计算二进制中1的个数

#include
using namespace std;
using ll = long long;
const int N = 2e5 + 9;
ll a[N];
 
int f(int x)
{
	int res = 0;
	while (x)
	{
		if (x & 1) res ++;//每次判断最后一位
		x >>= 1;//右移一位
	}
 
	return res;
}
 
 
int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n; cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		int x; cin >> x;
		cout << f(x) << ' ';
	}
 
 
	return 0;
}