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)

#![allow(unused_imports)]
use proconio::{
   fastout, input,
   marker::{Bytes, Isize1, Usize1},
};

const INF: usize = std::usize::MAX;
macro_rules !min {($a :expr $(,) *) =>{{$a } } ;($a :expr ,$b :expr $(,) *) =>{{std ::cmp ::min ($a ,$b ) } } ;($a :expr ,$($rest :expr ) ,+$(,) *) =>{{std ::cmp ::min ($a ,min !($($rest ) ,+) ) } } ;}

#[fastout]
fn main() {
   input! {
      n: usize,
      mut t: [usize; n],
      mut v: [usize; n]
   }
   for i in 0..n {
      t[i] *= 2;
      v[i] *= 2;
   }
   for i in 0..n - 1 {
      t[i + 1] += t[i];
   }

   let mut l = vec![INF; t[n - 1] + 1];
   for i in 0..n - 1 {
      l[t[i]] = i;
   }
   l[0] = 0;
   for tn in 1..=t[n - 1] {
      if l[tn] != INF {
         l[tn] = std::cmp::min(l[tn - 1] + 1, v[l[tn]]);
      } else {
         l[tn] = l[tn - 1] + 1;
      }
   }

   let mut r = vec![INF; t[n - 1] + 1];
   for i in 1..n {
      r[t[i - 1]] = i;
   }
   r[t[n - 1]] = 0;
   for tn in (0..t[n - 1]).rev() {
      if r[tn] != INF {
         r[tn] = std::cmp::min(r[tn + 1] + 1, v[r[tn]]);
      } else {
         r[tn] = r[tn + 1] + 1;
      }
   }

   let mut m = vec![INF; t[n - 1] + 1];
   let mut tr = 0;
   for i in 0..n {
      for tn in tr..=t[i] {
         m[tn] = v[i];
      }
      tr = t[i];
   }

   let mut ans = 0;
   for tn in 0..=t[n - 1] {
      ans += 2 * min!{l[tn], r[tn], m[tn]};
   }
   println!("{}", (ans as f64) / 8.0);
}