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: Wrong Answer)

import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int[] t = new int[n];
		int[] v = new int[n];
		int sumT = 0;
		for(int i = 0; i < n; i++){
		    t[i] = sc.nextInt();
		    sumT += t[i];
		}
		for(int i = 0; i < n; i++){
		    v[i] = sc.nextInt();
		}
		
		// 計算
		double result = 0;
		int[] minV = new int[sumT+1];
		int index = 0;
		for(int i = 0; i < n; i++){
		    for(int j = 0; j < t[i]; j++){
		        index++;
		        if(index == sumT) continue;
		        if(j < t[i]-1){
		            minV[index] = v[i];
		        }else{
		            minV[index] = min(v[i], v[i+1]);
		        }
		    }
		}
		int[] minA = new int[sumT+1];
		for(int i = 0; i < (sumT+2)/2; i++){
		    minA[i] = i;
		    minA[sumT-i] = i;
		}
		int[] r = new int[sumT+1];
		for(int i = 1; i < sumT+1; i++){
		    r[i] = min(r[i-1]+1, min(minV[i], minA[i]));
		}
		int[] l = new int[sumT+1];
		for(int i = 1; i < sumT+1; i++){
		    l[sumT-i] = min(l[sumT-i+1]+1, min(minV[sumT-i], minA[sumT-i]));
		}
		int[] res = new int[sumT+1];
		for(int i = 0; i < sumT+1; i++){
		    res[i] = min(r[i], l[i]);
		}
		
		for(int i = 1; i < sumT+1; i++){
		    result += (double)abs(res[i]-res[i-1])/2 + min(res[i], res[i-1]);
		}
		for(int i = 1; i < sumT+1; i++){
		    if(res[i] == res[i-1] && res[i] < minV[i] && res[i-1] < minV[i-1]) result += 0.25;
		}
		//System.out.println(Arrays.toString(minV));
		//System.out.println(Arrays.toString(minA));
		//System.out.println(Arrays.toString(r));
		//System.out.println(Arrays.toString(l));
		//System.out.println(Arrays.toString(res));
		
		// 出力
		System.out.println(result);
	}
}