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 <bits/stdc++.h>
using namespace std;
typedef long long int ll;
int n,t[110],v[110];
int main(){
    cin >> n;
    for(int i=0;i<n;i++) cin >> t[i];
    for(int i=0;i<n;i++) cin >> v[i];
    int sum = 0;
    for(int i=0;i<n;i++){
        sum += t[i];
        v[i] = min(v[i],sum);
    }
    sum = 0;
    for(int i=n-1;i>=0;i--){
        sum += t[i];
        v[i] = min(v[i],sum);
    }
    v[n] = 0;
    double ans = 0;
    double before = 0;
    for(int i=0;i<n;i++){
        if(v[i+1]<=v[i]){
            int tmp = (v[i]-before)+abs(v[i+1]-v[i]);
            if(tmp<=t[i]){
                ans += ((double)v[i]+before)*((double)v[i]-before)*0.5;
                ans += ((double)t[i]-tmp)*v[i];
                ans += abs((double)v[i+1]+v[i])*abs(v[i+1]-v[i])*0.5;
            }
            else {
                double x = ((double)t[i]+v[i+1]-before)*0.5;
                ans += ((double)before*2+x)*x*0.5;
                ans += ((double)before+x+v[i+1])*(t[i]-x)*0.5;
            }
            before = v[i+1];
        }
        else {
            int tmp = (v[i]-before);
            if(tmp<=t[i]){
                ans += ((double)v[i]+before)*tmp*0.5;
                ans += (double)v[i]*(t[i]-tmp);
                before = v[i];
            }
            else {
                ans += ((double)v[i]*2+t[i])*t[i]*0.5;
                before = before+t[i];
            }
        }
    }
    printf("%.9f\n",ans);
    return 0;
}