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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
using System.Linq;
using static MyIO;


public class D
{
	public static void Main()
	{
		int N = GetInt();
		double[] l = new double[N+2];	// l[0] = 0
		double[] r = new double[N+2];	// r[0] = 0;
		double T = 0.0;
		for(int i = 1; i <= N; i++)
		{
			l[i] = T;
			T += GetDouble();
			r[i] = T;
		}
		l[N + 1] = T;
		r[N + 1] = T;

		double[] v = new double[N+2];	// v[0] = v[N+1] = 0
		for(int i = 1; i <= N; i++)
			v[i] = GetDouble();

		double sum = 0;		
		double prevV = 0.0;
		for(double x = 0.5; x <= T; x += 0.5)
		{
			double currentV = 1000;
			for(int i = 0; i < N + 2; i++)
			{
				double tmpV;
				if(x < l[i])
					tmpV = v[i] + (l[i] - x);
				else if(x <= r[i])
					tmpV = v[i];
				else
					tmpV = v[i] + (x - r[i]);
				currentV = Math.Min(currentV, tmpV);
			}
			sum += 0.25 * (prevV + currentV);
			prevV = currentV;
		}

		Console.WriteLine($"{sum:F6}");
	}
}


public static class MyIO
{
	private static string[] args = null;
	private static int num = -1;
	private static int used = -1;

	private static string getArg()
	{
		if(used == num)
		{
			args = Console.ReadLine().Split(' ');
			num = args.Length;
			used = 0;
		}
		return args[used++];
	}

	public static int GetInt(){ return int.Parse(getArg()); }
	public static long GetLong(){ return long.Parse(getArg()); }
	public static double GetDouble(){ return double.Parse(getArg()); }
	public static string GetString(){ return getArg(); }
}