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 <bits/stdc++.h>

#define REP(i, n) for(int (i)=0;(i)<(n);++(i))
#define REPV(iter, v) for(auto (iter)=(v).begin(); (iter)!=(v).end();++(iter))
#define ALL(v) (v).begin(),(v).end()

#define MOD 1000000007

using namespace std;

typedef long long ll;

int main()
{
    ll N;
    cin >> N;
    vector<double> t(N), v(N);
    REP(i, N) cin >> t[i];
    REP(i, N) cin >> v[i];

    vector<double> tA(N), tE(N), tD(N), vStart(N);
    vStart[0] = 0;
    REP(i, N)
    {
        double vEndMax;
        if (i == N - 1) vEndMax = 0;
        else if (i == N - 2) vEndMax = min(v[i + 1], t[i + 1]);
        else vEndMax = min(v[i + 1], v[i + 2] + t[i + 1]);
        double vMax;
        for (int j = 0; 0.5 * j <= t[i]; ++j)
        {
            double ta = 0.5 * j;   
            double vT = vStart[i] + ta;
            if (vT < v[i] && vT < vEndMax + (t[i] - ta))
            {
                continue;
            }
            else
            {
                vMax = vT;
                break;
            }
        }
        tA[i] = vMax > vStart[i] ? (vMax - vStart[i]) : 0;
        tD[i] = vMax > vEndMax ? (vMax - vEndMax) : 0;
        tE[i] = t[i] - tA[i] - tD[i];
        if (i + 1 < N) vStart[i + 1] = min(vMax, vEndMax);
    }
    double ans = 0;
    REP(i, N)
    {
        ans += (vStart[i] + vStart[i] + tA[i]) * tA[i] / 2.0;
        ans += (vStart[i] + tA[i]) * tE[i];
        ans += (vStart[i] + tA[i] + vStart[i] + tA[i] - tD[i]) * tD[i] / 2.0;
    }

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