Answer:
- import java.util.Scanner;
-
- public class Main {
- public static void main(String[] args) {
- int num[] = new int[51];
- Scanner input = new Scanner(System.in);
- System.out.print("Number of input: ");
- int limit = input.nextInt();
- for(int i=0; i < limit; i++){
- System.out.print("Input a number (1-50): ");
- int k = input.nextInt();
- num[k]++;
- }
- for(int j=1; j < 51; j++){
- if(num[j] > 0){
- System.out.println("Number of occurrence of " + j + ": " + num[j]);
- }
- }
- }
- }
Explanation:
The solution is written in Java.
Firstly, create an integer array with size 51. The array will have 51 items with initial value 0 each (Line 5).
Create a Scanner object and get user entry the number of input (Line 6-7).
Use the input number as the limit to control the number of the for loop iteration to repeatedly get integer input from user (Line 9-13). Whenever user input an integer, use that integer, k, as the index to address the corresponding items in the array and increment it by one (LINE 11-12).
At last, create another for loop to iterate through each item in the array and check if there is any item with value above zero (this means with occurrence at least one). If so, print the item value as number of occurrence (Line 14-17).
Answer:
hi sorry for not knowing the answer
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Answer:
Aggregation
Explanation:
In Java, Aggregation is a relationshipthat exists between two classes that is best described as a "has-a"
Ex:
FOR Count= 1 TO 50
.......................
Body
......................
NEXT
Next keeps incrementing count by 1
Hope it helps
