Scratch
Here is the algorithm in variant 2 (see Java algorithm below) in Scratch. The output should be identical.
Java
Here is the algorithm in Java where I did comment the steps which should give you a step-by-step guide on how to do it in Scratch as well.
I have also implemented two variants of the algorithm to show you some considerations that a programmer often has to think of when implementing an algorithm which mainly is time (= time required for the algorithm to complete) and space (= memory used on your computer).
Please note: the following algorithms do not handle errors. E.g. if a user would enter a instead of a number the program would crash. It is easy to adjust the program to handle this but for simplicity I did not do that.
Variant 1: Storing all elements in array numbers
This variant stores all numbers in an array numbers and calculates the sum at the end using those numbers which is slower than variant 2 as the algorithm goes over all the numbers twice. The upside is that you will preserve all the numbers the user entered and you could use that later on if you need to but you will need storage to store those values.
public static void yourAlgorithm() {
// needed in Java to get input from user
var sc = new Scanner(System.in);
// print to screen (equivalent to "say"/ "ask")
System.out.print("How many numbers do you want? ");
// get amount of numbers as answer from user
var amount = sc.nextInt();
// create array to store all elements
var numbers = new int[amount];
// set iterator to 1
int iterator = 1;
// as long as the iterator is smaller or equal to the number of required numbers, keep asking for new numbers
// equivalent to "repeat amount" except that retries are possible if no number was entered
while (iterator <= amount) {
// ask for a number
System.out.printf("%d. number: ", iterator);
// insert the number at position iterator - 1 in the array
numbers[iterator - 1] = sc.nextInt();
// increase iterator by one
iterator++;
}
// calulate the sum after all the numbers have been entered by the user
int sum = 0;
// go over all numbers again! (this is why it is slower) and calculate the sum
for (int i = 0; i < amount; i++) {
sum += numbers[i];
}
// print average to screen
System.out.printf("Average: %s / %s = %s", sum, amount, (double)sum / (double)amount);
}
Variant 2: Calculating sum when entering new number
This algorithm does not store the numbers the user enters but immediately uses the input to calculate the sum, hence it is faster as only one loop is required and it needs less memory as the numbers do not need to be stored.
This would be the best solution (fastest, least space/ memory needed) in case you do not need all the numbers the user entered later on.
// needed in Java to get input from user
var sc = new Scanner(System.in);
// print to screen (equivalent to "say"/ "ask")
System.out.print("How many numbers do you want? ");
// get amount of numbers as answer from user
var amount = sc.nextInt();
// set iterator to 1
int iterator = 1;
int sum = 0;
// as long as the iterator is smaller or equal to the number of required numbers, keep asking for new numbers
// equivalent to "repeat amount" except that retries are possible if no number was entered (e.g. character was entered instead)
while (iterator <= amount) {
// ask for a number
System.out.printf("%d. number: ", iterator);
// get number from user
var newNumber = sc.nextInt();
// add the new number to the sum
sum += newNumber;
// increase iterator by one
iterator++;
}
// print average to screen
System.out.printf("Average: %s / %s = %s", sum, amount, (double)sum / (double)amount);
Variant 3: Combining both approaches
You could also combine both approaches, i. e. calculating the sum within the first loop and additionally storing the values in a numbers array so you could use that later on if you need to.
Expected output
