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)


#[allow(dead_code)]
const MOD: usize = 1_000000000 + 7;

fn main() {
    let mut cin = Scanner::new();
    let n = cin.next_usize();
    let mut t = (0..n).map(|_| cin.next_f64()).collect::<Vec<_>>();
    let mut v = (0..n).map(|_| cin.next_f64()).collect::<Vec<_>>();
    t.push(1.0);
    v.push(0.0);

    let mut ans = 0_f64;
    let mut cur_v = 0_f64;
    for i in 0..n {
        let mut acc_t = v[i] - cur_v;
        let mut dec_t = t[i];
        let mut sum_t = t[i];
        for j in i+1..t.len() {
            if dec_t > sum_t - (v[i] - v[j]) {
                dec_t = sum_t - (v[i] - v[j]);
            }
            sum_t += t[j];
        }
        if acc_t > dec_t {
            acc_t = (acc_t + dec_t) / 2.0;
            dec_t = acc_t;
        }
        if acc_t > t[i] {
            acc_t = t[i];
            dec_t = acc_t;
        }
        ans += cur_v * acc_t + acc_t * acc_t / 2.0;
        cur_v += acc_t;
        ans += cur_v * (dec_t - acc_t);
        ans += cur_v * (t[i] - dec_t) - (t[i] - dec_t) * (t[i] - dec_t) / 2.0;
        cur_v -= (t[i] - dec_t);
    }

    println!("{:.9}", ans);
}


struct Scanner {
    reader: std::io::Stdin,
    tokens: std::collections::VecDeque<String>,
}

impl Scanner {
    fn new() -> Self {
        Scanner {
            reader: std::io::stdin(),
            tokens: std::collections::VecDeque::new(),
        }
    }

    fn next(&mut self) -> String {
        if self.tokens.is_empty() {
            let mut s = String::new();
            loop {
                self.reader.read_line(&mut s).ok();
                let s = s.trim();
                if s.len() != 0 {
                    for it in s.split_whitespace() {
                        self.tokens.push_back(it.into())
                    }
                    break;
                }
            }
        }
        self.tokens.pop_front().unwrap()
    }

    fn next_generics<T>(&mut self) -> T where
        T: std::str::FromStr + std::marker::Copy,
        <T as std::str::FromStr>::Err: std::fmt::Debug
    {
        self.next().parse().unwrap()
    }
    
    #[allow(dead_code)] fn next_i32(&mut self) -> i32 { self.next_generics::<i32>() }
    #[allow(dead_code)] fn next_i64(&mut self) -> i64 { self.next_generics::<i64>() }
    #[allow(dead_code)] fn next_u64(&mut self) -> u64 { self.next_generics::<u64>() }
    #[allow(dead_code)] fn next_f64(&mut self) -> f64 { self.next_generics::<f64>() }
    #[allow(dead_code)] fn next_usize(&mut self) -> usize { self.next_generics::<usize>() }
    #[allow(dead_code)] fn next_isize(&mut self) -> isize { self.next_generics::<isize>() }
}