100763: [AtCoder]ABC076 D - AtCoder Express

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