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: Accepted)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let n = read::<usize>();
    let ts = read_vec::<i64>();
    let vs = read_vec::<i64>();

    let mut v_limits = vec![0; n];
    for i in (0..n - 1).rev() {
        v_limits[i] = std::cmp::min(vs[i], vs[i + 1]);
        v_limits[i] = std::cmp::min(v_limits[i], v_limits[i + 1] + ts[i + 1]);
    }
    let ts = ts.iter().map(|&x| x as f64).collect::<Vec<_>>();
    let vs = vs.iter().map(|&x| x as f64).collect::<Vec<_>>();
    let v_limits = v_limits.iter().map(|&x| x as f64).collect::<Vec<_>>();
    let mut cur_v = 0.0;
    let mut ans = 0.0;
    for i in 0..n {
        // accel whole span
        if cur_v + ts[i] <= v_limits[i] {
            ans += cur_v * ts[i] + 0.5 * ts[i] * ts[i];
            cur_v += ts[i];
        } else {
            // we need decel
            let cross_t = 0.5 * (v_limits[i] - cur_v + ts[i]);
            let cross_v = cur_v + cross_t;
            if cross_v <= vs[i] as f64 {
                ans += cur_v * cross_t;
                ans += 0.5 * cross_t * cross_t;
                ans += v_limits[i] * (ts[i] - cross_t);
                ans += 0.5 * (ts[i] - cross_t) * (ts[i] - cross_t);
            } else {
                let cross_t1 = vs[i] - cur_v;
                let cross_t2 = ts[i] - (vs[i] - v_limits[i]);
                ans += cross_t1 * cur_v;
                ans += 0.5 * cross_t1 * cross_t1;
                ans += (cross_t2 - cross_t1) * vs[i];
                ans += v_limits[i] * (ts[i] - cross_t2);
                ans += 0.5 * (ts[i] - cross_t2) * (ts[i] - cross_t2);
            }
            cur_v = v_limits[i];
        }
        // println!("{} {}", ans, cur_v);
    }
    println!("{}", ans);
}