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: Wrong Answer)

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

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).bigin(), (c).end())

int main()
{

    int n, tsum, t[110], v[110], l[110], r[110];
    cin >> n;
    v[0] = 0;
    tsum = 0;
    l[0] = 0;
    r[0] = 0;
    FOR(i, 1, n + 1)
    {
        cin >> t[i];
        l[i] = tsum;
        tsum += t[i];
        r[i] = tsum;
    }
    FOR(i, 1, n + 1)
    cin >> v[i];
    v[n + 1] = 0;
    l[n + 1] = tsum;
    r[n + 1] = tsum;

    REP(i, n + 2)
    cout << l[i] << " " << r[i] << endl;
    double ans;
    for (double x = 0.0; x < tsum; x += 0.5)
    {
        double minV1 = 1000000;
        double minV2 = 1000000;

        REP(j, n + 2)
        {
            double hoge;
            if (x < (double)l[j])
            {
                hoge = (double)v[j] + ((double)l[j] - x);
            }
            else if (x < (double)r[j])
            {
                hoge = (double)v[j];
            }
            else
            {
                hoge = (double)v[j] + (x - (double)l[j]);
            }

            minV1 = min(minV1, hoge);

            if (x + 0.5 < (double)l[j])
            {
                hoge = (double)v[j] + ((double)l[j] - x - 0.5);
            }
            else if (x + 0.5 < (double)r[j])
            {
                hoge = (double)v[j];
            }
            else
            {
                hoge = (double)v[j] + (x + 0.5 - (double)l[j]);
            }

            minV2 = min(minV2, hoge);
        }

        //cout << minV1 << " " << minV2 << endl;

        ans += (minV1 + minV2) * 0.25;
    }

    cout << fixed << setprecision(10) << ans << endl;

    return 0;
}