Problem:
Score : 600 points

Problem Statement

Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

Constraints

0 \leq N \leq 10^5

0 \leq A_i \leq 10^{8} (0 \leq i \leq N)

A_N \geq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 A_2 \cdots A_N

Output

Print the answer as an integer.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using P = pair<int,int>;
using T = tuple<int,int,int>;
using ll = long long;
using vec_ll = vector<long long>;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)

int charToInt(char c){
    char zero_num = '0';
    return (int)c - (int)zero_num;
}

int main(){
    int N; cin>>N;
    vec_ll A(N+1);
    vec_ll cumul_A(N+1);
    ll tot_A =0;
    rep(i,N+1){
     cin>>A.at(i);
     tot_A += A.at(i);
     cumul_A.at(i) = tot_A;
    }

    vec_ll avairable_edges(N+1); //たぶんここまでたくさんのノードのことは考えないでよいはず
    vec_ll avairable_nodes(N+1); //たぶんここまでたくさんのノードのことは考えないでよいはず


    if(A.at(0)!=0){
        cout<<-1<<endl;
        return 0;
    }

    ll the_depth = 0;
    avairable_edges.at(0) = 2;
    avairable_nodes.at(0) = 1;

    ll total_nodes = 1; //1は深さ0の文
    for(int i=1;i<N+1;i++){
        /*
        cout<<"i:"<<i<<endl;
        cout<<"A[i]"<<A.at(i)<<endl;
        cout<<"the depth:"<<the_depth<<endl;
        cout<<"avairable_nodes:"<<avairable_nodes.at(the_depth)<<endl;
        */


        ll temp_val = A.at(i);
        ll temp2 = 0;
        while(true){
            //the_nodeのedgeの数をA(i)本減らす
            //the_nodeのedgeの数が足りなければ次の層に移す
            if(temp_val<=avairable_edges.at(the_depth)){
                avairable_edges.at(the_depth) -= temp_val;
                //the_depthからi層までノードが伸びてくるので、i-the_depth個ノードが加算される
                total_nodes += (i-the_depth)*temp_val;
                break;
            }else{
                //cout<<"go to next node"<<the_depth<<endl;
                //次の層に行く
                temp_val -=avairable_edges.at(the_depth);
                temp2 += avairable_edges.at(the_depth);
                //ノードの数を足す
                total_nodes += (i-the_depth)*avairable_edges.at(the_depth);
                avairable_edges.at(the_depth) = 0;
                //次の層で使えるエッジの数を数える
                the_depth++;
                if(the_depth>i-1){
                    //条件を満たせない
                    cout<<-1<<endl;
                    return 0;
                }
                //この時点ですでにこの階層でcumul_A(i) - cumul_A(the_depth)
                //だけedgeを使っている。
                //そもそも、子供を作れるノードの最大数は
                avairable_nodes.at(the_depth) = avairable_nodes.at(the_depth-1)*2 - A.at(the_depth);
                //である。
                avairable_edges.at(the_depth) = avairable_nodes.at(the_depth)*2 - (cumul_A.at(i-1) - cumul_A.at(the_depth)-temp2);

            }
        }
    }
    cout<<total_nodes<<endl;

    return 0;
}