Problem:
Score : 800 points

Problem Statement

There are N lines in the xy-plane. The i-th line is represented by A_ix+B_iy=C_i.
Any two lines among the N+2 lines, the above N lines plus the x-axis and y-axis, cross each other at exactly one point.

For each pair 1 \leq i < j \leq N, there is a car at the cross point of the i-th and j-th lines.
Even where three or more lines intersect at a point, a car is individually placed for each pair of lines.
That is, there will be k(k-1)/2 cars placed at the intersection of k lines.

Those cars are already very old, and can only be moved parallel to the x-axis or y-axis.

Takahashi will hold an exhibition of antique cars at a place on the xy-plane.
In order to avoid damaging the half-broken cars too much, he will select the place of the exhibition so that the total distance covered will be minimized when all the cars are moved to the place.
If such a place is not uniquely determined, among the places that satisfy the condition above, the place with the minimum x-coordinate will be selected.
If the place is still not uniquely determined, among the places that satisfy the two conditions above, the place with the minimum y-coordinate will be selected.

Find the place of the exhibition that will be selected.

Constraints

2 \leq N \leq 4 × 10^4

1 \leq |A_i|,|B_i| \leq 10^4(1 \leq i \leq N)

0 \leq |C_i| \leq 10^4(1 \leq i \leq N)

No two given lines are parallel.

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
A_1 B_1 C_1
:
A_N B_N C_N

Outputs

Print the x-coordinate and y-coordinate of the place of the exhibition that will be selected, in this order, with a space in between. The output will be judged as correct when the absolute or relative error is at most 10^{-9}.

Sample Input 1

3
1 1 1
2 -1 2
-1 2 2

Sample Output 1

1.000000000000000 1.000000000000000

There is a car at each place shown by a blue circle in the figure. The place to be selected is shown by a purple circle.

Sample Input 2

4
1 1 2
1 -1 0
3 -1 -2
1 -3 4

Sample Output 2

-1.000000000000000 -1.000000000000000

Sample Input 3

7
1 7 8
-2 4 9
3 -8 -5
9 2 -14
6 7 5
-8 -9 3
3 8 10

Sample Output 3

-1.722222222222222 1.325000000000000

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::cmp::Ordering;

fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };
    let n: usize = sc.read();
    let abc: Vec<(f64, f64, f64)> = (0..n).map(|_| (sc.read(), sc.read(), sc.read())).collect();
    let bac = abc.iter().map(|&(a, b, c)| (b, a, c)).collect();
    let x = solve(abc);
    let y = solve(bac);
    println!("{} {}", x, y);
}

fn solve(mut abc: Vec<(f64, f64, f64)>) -> f64 {
    let n = abc.len();
    let total = n * (n - 1) / 2;
    let count = (total + 1) / 2;

    let mut high = 1e15 as f64;
    let mut low = -1e15 as f64;
    let mut prev_high = high;
    let mut prev_low = low;

    abc.sort_by_key(|&(a, b, c)| F64((-a * low + c) / b));

    loop {
        let x = (high + low) / 2.0;
        let mut v: Vec<(F64, usize)> = abc.iter().enumerate().map(|(i, &(a, b, c))| (F64((-a * x + c) / b), i)).collect();
        v.sort();

        let mut bit = fenwick_tree::FenwickTree::new(v.len(), 0);
        let mut result = 0;
        for &(_, i) in v.iter() {
            result += i - bit.sum_one(i + 1);
            bit.add(i, 1);
        }

        if result >= count { high = x; } else { low = x; }
        if prev_high == high && prev_low == low { break; }
        prev_high = high;
        prev_low = low;
    }
    (high + low) / 2.0
}

pub mod fenwick_tree {
    use std::ops::{AddAssign, Sub};

    /// `FenwickTree` is a data structure that can efficiently update elements

    /// and calculate prefix sums in a table of numbers.
    /// [https://en.wikipedia.org/wiki/Fenwick_tree](https://en.wikipedia.org/wiki/Fenwick_tree)
    pub struct FenwickTree<T> {
        n: usize,
        data: Vec<T>,
        init: T,
    }

    impl<T: Copy + AddAssign + Sub<Output=T>> FenwickTree<T> {
        /// Constructs a new `FenwickTree`. The size of `FenwickTree` should be specified by `size`.
        pub fn new(size: usize, init: T) -> FenwickTree<T> {
            FenwickTree {
                n: size + 1,
                data: vec![init; size + 1],
                init: init,
            }
        }

        pub fn add(&mut self, k: usize, value: T) {
            let mut x = k;
            while x < self.n {
                self.data[x] += value;
                x |= x + 1;
            }
        }

        /// Returns a sum of range `[l, r)`
        pub fn sum(&self, l: usize, r: usize) -> T {
            self.sum_one(r) - self.sum_one(l)
        }

        /// Returns a sum of range `[0, k)`
        pub fn sum_one(&self, k: usize) -> T {
            if k >= self.n {
                panic!("");
            }

            let mut result = self.init;
            let mut x = k as i32 - 1;
            while x >= 0 {
                result += self.data[x as usize];
                x = (x & (x + 1)) - 1;
            }

            result
        }
    }
}

#[derive(PartialOrd, PartialEq)]
struct F64(f64);

impl Ord for F64 {
    fn cmp(&self, other: &Self) -> Ordering {
        self.0.partial_cmp(&other.0).unwrap()
    }
}

impl Eq for F64 {}

pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n')
            .take_while(|&b| b != b' ' && b != b'\n')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}