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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)



import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	private static List<Long> getStrings(Scanner sc, int num) {
		List<Long> stringList = new ArrayList<>();
		while(num>0) {
			long n = sc.nextLong();
			stringList.add(n);
			num--;
		}
		return stringList;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		List<Long> treeList = getStrings(sc, n+1);
		long[] treeArray = new long[n+1];
		if (n == 0) {
			if (treeList.get(0) == 1) {
				System.out.println(1);
			} else {
				System.out.println(-1);
			}
			return;
		}
		treeArray[n] = treeList.get(n);
		for (int i = n-1; i >= 0; i--) {
			long max = (long)Math.pow(2.0, i);
			if (treeList.get(i) > max) {
				System.out.println(-1);
				return;
			}
			long tobeRoot = max - treeList.get(i);
			if (tobeRoot <=0) {
				System.out.println(-1);
				return;
			}
			long uppermax = 1;
			if (i != 0) {
				long upper = (long)Math.pow(2.0, i-1) - treeList.get(i-1);
				uppermax = upper*2;
			}
			treeArray[i] = Math.min(Math.min(treeArray[i+1] + treeList.get(i), max), uppermax);
		}
		for (int i = 1; i <= n; i++) {
			treeArray[i] = Math.min(treeArray[i], (treeArray[i-1]-treeList.get(i-1))*2);
		}
		long sum = 0;
		for (int i = 0; i <= n; i++) {
			sum+=treeArray[i];
			// comment out
//			System.out.print(treeArray[i] + ", " + Math.pow(2, i));
//			System.out.println();
		}
		System.out.println(sum);
	}

}