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>
#define REP(i, n) for(ll i = 0; i < (ll)n; i++)
#define FOR(i, a, b) for(ll i = (a); i < (ll)b; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1000000000000000
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;
typedef pair<ll, ll> p;
constexpr int MOD = 1000000007;
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

void print(const std::vector<int> &v) {
    std::for_each(v.begin(), v.end(), [](int x) { std::cout << x << " "; });
    std::cout << std::endl;
}

int main() {
    int N;
    cin >> N;
    vector<double> t(N), v(N);
    REP(i, N) { cin >> t[i]; }
    REP(i, N) { cin >> v[i]; }
    int T = accumulate(t.begin(), t.end(), 0);
    // maxv i t=iのときの最大
    vector<double> maxv(2 * T + 1, 10000000);
    int time = 1;
    int nowT = 0;
    REP(i, N) {
        for(int ti = 0; ti < t[i]; ti++) {
            int t1 = nowT + ti * 2;
            int t2 = nowT + ti * 2 + 1;
            maxv[t1] = min(maxv[t1], v[i]);
            maxv[t2] = min(maxv[t2], v[i]);
        }
        nowT += t[i] * 2;
        maxv[nowT] = min(maxv[nowT], v[i]);
    }
    maxv[0] = 0;
    maxv[2 * T] = 0;
    // cout << "T " << T << endl;
    for(int i = 0; i < 2 * T + 1; i++) {
        maxv[i + 1] = min(maxv[i + 1], maxv[i] + 0.5);
    }
    for(int i = 2 * T - 1; i >= 0; i--) {
        maxv[i] = min(maxv[i], maxv[i + 1] + 0.5);
    }
    double ans = 0;
    for(int i = 0; i < 2 * T; i++) {
        // cout << i << ":::" << maxv[i] << endl;
        ans += (maxv[i] + maxv[i + 1]) / 2 / 2;
    }
    printf("%.4f\n", ans);
    return 0;
}