lowbit(x)：返回x的最后一位1

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

例：
x = 1010    lowbit(x) = 10
x = 101000  lowbit(x) = 1000

#include <iostream>

using namespace std;

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

int main ()
{
    int n;
    cin >> n;

    while(n -- )
    {
        int x;
        cin >> x;

        int res = 0;
        while (x) x -= lowbit(x), res ++ ;

        cout << res << ' ';        
    }
    return 0;
}

x >> k & 1 x的二进制表示

int x = 10;
for (int k = 31; k >= 0 ; k -- ) cout << (x >> k & 1);  => 1010

int main()
{
    int n;
    cin >> n;

    while(n -- )
    {
        int x;
        cin >> x;

        int res = 0;

        for (int k = 31; k >= 0; k -- ) 
        {
            if (x >> k & 1) res ++ ;
        }
        cout << res << ' ';
    }

    return 0;
}