解法1：调用__builtin_popcount(unsigned int x)接口

#include <bits/stdc++.h>
using namespace std;

const int N=1e5+5;
int a[N];
int n;

int main(){
    memset(a, 0x00, sizeof a);
    cin>>n;
    for(int i=0; i<n; ++i) cin>>a[i];
    for(int i=0; i<n; ++i) cout<<__builtin_popcount(a[i])<<" ";

    return 0;
}

解法2：lowbit(unsigned int x)操作

#include <bits/stdc++.h>
using namespace std;

const int N=1e5+5;
int a[N];
int n;

// 每次减去lowbit(x), 计算减了多少次即x含有多少个1
int lowbit(int x){
    return x&-x;
}

int main(){
    memset(a, 0x00, sizeof a);
    cin>>n;
    for(int i=0; i<n; ++i) cin>>a[i];
    for(int i=0; i<n; ++i){
        int res=0;
        while(a[i]) a[i]-=lowbit(a[i]), res++;
        cout<<res<<" ";
    }

    return 0;
}