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 Kotlin (1.3.71) (judge verdict: Runtime Error)

import java.lang.Math.min
import java.lang.Math.pow
import kotlin.math.ceil

//

fun main() {
    nomura2020c()
}

fun nomura2020c() {
    val n = readLine()!!.toInt()
    val a = readLine()!!.split(" ").map { it.toLong() }
    // 深さnでちょうど終わらないとダメ
    // つまり最終のnの葉の数から逆算して二分木を構築しないとダメ
//    var total = 0L
//    var nodes = 1L
//    for (d in 0..n) {
//        if (a[d] <= nodes) {
//            total += nodes
//        } else {
//            println(-1)
//            return
//        }
//        nodes = (nodes - a[d]) * 2
//    }
//    println(total)
    if (n > 0 && a[0] > 0) {
        // ルートが葉でありうるのはn=0のときだけ
        println(-1)
        return
    }
//    var total = 0L
//    var nodes = 1L
//    var nodesLimit = 1L
//    val limit = LongArray(n + 1) { 0L }
//    for (d in 0 .. n) {
//        val temp = nodes * 2 + a[d]
//        if (temp > nodesLimit) {
//            println(-1)
//            return
//        }
//        limit[d] = temp
//        nodesLimit *= 2
//    }
//    nodes = 0L
//    for (d in n downTo 0) {
//        // 直前のノード数が限界いっぱいの場合、そもそもすべてのノードが枝でないと二分木を構築できないのでそのチェック
//        val check = ceil(nodes.toDouble() / 2L) + a[d]
//        if (check > limit[d]) {
//            println(-1)
//            return
//        }
//        nodes = min(limit[d], nodes + a[d])
//        total += nodes
//        nodesLimit /= 2
//    }
//    println(total)
    val nodesCount = mutableListOf<Long>()
    dfs(nodesCount, a, 0, n, 1L, 1L, 0L)
    println(nodesCount.max() ?: -1)
}

private fun dfs(nodesCount: MutableList<Long>, leafCount: List<Long>, depth: Int, n: Int, nodesLimit: Long, currentNode: Long, total: Long) {
    if (depth == n) {
        if (currentNode == leafCount[depth]) {
            nodesCount.add(total + leafCount[n])
        }
        return
    }
    if (currentNode + leafCount[depth] > nodesLimit) return
    val nextLimit = (nodesLimit - leafCount[depth]) * 2
    val nextTotal = total + currentNode + leafCount[depth]
    dfs(nodesCount, leafCount, depth + 1, n, nextLimit, currentNode * 2, nextTotal)
    dfs(nodesCount, leafCount, depth + 1, n, nextLimit, currentNode, nextTotal)
}