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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main()
{
	int i, N, t[101], v[101];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d", &(t[i]));
	for (i = 1; i <= N; i++) scanf("%d", &(v[i]));

	int j, k;
	double dp[101][102] = {}, tmp;
	for (i = 0; i <= 100; i++) {
		for (j = 0; j <= 101; j++) dp[i][j] = -1.0;
	}
	for (i = 1, dp[0][0] = 0.0; i <= N; i++) {
		for (j = 0; j <= v[i] && dp[i-1][j] >= 0.0; j++) {
			for (k = 0; j - k > t[i]; k++);
			for (; k <= v[i] && k - j <= t[i]; k++) {
				if (t[i] >= v[i] * 2 - j - k) tmp = (double)(v[i] + j) * (v[i] - j) / 2.0 + (double)(v[i] + k) * (v[i] - k) / 2.0 + (double)v[i] * (t[i] - v[i] * 2 + j + k);
				else tmp = ((double)(t[i] - j + k) / 2.0 + j * 2.0) * (double)(t[i] - j + k) / 4.0 + ((double)(t[i] - k + j) / 2.0 + k * 2) * (double)(t[i] - k + j) / 4.0;
				if (dp[i-1][j] + tmp > dp[i][k]) dp[i][k] = dp[i-1][j] + tmp;
			}
		}
	}

	printf("%f\n", dp[N][0]);
	fflush(stdout);
	return 0;
}