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 <iostream>
#include <vector>
#include <string.h>
#include <stack>
#include <queue>
#include <algorithm>
#include <climits>
#include <cmath>
#include <map>
#include <set>
#include <assert.h>
#include <stdio.h>
#define REP(i,n) for(ll i=0;i<(n);i++)
#define MOD 1000000007
#define int long long
#ifdef int
const long long INF = LLONG_MAX / 10;
#else
const int INF = 1010101010;
#endif
using namespace std;
typedef long long ll;
typedef vector<vector<ll> > mat;
typedef pair<int, int> P;
//typedef pair<double, double> P;
const int dx[9] = {-1, 0, 0, 1, -1, -1, 1, 1, 0};
const int dy[9] = {0, -1, 1, 0, -1, 1, -1, 1, 0};
 
int N;
int t[110], v[110];
double ul[40404]; //upper limit(0.5ずつ)

signed main()
{
  cin >> N;
  REP(i,N) cin >> t[i];
  REP(i,N) cin >> v[i];

  REP(i,40404) ul[i] = 99999.9;

  ul[0] = 0.0;
  int ulidx = 0;
  REP(i,N) {
    REP(j,2*t[i]+1) {
      ul[ulidx+j] = min(ul[ulidx+j], 1.0 * v[i]);
    }
    ulidx += t[i] * 2;
  }
  ul[ulidx++] = 0.0;

  cout << ulidx << endl;

  //左から
  double nowspeed = 0.0;
  REP(i,ulidx) {
    ul[i] = min(ul[i], nowspeed);
    nowspeed = ul[i];
    nowspeed += 0.5;
  }

  //右から
  nowspeed = 0.0;
  for (int i=ulidx-1; i>=0; i--) {
    ul[i] = min(ul[i], nowspeed);
    nowspeed = ul[i];
    nowspeed += 0.5;
  }

  // REP(i,30) cout << ul[i] << " ";

  double ans = 0.0;
  REP(i,ulidx-1) {
    ans += 0.5 * min(ul[i], ul[i+1]);
    if (abs(ul[i] - ul[i+1]) > 0.01) {
      ans += 0.5 * 0.5 / 2.0;
    }
  }

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