Case 3 and 4 Halloween Candy - Java

Question

I solved it but case 3 and 4 isn't solved. I thought that one of the cases is that "houses" can't be lower than 3, but I guess not
My code:

import java.util.Scanner;

public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);

int houses = 0;
while (houses < 3){
houses = input.nextInt();
}
//your code goes here

int bill = 2 *100;
int billChance = bill / houses;

houses = houses % 2;
billChance = billChance + houses ;

System.out.println(billChance);
}
}

Answer 1

Edited:
OK. Why you are changing given code..?
By your logic,
For ex: if Bill =2
House=3
200/3=66
If house =5
200/5=40
If house =4
200/4=50
Finally
Output : 67 41 50
40 has no fraction part but still rounded up to 41..
Am I right?
Edit: use a double type for dollar.
Use a ciel function to round up result..

Answer 2

Thanks, and I changed the upper code because it says that the int should be >= to 3 and I thought that that is one of the cases

Answer 3

This code works guys:
import java.util.Scanner;

public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int houses = input.nextInt();

//your code goes here
int chances = 200/houses;

if(200%houses!=0)
System .out .println (chances+1);
else
System .out .println (chances);
}
}