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;

#define FOR(i,a,b)  for(ll (i)=a;(i)<(b);++(i))
#define RFOR(i,a,b) for(ll (i)=a;(i)>=(b);--(i))
#define REP(i,n)    FOR(i,0,n)
#define RREP(i,n)   RFOR(i,n,0)
#define SZ(v)       (ll)(v).size()
#define ALL(v)      v.begin(), v.end()
#define UNIQ(v)     sort(ALL(v)); v.erase(unique(ALL(v)), v.end())
#define BIT(n)      (1LL<<(n))
#define PQ(T)       priority_queue<T, vector<T>, greater<T>>
#define SHOW(x)     cout << #x << " = " << x << endl

template<typename T>string join(const vector<T>& v, string d=" ")
{stringstream s;REP(i,SZ(v))s<<d<<v[i];return s.str().substr(d.length());}

template<typename T> ostream& operator<<(ostream& s, const vector<T>& v)
{if(SZ(v))s<<join(v,", ");return s;}

const ll inf = 1e15;
const ll mod = 1e9+7;
const double eps = 1e-10;

int dy[] = {0, 0, 1, -1};
int dx[] = {1, -1, 0, 0};

double dp[105][105];

int main() {
    ll N; cin >> N;
    vector<ll> T(N); REP(i, N) cin >> T[i];
    vector<ll> V(N); REP(i, N) cin >> V[i];
    
    fill((double *)dp, (double *)dp + 105 * 105, 0.0);
    dp[0][0] = 1;
    REP(i, N) REP(v, 105) if (dp[i][v] and v <= V[i]) {
        REP(vv, 105) if (vv <= V[i]) {
            ll a = V[i] - v;
            ll b = V[i] - vv;
            if (a + b <= T[i]) {
                double up = 1.0 * (v + V[i]) * a / 2;
                double stay = V[i] * (T[i] - a - b);
                double down = 1.0 * (vv + V[i]) * b / 2;
                double c = up + stay + down;
                dp[i+1][vv] = max(dp[i+1][vv], dp[i][v] + c);
            } else if (T[i] < abs(vv - v)) {
                
            } else {
                double x = 1.0 * (vv - v + T[i]) / 2;
                double y = 1.0 * (T[i] - vv + v) / 2;
                double vvv = v + x;
                
                double up = 1.0 * (v + vvv) * x / 2;
                double down = 1.0 * (vv + vvv) * y / 2;
                
                double c = up + down;
                dp[i+1][vv] = max(dp[i+1][vv], dp[i][v] + c);
            }
        }
    }
    cout << fixed << setprecision(9);
    cout << dp[N][0] - 1 << endl;;
    
    return 0;
}