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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

typedef long long ll;

ll a[100005],b[100005];
ll siz[100005],sum[100005];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	for (int i = 0; i <= n; i++)
	{
		cin >> a[i];
		if( i != 0 ) sum[i] = sum[i-1] + a[i];
	}
	ll sumx = 0;
	b[0] = 1;
	for (int i = 1; i <= n; i++)
	{
		if(b[i - 1] > 1e17) b[i-1] = 1e17;
		b[i] = b[i - 1] * 2;
	}
	for (int i = n; i >= 0; i --)
	{
		if( a[i] + sumx > b[i] ) 
		{
			cout << "-1" << endl;
			return 0;
		}
		sumx = (sumx + a[i] + 1) / 2;
	}
	if( a[0] != 0 )
	{
		cout << -1 << '\n';
		return 0;
	}
	ll ans = 1;
	ll temp = 0;
	siz[0] = 2;
	int index = 0;
	ll help = 0;
	for (int i = 1; i <= n; )
	{
		if( a[i] == 0 )
		{
			i ++;
			continue;
		}
		if( index >= i )
		{
			cout << -1 << '\n';
			return 0;
		}
		ll t = min(a[i],siz[index]);
		ans += t * (i - index);
		help += t;
		a[i] -= t;
		siz[index] -= t;
		if( siz[index] == 0 )
		{
			index ++;
			siz[index] = help - sum[index];
		}
		if( a[i] == 0 ) i++;
	}
	cout << ans << '\n';
	return 0;
}