1、位运算

AcWing 801. 二进制中1的个数

n>>k&1 //取出二进制第k位的数字

import java.util.*;

//ACWing
public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.erjingzhi1();
    }

    //求的是最后一位1
    //例如10010
    //补码=~10010+1=01110
    //两者做与运算得
    //00010
    int lowbit(int n){
        return n&-n;
    }
    //判断子序列
    void erjingzhi1() {
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt();
        int a[]=new int[n];
        for (int i = 0; i < n; i++) a[i]=scanner.nextInt();

        for (int i = 0; i < n; i++) {
            int res=0;
            while(a[i]>0){
                a[i]-=lowbit(a[i]);
                res++;
            }
            System.out.print(res+" ");
        }
    }
}