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>
using namespace std;

#define fs first
#define sc second
#define pb emplace_back
#define mp make_pair
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

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

  vector<int> t(n + 1, 0), v(n + 1, 0);
  for (int i = 1; i <= n; ++i) {
    cin >> t[i];
    t[i] += t[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
  }

  double ans = 0;
  double s = 0.0;
  for (int i = 1; i <= n; ++i) {
    if (v[i] > v[i + 1]) {
      double d = v[i] - s;
      double b = v[i] - v[i + 1];

      if (t[i - 1] + d > t[i] - b) {
        double m = (t[i - 1] + d + t[i] - b) / 2.0;
        ans += (m - t[i - 1]) * s + (m - t[i - 1]) * (m - t[i - 1]) / 2.0;
        ans += (t[i] - m) * (t[i] - m) / 2.0 + (t[i] - m) * v[i + 1];
        s = v[i + 1];
      } else {
        ans += s * d + d * d / 2.0 + (t[i] - b - (t[i - 1] + d)) * v[i] + v[i + 1] * b + b * b / 2.0;
        s = v[i + 1];
      }
    } else {
      double b = t[i] - t[i - 1];
      double d = v[i] - s;
      if (d > b) {
        ans += b * s + b * b / 2.0;
        s += b;
      } else {
        ans += s * d + d * d / 2.0 + (b - d) * v[i];
        s = v[i];
      }
    }
  }

  printf("%.10lf\n", ans);
}