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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long int ll;
typedef pair<int, int> ii;
#define DEBUG freopen("in.txt", "r", stdin);

struct fastio {
    fastio() {
        ios::sync_with_stdio(false);
        cout << setprecision(10) << fixed;
        cin.tie(0);
    }
};

fastio _fast_io;

const int N = 100 + 5;
int n;
int t[N], l[N], m[N], r[N];
double ans;
int cur;

double f(int i) {
    double lm = l[i];
    double rm = r[i];
    double ll = min(l[i], r[i]), rr = m[i];
    double t1, t2;
    while (ll + 1e-6 < rr) {
        double mm = (ll + rr) / 2;
        t1 = abs(mm - l[i]);
        t2 = abs(mm - r[i]);
        if (t1 + t2 <= t[i]) {
            ll = mm;
        } else {
            rr = mm - 1e-6;
        }
    }
    // cout << lm << ' ' << m[i] << ' ' << rm << " # " << ll << endl;
    t1 = abs(ll - lm);
    t2 = abs(ll - rm);
    return (t[i] - t1 - t2) * ll + (ll + l[i]) / 2 * t1 + (ll + r[i]) / 2 * t2;
}

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> t[i];
    }
    for (int i = 0; i < n; ++i) {
        cin >> m[i];
        l[i] = r[i] = m[i];
    }
    r[n - 1] = 0;
    for (int i = n - 1; i >= 0; --i) {
        if (i < n - 1) {
            r[i] = min(r[i], l[i + 1]);
        }
        l[i] = min(l[i], r[i] + t[i]);
    }
    l[0] = 0;
    for (int i = 0; i < n; ++i) {
        if (i > 0)
            l[i] = min(l[i], r[i - 1]);
        r[i] = min(r[i], l[i] + t[i]);
    }
    for (int i = 0; i < n; ++i) {
        ans += f(i);
    }
    cout << ans << endl;
    return 0;
}