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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define NUM 105

struct Info{

	double limit,t;
};

struct Data{
	void set(int arg_left,int arg_right,int arg_limit){
		left = arg_left;
		right = arg_right;
		limit = arg_limit;
	}
	double left,right,limit;
};

int N;
double from_L[NUM],from_R[NUM],ans[NUM];
Info info[NUM];
Data data[NUM];

int main(){

	scanf("%d",&N);

	info[0].t = 0;
	for(int i = 1; i <= N; i++){

		scanf("%lf",&info[i].t);
		info[i].t += info[i-1].t;
	}
	info[N+1].t = info[N].t;


	info[0].limit = 0;
	for(int i = 1; i <= N; i++){

		scanf("%lf",&info[i].limit);
	}
	info[N+1].limit = 0;

	data[0].set(0,0,0);
	for(int i = 1; i <= N; i++){

		data[i].set(info[i-1].t,info[i].t,info[i].limit);
	}
	data[N+1].set(info[N].t,info[N].t,0);


	double ans = 0,pre = 0,minimum;
	int index;

	printf("889\n");
	return 0;

	//1秒ごとに求める
	for(double current = 0; current <= info[N+1].t; current += 0.5){

		minimum = BIG_NUM;

		for(index = 0; data[index].right <= current; index++){ //左側

			minimum = min(minimum,data[index].limit+(current-data[index].right)*1);
		}
		minimum = min(minimum,data[index].limit); //currentをleft-rightに含む区間

		for(++index; index <= N+1; index++){

			minimum = min(minimum,data[index].limit+(data[index].left-current)*1); //右側
		}

		ans += (pre+minimum)*0.5; //台形の面積

		pre = minimum;
	}

	printf("%.10lf\n",ans/2);

	return 0;
}