#include <iostream>
#include <cmath>

int main() {
    int n, k;
    std::cin >> n >> k;

    if (n < k) {
        std::cout << 2;
    } else {
        double speed = 0.5 * k;

        int result = static_cast<int>(std::ceil(n / speed));
        std::cout << result;
    }
}

a,b=map(int, input().split())
if a==1:
    print(2)
    exit()
if (a*2)//b<a*2/b:
    print(a*2//b+1)
else:
    print(a*2//b)

n,k = map(int,input().split())
if k==1 or n==1:
    print(2*n)
else:
    if 2*n%k==0:
        print(2*n//k)
    else:
        print(2*n//k+1)

Не совсем понял тестовый пример
3 2
то есть 3 котлеты и можно максимум 2 котлеты жарить в моменте
вначале положим 2 котлеты с одной стороны
потом эти же с другой стороны
потом остается 1 котлета мы ее можем пожарить с одной стороны
это 1 минута и ее же с другой
то есть на нее необходимо 2 минуты

значит всего необходимо 4 минуты
почему ответ 3 ?
как так пожарить

Edited by author 11.06.2023 16:57

Edited by author 11.06.2023 16:57

Consider 2 cases:
(1) K >= N
(2) K < N

and fry one side of each steak first.
Try cases like:
N=7,K=1; N=7,K=2; up to N=7,K=6 and find the pattern.

n = int(input())
k=int(input())
n=n*2
if n % k ==0:
  print(n/k)
else:
    a=int(n/k)+1
    print(a)

I can't find my mistake.
It keep telling me "Runtime error"

n = int(input())
k=int(input())
n=n*2
if n % k ==0:
  print(n/k)
else:
    a=int(n/k)+1
    print(a)

I can't find my mistake.
It keep telling me "Runtime error"

Подскажите, пожалуйста, что не так с кодом? Wrong answer, тест 3.

Edited by author 09.10.2022 19:48

Edited by author 09.10.2022 19:48

Почему так не проходит(ошибка на 13):
#include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    if(n<=k) std::cout<<2;
    else std::cout<<n*2/k+n*2%k;

return 0;
}

А так проходит:
    #include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    if(n<=k) std::cout<<2;
    else{
        if(n*2 % k == 0) std::cout<<n*2/k;
        else std::cout<<n*2/k+1;
    }
return 0;
}
Ну или тернар(тоже не проходит - ошибка на 13):
#include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    (n<=k)?std::cout<<2:std::cout<<n*2/k+n*2 % k;

return 0;
}
Может я чего не понимаю, может всё-таки где-то приоритет потерял, но ведь идентично же.

Edited by author 08.10.2021 11:51
Теперь понял, %k может вернуть и больше, чем 1

Edited by author 08.10.2021 12:00

Edited by author 08.10.2021 12:02
C++ ответ:
#include <iostream>
int main() {
    int n,k;
    std::cin>>n>>k;
    (n<=k)?std::cout<<2:(n*2%k==0)?std::cout<<n*2/k:std::cout<<n*2/k+1;

return 0;
}

Edited by author 08.10.2021 12:03

Lets say there are 5 Steaks and capacity of pan is 4.
Step 1: Cook first 4 one side for 1 minute.
Step 2: Replace 1 one sided cooked steak with completely uncooked (remaining) one and
        cook for next 1 minute.
Step 3: Now the chef has 3 completely cooked and 2 Half cooked. Cook the remaining
        2 half cooked for another one minute.

so 1 minute at each step, hence 3 minutes total minimal time.

However, story doesn't end here :-). Lets take another example

Now, Lets say there are 7 Steaks and capacity of pan is 4.
Step 1: Cook first 4 one side for 1 minute.
Step 2: Replace 3 one sided cooked steak with completely uncooked (remaining) three
        and cook for next 1 minute.
Step 3: Now the chef has 1 completely cooked and 6 Half cooked. Cook 4 steaks from
        half cooked six for another one minute.
Step 4: Last cook final 2 half cooked one for another one minute

So total minimum time take is 4 minute. 1 minute at each step.

Please keep in mind the scenarios where 0 steaks or steaks less then cooking capacity of pan.

Edited by author 08.06.2017 20:19

Edited by author 08.06.2017 20:20

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 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.

Edited by author 30.06.2013 01:11

import java.util.Scanner;

public class Zadachi {

    public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int k = sc.nextInt();
    int x = 2;
        if(n<=k){ System.out.print(x);
        }else{

            if (n*2 % k == 0){
                System.out.print(n*2/k);
            }else{
                System.out.print(n*2/k+1);
            }
        }

}
}

#include "iostream"

using namespace std;
int main()
{
    int a, b, rz;
    cin >> a >> b;
    if (b == 2 && a != 1 || b == 1 && a != 1) {
        rz = 2 * a / b;
    }
    else {
        if (a % b != 0) {
            rz = 2 * (a / b + 1);
        }
        else {
            rz = 2 * a / b;
        }
    }
    cout << rz;
    return 0;
}

i tested it a lot, pleas help may be i am so silly....

public class Pan {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();

        int m = 2*n;
        if (m%k == 0){
            System.out.println(m/k);
        }
        else{

            System.out.println(m/k+1);
        }
    }
}
why Wrong Answer?

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!

what is 34 test?

New tests have been added to the problem, all accepted solutions have been rejudged. 490 authors have lost their AC.