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)

// #pragma GCC optimize("unroll-loops", "omit-frame-pointer", "inline")
// #pragma GCC option("arch=native", "tune=native", "no-zero-upper")
// #pragma GCC
// target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("tree-vectorize","openmp","predictive-commoning")
// #pragma GCC option("D_GLIBCXX_PARALLEL","openmp")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<ll> vll;
// #define int long long
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 2000000000               // 2e9
#define LLINF 2000000000000000000ll  // 2e18 (llmax:9e18)
#define fi first
#define sec second
#define all(x) (x).begin(), (x).end()
#define sq(x) ((x) * (x))
#define dmp(x) cerr << #x << ": " << x << endl;

template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}

template <class T>
using MaxHeap = priority_queue<T>;
template <class T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <class T>
vector<T> vect(int len, T elem) {
  return vector<T>(len, elem);
}

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << p.fi << ',' << p.sec;
  return os;
}
template <class T, class U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.fi >> p.sec;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i];
    if (i + 1 < vec.size()) os << ' ';
  }
  return os;
}
template <class T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) is >> vec[i];
  return is;
}
void fastio() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
}

#define endl "\n"

void solve() {
  int N;
  cin >> N;
  vector<int> t(N), v(N);
  cin >> t;
  cin >> v;
  auto dp = vect(N + 1, vect(101, -1.0));
  dp[0][0] = 0.0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= v[i]; j++) {
      if (dp[i][j] < 0.0) continue;
      for (int k = max(0, j - t[i]); k <= min(v[i], j + t[i]); k++) {
        if (j <= k) {
          double add = 0.0;
          double d = abs(j - k);
          double rem = (double)t[i] - d;
          double h = (double)(v[i] - k);
          add += (double)(j + k) * d / 2.0 + (double)k * rem;
          if (rem <= 2.0 * h) {
            add += rem * rem / 4.0;
          } else {
            add += h * h + (rem - h * 2.0) * h;
          }
          chmax(dp[i + 1][k], dp[i][j] + add);
        } else {
          double add = 0.0;
          double d = abs(j - k);
          double rem = (double)t[i] - d;
          double h = (double)(v[i] - j);
          add += (double)(j + k) * d / 2.0 + (double)j * rem;
          if (rem <= 2.0 * h) {
            add += rem * rem / 4.0;
          } else {
            add += h * h + (rem - h * 2.0) * h;
          }
          chmax(dp[i + 1][k], dp[i][j] + add);
        }
      }
    }
  }
  cout << dp[N][0] << endl;
  return;
}

signed main() {
  fastio();
  solve();
  // int t; cin >> t; while(t--)solve();

  // int t; cin >> t;
  // for(int i=1;i<=t;i++){
  //   cout << "Case #" << i << ": ";
  //   solve();
  // }
  return 0;
}