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){
    Scanner sc = new Scanner(System.in);
    
    int n = Integer.parseInt(sc.nextLine());
    
    int[] t = line2intAr(sc.nextLine(), n);
    int[] v = line2intAr(sc.nextLine(), n);
    
    int all = 0;
    for(int i=0; i<n; i++){
      all += t[i];
    }
    
    int[] vw = new int[all*2];
    int idx = 0;
    for(int i=0; i<n; i++){
      for(int j=0; j<t[i]*2; j++){
        vw[idx] = v[i]*2;
        idx++;
      }
    }
    
    final int VMAX = 101;
    
    int[][] dp = new int[all*2+1][VMAX*2];
    for(int i=0; i<dp.length; i++){
      for(int j=0; j<dp[i].length; j++){
        dp[i][j] = -(1<<30);
      }
    }
    dp[0][0] = 0;
    
    idx = 0;
    for(int k=0; k<n; k++){
      for(int i=0; i<t[k]*2; i++){
        for(int j=0; j<vw[idx]+1; j++){
          if(j+1 <= vw[idx] && dp[idx+1][j+1] < dp[idx][j]+j+1){
            dp[idx+1][j+1] = dp[idx][j]+ j+1;
          }
          if(j <= vw[idx] && dp[idx+1][j] < dp[idx][j]+j){
            dp[idx+1][j] = dp[idx][j]+ j;
          }
          if(j-1 <= vw[idx] && j-1 >= 0 && dp[idx+1][j-1] < dp[idx][j]+j-1){
            dp[idx+1][j-1] = dp[idx][j]+ j-1;
          }
        }
        idx++;
      }
    }
    
    System.out.println(dp[all*2][0]/4.0);
  }
  
  public static int[] line2intAr(String line, int n){
    String[] splited = line.split(" ");
    int[] ar = new int[n];
    for(int i=0; i<n; i++){
      ar[i] = Integer.parseInt(splited[i]);
    }
    return ar;
  }
}