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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

//package com.company;

import java.io.*;
import java.util.*;


public class Main {
    static int N;
    static int totalt;
    static int[] t;
    static int[] v;
    static int[] maxv;
    public static void main(String[] args) throws IOException {
        MyScanner sc = new MyScanner(System.in);
        int n = sc.nextInt();
        t = new int[n];
        v = new int[n];
        for (int i = 0; i < n; i++) {
            t[i] = sc.nextInt();
            totalt += t[i];
        }
        for (int i = 0; i < n; i++) {
            v[i] = sc.nextInt();
        }
        if (n == 1 && t[0] == 1 && v[0] > 0){
            System.out.println(0.25);
            return;
        }
        maxv = new int[totalt + 1];
        int[] ctr = new int[totalt + 1];
        Arrays.fill(maxv, -1);
        maxv[0] = 0;
        maxv[totalt] = 0;
        int tidx = n - 1;
        int mov = 0;
        ctr[totalt] = v[tidx];
        for (int i = totalt - 1; i > 0; i--) {
            mov++;
            maxv[i] = Math.min(maxv[i + 1] + 1, v[tidx]);
            ctr[i] = v[tidx];
            if (mov == t[tidx]){
                mov = 0;
                tidx--;
            }
        }
        long size = 0;
        long btt = 0;
        int prevHei = 0;
        int ttt = 0;
        int curHei = 0;
        int perform = 0; //0--up, 1---straight, 2---down
        for (int i = 1; i < totalt + 1; i++) {
            int mv = maxv[i];
            switch (perform){
                case 0:
                    if (mv >= curHei + 1){
                        curHei++;
                        ttt++;
                    } else {
                        size += (curHei + prevHei) * ttt;
                        prevHei = curHei;
                        ttt = 0;
                        if (mv == curHei){
                            if (maxv[i + 1] == curHei - 1 && ctr[i + 1] > curHei){
                                btt += 1;
                            }
                            perform = 1;
                        } else {
                            perform = 2;
                        }
                        i--;
                    }
                    break;
                case 1:
                    if (mv == curHei){
                        ttt++;
                    } else {
                        size += curHei * ttt * 2;
                        ttt = 0;
                        prevHei = curHei;
                        if (mv > curHei && maxv[i + 1] > curHei){
                            perform = 0;
                            size += curHei * 2;
                            i++;
                        } else {
                            if (mv > curHei){
                                size += curHei;
                                i++;
                                btt++;
                            }
                            perform = 2;
                        }
                        i--;
                    }
                    break;
                case 2:
                    if (mv == curHei - 1){
                        curHei--;
                        ttt++;
                    } else {
                        size += (prevHei + curHei) * ttt;
                        prevHei = curHei;
                        ttt = 0;
                        if (mv == curHei){
                            perform = 1;
                        } else {
                            perform = 0;
                        }
                        i--;
                    }
                    break;
            }
        }
        size += (prevHei + curHei) * ttt;
        System.out.println((size+0.0)/ 2 + (btt+0.0)/4);
    }

    static class MyScanner {
        StringTokenizer st;
        BufferedReader br;

        public MyScanner(InputStream s){  br = new BufferedReader(new InputStreamReader(s));}

        public MyScanner(FileReader s) throws FileNotFoundException {br = new BufferedReader(s);}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {return Integer.parseInt(next());}

        public long nextLong() throws IOException {return Long.parseLong(next());}

        public String nextLine() throws IOException {return br.readLine();}

        public double nextDouble() throws IOException { return Double.parseDouble(next()); }

        public boolean ready() throws IOException {return br.ready();}
    }
}