50 49 8.4 5.5
//1
50 49 8.4 5
//25
50 49 8.4 4
//28
50 49 8.4 2.0
//37
50 49 8.4 0.2
//48

UP! Can someone explain why it happened?

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Lift {
    public static final float LIFT_DELAY = 15;
    public static final float PETYA_DELAY = 5;

    public static void main(String[] args) throws IOException {

        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(reader.readLine());

        float petyaFloor = Float.parseFloat(st.nextToken());
        float liftFloor = Float.parseFloat(st.nextToken());

        float petyaSpeed = Float.parseFloat(st.nextToken());
        float liftSpeed = Float.parseFloat(st.nextToken());

        float timeByFootToTheFirstFloor = ((petyaFloor - 1) * petyaSpeed); //время, если идти до 1го этажа пешком
        float firstLiftDescentTime = ((liftFloor - 1) * liftSpeed) + LIFT_DELAY; //время, затраченное лифтом при спуске и простое на 1м этаже

        Float numOfFloorToGoByFoot = firstLiftDescentTime / petyaSpeed; //количество этажей, которые может успеть пройти Петя за время спуска лифта

        int numOfFloors = numOfFloorToGoByFoot.intValue(); //см. предыдущее

        Float numOfSeekingFloor = petyaFloor - numOfFloors; //искомый этаж

        float timeLiftWithPetya = (((numOfSearchingFloor - 1) * liftSpeed) * 2) + firstLiftDescentTime + PETYA_DELAY; //время на спуск, подъём к Пете на новый этаж и спуск с ним до 1го этажа

        if (timeByFootToTheFirstFloor < timeLiftWithPetya) {
            System.out.println("1");
        } else if (timeByFootToTheFirstFloor == timeLiftWithPetya) {
            String s = numOfSearchingFloor.longValue() + "";
            System.out.println(s);
        } else {
            String s = numOfSearchingFloor.longValue() + "";
            System.out.println(s);
        }
    }
}

Thank you very much. You really helped me: I did the same thing and got AC.

for wa7 too work.. thx

Edited by author 11.08.2014 17:44

It's not necessary now, I've got AC long time ago.

I passed this test, when I understood that, if he has to go to the first floor using his legs, he mustn't enter the elevator. So, when calculating time, when he goes to the first floor, you shouldn't add 5 seconds to the time.

I agree, double is all right in case you use comparison like (a+1e-9 < b).

Oh! I think that you are not right at r. Because the elevator goes down and then waits for 15 minute, after that it goes up. So your fomular computes r is wrong! ^_^.

Edited by author 07.11.2008 05:38

read problem more attentively. when you find pos, you way from 2 to N, but right way from N downto 2... and you get AC :)

Edited by author 06.07.2008 02:43

Your source is right. But please don't remember that "If there are several equivalent variants, output such one in which Petr will go by foot the smaller distance."

if you compare two doubles use
a - b > 0.000000001 instead of a > b

When comparing calculate time _t_ to the current _best_ time, where both variables are doubles, use e.g.
if (t + 1e-9 < best)

You are right, the answer is wrong. Right answer for the input is 76.

The man should be at the floor before he call for the elevator.

My English is so poor....

I understand too...


Edited by author 25.03.2008 23:50