题目描述

blablabla

样例

blablabla

算法1

$O(n^logn)$

返回n的最后一位1：lowbit(n) = n & -n
-n = ~n + 1 //(x取反加1)

Java 代码

import java.util.*;
import java.io.*;

public class Main {
    final static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    final static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out)); 
    public static void main(String[] args) throws IOException{
        String[] str = reader.readLine().split("\\s+");
        int n = Integer.parseInt(str[0]);
        int[] a = new int[n];
        str = reader.readLine().split("\\s+");
        for(int i = 0; i < n; i++){
            int val = Integer.parseInt(str[i]);
            System.out.print(findOne(val) + " ");
        }
    }

    private static int findOne(int val){
        int res = 0;
        int k = 1;
        while(val > 0){
            val -= lowbit(val);
            res ++;
        }
        return res;
    }

    public static int lowbit(int x){
        return x & -x;
    }

}