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;

int main() {
  int N;
  cin >> N;
  vector<int> A(N+1);
  for(int i = 0; i < N+1; ++i){
    cin >> A.at(i);
  }
  if(N == 0){
    if(A.at(0) == 1){
      cout << 1 << endl;
    } else {
      cout << -1 << endl;
    }
    return 0;
  }
  if(A.at(0) == 1){
    cout << -1 << endl;
    return 0;
  }

  vector<int> maxChild(N+1);
  maxChild.at(0) = 2;
  maxChild.at(N) = 0;

  for(int i = 1; i < N; ++i){
    maxChild.at(i) = (maxChild.at(i-1) - A.at(i)) * 2;
  }

  maxChild.at(N-1) = A.at(N);
  for(int i = N-2; 0 <= i; --i){
    maxChild.at(i) = min(maxChild.at(i),maxChild.at(i+1) + A.at(i+1));
  }

  int cnt = 1;
  for(int i = 0; i < N; ++i){
    // cout << maxChild.at(i) << endl;
    if(maxChild.at(i) <= A.at(i+1) && i != N-1 || maxChild.at(i) != A.at(i+1) && i == N-1){
      cout << -1 << endl;
      return 0;
    }

    cnt += maxChild.at(i);
  }

  cout << cnt << endl;
}