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;
import java.util.Arrays;

public class Main{
  static final int INF = Integer.MAX_VALUE / 2;
  static int n;
  static int[] t, v;
  static final boolean DEBUG = true;

  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);

    n = sc.nextInt();
    t = new int[n];
    for(int i = 0; i < n; i++){
      t[i] = sc.nextInt();
    }
    v = new int[n];
    for(int i = 0; i < n; i++){
      v[i] = sc.nextInt();
    }

    int[] st = new int[n + 1];
    for(int i = 1; i <= n; i++){
      st[i] = st[i - 1] + t[i - 1];
    }

    double[] mv = new double[2 * st[n] + 1];
    for(int i = 0; i <= 2 * st[n]; i++){
      double k = 0.5 * i;
      double min = Math.min(k, (double)st[n] - k);
      for(int j = 0; j < n; j++){
        min = Math.min(min, f(j, st, k));
      }
      mv[i] = min;
    }

    double sum = 0;
    for(int i = 0; i < 2 * st[n]; i++){
      sum += 0.25 * (mv[i] + mv[i + 1]);
    }

    System.out.println(sum);
  }

  static double f(int i, int[] st, double k){
    if(k < st[i]){
      return (double)v[i] + ((double)st[i] - k);
    }
    else if(k > st[i + 1]){
      return (double)v[i] + (k - (double)st[i + 1]);
    }
    else{
      return v[i];
    }
  }
}