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 <iomanip>
#include <vector>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <string>
#define _repargs(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _repargs(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define inf 2000000007
#define mp make_pair
#define pb push_back
typedef long long ll;
using namespace std;
template <typename T>
inline void output(T a, int p = 0) {
    if(p) cout << fixed << setprecision(p)  << a << "\n";
    else cout << a << "\n";
}
// end of template

template <typename T> inline void voutput(T &v){
    rep(i, v.size()){
        if (i) cout << " " << v[i];
        else cout << v[i];
    }
    cout << endl;
}

double calc(ll t, ll v1, ll v2, ll vm) {
    vector<ll> V(t + 1);
    V[0] = v1, V[t] = v2;
    rep(i, 1, t){
        V[i] = min({V[i - 1] + 1, vm, v2 + t - i});
    }
    double ret = 0;
    rep(i, t){
        if(V[i] == V[i + 1] && V[i] < vm){
            ret += (double)V[i] + 0.25;
        }
        else{
            ret += (double)(V[i] + V[i + 1]) / 2.0;
        }
        
    }
    return ret;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    // source code
    int N;
    cin >> N;
    vector<ll> T(N), V(N);
    rep(i, N) cin >> T[i];
    rep(i, N) cin >> V[i];
    vector<ll> TT(N);
    TT[0] = T[0];
    rep(i, 1, N) TT[i] = TT[i - 1] + T[i];
    
    vector<ll> VV(N + 1);
    VV[0] = VV[N] = 0;
    rep(i, 1, N){
        VV[i] = min(TT[i - 1], TT[N - 1] - TT[i - 1]);
        VV[i] = min(VV[i], min(V[i - 1], V[i]));
    }
    double ret = 0;
    rep(i, N){
        ret += calc(T[i], VV[i], VV[i + 1], V[i]);
    }
    
    output(ret, 10);
    
    
                    

    
    return 0;
}