#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

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

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.

H = 0 or W = 0

#include <iostream>
#include <algorithm>

int main()
{
    int x, y;
    std::cin >> x >> y;
    if ((x + y) & 1)
    {
        std::swap(x, y);
    }
    std::cout << x << " " << y << "\n";
    return 0;
}