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: Wrong Answer)

macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;let mut next =||{iter .next () .unwrap () } ;input_inner !{next ,$($r ) *} } ;($($r :tt ) *) =>{let stdin =std ::io ::stdin () ;let mut bytes =std ::io ::Read ::bytes (std ::io ::BufReader ::new (stdin .lock () ) ) ;let mut next =move ||->String {bytes .by_ref () .map (|r |r .unwrap () as char ) .skip_while (|c |c .is_whitespace () ) .take_while (|c |!c .is_whitespace () ) .collect () } ;input_inner !{next ,$($r ) *} } ;}
macro_rules !input_inner {($next :expr ) =>{} ;($next :expr ,) =>{} ;($next :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($next ,$t ) ;input_inner !{$next $($r ) *} } ;}
macro_rules !read_value {($next :expr ,($($t :tt ) ,*) ) =>{($(read_value !($next ,$t ) ) ,*) } ;($next :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($next ,$t ) ) .collect ::<Vec <_ >>() } ;($next :expr ,chars ) =>{read_value !($next ,String ) .chars () .collect ::<Vec <char >>() } ;($next :expr ,usize1 ) =>{read_value !($next ,usize ) -1 } ;($next :expr ,$t :ty ) =>{$next () .parse ::<$t >() .expect ("Parse error" ) } ;}

fn main() {
   input! {
      n: usize,
      t: [usize; n],
      v: [usize; n],
   }
   let mut v = v;
   v.push(0);
   for i in (0..n).rev() {
      if v[i + 1] < v[i] {
         v[i] = std::cmp::min(v[i], v[i + 1] + t[i]);
      }
   }
   let mut vn = 0;
   let mut ans = 0.0f64;
   for i in 0..n {
      if v[i] <= v[i + 1] {
         for _ in 0..t[i] {
            if vn < v[i] {
               ans += ((vn + vn + 1) as f64) / 2.0;
               vn += 1;
            } else {
               ans += vn as f64;
            }
         }
      } else {
         for tn in 0..t[i] {
            if vn == v[i+1] + t[i] - tn {
               ans += ((vn + vn - 1) as f64) / 2.0;
               vn -= 1;
            } else if vn == v[i] {
               ans += vn as f64;
            } else {
               ans += ((vn + vn + 1) as f64) / 2.0;
               vn += 1;
            }
         }
      }
   }
   println!("{}", ans);
}