AcWing 1240. 完全二叉树的权值（二叉树）

Problem

第N次取2^N - 1^个数，判断哪次取数的累加最大。
分析一下就很简单了。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Queue;

class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(System.out);
    static int time = 1, n;
    static int N = 100010;
    static long point[][] = new long[N][2];
    static Queue<Integer> queue = new ArrayDeque<>();

    public static void main(String[] args) throws Exception {
        String s[] = br.readLine().split(" ");
        n = Integer.parseInt(s[0]);
        s = br.readLine().split(" ");
        for (int i = 0; i < n; i++) queue.offer(Integer.parseInt(s[i]));
        long max = Long.MIN_VALUE;
        int k = 0;
        while (!queue.isEmpty()) {
            long sum = 0;
            for (int i = 1; i <= time && !queue.isEmpty(); i++) {
                sum += queue.poll();
            }
            if (sum > max) {
                max = sum;
                point[k][0] = log2(time) + 1;
                point[k++][1] = sum;
            }
            time <<= 1;
        }

        long res = 0;
        long mmax = Long.MIN_VALUE;
        for (int i = 0; i < k; i++) {
            if (point[i][1] > mmax) {
                res = point[i][0];
                mmax = point[i][1];
            }
        }
        pw.print(res);
        pw.flush();
        pw.close();
        br.close();
    }

    public static int log2(int N) {
        return (int) (Math.log(N) / Math.log(2));//Math.log的底为e
    }

}