numbers that I enter for n1, and average them out, how do i refer to all these numbers? I am going to put them in the IF statement, so if a negative number is entered, the program stops and posts their average.
This is the pseudocode you need to do this task (pseudo-code since it looks suspiciously like homework/classwork and you'll become a better developer if you nut out the implementation yourself).
Because you don't need the numbers themselves to work out the average, there's no point in storing them. The average is defined as the sum of all numbers divided by their count, so that's all you need to remember. Something like this should suffice:
total = 0
count = 0
n1 = get_next_number()
while n1 >= 0:
total = total + n1
count = count + 1
n1 = get_next_number()
if count == 0:
print "No numbers were entered.
else:
print "Average is ", (total / count)
A couple of other points I'll mention. As it stands now, your for statement will exit at the first non-positive number (<= 0), making the if superfluous.
In addition, you probably want any zeros to be included in the average: the average of {1,2,3} = 2 is not the same as the average of {1,2,3,0,0,0} = 1.
You can do this in the for statement itself with something like:
for (int n1 = in.nextInt(); n1 >= 0; n1 = in.nextInt())
and then you don't need the if/break bit inside the loop at all, similar to my provided pseudo-code.
An outline of what you will need to do: Create a variable sum to add up all of the values and create another variable called count that will be used to store the number of non-negative numbers in your list, aList. Then divide the sum by the count to find the average.
Another way to do running average is:
new_average = old_average + (new_number - old_average ) / count
If you ever hit max for a variable type, you would appreciate this formula.
Related
I'm trying to make a game show program where I enter the number of contestants, how fast they hit the buzzer, the fastest and slowest times for hitting the buzzer, and the number of people that hit the buzzer faster than average.
Everything goes well with my program when I have 5 contestants. But if I enter 4 contestants or anything else less than 5, it messes up some things. When I have 4 contestants, my fastest time is 0, even when I'm not putting in a 0 as a time, and the number of contestants that hit the buzzer faster than average is also messed up, having one extra than it should, because it's putting in a 0.
How do I stop this program from adding in a 0 when I don't have 5 contestants?
import java.util.Scanner;
public class FastestFingers {
public static void main(String[] args) {
//declare varialbes
int contestants;
int [] milisecs = new int [6];
int fastest, slowest, i;
int faster = 0;
double average;
Scanner scanInt = new Scanner (System.in);
//make user enter number of contestants
System.out.println ("Enter # of contestants: ");
contestants = scanInt.nextInt();
//make user enter times
for(i=1;i<=contestants;i++)
{
System.out.println ("Time (ms): ");
milisecs[i] = scanInt.nextInt();
}
//calculate fastest time
fastest = milisecs[1];
for(i=1;i<milisecs.length;i++)
{
if(milisecs[i] < fastest)
{
fastest = milisecs[i];
}
}
System.out.println ("Fastest: " + fastest);
//calculate slowest time
slowest = milisecs[1];
for(i=1;i<milisecs.length;i++)
{
if(milisecs[i] > slowest)
{
slowest = milisecs[i];
}
}
System.out.println ("Slowest: " + slowest);
//tell program how to find average
int total = 0;
for(i=1;i<milisecs.length;i++)
{
total = total + milisecs[i];
}
average = total/contestants;
//find numbers faster than the average
int count = 0;
for(i=1;i<milisecs.length;i++)
{
if(milisecs[i]<average)
{
count++;
}
}
System.out.println ("Faster than average: " + count);
}
}
You create an int[] of fixed size 6. Then you compare the last 5 values to find the lowest. All int values are default 0. If you only set 4 values to anything larger than 0 the fifth value will always be 0 and therefore your lowest value.
You can fix this problem by setting the size of your array to the number of contestants.
Also remember that arrays start counting at 0 not at 1. Your loops should therefore be:
for(i=0;i<milisecs.length;i++), etc.
The problem is that you initialize the array to have six elements.
int [] milisecs = new int [6];
By default all of these values will be zero. Then in your loop you only initialize four of them, meaning that two will still be zero. Change your code to:
System.out.println ("Enter # of contestants: ");
contestants = scanInt.nextInt();
int [] milisecs = new int [contestants];
This will ensure that you will only have as many time slots as you have contestants.
Also array indexes start at zero, so you should change
fastest = milisecs[1];
to
fastest = milisecs[0];
And in your loops, start your variable at 0 instead of 1.
Array Index should start with 0. Please be careful with this. You code reads input and assigns to the milisecs starting from index 1.
