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>
#include <stdlib.h>

int main (void){
	int	N;
	int	i,j;
	int	t[250],v[150];	/*tの値とvの値を格納*/
	double	gragh[1000000];	
	int	sumt;
	double	tmp;

	scanf("%d",&N);

	t[0] = 0;

	/*tの入力*/
	for(i=0;i<N;i++)
    	scanf("%d",&t[i+1]);

	/*tの配列を累積時間にする*/
	for (i = 2;i < N+1;i++)
 		t[i] += t[i-1];

	/*累積最終時間が終了時間*/
	sumt = t[N];

	/*vの入力*/
	for(i=0;i<N;i++){
 		scanf("%d",&v[i]);
	}

	/*graghの最初の状態をつくる*/
	for (i=1;i < sumt*2+1;i++){ 

		if (i < sumt+1){
 			gragh[i] = (double)i/2;
 		}

		else{
 			gragh[i] = gragh[i-1] - 0.5;
 		}

	}

	for(j=0;j<N;j++){
	for(i=1;i<sumt*2+1;i++){
		if (t[j]*2 > i){
			tmp = ((double)t[j]+(double)v[j]-(double)i/2);
			if (gragh[i] > tmp){
				gragh[i] = tmp;
			}
		}
		else{if ((t[j]*2 <= i) && (i <= (t[j+1]*2))){
			if (gragh[i] > (double)v[j]){
				gragh[i] = (double)v[j];
			}
		}
		else {
			tmp = ((double)v[j]+0.5*(double)(i-t[j+1]*2));
			if (gragh[i] > tmp){
				gragh[i] = tmp;
			}
		}
		}
	}
	}

	double val = 0.0;
	for(i=0;i<sumt*2;i++){
		if (gragh[i] != gragh[i+1]){
			val += ((gragh[i]+gragh[i+1]) * 0.25);
		}
		else{
			val += ((gragh[i]) * 0.5);
		}
	}

	/*答えの出力*/
	printf("%lf\n",val);
	return 0;
}