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 JavaScript (node.js v5.12) (judge verdict: Accepted)

process.stdin.resume();
process.stdin.setEncoding('utf8');

var g_input = '';
process.stdin.on('data', function(chunk) {
    g_input += chunk;
});

process.stdin.on('end', function() {
    Main(g_input.split(/\r?\n/));
});

function Main(input) {

    var n;
    var t;
    var v;
    var i;
    var j;
    var l;
    var r;
    var x;
    var minV;
    var distance;

    n = parseInt(input[0], 10);
    tvals = input[1].split(' ');
    vvals = input[2].split(' ');

    v = new Array(n + 2);
    l = new Array(n + 2);
    r = new Array(n + 2);

    v[0] = 0;
    l[0] = 0;
    r[0] = 0;

    for (i = 0; i < n; i++) {
        t = parseInt(tvals[i], 10);
        v[i + 1] = parseInt(vvals[i], 10);

        l[i + 1] = r[i];
        r[i + 1] = l[i + 1] + t;
    }

    v[n + 1] = 0;
    l[n + 1] = r[n];
    r[n + 1] = r[n];

    minV = new Array(r[n + 1] * 2 + 1);
    for (i = 0; i < r[n + 1] * 2 + 1; i++) {
        minV[i] = 100;
        x = i * 0.5;
        for (j = 0; j < n + 2; j++) {
            minV[i] = Math.min(minV[i], f(x, l[j], r[j], v[j]));
        }
    }

    distance = 0;
    for (i = 0; i < r[n + 1] * 2; i++) {
        distance += (minV[i] + minV[i + 1]) * 0.25;
    }

    console.log(distance);

    return 0;
}

function f(x, l, r, v) {

    if (x < l) {
        return -x + v + l;
    }
    else if (x > r) {
        return x + v - r;
    }
    else {
        return v;
    }
}