I am creating a program that prints the sum of the even numbers from a range of 0 to the number that the user entered. For example, if the user entered the number 20, the program would calculate the sum of all of the even numbers between 0 and 20.
When I tested the program out with the number 10, it worked. But I tried using a different number, 35, and it was just stuck in an infinite loop. I would appreciate any and all help. The code will be posted below:
(Edit) Thanks for the feedback everyone! After talking with a friend, we realized that the solution is actually pretty simple. We were just making it complicated. Still, thanks for all of the suggestions.
//**************************************************************
// Prints the sum of the even numbers within a range of 0
// and the integer that the user enters.
//
// #me
// #version_1.0_11.7.17
//**************************************************************
import java.util.Scanner;
public class EvenNumbersSum
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int user_num = 2; // variable that stores the user's number
int sum; // stores the sum of the needed values
System.out.print("Enter an integer greater than or equal to, 2: "); // prompt user for input
user_num = input.nextInt();
// checks to see if the value entered is valid or not.
while (user_num < 2)
{
System.out.println("Invalid entry. Must enter an integer greater than or equal to, 2.\n");
System.out.print("Enter an integer greater than or equal to, 2: ");
user_num = input.nextInt();
}
// starts adding the values
for (sum = 0; sum <= user_num;)
{
if (user_num % 2 == 0) // checks if the number is even
sum+=user_num; // add the number to sum
else
continue; // I thought that I might need this, but ended up changing nothing.
}
System.out.println(); // extra line for cleanliness
System.out.printf("The sum of the even numbers between 0 and %d is %d.", user_num, sum); // prints the result
}
}
Why are you writing loop for this, there are efficient way to do it.
Sum of numbers between 1-N = (N(N+1))/2
Sum of even numbers between 1-N = (N(N+2))/4
where N = user given input number till which you would like to add even numbers
NOTE: you can add validation on input number that it’s even by (n%2 == 0) and return error if it’s not
The variable you have used in condition (i.e. sum & user_num) no one changes in case of odd number and your code stuck in never-ending loop.
You should use counter variable ( e.g. i from 1 to user_num) and use that number in the condition. Example:
// starts adding the values
sum = 0;
for (int i = 0; i <= user_num; i++)
{
if (i % 2 == 0) // checks if the number is even
sum+=i; // add the number to sum
}
Your for loop should be like this.
int total_sum = 0;
for (int sum = 0; sum <= user_num; sum++)
{
if (sum % 2 == 0) // checks if the number is even
total_sum+=sum; // add the number to total sum
else
continue; // I thought that I might need this, but ended up changing nothing.
}
// note print total sum
System.out.println(totalsum);
your initial program just kept on checking entered number is even or odd.
And the entered number to sum.
So sum was always double of the entered number is even.
If entered number is odd it would go to infinite loop as entered_num(odd) % 2 == 0 always false and execute else statement.
Firstly, I'm taking AP Computer Science this year, and this question is related to an exercise we were assigned in class. I have written the code, and verified that it meets the requirements to my knowledge, so this is not a topic searching for homework answers.
What I'm looking for is to see if there's a much simpler way to do this, or if there's anything I could improve on in writing my code. Any tips would be greatly appreciated, specific questions asked below the code.
The exercise is as follows: Write a program called ProcessingNumbers that does:
Accepts a user input as a string of numbers
Prints the smallest and largest of all the numbers supplied by the user
Print the sum of all the even numbers the user typed, along with the largest even number typed.
Here is the code:
import java.util.*;
public class ProcessingNumbers {
public static void main(String[] args) {
// Initialize variables and objects
Scanner sc = new Scanner(System.in);
ArrayList<Integer> al = new ArrayList();
int sumOfEven = 0;
// Initial input
System.out.print("Please input 10 integers, separated by spaces.");
// Stores 10 values from the scanner in the ArrayList
for(int i = 0; i < 10; i++) {
al.add(sc.nextInt());
}
// Sorts in ascending order
Collections.sort(al);
// Smallest and largest values section
int smallest = al.get(0);
int largest = al.get(al.size() - 1);
System.out.println("Your smallest value is " + smallest + " and your largest value is " + largest);
// Sum of Even numbers
int arrayLength = al.size();
for (int i = 0; i < al.size(); i++) {
if (al.get(i) % 2 == 0) {
sumOfEven += al.get(i);
}
}
System.out.println("The sum of all even numbers is " + sumOfEven);
// Last section, greatest even number
if (al.get(arrayLength - 1) % 2 == 0) {
System.out.println("The greatest even number typed is " + al.get(arrayLength - 1));
} else {
System.out.println("The greatest even number typed is " + al.get(arrayLength - 2));
}
sc.close();
}
}
Here are specific questions I'd like answered, if possible:
Did I overthink this? Was there a much simpler, more streamlined way to solve the problem?
