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 Python (3.8.2) (judge verdict: Accepted)

import sys
from heapq import heapify, heappop
from bisect import bisect_left, bisect_right, insort_left
read=sys.stdin.read
readline=sys.stdin.readline

def main():
    n=int(readline())
    t,v=(list(map(int,l.split())) for l in read().splitlines())
    vi=[(x,i+1) for i,x in enumerate(v)]
    x=[0]*(n+2)
    y=[0]*(n+1)
    x[0]=0
    y[0]=0
    y[-1]=0
    for i in range(1,n+1):
        x[i]=x[i-1]+t[i-1]
    x[n+1]=x[n]
    heapify(vi)
    ans=0
    v=[0]+v+[0]
    memo=[0,n+1]
    while vi:
        e=heappop(vi)
        insort_left(memo,e[1])
        idx=bisect_left(memo,e[1])
        _i=memo[idx-1]
        i_=memo[idx+1]
        i=memo[idx]
        sx=x[_i]
        gx=x[i_-1]
        sy=y[_i]
        gy=y[i_-1]
        h=e[0]
        y[i]=min(h,gx-x[i]+gy,x[i]-sx+sy)
        y[i-1]=min(h,gx-x[i-1]+gy,x[i-1]-sx+sy)
        dw=gx-sx
        dh=abs(gy-sy)
        be=max(0,dw-(2*max(v[_i],v[i_])-sy-gy))
        ht=min(h-v[_i],h-v[i_],be/2)
        te=max(0,be-2*ht)
        ans+=(te+be)*ht/2
    print(ans)

if __name__=='__main__':
    main()