代码

边输入，边处理
时间复杂度$$O(n)$$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
int main()
{
    cin >> n;
    ll max1 = LONG_LONG_MIN, max2 = 0;
    int flag = 1, now = 1;
    for (int i = 1; i <= n; i++)
    {
        int t;
        cin >> t;
        max2 += t;
        if ((int)pow(2, flag) - 1 == i) //如果当前是第2^flag-1个，代表flag这是flag层的最后一个元素
        {
            if (max1 < max2)
            {
                now = flag; //记录每次最大值的层号
                max1 = max2;
            }
            flag++;
            max2 = 0;
        }
    }
    if (max2 > max1) //最后来一次特判，因为题目给的数据并没有覆盖到整个完全二叉树
    {
        now = flag;
    }
    cout << now;
    return 0;
}

刚开始用的BFS，很遗憾，超时

#include<bits/stdc++.h>

using namespace std;
const int N = 1e5 + 10;
typedef long long ll;
int a[N];
int visited[N];
int father[N];
int n;
ll sum[N];

void bfs()
{
    queue<int> q;
    q.push(0);
    visited[0] = 1;
    int head = 0, tail;
    int cnt = 0;
    while (!q.empty())
    {
        int now = q.front();
        sum[cnt] += a[now];
        q.pop();
        for (int i = 0; i < n; i++)
        {
            if (!visited[i] && father[i] == now + 1)
            {
                q.push(i);
                visited[i] = 1;
                tail = i;
            }
        }
        if (head == now)
        {
            cnt++;
            head = tail;
        }
    }
}

int main()
{
    cin >> n;
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        father[i] = cnt;
        if ((i + 1) % 2 == 1)
            cnt++;
    }
    father[0] = -1;
    bfs();
    int max = 0;
    for (int i = 0; i < cnt; i++)
        if (sum[max] < sum[i])
            max = i;
    cout << max + 1;
    return 0;
}