#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;
}

Precision problem: try rounding intersection points or using epsilon when comparing points.

brute force

Edited by author 18.08.2024 01:04

Problem is almost equal to 1203, there is really trivial solution if you now how to solve 1203.

But rating of this problem is 711 and of 1203 is 82. It's strange)

Just bruteforce

First 8 tests all cords are from -1000 to 1000, maybe it can help somebody)

#include<iostream>

using namespace std;
 int main() {
 string s;
 cin >> s;
 for(auto now : s) cout << 1;
}

Edited by author 14.01.2026 22:14

How to do without palindromic tree???

Don't forget about the stars)

ans >= 2^31

This task has some problems with accurancy, but 878 is too big rating for it.

It's strange that 1377 has 295 rating while 1364 has 953 rating. Solutions are almost same)

Given weighted undirected graph, every vertex has its country "C[v]" and money "V[v]". Let's call vertex "v" "responsible" if there exist at least one edge (v, u, cost) where C[v] == C[u]. Also, you can do this operation infinitely many times: Choose edge (v, u, cost), delete it, and add edge (k, u, cost), where C[u] == C[k] and u != k. After this operation make subtraction V[k] -= cost (of course after this operation V[k] must be >= 0). You need to maximize number of responsible vertices

Edited by author 14.08.2024 13:04

What it ask us to do?

Who can explain me the work.

Thanks.