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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
	public static void main(String[] args){
		Main main = new Main();
		main.solveD();
	}

	private void solveD() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] t = new int[N];
		int sum_t = 0;
		for (int i = 0; i < N; i++) {
			t[i] = sc.nextInt() * 2;
			sum_t += t[i];
		}
		int[] v = new int[N];
		for (int i = 0; i < N; i++) {
			v[i] = sc.nextInt() * 2;
		}
		int tmp_v = 0;
		int tmp_t = 0;
		int[] max_v = new int[sum_t + 1];
		for (int i = 0; i < N; i++) {
			max_v[tmp_t] = v[i] < tmp_v ? v[i] : tmp_v;
			tmp_v = max_v[tmp_t];
			for (int t_ind = 0; t_ind < t[i]; t_ind++) {
				max_v[++tmp_t] = v[i] < tmp_v + 1 ? v[i] : tmp_v + 1;
				tmp_v = max_v[tmp_t];
			}
		}
		tmp_v = 0;
		for (int i = N - 1; i >= 0; i--) {
			max_v[tmp_t] = v[i] < tmp_v ? v[i] : tmp_v;
			tmp_v = max_v[tmp_t];
			for (int t_ind = 0; t_ind < t[i]; t_ind++) {
				int back_v = v[i] < tmp_v + 1 ? v[i] : tmp_v + 1;
				--tmp_t;
				max_v[tmp_t] = max_v[tmp_t] < back_v ? max_v[tmp_t] : back_v;
				tmp_v = max_v[tmp_t];
			}
		}
		double answer = 0;
		for (int i = 0; i < sum_t; i++) {
			answer += (max_v[i] + max_v[i + 1]);
		}
		System.out.println(answer * 0.125);
	}
}