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 <cstdio>
#include <random>
#include <cassert>
#include <algorithm>    
#include <iostream>
#include <chrono>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <cstring>
#include <cmath>
#include <bitset>
#define pb push_back
#define mk make_pair
#define eb emplace_back
#define eps 1e-8
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
// #define int long long


using namespace std;
typedef long double ld;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef tuple<int, int, int> tiii;
typedef vector<int> vii;
typedef vector<long double> vd;
const int inf = 1e9;
const long long INF = 1e18;
const int M = 1e9 + 7;





signed main()
{
#ifndef ONLINE_JUDGE
	// freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
#endif
	int n;
	ios::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	int tot = 0;
	const int maxn = 50000;
	static int num[maxn];
	static double dp[maxn][200];
	static bool vis[maxn][200];
	int t[202], v[202];
	for(int i = 0; i < n; ++i) {
		cin >> t[i];
		for(int j = tot + 1; j <= tot + 2 * t[i]; ++j) {
			num[j] = i;
		}
		tot += 2 * t[i];
	}
	for(int i = 0; i < n; ++i) {
		cin >> v[i];
		v[i] *= 2;
	}	
	vis[0][0] = 1;
	dp[0][0] = 0;
	for(int i = 0; i <= tot; ++i) {
		int k = num[i];
		for(int j = 0; j <= v[k]; ++j) {
			if(vis[i][j]) {
				int tk = num[i + 1];
				if(j + 1 <= v[tk]) {
					dp[i + 1][j + 1] =  max(dp[i + 1][j + 1], dp[i][j] + (j + j + 1) / 2.);
					vis[i + 1][j + 1] = 1;
				}
				if(j <= v[tk]) {
					dp[i + 1][j] = max(dp[i][j] + j, dp[i + 1][j]);
					vis[i + 1][j] = 1;
				}
				if(j > 0 && j <= v[tk]) {
					dp[i + 1][j - 1] = max(dp[i + 1][j - 1], dp[i][j] + (j + j - 1) / 2.);
					vis[i + 1][j - 1] = 1;
				}
			}
		}
	}
	printf("%.5f\n", dp[tot][0] / 4);


    
	return 0;
}