Problem:
﻿Score : 400 points

Problem Statement

In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.

In the plan developed by the president Takahashi, the trains will run as follows:

A train will run for (t_1 + t_2 + t_3 + ... + t_N) seconds.

In the first t_1 seconds, a train must run at a speed of at most v_1 m/s (meters per second). Similarly, in the subsequent t_2 seconds, a train must run at a speed of at most v_2 m/s, and so on.

According to the specifications of the trains, the acceleration of a train must be always within ±1m/s^2. Additionally, a train must stop at the beginning and the end of the run.

Find the maximum possible distance that a train can cover in the run.

Constraints

1 \leq N \leq 100

1 \leq t_i \leq 200

1 \leq v_i \leq 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 t_2 t_3 … t_N
v_1 v_2 v_3 … v_N

Output

Print the maximum possible that a train can cover in the run.

Output is considered correct if its absolute difference from the judge's output is at most 10^{-3}.

Sample Input 1

1
100
30

Sample Output 1

2100.000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 30 seconds, it accelerates at a rate of 1m/s^2, covering 450 meters.

In the subsequent 40 seconds, it maintains the velocity of 30m/s, covering 1200 meters.

In the last 30 seconds, it decelerates at the acceleration of -1m/s^2, covering 450 meters.

The total distance covered is 450 + 1200 + 450 = 2100 meters.

Sample Input 2

2
60 50
34 38

Sample Output 2

2632.000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 34 seconds, it accelerates at a rate of 1m/s^2, covering 578 meters.

In the subsequent 26 seconds, it maintains the velocity of 34m/s, covering 884 meters.

In the subsequent 4 seconds, it accelerates at a rate of 1m/s^2, covering 144 meters.

In the subsequent 8 seconds, it maintains the velocity of 38m/s, covering 304 meters.

In the last 38 seconds, it decelerates at the acceleration of -1m/s^2, covering 722 meters.

The total distance covered is 578 + 884 + 144 + 304 + 722 = 2632 meters.

Sample Input 3

3
12 14 2
6 2 7

Sample Output 3

76.000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 6 seconds, it accelerates at a rate of 1m/s^2, covering 18 meters.

In the subsequent 2 seconds, it maintains the velocity of 6m/s, covering 12 meters.

In the subsequent 4 seconds, it decelerates at the acceleration of -1m/s^2, covering 16 meters.

In the subsequent 14 seconds, it maintains the velocity of 2m/s, covering 28 meters.

In the last 2 seconds, it decelerates at the acceleration of -1m/s^2, covering 2 meters.

The total distance covered is 18 + 12 + 16 + 28 + 2 = 76 meters.

Sample Input 4

1
9
10

Sample Output 4

20.250000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 4.5 seconds, it accelerates at a rate of 1m/s^2, covering 10.125 meters.

In the last 4.5 seconds, it decelerates at the acceleration of -1m/s^2, covering 10.125 meters.

The total distance covered is 10.125 + 10.125 = 20.25 meters.

Sample Input 5

10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70

Sample Output 5

20291.000000000000

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::{BufRead, Write};
use std::io;
use std::iter;

fn solve<R: BufRead, W: Write>(mut input: R, mut output: W) {
    let mut buf = String::new();
    input.read_line(&mut buf).unwrap();
    let n: usize = buf.trim_right().parse().unwrap();
    buf.clear();
    input.read_line(&mut buf).unwrap();
    let t: Vec<usize> = buf.split_whitespace().map(|x| x.parse().unwrap()).chain(iter::once(0)).collect();
    buf.clear();
    input.read_line(&mut buf).unwrap();
    let v: Vec<usize> = buf.split_whitespace().map(|x| x.parse().unwrap()).chain(iter::once(0)).collect();
    assert!(n == t.len() - 1 && n == v.len() - 1);
    let mut result = 0.;
    let mut current_speed = 0.;
    for i in 0..n {
        let mut accu = 0;
        let target_speed = (i + 1..n + 1).map(|j| {
            let ret = v[j] + accu;
            accu += t[j];
            ret
        }).min().unwrap();
        let max_speed = (v[i] as f64).min((t[i] as f64 + current_speed + target_speed as f64) / 2.); // derived from (time-diff)/2+max
        let goal_speed = max_speed.min(target_speed as f64);
        let accel = max_speed - current_speed;
        let decel = max_speed - goal_speed;
        let keep = t[i] as f64 - accel - decel;
        result += (current_speed + max_speed) * accel / 2. + keep * max_speed + (goal_speed + max_speed) * decel / 2.;
        current_speed = goal_speed;
    }
    writeln!(output, "{}", result).unwrap();
}

fn main() {
    let stdin = io::stdin();
    let stdout = io::stdout();
    solve(stdin.lock(), stdout.lock());
}

#[cfg(test)]
mod tests {
    fn test(input: &str, output: &str) {
        let mut v: Vec<u8> = Vec::new();
        ::solve(input.as_bytes(), &mut v);
        assert_eq!(String::from_utf8(v).unwrap(), output);
    }

    #[test]
    fn case1() {
        test(
            "1
100
30\n",
            "2100.000000000000000\n",
        );
    }

    #[test]
    fn case2() {
        test(
            "2
60 50
34 38\n",
            "2632.000000000000000\n",
        );
    }

    #[test]
    fn case3() {
        test(
            "3
12 14 2
6 2 7\n",
            "76.000000000000000\n",
        );
    }

    #[test]
    fn case4() {
        test(
            "1
9
10\n",
            "20.250000000000000\n",
        );
    }

    #[test]
    fn case5() {
        test(
            "10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70\n",
            "20291.000000000000000\n",
        );
    }
}