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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
using ll=long long;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
using pq_int=priority_queue<int,vector<int>,greater<int>>;
const ll LINF=0x3fffffffffffffff;
const ll MOD=1000000007;
const ll MODD=0x3b800001;
const int INF=0x3fffffff;
#define _overload4(_1,_2,_3,_4,name,...) name
#define _overload3(_1,_2,_3,name,...) name
#define _rep1(n) _rep2(i,n)
#define _rep2(i,n) _rep3(i,0,n)
#define _rep3(i,a,b) for(ll i=a;i<b;++i)
#define _rep4(i,a,b,c) for(ll i=a;i<b;i+=c)
#define rep(...) _overload4(__VA_ARGS__,_rep4,_rep3,_rep2,_rep1)(__VA_ARGS__)
#define _rrep1(n) _rrep2(i,n)
#define _rrep2(i,n) _rrep3(i,0,n)
#define _rrep3(i,a,b) for(ll i=b-1;i>=a;i--)
#define _rrep4(i,a,b,c) for(ll i=a+(b-a-1)/c*c;i>=a;i-=c)
#define rrep(...) _overload4(__VA_ARGS__,_rrep4,_rrep3,_rrep2,_rrep1)(__VA_ARGS__)
#define each(i,a) for(auto&& i:a)
#define sum(...) accumulate(range(__VA_ARGS__),0)
#define _range(i) (i).begin(),(i).end()
#define _range2(i,k) (i).begin(),(i).begin()+k
#define _range3(i,a,b) (i).begin()+a,(i).begin()+b
#define range(...) _overload3(__VA_ARGS__,_range3,_range2,_range)(__VA_ARGS__)
#define Yes(i) out(i?"Yes":"No")
#define YES(i) out(i?"YES":"NO")
//#define START auto start=system_clock::now()
//#define END auto end=system_clock::now();cerr<<duration_cast<milliseconds>(end-start).count()<<" ms\n"
#define elif else if
#define unless(a) if(!(a))
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define STR(...) string __VA_ARGS__;in(__VA_ARGS__)
#define CHR(...) char __VA_ARGS__;in(__VA_ARGS__)
#define DBL(...) double __VA_ARGS__;in(__VA_ARGS__)
#define vec(type,name,...) vector<type> name(__VA_ARGS__)
#define VEC(type,name,...) vector<type> name(__VA_ARGS__);in(name)
#define vv(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))
__attribute__((constructor)) void SETTINGS(){cin.tie(0); cout.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(15);};
template<class T>
inline constexpr T gcd (T a,T b) {if(a==b)return a;else return gcd(b,(a-1)%b+1);}
template<class T>
inline constexpr T min(vector<T>& v){return *min_element(range(v));}
inline char min(string& v){return *min_element(range(v));}
template<class T>
inline constexpr T max(vector<T>& v){return *max_element(range(v));}
inline char max(string& v){return *max_element(range(v));}
template<typename T>
inline bool update_min(T& mn,const T& cnt){if(mn>cnt){mn=cnt;return 1;}else return 0;}
template<typename T>
inline bool update_max(T& mx,const T& cnt){if(mx<cnt){mx=cnt;return 1;}else return 0;}
inline void in() {}
template<class T>
istream& operator >> (istream& is, vector<T>& vec);
template<class T,size_t size>
istream& operator >> (istream& is, array<T,size>& vec);
template<class T,class L>
istream& operator >> (istream& is, pair<T,L>& p);
template<class T>
ostream& operator << (ostream& os, vector<T>& vec);
template<class T,class L>
ostream& operator << (ostream& os, pair<T,L>& p);
template<class T>
istream& operator >> (istream& is, vector<T>& vec){for(T& x: vec) is >> x;return is;}
template<class T,class L>
istream& operator >> (istream& is, pair<T,L>& p){is >> p.first;is >> p.second;return is;}
template<class T>
ostream& operator << (ostream& os, vector<T>& vec){os << vec[0];rep(i,1,vec.size()){os << ' ' << vec[i];}return os;}
template<class T>
ostream& operator << (ostream& os, deque<T>& deq){os << deq[0];rep(i,1,deq.size()){os << ' ' << deq[i];}return os;}
template<class T,class L>
ostream& operator << (ostream& os, pair<T,L>& p){os << p.first << " " << p.second;return os;}
template <class Head, class... Tail>
inline void in(Head&& head,Tail&&... tail){cin>>head;in(move(tail)...);}
template <class T>
inline void out(T t){cout<<t<<'\n';}
inline void out(){cout<<'\n';}
template <class Head, class... Tail>
inline void out(Head head,Tail... tail){cout<<head<<' ';out(move(tail)...);}
template <class T>
inline void err(T t){cerr<<t<<'\n';}
inline void err(){cerr<<'\n';}
template <class Head, class... Tail>
inline void err(Head head,Tail... tail){cerr<<head<<' ';out(move(tail)...);}




signed main(){
    INT(n);
    vector<int> t(n+1),v(n);
    rep(i,1,n+1)in(t[i]);
    in(v);
    partial_sum(range(t),t.begin());
    each(i,t)i*=2;
    each(i,v)i*=2;
    vector<int> r(t[n]+1),l(t[n]+1);
    int at=0;
    rep(i,1,t[n]+1){
        r[i]=min(r[i-1]+1,v[at]);
        if(t[at+1]<=i)at++;
    }
    rrep(i,0,t[n]){
        l[i]=min(l[i+1]+1,v[at-1]);
        if(t[at-1]>=i)at--;
    }
    rep(t[n]+1)update_min(r[i],l[i]);
    int ans=0;
    rep(t[n])ans+=r[i]+r[i+1];
    out(ans/8.0);
}