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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;
class Main{
 
public static void  main(String args[]){
int n;
int t[]= new int[102];
int v[] = new int[101];
int a[] = new int[20100];
int b[] = new int[20100];
int s = 0;
double ans = 0;
 
Scanner in = new Scanner (System.in);	
n = Integer.parseInt(in.next());
	for(int i=1; i<=n; ++i){
		t[i]=Integer.parseInt(in.next());
		s += t[i];
		t[i] += t[i-1];
	}
	++t[n+1];
	t[n+1] += t[n];
//	for(int i=0; i<=n+1; ++i) printf("%d\t", t[i]);
//	printf("\n");
	for(int i=1; i<=n; ++i){
		v[i]=Integer.parseInt(in.next());
		for(int j=t[i-1]; j<t[i]; ++j) b[j] = v[i];
	}
	for(int i=0; i<=s; ++i) a[i] = Math.min(i, s-i);
//	for(int i=0; i<=s; ++i) printf("%d\t", a[i]);
//	printf("\n");
	for(int i=1; i<=n; ++i){
		for(int j=t[i-1]; j<=t[i]; ++j) a[j] = Math.min(a[j], v[i]);
		for(int j=t[i-1]; j>=0; --j) a[j] = Math.min(a[j], v[i]+t[i-1]-j);
		for(int j=t[i]; j<=s; ++j) a[j] = Math.min(a[j], v[i]+j-t[i]);
	}
//	for(int i=0; i<=s; ++i) printf("%d\t", a[i]);
//	printf("\n");
	for(int i=1; i<s; ++i) ans += (float)a[i];
	for(int i=1; i+3<s; ++i){
		if(a[i]+1==a[i+1] && a[i+1]==a[i+2] && a[i+2]-1==a[i+3] && b[i+1]>a[i+1] && b[i+2]>a[i+2])
			ans += 0.25;
	}
	System.out.println(ans);
//	for(int i=1; i<s; ++i) printf("%d %d\n", a[i], b[i]);
}}