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<stdio.h>
int main()
{
	int n;
	int a[100002],l[100002];  //共n+1个数;l[i]存储第i层的结点总数 
	int i,ans=0,mi=2;  //mi代表第n层最节点数 
	scanf("%d",&n);
	scanf("%d",&a[0]);
	if((n==0&&a[0]>1)||(n!=0&&a[0]>=1))
	{
		printf("-1");
		return 0;
	}
	for(i=1;i<=n-1;i++)
	{
		scanf("%d",&a[i]);
		if(a[i]>=mi||a[i]>mi-2*a[i-1])
		{
			printf("-1");
			return 0;
		}
		mi*=2;
	}
	if(n!=0)
	{
		scanf("%d",&a[n]);
		if(a[n]>mi||a[n]>mi-2*a[n-1])
		{
			printf("-1");
			return 0;
		}
	}
	l[n]=a[n];
	mi/=2;  //第n-1层 
	ans+=l[n];
	for(i=n-1;i>=0;i--)
	{
		if(mi<=a[i]+l[i+1]&&mi<=mi-2*a[i-1])
		{
			l[i]=mi;
		}
		if(a[i]+l[i+1]<=mi&&a[i]+l[i+1]<=mi-2*a[i-1])
		{
			l[i]=a[i]+l[i+1];
		}
		if(mi-2*a[i-1]<=mi&&mi-2*a[i-1]<=a[i]+l[i+1])
		{
			l[i]=mi-2*a[i-1];
		}
		if(l[i]*2<l[i+1])
		{
			printf("-1");
			return 0;
		}
		ans+=l[i];
		mi/=2;
	}
	printf("%d",ans);
	return 0;
}