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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include <iomanip>
#include<vector>
#include<algorithm>
#include<string>
#include <cmath>
#include <limits>
#include <utility>
#include <map>


#define rep(p,q) for(int i = p;i < q;i++)
#define ll long long

using namespace std;

const int cons = 1e9 + 7;

int n;
vector<double> v;
vector<double> l;
vector<double> r;
double speed(double t) {
	double maxV = 100;
	rep(0, n + 2) {
		double reti = 0;
		if (t <= l[i]) reti = v[i] + l[i] - t;
		else if (r[i] <= t) reti = v[i] + t - r[i];
		else reti = v[i];
		maxV = min(maxV, reti);
	}
	return maxV;
}

//D

int main(void) {
	
	cin >> n;	
	v = vector<double>(n + 2);
	l = vector<double>(n + 2);
	r = vector<double>(n + 2);
	v[0] = 0; l[0] = 0; r[0] = 0;
	v[n + 1] = 0;
	int T= 0;
	rep(1, n+1) {
		int ti = 0;
		cin >> ti;
		r[i] = r[i - 1] + ti;
		l[i] = r[i - 1];
		T += ti;
	}
	l[n] = r[n - 1];
	rep(1, n + 1) {
		cin >> v[i];
	}
	l[n + 1] = T; r[n + 1] = T;
	double ans = 0;
	for (double i = 0; i < T; i += 0.5) {
		double s1 = speed(i);
		double s2 = speed(i + 0.5);
		ans += 0.25*(s1 + s2);
	}
	cout << setprecision(20) << ans << endl;
}