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: Wrong Answer)

using System;

namespace ABC076D
{
    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            int[] t = new int[N + 1];
            int[] v = new int[N + 1];

            string[] bufs = Console.ReadLine().Split(' ');
            for (int i = 0; i < N; i++)
            {
                t[i] = int.Parse(bufs[i]);
            }
            t[N] = 999;

            bufs = Console.ReadLine().Split(' ');
            for (int i = 0; i < N; i++)
            {
                v[i] = int.Parse(bufs[i]);
            }
            v[N] = 0;
            // END Initialize //

            // Recurrsive Speed Limit
            for (int i = N - 1; i >= 0; i--)
            {
                if (v[i] > v[i + 1] + t[i])
                {
                    v[i] = v[i + 1] + t[i];
                }
            }


            double curv = 0.0;
            int curt = 0;
            double dist = 0.0;

            for (int i = 0; i < N; i++)
            {
                double upTime = v[i] - curv;
                double downTime = v[i] - v[i + 1];
                double flatTime = t[i] - upTime - Math.Max(downTime, 0.0);

                if (flatTime < 0)
                {
                    flatTime = 0;
                    double cross = ((v[i + 1] + t[i]) - curv) / 2.0;
                    upTime = cross;
                    downTime = t[i] - cross;
                }

                if (upTime > 0)
                {
                    dist += upTime * ( 2 * curv + upTime ) / 2.0;
                    curv += upTime;
                }
                dist += flatTime * v[i];
                if (downTime > 0) {
                    dist += downTime * (curv + v[i + 1] ) / 2.0;
                    curv = v[i + 1];
                }

                curt += t[i];
            }


            Console.WriteLine(dist.ToString("F4"));
        }
    }
}