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 (1.42.0) (judge verdict: Accepted)

macro_rules! get {
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
            get!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
            get!($($t),*)
        ).collect::<Vec<_>>()
    };
}

use std::cmp;

fn main() {
    let n = get!(usize);
    let mut ts = get!(String).split_whitespace().map(|i| i.parse::<usize>().unwrap() * 2).collect::<Vec<_>>();
    for i in 1..n {
        ts[i] += ts[i-1];
    }
    let vs = get!(String).split_whitespace().map(|i| i.parse::<usize>().unwrap() * 2).collect::<Vec<_>>();

    let &t_sum = ts.last().unwrap();
    let &v_max = vs.iter().max().unwrap();

    let mut tt = vec![v_max+1; t_sum+1];

    for i in 0..n {
        let s = if i == 0 { 0 } else { ts[i-1]+1 };
        for j in s..ts[i]+1 {
            tt[j] = cmp::min(tt[j], vs[i]);
        }
    }

    for i in 0..cmp::min(t_sum, v_max)+1 {
        tt[i] = cmp::min(tt[i], i);
    }
    for i in 0..n {
        let mut j = ts[i];
        let mut k = vs[i];
        loop {
            if j > t_sum || k > v_max {
                break;
            }
            tt[j] = cmp::min(tt[j], k);
            j += 1;
            k += 1;
        }
    }
    let mut j = t_sum;
    let mut k = 0;
    loop {
        tt[j] = cmp::min(tt[j], k);
        if j == 0 || k == v_max {
            break;
        }
        j -= 1;
        k += 1;
    }
    for i in 0..n-1 {
        let mut j = ts[i];
        let mut k = vs[i+1];
        if k > v_max {
            break;
        }
        loop {
            tt[j] = cmp::min(tt[j], k);
            if j == 0 || k == v_max {
                break;
            }
            j -= 1;
            k += 1;
        }
    }

    let mut s = 0;
    for i in 0..t_sum+1 {
        s += tt[i];
    }
    println!("{:.12}", (s as f64) / 4.0);
}