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 REP(i,n) for (int i=0;i<(n);i++)
//#define int long long  
int main(){
    int N;
    int t[104];
    int v[104];
    int Tm[104]={};
    int Ans=0;
    int lim;
    int tt[20100]={};
    cin >> N;
    REP(i,N){cin >> t[i];}
    Tm[0]=t[0];
    REP(i,N){Tm[i+1]+=t[i+1]+Tm[i];}
    REP(i,N){cin >> v[i];}
    if(N==1){
        if(t[0]%2==1){
            printf("%f\n",((double)t[0]*(double)t[0])/4);
            return 0;
        }else{
            REP(i,t[0]){tt[i]=v[0];}
        }
    }else{
        for(int i=0;i<=t[0];i++){
            tt[i]=v[0];
        }
        REP(j,N){
            if(v[j]==v[j+1]){
                //cout << "==" <<endl;
                for(int i=Tm[j];i<Tm[j+1];i++){
                    tt[i]=v[j];
                }
            }else if(v[j]<v[j+1]){
                //cout << "<" <<endl;
                for(int i=Tm[j];i<Tm[j+1];i++){
                    tt[i]=min(v[j]+i-Tm[j],v[j+1]);
                }
            }else{
                //cout << ">" <<endl;
                for(int i=Tm[j];i<Tm[j+1];i++){
                    tt[i]=v[j+1];
                }
                for(int i=Tm[j];i>=0;i--){
                    if(tt[i]>=tt[Tm[j]]+(Tm[j]-i)){
                        tt[i]=tt[Tm[j]]+(Tm[j]-i);
                    }else{
                        break;
                    }
                }
            }
        }
    }
    //REP(i,N){cout << Tm[i] << " ";}
    REP(i,Tm[N-1]){
        lim=min(Tm[N-1]-i,i);
        Ans+=min(tt[i],lim);
        //cout << i << ":" <<min(tt[i],lim) << endl;
    }
    //cout <<endl;
    cout << Ans<<endl;
	return 0;
}