It works on my laptop! I have Pascal ABC.

var a,b,c:integer;
begin
readln(a,b);
if a>b then c:=ceil(2*a/b) else c:=2;
write(c);
end.

wrong answer test 13
why

using System;


namespace ConsoleApplication5
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] A = Console.ReadLine().Split(' ');
          float a = int.Parse(A[0]);
          float b = int.Parse(A[1]);
          if (a <= b)
              Console.WriteLine(2);
          else
              Console.WriteLine("{0:#}",a/b*2);
        }
    }
}

i am doing (n%k)+(n/k) after putting restriction on k(k<=1000) and n(n>=1). but my answer is  said to be wrong.please help.

import java.util.*;
public class Bifshteks      //zadacha -1820
{ int n; int k; int clock;
   public static void main(String args[])
    {
Bifshteks b = new Bifshteks();
Scanner in = new Scanner(System.in);
System.out.println("Skolko bifshteksov zakazat? - ");
b.n = in.nextInt();
while(b.n > 1000 || b.n<0 )
 {System.out.println("Skajite zanovo . Vam Skolko bifshteksov zakazat? - ");
    b.n = in.nextInt();
     }
System.out.println("Skolko bifshteksov mojet gotovitsia odnoremenno? - ");
    b.k = in.nextInt();
      int y = b.n/b.k;
   int x = b.n-(y*b.k);
   if(x==0) y+=0;
   else y+=1;
   System.out.println("Vaw zakaz budet gotov cherez "+y*2+" minut.");
    }
}

using System;

namespace console_aplication54
{
    internal class Program
    {
        private static void Main(string[] args)
        {
            string[] nm = Console.ReadLine().Split(' ');
            if (int.Parse(nm[0]) >= 1 && int.Parse(nm[1]) <= 1000)
            {
                double x,y;
                x = double.Parse(nm[0]);
                y = double.Parse(nm[1]);
                Console.Write(Convert.ToInt32(Math.Ceiling(x*2)/y));
            }
        }
    }
}

#include <iostream>


using namespace std;
int main()
{
   int a,b,i=0,x=0,y=0;
   cin>>a>>b;

       while(x+b<=a)
       {
           x=x+b;
           i++;
       }
       y=a-x;
       if(y>0)
       {
            i++;
       }
       while((y+b)<a)
       {
            y=y+b;
            i++;
       }
       i++;

   cout<<i;


return 0;

}

Пример
исходные данные    результат
3 2            3

||| - 3 бифштекса, 2 на одной сковороде

|| - 2 минуты
|  - 2 минуты

итого 4 минуты, а не 3

my program work. but timus don't accept answer.

var n,k,m:word;
begin
read(n,k);
if 2*n mod k=0 then m:=2*n div k
else m:=(2*n div k)+1;
write(m);
end.

I donno how to make it work
somebody give me hint

test   #17   ???

There is my code on C. All test I found are passed. For example:
hot (3,2) == 3);
hot (3,10) == 2);
hot (1,5) == 2);
hot (5,3) == 4);
hot (10,3) == 7);
hot (7,3) == 5);
hot (9,5) == 4);
But server says that anyway test 1 - wrong. What is wrong?
Please give me any test, which my program can't decide corectly.

int hot (int n1, int k) {
   int n2, minutes = 0;

   //Жарим первую сторону по максимальному числу бифштексов
    minutes += n1/k;
    //Пожаренные первые стороны становятся недожаренными вторыми
    n2 = minutes * k;
    //Ищем недожаренные первые стороны
    n1 -= minutes*k;
    //Докладываем на сковороду n2, если есть
    if (n2 > 0) {
        n2 -= (k-n1);
        if (n2<0) n2=0;
    }
    //Жарим остатки n1 и то, что доложили
    minutes++;
    //Пожаренное n1 переходит в n2
    n2 += n1; n1 = 0;
    //Жарим вторую сторону
    minutes += (int) ceil((float)n2/k);

    return minutes;
}

int main() {
    int n,k;

    scanf("%d %d",&n, &k);

    printf("%d", hot(n,k));

    return (EXIT_SUCCESS);
}

Edited by author 01.10.2012 09:26

#include <iostream>
int main()
{
int n,k,a;
std::cin>>n>>k;
if ((2*n)%k==0)
a=(2*n-(2*n)%k)/k;
else  a=(2*n-(2*n)%k)/k+1;

std::cout<<a;
return 0;
}

wrong answer
po4emu ne pravilno, ved algoritm verni?

Edited by author 25.10.2012 22:40

Edited by author 25.10.2012 22:40

Edited by author 21.10.2012 22:36

Edited by author 21.10.2012 22:39

The cook must cook the one steak during 2 minutes (each side during 1 minute). Right?

In example n = 3, k = 2, then:
* at first step he cook 2 stake during 2 minutes;
* at second step he cook 1 stake during 2 minutes;
Answer = 2 + 2 (minutes), but in example 3?

If try to send solution based on above logic, checker answered with mistake.

Probably checking occurs on this way (all test passed):

answer == (k >= n) ? 2 : n * 2 / k + ((n*2 % k == 0) ? 0 : 1)

but, more correctly (in my opinion) variant:

answer == (k >= n) ? 2 : n * 2 / k + ((n % k == 0) ? 0 : 1)

what is in the test #7?

here is my program

var n,k:longint;
begin
readln(n,k);
  if (n<=k) then write(2*n)
  else if (n mod k=0) then write(trunc(2*n/k))
       else write(trunc(1+(n-(n mod k))/k+((n-abs(k-(n mod k)))-((n-abs(k-(n mod k))) mod k))/k+((n-abs(k-(n mod k))) mod k)/k));
end.

what is in 13 test?

#include <iostream>
#include <stdio.h>

int main()
{
    long n,k,t;
    scanf("%ld%ld",&n,&k);
    t=2*(n/k)+(n%k!=0)+((n%k>k/2.0)||(n<k));
    printf("%ld\n",t);
    return 0;
}

Edited by author 05.05.2012 14:45

Edited by author 05.05.2012 14:45

take me idea what is wrong:
var a,b,c:integer;
begin
read(a,b);
c:= (2*a) div b;
if c=2*a/b then write(c)else write (c+1);
end.
p.s sorry for my english