Was the use of an ArrayList mostly necessary? We haven't learned about them yet, I did get approval from my teacher to use them though.
How could I possibly code it so that there is no 10 integer limit?
This is my first time on Stackoverflow in quite some time, so let me know if anything's out of order.
Any advice is appreciated. Thanks!
Usage of the ArrayList wasn't necessary, however it does make it much simpler due to Collections.sort().
To remove the 10 integer limit you can ask the user how many numbers they want to enter:
int numbersToEnter = sc.nextInt();
for(int i = 0; i < numbersToEnter; i++) {
al.add(sc.nextInt());
}
Another note is that your last if-else to get the highest even integer doesn't work, you want to use a for loop, something like this:
for (int i = al.size() - 1; i >= 0; i--) {
if (al.get(i) % 2 == 0) {
System.out.println("The greatest even number typed is " + al.get(i));
break;
}
I wouldn't say so. Your code is pretty straightforward and simple. You could break it up into separate methods to make it cleaner and more organized, though that isn't necessary unless you have sections of code that have to be run repeatedly or if you have long sections of code cluttering up your main method. You also could have just used al.size() instead of creating arrayLength.
It wasn't entirely necessary, though it is convenient. Now, regarding your next question, you definitely do want to use an ArrayList rather than a regular array if you want it to have a variable size, since arrays are created with a fixed size which can't be changed.
Here's an example:
int number;
System.out.print("Please input some integers, separated by spaces, followed by -1.");
number = sc.nextInt();
while (number != -1) {
al.add(number);
number = sc.nextInt();
}
Here is a solution that:
Doesn't use Scanner (it's a heavyweight when all you need is a line of text)
Doesn't have a strict limit to the number of numbers
Doesn't need to ask how many numbers
Doesn't waste space/time on a List
Handles the case when no numbers are entered
Handles the case when no even numbers are entered
Fails with NumberFormatException if non-integer is entered
Moved actual logic to separate method, so it can be mass tested
public static void main(String[] args) throws Exception {
System.out.println("Enter numbers, separated by spaces:");
processNumbers(new BufferedReader(new InputStreamReader(System.in)).readLine());
}
public static void processNumbers(String numbers) {
int min = 0, max = 0, sumOfEven = 0, maxEven = 1, count = 0;
if (! numbers.trim().isEmpty())
for (String value : numbers.trim().split("\\s+")) {
int number = Integer.parseInt(value);
if (count++ == 0)
min = max = number;
else if (number < min)
min = number;
else if (number > max)
max = number;
if ((number & 1) == 0) {
sumOfEven += number;
if (maxEven == 1 || number > maxEven)
maxEven = number;
}
}
if (count == 0)
System.out.println("No numbers entered");
else {
System.out.println("Smallest number: " + min);
System.out.println("Largest number: " + max);
if (maxEven == 1)
System.out.println("No even numbers entered");
else {
System.out.println("Sum of even numbers: " + sumOfEven);
System.out.println("Largest even number: " + maxEven);
}
}
}
Tests
Enter numbers, separated by spaces:
1 2 3 4 5 6 7 8 9 9
Smallest number: 1
Largest number: 9
Sum of even numbers: 20
Largest even number: 8
Enter numbers, separated by spaces:
1 3 5 7 9
Smallest number: 1
Largest number: 9
No even numbers entered
Enter numbers, separated by spaces:
-9 -8 -7 -6 -5 -4
Smallest number: -9
Largest number: -4
Sum of even numbers: -18
Largest even number: -4
Enter numbers, separated by spaces:
No numbers entered
NOTE: (I do not need anyone to write the whole program for me, I only need the algorithm!)
I need to create a program that prompts the user to enter two integers. The program then needs to list all the even numbers in between the two inputed integers and output the sum. And then the same for the odd numbers. (using While loops)
I will then need to rewrite the code to use a do-while loop, and then rewrite it AGAIN using a for loop.
Here is an example of what the result should look like:
Enter an integer: 3
Enter another integer larger than the first: 10
Even Numbers: 4, 6, 8, 10
Sum of even numbers = 28
Odd Numbers: 3, 5, 7, 9
Sum of odd numbers = 24}
I tried starting off with the even numbers with something like this, but it just gets stuck at the first number, even if the first number is even.
import java.util.Scanner;
public class EvenOddSum_While {
public static void main(String[] args){
Scanner keyboard = new Scanner(System.in);
System.out.println("Please enter a number: ");
int num1 = keyboard.nextInt();
System.out.println("And another: ");
int num2 = keyboard.nextInt();
while (num1 < num2){
while (num1 %2 == 0){
System.out.print(num1 + ", ");
num1++;
}
}
}
}
The inner while has no end criteria, you need if there instead. Also, your num1++ Statement must be in the outer while loop, not the inner one.
