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)

use std::cmp;

fn main() {
    let mut sc = Scanner::new();
    let n: usize = sc.read();
    let t: Vec<usize> = (0..n).map(|_| sc.read()).collect();
    let v: Vec<usize> = (0..n).map(|_| sc.read()).collect();

    let t_sum = t.iter().sum();
    let mut max_v = vec![0; t_sum + 1];

    let mut pos = 0;
    for i in 0..n {
        let ti = t[i];
        let vi = v[i];
        max_v[pos] = cmp::min(max_v[pos], vi);
        for _ in 0..ti {
            max_v[pos + 1] = cmp::min(max_v[pos] + 1, vi);
            pos += 1;
        }
    }

    pos = t_sum;
    max_v[t_sum] = 0;
    for i in (0..n).rev() {
        let ti = t[i];
        for _ in 0..ti {
            max_v[pos - 1] = cmp::min(max_v[pos] + 1, max_v[pos - 1]);
            pos -= 1;
        }
    }

    let mut ans = 0.0;
    pos = 0;
    for i in 0..n {
        let ti = t[i];
        let vi = v[i];
        for _ in 0..ti {
            let cur = max_v[pos] as f64;
            let next = max_v[pos + 1] as f64;
            ans += cur + (next - cur) / 2.0;
            if cur == next && vi as f64 > cur {
                ans += 0.25;
            }
            pos += 1;
        }
    }

    println!("{}", ans);
}

struct Scanner {
    ptr: usize,
    length: usize,
    buf: Vec<u8>,
    small_cache: Vec<u8>,
}

impl Scanner {
    fn new() -> Scanner {
        Scanner { ptr: 0, length: 0, buf: vec![0; 1024], small_cache: vec![0; 1024] }
    }

    fn load(&mut self) {
        use std::io::Read;
        let mut s = std::io::stdin();
        self.length = s.read(&mut self.buf).unwrap();
    }

    fn byte(&mut self) -> u8 {
        if self.ptr >= self.length {
            self.ptr = 0;
            self.load();
            if self.length == 0 {
                self.buf[0] = b'\n';
                self.length = 1;
            }
        }

        self.ptr += 1;
        return self.buf[self.ptr - 1];
    }

    fn is_space(b: u8) -> bool { b == b'\n' || b == b'\r' || b == b'\t' || b == b' ' }

    fn read<T>(&mut self) -> T where T: std::str::FromStr, T::Err: std::fmt::Debug, {
        let mut b = self.byte();
        while Scanner::is_space(b) {
            b = self.byte();
        }

        for pos in 0..self.small_cache.len() {
            self.small_cache[pos] = b;
            b = self.byte();
            if Scanner::is_space(b) {
                return String::from_utf8_lossy(&self.small_cache[0..(pos + 1)]).parse().unwrap();
            }
        }

        let mut v = self.small_cache.clone();
        while !Scanner::is_space(b) {
            v.push(b);
            b = self.byte();
        }
        return String::from_utf8_lossy(&v).parse().unwrap();
    }
}