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;

#define LCM(a,b)                (a / __gcd(a,b) ) *b
#define GCD(a,b)                __gcd(a,b)
#define dist(x1,y1,x2,y2)       sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1))
#define distl(x1,y1,x2,y2)      sqrtl((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1))
#define fRead(x)                freopen(x,"r",stdin)
#define fWrite(x)               freopen (x,"w",stdout)
#define LL                      long long
#define ULL                     unsigned long long
#define pb                      push_back
#define tcase                   cout<<"Case "<<C++<<": ";
#define all(a)                  a.begin(),a.end()
#define Unique(a)               sort(all(a)),a.erase(unique(all(a)),a.end
#define ABS(x)                  ((x)<0?-(x):(x))
#define min3(a,b,c)             min(a,min(b,c))
#define min4(a,b,c,d)           min(a,min(b,min(c,d)))
#define max3(a,b,c)             max(a,max(b,c))
#define max4(a,b,c,d)           max(a,max(b,max(c,d)))
#define PI                      acos(-1.0)
#define ff                      first
#define ss                      second
#define SP(x,y)                 fixed<<setprecision((y))<<(x)
#define FastRead                ios_base::sync_with_stdio(0);cin.tie(NULL);
#define FOR(x, l, r)            for (int x = l; x<=r; ++x)
#define REP(x,r)                for (int x = 0; x <r; ++x)
#define NN                      cout<<"\n";
#define NNN                     cout<<"\n------------------------------\n";
#define SORT(v) sort(v.begin(),v.end())
#define REV(v) reverse(v.begin(),v.end())
//***********************************************************************//
//                      SHAHED AHMED, CSE'15, IUT                        //
//***********************************************************************//


LL lim[200020];
LL fix[200020];
LL val[200020];


int main()
{
    LL n;

    cin>>n;

    vector<LL> t(n);
    vector<LL> v(n);

    for(LL i=0;i<n;i++) cin>>t[i];
    for(LL i=0;i<n;i++) cin>>v[i];


    LL pos = 0;

    for(LL i=0;i<t.size();i++)
    {
        lim[pos] = min(lim[pos],v[i]);
        while(t[i]--)
        {
            lim[++pos]= v[i];
        }

    }




    fix[0]=1;
    fix[pos]=1;



    LL increase = 111;

    while(increase--)
    {
        for(LL i=1;i<=pos;i++)
        {
            if(val[i]==lim[i]) fix[i]=1;
            if(val[i]==val[i-1]+1 && fix[i-1]==1) fix[i]=1;
        }

        for(LL i=pos;i>=1;i--)
        {
            if(val[i]==val[i+1]+1 && fix[i+1]==1) fix[i]=1;
        }

        for(LL i=1;i<=pos;i++)
        {
            if(fix[i]!=1)
            {
                val[i]++;
            }
        }
    }

    double tot=0;
    for(LL i=1;i<=pos;i++)
    {
        //cout<<val[i]<<" ";
        tot+=val[i];
    }

    for(LL i=1;i<=pos;i++)
    {
        if(val[i]==val[i-1] && (val[i]<max(lim[i],lim[i-1])))
        {
            tot+=0.25;
        }

    }
//
//    cout<<"\n";
//    for(LL i=1;i<=pos;i++)
//    {
//        cout<<val[i]<<" ";
//    }
//    cout<<"\n";

    cout<<SP(tot,18)<<"\n";


}