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; i++){
    cin >> A[i];
  }
  if (A[0] != 0){
    if (N != 0){
      cout << -1 << endl;
    } else {
      cout << 1 << endl;
    }
  } else {
    vector<long long> sum(N + 1, 0);
    sum[N] = 0;
    for (int i = N; i >= 1; i--){
      sum[i - 1] = sum[i] + A[i];
    }
    /*
    for (int i = 0; i <= N; i++){
      cout << sum[i] << ' ';
    }
    cout << endl;
    */
    vector<long long> B(N + 1, 0);
    B[0] = 1;
    for (int i = 0; i < N; i++){
      long long tmp = B[i] * 2 - A[i + 1];
      B[i + 1] = min(tmp, sum[i + 1]);
    }
    /*
    for (int i = 0; i <= N; i++){
      cout << B[i] << ' ';
    }
    cout << endl;
    */
    bool ok = true;
    for (int i = 0; i <= N; i++){
      if (B[i] < 0){
        ok = false;
      }
    }
    if (!ok){
      cout << -1 << endl;
    } else {
      long long ans = 0;
      for (int i = 0; i <= N; i++){
        ans += A[i];
        ans += B[i];
      }
      cout << ans << endl;
    }
  }
}