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 Ruby (2.3.3) (judge verdict: Accepted)

require 'prime'
require 'set'
require 'tsort'
include Math
ALP = ('a'..'z').to_a
INF = 0xffffffffffffffff
def max(a,b);              a > b ? a : b                              end
def min(a,b);              a < b ?  a : b                             end
def swap(a,b);             a, b = b, a                                end
def gif;                   gets.to_i                                  end
def gff;                   gets.to_f                                  end
def gsf;                   gets.chomp                                 end
def gi;                    gets.split.map(&:to_i)                     end
def gf;                    gets.split.map(&:to_f)                     end
def gs;                    gets.chomp.split.map(&:to_s)               end
def gc;                    gets.chomp.split('')                       end
def pr(num);               num.prime_division                         end
def pr?(num);              Prime.prime?(num)                          end
def digit(num);            num.to_s.length                            end
def array(s,ini=nil);      Array.new(s){ini}                          end
def darray(s1,s2,ini=nil); Array.new(s1){Array.new(s2){ini}}          end
def rep(num);              num.times{|i|yield(i)}                     end
def repl(st,en,n=1);       st.step(en,n){|i|yield(i)}                 end

N = gif
t = gi
v = gi
repl 1, N-1 do |i|
  t[i] += t[i-1]
end

t.map!{|i| i * 2 }
t << 0

x = [0] * (t[-2] + 1)

rep N do |i|
  if v[i] < x[t[i - 1]]
    x[t[i - 1]] = v[i]
    repl t[i - 1] - 1, t[i - 2], -1 do |j|
      x[j] = [x[j + 1] + 0.5, x[j]].min
    end
  end

  repl t[i - 1] + 1, t[i] do |j|
    x[j]=[x[j-1]+0.5, v[i]].min
  end
end

x[-1]=0
repl t[-2], 1, -1 do |i|
  x[i-1] = [x[i-1], x[i]+0.5].min
end

puts x.each_cons(2).map{|l, r|
  (l + r)*0.5/2.0
}.inject(:+)