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.Linq;
using static System.Math;
using static System.Console;
class Z { static void Main() => new K(); }
class K
{
	int[] G() => ReadLine().Split().Select(int.Parse).ToArray();
	public K()
	{
		var N = int.Parse(ReadLine());
		var t = G();
		var v = G();
		var V = v.Max();
		// dp[n, v] = t[0]+...+t[n-1] を走り終えたときに速度が v/2 である場合の、そこまでの最大の走行距離 * 16
		// その速度に到達できなければ dp[n, v] = -1 とする
		// 答えは dp[N, 0] / 16
		var dp = new int[N + 1, 2 * V + 1];
		dp[0, 0] = 0;
		// dp[n, v] >= 0 となる v の最大値を M[n] とする
		var M = new int[N + 1];
		for (var i = 1; i <= 2 * V; i++) dp[0, i] = -1;
		for (var n = 0; n < N; n++)
		{
			M[n + 1] = Min(M[n] + 2 * t[n], 2 * v[n]);
			for (var w = M[n + 1] + 1; w <= 2 * V; w++) dp[n + 1, w] = -1;
			for (var w = 0; w <= M[n + 1]; w++)
			{
				for (var u = 0; u <= M[n]; u++)
				{
					// dp[n, u] => dp[n+1, w]
					if (u > 2 * v[n] || u - 2 * t[n] > w || u + 2 * t[n] < w) continue;
					int au = 2 * v[n] - u,  bw = 2 * v[n] - w, run;
					// 時間に余裕があれば、加速して速度キープして減速
					if (au + bw <= 2 * t[n]) run = 2 * (8 * t[n] * v[n] - au * au - bw * bw);
					// 加速して減速、最大速度は x/4 (< v[n]) までしか到達できない
					else
					{
						var x = 2 * t[n] + u + w;
						run = x * x - 2 * (u * u + w * w);
					}
					dp[n + 1, w] = Max(dp[n + 1, w], dp[n, u] + run);
				}
			}
		}
		WriteLine(dp[N, 0] / 16m);
	}
}