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<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
#include<iomanip>
#include<tuple>
using namespace std;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,int> LP;
const int INF=1<<30;
const LL MAX=1e9+7;

void array_show(int *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%d%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(LL *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%lld%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(vector<int> &vec_s,int vec_n=-1,char middle=' '){
	if(vec_n==-1)vec_n=vec_s.size();
	for(int i=0;i<vec_n;i++)printf("%d%c",vec_s[i],(i!=vec_n-1?middle:'\n'));
}
void array_show(vector<LL> &vec_s,int vec_n=-1,char middle=' '){
	if(vec_n==-1)vec_n=vec_s.size();
	for(int i=0;i<vec_n;i++)printf("%lld%c",vec_s[i],(i!=vec_n-1?middle:'\n'));
}

int v[200005];
int t[105];
int vv[205];

int main(){
	int n,m;
	int i,j,k;
	int a,b,c;
	cin>>n;
	m=0;
	for(i=0;i<n;i++){
		cin>>t[i];
		m+=t[i];
	}
	for(i=0;i<=m;i++){
		v[i]=min(i,m-i);
	}
	b=0;
	for(i=0;i<n;i++){
		cin>>a;
		for(j=0;j<=m;j++){
			if(j<b)c=a+b-j;
			else if(j<=b+t[i])c=a;
			else c=a+j-(b+t[i]);
			v[j]=min(v[j],c);
		}
		vv[i]=a;
		b+=t[i];
	}
	double s=0;
	b=0;
	//array_show(v,100);
	for(i=0;i<n;i++){
		for(j=b;j<b+t[i];j++){
			s+=(v[j]+v[j+1])/2.;
			if(v[j]==v[j+1] && v[j]!=vv[i])s+=1./4;
		}
		b+=t[i];
	}
	cout<<s<<endl;
}