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)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll MOD=(ll)1e9+7;

int main(){
    //縦横2倍する
    int N; cin>>N;
    vector<int> t(N),v(N);
    vector<int> point(N+1,0);
    for (int i = 0; i < N; ++i) {
        cin>>t[i]; t[i]*=2;
        point[i+1]=point[i]+t[i];
    }
    for (int i = 0; i < N; ++i) {
        cin>>v[i]; v[i]*=2;
    }
    vector<vector<int>> limit(N+2,vector<int> (point[N]+1,0));

    //初速0制限
    for(int i=0;i<=point[N];++i){
        limit[0][i]=i;
    }
    //最終0制限
    for(int i=0;i<=point[N];++i){
        limit[N+1][point[N]-i]=i;
    }

    for (int i = 0; i < N; ++i) {
        for (int j = 0; j <=point[N]; ++j) {
            if(j<point[i]) limit[i+1][j]=v[i]+point[i]-j;
            else if(j>point[i+1]) limit[i+1][j]=v[i]+j-point[i+1];
            else limit[i+1][j]=v[i];
        }
    }

    vector<int> speed(point[N]+1,(int)1e9);
    for (int i = 0; i <=point[N]; ++i) {
        for (int j = 0; j <=N+1; ++j) {
            speed[i]=min(speed[i],limit[j][i]);
        }

    }

    double ans=0.0;
    for(int i=0;i<point[N];++i){
        ans+=(speed[i]+speed[i+1])/2.0;
        //printf("%d : %d\n",i,speed[i]);
    }
    ans/=4;

    printf("%.5f \n",ans);

    return 0;
}