i also had tried forward and backward labels, i think my program work correct

How to solve it without hashes?

Actually, u can write double-hashes and easy pass these anti-hash tests.

why?

It's seems your answer on first test is wrong. Point is that my AC program answer is
00:00:00
00:00:18
00:00:38
00:00:42
00:00:07
00:00:17
00:00:27
00:00:37

Your test 2 inconsistentб because
"The times in each list are pairwise different and are given in ascending order."
1 2 3 1
5
00:00:01
00:00:05
00:00:07
00:00:08
00:00:09
5
00:00:02
00:00:03
00:00:05
00:00:07
00:00:10

Answer
00:00:01
00:00:13
00:00:14
00:00:15
00:00:16
00:00:03
00:00:04
00:00:05
00:00:07
00:00:10


Edited by author 24.01.2012 22:39

OUCH

Please do not writeln your code there.

> This code Accepted so
So no run-time checks if a, b, c are in [0-1] interval.

Example:
[code deleted]

Edited by author 25.05.2016 18:45

Edited by moderator 24.11.2019 13:45

nuuu.. che skazat nu krasavchik.

умный, молодец

your english are good)
i got it.
may be i got it because my solution is similar yours=)

Your english good enouth to speak with Russians)

In the "For" can not be used "int64"

new Busket[100] means indices 0..99. Use [101] for 0..100 indices.

Edited by author 24.05.2016 14:33

Another hint: imagine if you got n points and you could cheat and make them all respect each other by stretching or compressing the coordinate grid (while keeping the points static); what would be the simplest way to transform the grid? In this problem you can't cheat like that, but on the other hand you only need to have n/5 points work together.

input
0 30
-25.9807621135 -15
25.9807621135 -15
output
-8.2902573694809689e-012 -4.7855053253442748e-012
8.0829037686437069 13.999999999999361
16.165807537287968 -3.1885605267234496e-013
-1.6580514738954005e-011 -29.999999999980858
-25.980762113541445 15.000000000004786
25.980762113533153 15.000000000019146




input
0 0
5 8.6602540378443864676372317075294
10 0
output
5 2.8867513459481291
3.9106836025229588 0.99999999999999989
6.0893163974770417 0.99999999999999989
10 5.7735026918962582
0 5.7735026918962582
5 -2.8867513459481291

WA1 could also be too low output precision!

cin/cout with "ios_base::sync_with_stdio (false);" got TLE on 9 for me.

But these lines are faster:

scanf("%s", input);
.....
puts(answer);

where input and answer are char arrays.

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);
        }
    }
}

For some tasks, they had a bigger time limit before, but after it changed, old solutions might have not been rejudged for whatever reasons.