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.

what is EOL, EOF symbols???????

check whether you output EOF. If you output it you will get WA. I got
really angry when I understood that my program didn't work, because I
print EOF!!!!

thnx man .. that should help

5 years later, same trap grabbing me :(

В администрации есть кто живой?

Nevermind, this test case helped me find my mistake: "ababbaa"