First code gives wrong answer test #13. In the next code I changed

   System.out.print(2*n/k+2*n%k);

and used Math.ceil()

   System.out.print((int)Math.ceil(2*(float)n/(float)k));

The code was accepted. What's wrong?

Edited by author 03.08.2015 07:16

Edited by author 03.08.2015 07:17

Edited by author 03.08.2015 07:17

Edited by author 03.08.2015 07:17

Edited by author 03.08.2015 07:18

Edited by author 03.08.2015 07:18

Edited by author 03.08.2015 07:22

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

//URAL STEAKS
public class P {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        String line;
        int ans;
        line = br.readLine().trim();
        String str[];
        str = line.split(" ");
        int N = Integer.parseInt(str[0]);
        int K = Integer.parseInt(str[1]);
        if(2*N%K>1){
            ans = (2*N/K)+1;
        }else{
            ans = (2*N/K);
        }
        out.print(ans);
        out.flush();
    }
}

import java.io.*;
import java.util.*;

public class task {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int k = in.nextInt();
        int t;

        if(n>=1 && k<=1000) {
            if((int) (n * 2.0 / k) == (n * 2.0 / k)) t = (int) (n * 2.0 / k);
            else t = 1 + (int) (n * 2.0 / k);
            out.print(t);

        }
        out.flush();
   }
}

i understand problem !!!

Edited by author 03.06.2015 12:14

var n,k,l:real;
begin
read(n,k);
l:=(n*2/k);
if trunc(l)=0 then write(2)
else if frac(l)=0 then write(l)
else write(trunc(l)+1);
end.

Whoever thought of this problem haven't fried steaks in real life.

The only way to get an acceptable solution to 1820 is to cook the steaks somewhat similar to this order:
Minute 1: St1A, St2A
Minute 2: St1B, St3A
Minute 3: St2B, St3B
(where there are three steaks: St1, St2 and St3, and each having side A and side B).

The problem with this is:
After St2A side is cooked, it will be taken off the heat and cool down for one minute before cooking the other side, thus in the end the steak will be less cooked than stake 1 and 3. That's all sorts of wrong, you'll be ruining a good steak!

:-D

import static java.lang.Math.*;
import java.util.*;

public class UralSteaks {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n,k;
        n=in.nextInt();
        k=in.nextInt();
        int t=(int)ceil(2*(float)n / (float)k);
        System.out.print(t);
    }

}
help plz!

Pay attention to your data types, and convert before ceil if needed.

what is the difference?? to count as whole stakes or to count it with side.
as i think this is the same problem but we get different answers to this question.

my first solution got ac, but with my test 3 78 it wrote 1
i think you should add some tests

1.you should consider total sides that must be fried instead the number of steaks.
2.with every pan we can fry one side of steak independently. for example if we have only 1 steak and 2 pan we can fry it in 1 minute or for 5 steaks and 6 pan we can do it for 1 minute.

Edited by author 07.07.2014 17:30

Edited by author 07.07.2014 17:32

[code deleted by moderator]

You are interested in the number of sides, not steaks as total. First you calculate number of sides, then you calculate the remainder between it and the maximum number of steaks that can be cooked in 1 minute. If the remainder is bigger than 1(not equal, if equal you will get wrong answer ex: 7 and 4 yields answer 3 with actually is 4) then you write the number of sides divided by the maximum number(of sides) in the pan plus the extra that remains. Else you do the same without adding the extra. If k>n, then the maximum is 2 minutes(1 minute for each side) because all steaks fit in the pan. It's a simple and quick solution. Some examples:
1. 4 steaks and maximum 3 in a pan. You have 8 sides and a maximum of 3 in a pan. 8%3=1, that means that you have 2 sides extra. Answer is 8/3 + 1 =3
2. 7 steaks and maximum 4 in a pan. You have 14 sides and maximum 4 in a pan. 14%4=2, that is why you need to put 2*n%k>=1 and not 2*n%n==1. The answer is 14%4 + 1 =3+1=4
I hope that it helped!

Edited by moderator 21.07.2014 01:04

если жарить 3 бифштекса на сковороде, где умещается всего 2, то потребуется не 3, а 4 минуты, поскольку последний бифштекс на 2 части не разделить и не положить на сковороду одновременно обоими сторонами

if n=1(k any) we need 2 minuten, but when n=2 , k=5 we need 1 minutes. Why???

Give me please test 8.

I have solved this problem using heap:)
Please, give me advise, how to solve this problem without structures and etc.

Module Module1

    Sub Main()
        Dim a() As String = Console.ReadLine.Split(" ")
        Dim n, k As Integer
        n = 2 * CInt(a(0))
        k = CInt(a(1))
        Dim rez As Integer = Math.Ceiling(n / k)
        Console.WriteLine(rez)
    End Sub

End Module

Just think about it. In the example there are three steaks, and only two can be cooked at the same time. It would take two minutes to cook both sides of the first pair and another two to prepare the last steak. In total it makes four. I'm just curious, did the author mistype or is there an actual mistake in the program.

My solution failed in test 30. How I can find what is wrong? Help me please

Edited by author 01.09.2013 23:49