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 <fstream>
#include <string> 
#include <cmath>  
#include <cstdlib>
#include <ctime>
#include <vector>
#include <algorithm>
#include <numeric>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <functional>

using namespace std;
using ll = long long;
using ld = long double;

#define FOR(i, m, n) for(int i = (m);i < (n);i++)
#define REP(i,n) for(int i = 0; i < int(n); i++)
#define VI vector<int>
#define VVI vector<vector<int>>
#define VVVI vector<vector<vector<int>>>
#define VL vector<ll>
#define VB vector<bool>
#define VVB vector<vector<bool>>
#define PAIR pair<int,int>
#define MP make_pair
#define VP vector<pair<int,int>>
#define VS vector<string>
#define QUE queue<int>
#define DEQ deque<int>
#define PQUE priority_queue<int> //5,5,4,3,3,2,...
#define REPQUE priority_queue<int, vector<int>, greater<int>> //1,1,2,3,4,4,5,...
#define SUM(obj) accumulate((obj).begin(), (obj).end(), 0)
#define SORT(obj) ((obj).begin(), (obj).end()) // 1,2,3,4,5...
#define RESORT(obj) ((obj).begin(), (obj).end(), greater<int>()) // 5,4,3,2,1...
#define UB(obj,n) upper_bound((obj).begin(), (obj).end(), n) //itr > n
#define LB(obj,n) lower_bound((obj).begin(), (obj).end(), n) //itr>= n

const ll MOD = (ll)1e9 + 7;

int main() {
	int N;
	cin >> N;
	VI ti(N+1), vi(N+1);
	REP(i, N) cin >> ti[i];
	REP(i, N) cin >> vi[i];
	ti[N] = 0;
	vi[N] = 0;

	int t = 0, v = 0, v_ = 0;
	ld x = 0;
	REP(i, N) {
		REP(j, ti[i]) {
			v_ = v;
			int dv = vi[i + 1] - v_;
			int dt = ti[i] - j;
			t++;
			if (dv < 0) {
				if (dt <= -dv) v--;
				if (dt > -dv) if (dt > 1) if (v_ < vi[i]) v++;
			}
			if (dv == 0) {
				if (dt > 1) if (v_ < vi[i]) v++;
			}
			if (dv > 0) {
				if (v_ < vi[i]) v++;
			}

			x += (v_ == v) ? (ld)v : (ld)(v + v_) / 2.;
		}
	}
	cout << x << endl;
	return 0;
}