New tests have been added. 95 authors have lost their AC.

New tests have been added. 86 authors have lost their AC.

use int instead of double
Good Luck

There exist a closed form for the answer. hence the solution is O(K).

The most important functions are listed below:

// Judge if there exist a parabola passing P[a] and P[b]
// The validation of direction of the parabola is included in the positive sign
bool Valid(int a, int b)
{
    if (P[a].x == P[b].x)
        return false;
    else return (P[a].x - P[b].x) * (P[a].x * P[b].y - P[b].x * P[a].y) > 0;
}

// Judge if the parabola passing P[a] and P[b] also passes P[c]
// Using the determinant for judgement
bool Pass(int a, int b, int c)
{
    return (P[a].x * P[a].x) * P[b].x * P[c].y +
        P[a].x * P[b].y * (P[c].x * P[c].x) +
        P[a].y * (P[b].x * P[b].x) * P[c].x -
        (P[a].x * P[a].x) * P[b].y * P[c].x -
        P[a].x * (P[b].x * P[b].x) * P[c].y -
        P[a].y * P[b].x * (P[c].x * P[c].x) == 0;
}

Note: using long integers (64-bit) to avoid overflow.

Good luck.

I've got AC then complete test for a<0 && b!=0:
1 999
2 999
3 0
4 1
5 2
answ 3...
Good Luck!

Edited by author 23.10.2011 01:27

Edited by author 23.10.2011 04:20

What's test 8 case?

For each pair of points, consider the family of lines parallel to the line that connects the points. Project all the points on the line perpendicular to the family, sort the coordinate, and find the smallest coordinate distance that covers k points.

EDIT: Finally passed with Python in 0.109s. deque() really works for this problem. The key insight is to recognise how we design the pop() and popleft() to make sure the first element of the list is the largest element which has not expired. Whether or not the middle elements have expired does not matter.

Edited by author 01.04.2018 15:32

Hi. Can anyone help me?
5 4
4 1 4 1 2

Why answer 3 2 ?? why it isn't 3 1?
i can cast power = 2, then i'll destroy three coin (1, 1, 2) which is equal to 4, and i can survive.
But i can't cast with power 4, because then i take (4,4) = 8 damage.



Edited by author 31.03.2018 04:18

Edited by author 31.03.2018 04:18

Edited by author 31.03.2018 04:20

help me

Так до сих пор и не выяснилось, что делать с окончанием ввода?
Или измените условия так, чтобы окончание ввода оканчивалось явно - например пустой строкой или -1, или же уточните, нужно ли это делать с помощью файла.
Перенаправление ввода из файла, например в bash, не воспринимает символ конца файла EOF. А ваш сайт при тесте делает расчет на то, что ввод окончится EOFом. Думаю по этой проблеме так мало успешных попыток

Could anyone give me some tests for this one?

One of the solutions would involve sorting and stacks. I've tried to use reb-black tree, but it seems to have problems with dublicate-keys.

Edited by author 14.08.2015 22:05

HINT1: solution always exist
HINT2: try to solve this problem with end.
HINT3: this problem can solve <= O(n logn)

I tried implementing mergesort which probably had recursion depth runtime error; bisect insertion to save some memory, but too slow for python list (perhaps a good alternative if I have access to pointer?); heapq is much faster but instable. What are other options?

import java.io.InputStreamReader;
import java.text.DecimalFormat;
import java.util.Scanner;

public class Reverse {

    private static void printer(double value){
        String pattern = "###0.0000";
        DecimalFormat myFormatter = new DecimalFormat(pattern);
        String output = myFormatter.format(value);
        System.out.println(output);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        long a,b,c,d;

        a = sc.nextLong();
        b = sc.nextLong();
        c = sc.nextLong();
        d = sc.nextLong();

        printer(Math.sqrt(d));
        printer(Math.sqrt(c));
        printer(Math.sqrt(b));
        printer(Math.sqrt(a));
    }
}

It's very easy problem, but many people don't understand how to solve it ... :)

7 lines :

#include <iostream>
using namespace std;
int main(){
unsigned long long N;
cin >> N;
cout << N*N << endl << N;
return 0; }