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 (3.4.3) (judge verdict: Wrong Answer)

N=int(input())
t=list(map(int, input().split()))
v=list(map(int, input().split()))
T=sum(t)
vmax=[0]*(T+1)
from collections import deque
q=[deque() for i in range(101)]
s=0
for i in range(N):
    vmax[s]=min(vmax[s-1],v[i])
    q[vmax[s]].append(s)
    for j in range(t[i]):
        vmax[s+j+1]=v[i]
    s+=t[i]
vmax[-1]=0
q[0].append(T)
vans=[-1]*(T+1)
for s in q[0]:
    vans[s]=0
for i in range(101):
    while(len(q[i])>0):
        s=q[i].popleft()
        if s>0 and vans[s-1]==-1:
            if s-1 not in q[i] and vmax[s-1]>i:
                q[i+1].append(s-1)
                vans[s-1]=i+1
            else:
                q[i].append(s-1)
                vans[s-1]=i
        if s<T and vans[s+1]==-1:
            if s+1 not in q[i] and vmax[s+1]>i:
                q[i+1].append(s+1)
                vans[s+1]=i+1
            else:
                q[i].append(s+1)
                vans[s+1]=i
ans=0
for i in range(T):
    if vans[i+1]==vans[i]+1:
        ans+=vans[i]+0.5
    elif vans[i+1]==vans[i]-1:
        ans+=vans[i+1]+0.5
    elif vmax[i+1]>vans[i+1] or vmax[i]>vans[i]:
        ans+=vans[i]+0.25
    else:
        ans+=vans[i]
print(ans)