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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vint;
typedef pair<int,int> pint;
typedef vector<pint> vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=n-1;i>=(0);i--)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define all(v) (v).begin(),(v).end()
#define eall(v) unique(all(v), v.end())
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll INFF = 1e18;

int N;
int t[110], v[110];
int sum[200010];
int memo[20010];
int tmpl[200010], tmpr[200010];

int main(void) {
	cin >> N;
	rep(i, N) cin >> t[i];
	rep(i, N) cin >> v[i];
	sum[0] = 0;
	rep(i, N) sum[i + 1] += sum[i] + t[i];
	rep(i, 20010) memo[i] = INF;
	memo[0] = memo[sum[N]] = 0;
	rep(i, N) {
		reps(j, sum[i], sum[i + 1] + 1) chmin(memo[j], v[i]);
		// reps(j, sum[i], sum[i + 1] + 1) tmp[j] = v[i];
		reps(j, sum[i] + 1, sum[i + 1]) tmpl[j] = tmpr[j] = v[i];
		tmpr[sum[i]] = tmpl[sum[i + 1]] = v[i];
	}
	tmpl[0] = tmpr[0] = tmpl[sum[N]] = tmpr[sum[N]] = 0;

	/*
	printf("sum[N] %d\n", sum[N]);
	rep(i, sum[N] + 1) printf("%d ", memo[i]);
	printf("\n");
	*/

	rep(i, N + 1){
		for (int k = sum[i]; k >= 0; --k) {
			int dist = sum[i] - k;
			int dv = tmpr[sum[i]] + dist;
			if(dv > 100) break;
			chmin(memo[k], dv);
		}
	}
	/*
	printf("\n");
	rep(i, sum[N] + 1) printf("%d ", memo[i]);
	printf("\n");
	*/

	rep(i, N + 1){
		// printf("sum %d tmp %d\n", sum[i], tmpl[sum[i]]);
		for (int k = sum[i]; k <= sum[N]; ++k) {
			int dist = k - sum[i];
			int dv = tmpl[sum[i]] + dist;
			if(dv > 100) break;
			chmin(memo[k], dv);
		}
	}
	/*
	rep(i, sum[N] + 1) printf("%d ", memo[i]);
	printf("\n");

	printf("tmpl\n");
	rep(i, sum[N] + 1) printf("%d ", tmpl[i]);
	printf("\n");
	printf("tmpl\n");
	rep(i, sum[N] + 1) printf("%d ", tmpr[i]);
	printf("\n");
	*/

	double ans = 0;
	rep(i, sum[N] + 1) ans += memo[i];
	rep(i, sum[N]) if(memo[i] == memo[i + 1] && memo[i] < tmpr[i] && memo[i + 1] < tmpl[i + 1]) ans += 0.250;
	printf("%.6f\n", ans);
	return 0;
}