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;

#ifdef LOCAL
#include "../../dump.hpp"
#else
#define dump(...)
#endif

#define rep(i,n) for(int i=0,i##_cond=(n);i<i##_cond;i++)
#define FOR(i,a,b) for(int i=(a),i##_cond=(b);i<i##_cond;i++)
#define ROF(i,a,b) for(int i=(a)-1,i##_cond=(b);i>=i##_cond;i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend() //sortで大きい順
#define UNIQUE(v) v.erase(unique(all(v)),v.end())
#define SUM(a) accumulate(all(a),0)
#define sz(x) ((int)(x).size())
#define pb push_back
#define fst first
#define snd second
#define mp make_pair

typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<ll,ll> pii;

const ll inf = 1ll<<62;
const ll mod = 1e9+7;

#define int ll

struct func{
  int l,r,v;
};

main(){
  int n; cin >> n;
  vector<struct func> f(n+2);
  int t = 0;
  vi a(n),b(n);
  rep(i,n) cin >> a[i];
  rep(i,n) cin >> b[i];
  rep(i,n){
    f[i].l = t;
    t += 2 * a[i];
    f[i].r = t;
    f[i].v= 2 * b[i];
  }
  f[n].l = f[n].r = f[n].v = f[n+1].v = 0;
  f[n+1].l = f[n+1].r = t;
  //rep(i,n+2) cout << f[i].l << " " << f[i].r << " " << f[i].v << endl;

  int before = 0, now;
  int ans = 0;
  FOR(i,1,t+1){
    now = inf;
    rep(j,n+2){
      if(i < f[j].l){
	now = min(now, f[j].v + (f[j].l - i));
      }else if(f[j].r < i){
	now = min(now, f[j].v + (i - f[j].r));
      }else{
	now = min(now, f[j].v);
      }
    }
    //cout << i << " " << before << " " << now << endl;
    assert(abs(now - before) <= 1);
    ans += 4 * min(before, now);
    if(before != now) ans += 2;
    before = now;
  }
  double x = ans;
  x /= 16;
  cout << x << endl;
}