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;

#define fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define repr(i,a,b) for(int i=(int)(a-1);i>=b;i--)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout << #x" = " << x << endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<double, double> P;

#define INF INT_MAX/3

int main(){
    cin.sync_with_stdio(false);
    int n;
    cin >> n;
    double t[n], v[n + 2];
    rep(i, n) cin >> t[i];
    repl(i, 1, n + 1) cin >> v[i];
    v[0] = v[n + 1] = 0;
    repr(i, n + 1, 1) minch(v[i], v[i + 1] + t[i - 1]);
    repl(i, 1, n + 1) minch(v[i], v[i - 1] + t[i - 1]);
    double len = 0, vv = 0, tt;
    repl(i, 1, n + 1) {
        if (v[i] <= v[i + 1]) {
            if (v[i] - vv < t[i - 1]) {
                tt = v[i] - vv;
                len += ((vv * tt + 0.5 * tt * tt) + v[i] * (t[i - 1] - tt));
                vv = v[i];
            } else {
                tt = t[i - 1];
                len += (vv * tt + 0.5 * tt * tt);
                vv += tt;
            }
        } else {
            if ((v[i] - vv) + (v[i] - v[i + 1]) > t[i - 1]) {
                tt = (v[i + 1] - vv + t[i - 1]) / 2.0;
                double vvv = vv + tt, ttt = t[i - 1] - tt;
                len += ((vv * tt + 0.5 * tt * tt) + (vvv * ttt - 0.5 * ttt * ttt));
            } else {
                tt = v[i] - vv;
                double vvv = v[i], ttt = v[i] - v[i + 1];
                len += ((vv * tt + 0.5 * tt * tt) + (vvv * ttt - 0.5 * ttt * ttt) + v[i] * (t[i - 1] - tt - ttt));
            }
            vv = v[i + 1];
        }
    }
    printf("%.8f\n", len);
    return 0;
}