does anyone know what is this test?
thanks

n^3 is suitable for which n<=200
but pay attention to the collinear problem

Edited by author 17.05.2019 14:43

И все же...
Что там за набор такой?

6
0 3
0 0
3 0
0 3
1 2
2 1

answer 5 (0 0 is not)

16
15 15
16 17
17 19
18 21
19 23
75 19
16 62
13 69
10 76
7 83
4 90
1 97
-2 104
-5 111
-8 118
-11 125

answer: 10 (last 10)

13
1 1
1 1
1 1
16 62
13 69
10 76
7 83
4 90
1 97
-2 104
-5 111
-8 118
-11 125

answer: 10 (last 10 your program could get 10 in prev test and get 13 here)


8
1 1
1 1
1 1
0 0
0 0
0 0
0 0
2 2

answer 8

5
4 4
4 4
4 4
4 4
4 4

answer 5

I hope, it will help you (P.S. you can click on my nickname and write me, i with big pleasure help you)

Edited by author 03.03.2017 22:18

What's on test 6? help, please!

#include <cstdio>
#include <vector>
#include <math.h>
#include <cstdlib>
#include <algorithm>
using namespace std;
double line_k(double x1, double y1, double x2, double y2)
{
    double k = (y2 - y1)/(x2 - x1);
    return k;
}
double line_b(double x1, double y1, double x2, double y2)
{
    double b = y2 - (y2 - y1) * x2 / (x2 - x1);
    return b;
}
bool is_on_line(double k, double b, double x, double y)
{
    if (y <= x * k + b + 0.01 && y >= x * k + b - 0.01) return true;
    return false;
}
struct point{
    double x,y;
};
int main()
{
    int n, counter = 2, max_zerosx = 0, max_zerosy = 0;
    double x, y, k, b;
    vector <point> koord;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%lf %lf", &x, &y);
        if (x == 0) max_zerosx++;
        if (y == 0) max_zerosy++;
        {
            koord.push_back(point());
            koord[i].x = x;
            koord[i].y = y;
        }
    }
    int maximal = max(max_zerosx, max_zerosy);
    for (int i = 0; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            k = line_k(koord[i].x, koord[i].y, koord[j].x, koord[j].y);
            b = line_b(koord[i].x, koord[i].y, koord[j].x, koord[j].y);
            for (int l = j + 1; l < n; l++)
                 if (is_on_line(k, b, koord[l].x, koord[l].y))
                     counter++;
            if (counter > maximal) maximal = counter;
            counter = 2;
        }
    }
    printf("%d", maximal);
    return 0;
}

I solved it in 0.015 with g++, but i can't optimize anymore

Edited by author 22.06.2015 13:32

Not understand the algorithm. What is it? Please explain the problem.

Edited by author 25.03.2012 18:41

Some new tests were added. 234 authors lost AC. Maybe we'll add more tests soon.

In my solution I read values like this:

        scanf("%d %d", &px[i], &py[i]);
        px[i] += 1000;
        py[i] += 1000;
        board[px[i]][py[i]] = 1;

and get access violation on test #4; however, if I add right before accessing the array:

        if (px[i] < 0 || py [i] < 0 || px[i] > 2000 || py[i] > 2000) {
            while (1);
        }

I get TLE on the same test.

Are all the values of test #4 within the bounds -1000 <= x,y <= 1000 ?!

Edited by author 11.08.2011 18:49

What test6? please help me.

Edited by author 12.05.2011 11:10

I have understood why you could have some strange WA2.
my program:

while (scanf("%d", &n) >= 1) {
    ...


}
if I write
while (scanf("%d", &n) >= 1) {
     if (!n)
        break;
    ...


}
I'll get AC..(though n >= 2=))
they have some trash at the end of the file...

if (pts[m].y == (int)(Math.Round(k * (double)pts[m].x + b)))

this i changed on

if (Math.Abs(pts[m].y - (k * (double)pts[m].x + b))<0.000000001)

and now I have AC

Edited by author 08.01.2010 03:08

what's test2? my program has passed test 1 and 3 as i know
thx

Why does N == 0 in test 2
?
An input contains an integer N (2 ≤ N ≤ 200) ...