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<cstdio>
#include<algorithm>
#include<vector>
#include<string>
#include<string.h>
#include<set>
#define p pair<int,int>
#define P pair<int,p>
#define Q pair<p,int>
using namespace std;

set<p>S;//time speed
P a[100];
signed main() {
	int n; cin >> n;
	int sum = 0;
	for (int i = 0; i < n; i++) {
		int x; cin >> x;
		a[i].second = { sum,sum + x };
		sum += x;
	}
	for (int i = 0; i < n; i++) {
		cin >> a[i].first;
	}
	double ans = 0;
	sort(a, a + n);
	S.insert({ 0,0 });
	S.insert({ sum,0 });
	for (int i = 0; i < n; i++) {
		auto k = S.upper_bound({ a[i].second.second,-1 });
		int r = k->second + (k->first - a[i].second.second);
		k--;
		int l = k->second + (a[i].second.first - k->first);
		l = min(l, a[i].first);
		int m = min(l, a[i].first);
		for (int j = a[i].second.first; j < a[i].second.second; j++) {
			if (m + 1 <= a[i].first&&m + 1 <= r + a[i].second.second - (j + 1)) {
				ans += m + 0.5;
				m++;
			}
			else if (m + 1 <= a[i].first&&m <= r + a[i].second.second - (j + 1)) {
				ans += m + 0.25;
			}
			else if (m <= r + a[i].second.second - (j + 1)) {
				ans += m;
			}
			else {
				ans += m - 0.5;
				m--;
			}
		}
		if (k->first != a[i].second.first) {
			S.insert({ a[i].second.first,min(l,a[i].first) });
		}
		k++;
		if (k->first != a[i].second.second) {
			S.insert({ a[i].second.second,m });
		}
	}
	printf("%.10lf\n", ans);
}