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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.io.PrintWriter

fun main(args: Array<String>) {
    PrintWriter(System.out).use { out ->
        solve(out)
        out.flush()
    }
}

fun solve(out: PrintWriter) {
    val n = readInteger()
    val tList = readIntegerList()
    val vList = readDoubleList()

    val times = tList.sum() * 2 + 1
    val maxVs = DoubleArray(times) { Double.MAX_VALUE }
    maxVs[0] = 0.0
    maxVs[times - 1] = 0.0
    var curT = 0
    for ((i, t) in tList.withIndex()) {
        for (j in 0..(2 * t)) {
            if (curT + j in 0 until times) {
                maxVs[curT + j] = Math.min(maxVs[curT + j], vList[i]) // 境界値難しい...
            }
        }
        curT += 2 * t
    }

    for (i in 0 until (times - 1)) {
        maxVs[i + 1] = Math.min(maxVs[i + 1], maxVs[i] + 0.5)
    }

    for (i in (times - 1) downTo 1) {
        maxVs[i - 1] = Math.min(maxVs[i - 1], maxVs[i] + 0.5)
    }

    var ans = 0.0
    var preV = maxVs.first()
    for (v in maxVs.drop(1)) {
        ans += preV + v
        preV = v
    }
    println(ans / 4)
}

fun readInteger() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readStringList() = readLine()!!.split(" ")
fun readIntegerList() = readStringList().map(String::toInt)
fun readLongList() = readStringList().map(String::toLong)
fun readDoubleList() = readStringList().map(String::toDouble)