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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
using System.Numerics;

namespace AtCoder
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static long Mod = (long)1e9 + 7;
        static void Main(string[] args)
        {
            SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
            Solve();
            Out.Flush();
            Read();
        }
        static void Solve()
        {
            var N = Sarray()[0];
            var Ti = Sarray();
            var Vi = Sarray();
            Vi = Vi.Concat(new long[] { 0 }).ToArray();
            for (var i = N - 1; 0 <= i; --i)
                Vi[i] = Min(Vi[i], Vi[i + 1] + Ti[i]);

            double Vl = 0, Vr = 0, sum = 0;
            for (var i = 0; i < N; ++i)
            {
                double men = 0;
                var t = Ti[i];
                if (Vi[i] + 0.1 < Vl) throw new Exception();
                if (Vi[i] <= Vi[i + 1])
                {
                    Vr = Min(Vl + t, Vi[i]);
                    var Adt = Vr - Vl;
                    var Cst = t - Adt;
                    men = (Vl + Vr) * Adt / 2.0 + Cst * Vr;
                }
                else
                {
                    Vr = Min(Vi[i + 1], Vl + t);
                    double Vm = Vi[i];
                    double Adt = Vm - Vl;
                    double Dst = Vm - Vr;
                    if (t < Adt + Dst)
                    {
                        Adt = (t + Vr - Vl) / 2.0;//Vl+Adt'=Dst'+Vr,Adt'+Dst'=t
                        Dst = t - Adt;
                        Vm = Vl + Adt;
                    }
                    men = (Vl + Vm) * Adt / 2.0 + Vm * (t - Adt - Dst) + (Vm + Vr) * Dst / 2.0;
                }
                Vl = Vr;
                sum += men;
            }
            WriteLine($"{sum:f5}");
        }
    }
}