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: Accepted)

#include<stdio.h>

double min(double x,double y){
  if(x<y) return x;
  else return y;
}

int main(){
  int n,t[100],sum[100],i,j;
  double dis=0,vl[40002],vv[40002],v[100];
  scanf("%d",&n);
  scanf("%d",&t[0]);
  sum[0]=t[0];
  for(i=1;i<n;i++){
    scanf("%d",&t[i]);
    sum[i]=sum[i-1]+t[i];
  }
  int max=sum[n-1];
  for(i=0;i<=2*max;i++)
    vl[i]=100.0;
  for(i=0;i<200;i++)
    vl[2*max-i]=(double)i/2;
  for(i=0;i<n;i++){
    scanf("%lf",&v[i]);
  }
  for(i=n-1;i>0;i--){
    for(j=2*sum[i-1];j<=2*sum[i];j++){
      vl[j]=min(vl[j],v[i]);}
    for(j=0;(double)j/2+v[i]<=100&&2*sum[i-1]-j>=0;j++)
      vl[2*sum[i-1]-j]=min(vl[2*sum[i-1]-j],v[i]+((double)j/2));
  }
  for(j=0;j<=2*sum[0];j++){
      vl[j]=min(vl[j],v[i]);}
  vv[0]=0.0;
  for(i=0;i<=2*max;i++){
    if(vv[i]<vl[i+1]) vv[i+1]=vv[i]+0.5;
    else if(vv[i]>vl[i+1]) vv[i+1]=vv[i]-0.5;
    else vv[i+1]=vv[i];
    dis+=vv[i+1]/2;
  }
  printf("%lf\n",dis);
  
}