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: Accepted)

#include <bits/stdc++.h>
#include <cmath>


#include <assert.h>    // LON
#include <math.h>    // sqrt()


using namespace std;
#define rep(i, n) for (ll i = 0; i < (n); ++i)
//#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) ll(x.size())
typedef long long ll;
//typedef pair<int, int> P;
//typedef pair<ll, ll> P;
//const double INF = 1e10;
//const ll INF = 10e15;
const ll MINF = -10e10;
const int INF = INT_MAX;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)


//ifstream myfile("C:\\Users\\riku\\Downloads\\0_00.txt");
//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);

//typedef priority_queue<P, vector<P>, greater<P>> PQ_ASK;
const int mod = 1000000007;

int main() {
    ll n;
    cin >> n;

    vector<ll> over_lines;

    {
        vector<ll> ts(n);
        vector<ll> vs(n);
        rep(i, n) {
            cin >> ts[i];
        }
        rep(i, n) {
            cin >> vs[i];
        }

        ll ts_sum = 0;
        for (ll i : ts) {
            ts_sum += i;
        }
        over_lines.resize(ts_sum + 2, 0);

        ll index = 1;
        for (ll i = 0; i < n; i++) {
            for (ll j = 0; j < ts[i]; j++) {
                over_lines[index] = vs[i];
                index++;
            }
        }

    }
    for (ll i = 1; i < over_lines.size(); i++) {
        over_lines[i] = min(over_lines[i], over_lines[i - 1] + 1);
    }

    for (ll i = over_lines.size() - 2; i >= 0; i--) {
        over_lines[i] = min(over_lines[i], over_lines[i + 1] + 1);
    }

    ll ans = 0;

    for (ll i = 1; i < over_lines.size() - 1; i++) {
        ll a = over_lines[i - 1];
        ll b = over_lines[i];
        ll c = over_lines[i + 1];
        assert(b > 0);
        assert(abs(a - b) <= 1);
        assert(abs(b - c) <= 1);
        if (a == b && b == c) {
            ans += b * 100ll;
        } else if (a > b && b < c) {
            ans += b * 100ll;
        } else if ((a > b && b == c) || (a == b && b < c)) {
            ans += b * 100ll;
        } else if ((a == b && b > c) || (a < b && b == c)) {
            ans += ((b - 1) * 100ll) + 50;
        } else if (a < b && b > c) {
            ans += ((b - 1) * 100ll) + 25;
        } else if ((a < b && b < c) || (a > b && b > c)) {
            ans += ((b - 1) * 100ll) + 50;
        } else {
            assert(false);
        }
    }

    cout << setprecision(10) << ((long double) ans) / 100 << endl;
}