halloween candy solution!

Question

anyone have a better solution? Mine works but it feels janky:
houses = int(input())

#your code goes here
dollarchance = 2 / houses
percent = (dollarchance * 100) + .5
print(round(percent))

Answer 1

Here you go. In Java.
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

double percent =200;
//your code goes here
if(houses>=3){
percent /=houses;
System.out.println((int)Math.ceil(percent));
}
}
}

Answer 2

Brenner Pieszak has added 0.5 as he is using Math.round
The 0.5 will tip any random fraction to over 0.5, but under 1.5 which will always be rounded to 1

When we compare with Ore, the formula used here is Math.ceil which in itself does the work of adding 0.5.

Hope what I said is comprehensible 😊

Answer 3

0.5 is used because if the round function is used on its own, if the number after the decimal is less than 5 (such as 30.4) then it will return 30, whereas we want it to return 31, using "+0.5" turns "30.4 into "30.9" which can then be used by the round function to make 31. Hope that helps :)

Answer 4

print(math.ceil((2 / houses ) * 100))

Answer 5

houses = int(input())

#your code goes here

print (round(2 / houses * 100 + 0.5) )

Answer 6

sorry, can someone explain to me why add `+0.5` to percent?

Answer 7

this is for the code coach halloween candy btw!

Answer 8

Awesome, thanks!

Answer 9

import math
houses = int(input())
print(math.ceil((100/houses) * 2))

Answer 10

Cpp?? Anyone???

Answer 11

This is for you C# bros out there.😎

using System;

double houses;
double percent;

houses = Convert.ToInt32(Console.ReadLine());

percent = Convert.ToInt32((2 / houses) * 100 + 0.5);

Console.WriteLine(percent);