## solution

$x$軸と$y$軸は独立。$x$座標$y$座標のそれぞれについて交点の座標の中央値を取ればよい。$O(N^2)$。定数倍最適化。

## 知見

• #pragma clang loop vectorize(enable) は、付けなくてもvectorizeされているためあまり速度は変わらないだが、
• vector<T>.push_back をloop内でするとvectorizeできないので、flagを作って外に出すテク

## 他

• tanakhさんの提出をとても参考にした。ほとんど同じと言ってよい。
• $23$WA $2$RE
• 速度は足りてたが普通にバグ

## implementation

#include <algorithm>
#include <cassert>
#include <cstdio>
#include <random>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define whole(x) begin(x), end(x)
using namespace std;

int main() {
// input
int n; scanf("%d", &n);
vector<int16_t> a(n), b(n), c(n);
repeat (i, n) scanf("%hd%hd%hd", &a[i], &b[i], &c[i]);

// solve

// // find the windows
constexpr int sample_size = 500000;
vector<double> samples_x(sample_size);
vector<double> samples_y(sample_size);
mt19937 gen;
repeat (iteration, sample_size) {
int i, j;
do {
i = uniform_int_distribution<int>(0, n - 1)(gen);
j = uniform_int_distribution<int>(0, n - 1)(gen);
} while (i == j);
double den = a[i] * b[j] - a[j] * b[i];
double x = (b[j] * c[i] - b[i] * c[j]) / den;
double y = (a[j] * c[i] - a[i] * c[j]) / - den;
samples_x[iteration] = x;
samples_y[iteration] = y;
}
sort(whole(samples_x));
sort(whole(samples_y));
double dlx = samples_x[sample_size / 2 - window_radius];
double drx = samples_x[sample_size / 2 + window_radius];
double dly = samples_y[sample_size / 2 - window_radius];
double dry = samples_y[sample_size / 2 + window_radius];
bool determined_x = (dlx == drx);
bool determined_y = (dly == dry);
float lx = dlx;
float rx = drx;
float ly = dly;
float ry = dry;

// // get points in them
int low_count_x = 0;
int low_count_y = 0;
vector<double> bucket_x;
vector<double> bucket_y;
vector<int> flag(n);
repeat (j, n) {
{
int aj = a[j];
int bj = b[j];
int cj = c[j];
repeat (i, j) {
float den = a[i] * bj - aj * b[i];
float x = (bj * c[i] - b[i] * cj) / den;
float y = (aj * c[i] - a[i] * cj) / - den;
low_count_x += (x < lx);
low_count_y += (y < ly);
flag[i] = (int(lx <= x and x <= rx) << 1) | int(ly <= y and y <= ry);
}
}
{
double aj = a[j];
double bj = b[j];
double cj = c[j];
if (not determined_x) repeat (i, j) if (flag[i] & 2) {
double x = (bj * c[i] - b[i] * cj) / (a[i] * bj - aj * b[i]);
bucket_x.push_back(x);
}
if (not determined_y) repeat (i, j) if (flag[i] & 1) {
double y = (aj * c[i] - a[i] * cj) / (aj * b[i] - a[i] * bj);
bucket_y.push_back(y);
}
}
}

// // make results
int half = (n * (n - 1) / 2 - 1) / 2;
double x;
if (determined_x) {
x = dlx;
} else {
int ix = half - low_count_x;
assert (0 <= ix and ix < bucket_x.size());
nth_element(bucket_x.begin(), bucket_x.begin() + ix, bucket_x.end());
x = bucket_x[ix];
}
double y;
if (determined_y) {
y = dly;
} else {
int iy = half - low_count_y;
assert (0 <= iy and iy < bucket_y.size());
nth_element(bucket_y.begin(), bucket_y.begin() + iy, bucket_y.end());
y = bucket_y[iy];
}

// output
printf("%.15lf %.15lf\n", x, y);
return 0;
}