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.io.*;
import java.util.*;

class Main {

	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] t = new int[N];
		int T = 0;
		for (int i = 0; i < N; i++) {
			t[i] = sc.nextInt();
			T += t[i];
		}
		long[] v = new long[N];
		for (int i = 0; i < N; i++) {
			v[i] = sc.nextLong();
		}

		double[] maxA = new double[T * 2 + 1];
		double[] maxB = new double[T * 2 + 1];

		int passed = 0;
		for (int i = 0; i < N; i++) {
			for (int j = 0; j <= t[i]; j++) {
				int time = passed + j;
				maxA[time * 2] = Math.min(v[i],
						(time == 0) ? 0 : maxA[time * 2 - 1] + 0.5);
				if (j == 0 && i > 0) maxA[time * 2] = Math.min(maxA[time * 2], v[i - 1]);
				if (j == t[i]) continue;
				maxA[time * 2 + 1] = Math.min(v[i], maxA[time * 2] + 0.5);
			}
			passed += t[i];
		}

		passed = 0;
		for (int i = N - 1; i >= 0; i--) {
			for (int j = 0; j <= t[i]; j++) {
				int time = T - (passed + j);
				maxB[time * 2] = Math.min(v[i], (time == T) ? 0 : maxB[time * 2 + 1] + 0.5);
				if (j == 0 && i < N - 1) {
					maxB[time * 2] = Math.min(maxB[time * 2], v[i + 1]);
				}
				if (j == t[i]) continue;
				maxB[time * 2 - 1] = Math.min(v[i], maxB[time * 2] + 0.5);
			}
			passed += t[i];
		}

		double answer = 0;
		for (int i = 0; i < T * 2 + 1; i++) {
			answer += Math.min(maxA[i], maxB[i]) * 0.5;
		}

		System.out.println(answer);
	}
}