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"
#define rep(i,n) for(int i = 0; i < n; i++)
typedef long long ll;
using namespace std;
#define llMAX numeric_limits<long long>::max()
#define intMAX numeric_limits<int>::max()
#define d_5 100000
#define d9_7 1000000007
#define vll vector<vector<long long>>
#define vl vector<long long>
#define vi vector<int>
#define vii vector<vector<int>>
#define pb push_back
#define pf push_front
int main(void){
  int n;
  cin>>n;
  long double sumt=0;
  vector<long double> t(n),v(n),accumt(n);
  rep(i,n){
    cin>>t[i];
    sumt+=t[i];
  }
  rep(i,n){
    cin>>v[i];
  }
  accumt[0]=0;
  rep(i,n-1){
    accumt[i+1]=accumt[i]+t[i];//以前まで何秒立ったか
  }
  vi posnum(2*(int)(sumt)+1,-1);
  int now=0;
  rep(i,n){
    rep(j,t[i]*2){
      posnum[now]=i;
      now++;
    }
  }
  posnum[2*sumt]=n-1;
  posnum[2*sumt-1]=n-1;
  /*cout<<"posnum ";
  rep(i,2*(int)(sumt)+1){
    cout<<posnum[i]<<" ";
  }
  cout<<endl;*/
  vector<long double> points(2*(int)(sumt)+1,99999);
  rep(i,2*sumt+1){
    long double tmp=min(((long double)(i))*0.5,(long double)(sumt)-((long double)(i))*0.5);
    rep(j,n){
      if(posnum[i]==j){
        tmp=min(v[j],tmp);
        //cout<<" == "<<i<<" p "<<posnum[i];
      }else{
        if(((long double)(i))*0.5<=accumt[j]){
          tmp=min(tmp,v[j]+(accumt[j]-((long double)(i))*0.5));
        }
        if(((long double)(i))*0.5>=accumt[j]+t[j]){
          tmp=min(tmp,v[j]+(((long double)(i))*0.5-(accumt[j]+t[j])));
        }
      }
    }
    points[i]=tmp;
  }
  long double ans=0;
  rep(i,2*sumt){
    ans+=(points[i]+points[i+1])*0.25;
  }
  /*rep(i,2*(int)(sumt)+1){
    cout<<points[i]<<" ";
  }
  cout<<endl;*/
  cout<<ans<<endl;
  return 0;
}