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 <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <array>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <unordered_map>
#include <unordered_set>
*/

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define SORT(c) sort((c).begin(),(c).end())

typedef long long ll;
const ll INF = LLONG_MAX - 100;
const ll mod = 1e9 + 7;
const int MAX_N = 5e5 + 5;
int dx[] = { -1LL,0,1LL,0 }, dy[] = { 0,1LL,0,-1LL };
vector<ll> prime;

ll inv[MAX_N], fac[MAX_N];

inline ll GCD(ll a, ll b) { ll c; while (b != 0) { c = a % b; a = b; b = c; }return a; }
inline ll LCM(ll a, ll b) { return a * b / GCD(a, b); }
inline ll POW(ll a, ll b) { ll c = 1LL; while (b > 0) { if (b & 1LL) { c = a * c%mod; }a = a * a%mod; b >>= 1LL; }return c; }
inline void _nCr() { fac[0] = 1LL; for (int i = 1LL; i < MAX_N; i++) { fac[i] = fac[i - 1LL] * i%mod; }for (int i = 0; i < MAX_N; i++) { inv[i] = POW(fac[i], mod - 2); } }
inline ll nCr(ll n, ll r) { return (fac[n] * inv[r] % mod)*inv[n - r] % mod; }
inline void PRI(ll n) { bool a[n + 1LL]; for (int i = 0; i < n + 1LL; i++) { a[i] = 1LL; }for (int i = 2; i < n + 1LL; i++) { if (a[i]) { prime.pb(i); ll b = i; while (b <= n) { a[b] = 0; b += i; } } } }

int cnt[26];

signed main() {
	int n;
	cin >> n;
	int t[n];
	int v[n+2];
	REP (i,n) cin >> t[i];
	FOR (i,1,n) cin >> v[i];
	v[0] = 0;v[n+1] = 0;
	int lim = 0;
	for (int i = n;i >= 1;i--) {
		if (lim + t[i-1] <= v[i]) v[i] = lim + t[i-1];
		lim = v[i];
	}
	lim = 0;
	FOR (i,1,n+1) {
		if (lim + t[i-1] <= v[i]) v[i] = lim + t[i-1];
		lim = v[i];
	}
	REP (i,n+2) cout << v[i] << endl;
	int now = 0;
	float ans = 0;
	FOR (i,1,n+1) {
		now = v[i];
		if (i == 1) now = min(v[i],v[i-1]);
		if (now == v[i+1]) {
			ans += (float)t[i-1] * (float)t[i-1] / 4;
		} else if (now < v[i+1]){
			ans += (float)(v[i] - now) * (float)(v[i] - now) / 2 + (float)(t[i-1]-v[i]+now) * (float)v[i];
		} else {
			ans += (float)(now - v[i]) * (float)(now - v[i]) / 2 + (float)(t[i-1]-now+v[i]) * (float)now;
		}
	}
	cout << ans << endl;
}