Also, there is no real algorithm here, you're struggling with the language itself ;)
General advice: either run through your code with a step-by-step debugger, virtually every IDE has one OR place excessive log /System.out.println statements in your code to understand what it's doing
When you said 'in between' did you mean including the two integers? Because you did include them. Okay. So do this.
int x = 0;
int y = 0;
int even = 0;
int odd = 0;
int evenx = 0;
int oddx = 0;
int evena = 0;
int odda = 0;
Scanner scan = new Scanner(System.in);
//Prompts the user to input the first number
x = scan.nextInt();
//Prompts the user to input the second number
y = scan.nextInt();
for(int i = x;i<y;i++,x++;){
if(x%2 = 0){
even = even + x;
evenx++;
}
if(x%2 = 1){
odd = odd + x;
oddx++;
}
}
evena = even/evenx;
odda = odd/oddx;
//print it out. There. The algorithm.
God. Do this site have auto-format?
I need to write a program that finds an average of all values in the array and then returns ones larger than the average.
import java.util.*;
public class AboveAverage {
public static void main(String args[]) throws Exception {
Scanner kbd = new Scanner(System.in);
int count=0, num;
int nums[] = new int[1000];
double sum = 0;
double average = 0;
System.out.println("Input +ve integers, to stop type 0");
num = kbd.nextInt();
while( num > 0 ) {
nums[count] = num;
count = count + 1;
num = kbd.nextInt();
}
int d = 0;
while ( d < 1000 ) {
sum += nums[d];
d++;
average = sum / nums[d];
}
for(int i=0; i < count ; i=i+1) {
System.out.print(nums[i] + " ");
}
System.out.println(average);
}
}
In my opinion problem is in this line
I tried the one below which gives an exception
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1000
at AboveAvarage.main(AboveAvarage.java:23)
average = sum / nums[d];
I also tried
average = sum / nums.length;
Which gave me the sum instead of the average.
Remember that array indices in Java go from zero? This means the allowed indices in your nums array are 0 through 999.
Now look at this loop:
int d = 0;
while ( d < 1000 ) {
sum += nums[d];
d++;
average = sum / nums[d];
}
First, it adds the current number to the sum. This is correct.
Then, it increments the number.
Then it uses the incremented number to access nums[d] to calculate the average.
Now, imagine that you are at the last round. d is 999.
First, it adds the current number to the sum.
Then it increments d. Now it is 1000.
Then it uses the incremented number to access nums[d]. But this means nums[1000]. And that's an illegal index.
So this explains why you have an ArrayIndexOutOfBoundsException. You should never use the index again after you incremented it, because the while only guaranteed that its previous value was less than 1000. If you change the value, you need to test again. This is why normally the increment step is the last in the loop.
As for your logic:
To calculate an average, you first need to know the sum. After you know the sum, you can divide by the number of items you are averaging. So you need to divide the sum by the number of items you read.
So:
You should not do a loop up to 1000. If you entered 0 after 5 numbers when you inputted the data, then there will be no values in the rest of the array (or rather, there will be zeros there).
You can calculate the sum, as another answer told you, in the first loop. No need to do that in the second loop.
You don't calculate the sum and the average in the same step. You first have to complete calculating the sum (finish the loop, be it the first or the second), and only then you can divide by the number of items (which you saved in count).
Then you have to go through another loop, that prints the numbers that are larger than the average. Your print loop prints all the numbers. You should check each number, see if it is greater than the average you calculated, and only if it is, print it.
Hint: there should really only be two loops: One that reads the numbers and calculates the count and the sum. Then you calculate the average, but that's not a repeating action, so it should not be in a loop. The second loop is for printing the numbers that are above average.
In the while loop that is gathering values, you are using count to track how many values are provided.
In the subsequent loop to calculate the average, you should only look at count items of the array:
int d = 0;
while(d<count) {
sum += nums[d];
}
average = sum/count;
Note you don't need to calculate average within the loop - do it once you've summed the values.
You don't need the other while loop, add up sum upon input. When done, just divide by count
while( num > 0 ) {
nums[count] = num;
count = count + 1;
sum += num;
num = kbd.nextInt();
}
average = sum / count;