Count SET Bits in a number | BIT Manipulation | C++

 How to get /count the number of ones in binary form of a number using BIT Manipulation?

 So in this blog, we will see how we can count the number of ones in a number in its binary form.

As we know that if we take AND of a number with 1 it will give 0 if the number is 0 and 1 if the number is one.

So we will take the AND every time with the LSB of the number and then right-shift the number.

Here is the code of the program in C++. The code starts in the Solve function.

#include <bits/stdc++.h>

using namespace std;

#define int long long int

#define F first

#define S second

#define pb push_back

#define si set<int>

#define vi vector<int>

#define pii pair<int, int>

#define vpi vector<pii>

#define vpp vector<pair<int, pii>>

#define mii map<int, int>

#define mpi map<pii, int>

#define spi set<pii>

#define endl "\n"

#define sz(x) ((int)x.size())

#define all(p) p.begin(), p.end()

#define double long double

#define que_max priority_queue<int>

#define que_min priority_queue<int, vi, greater<int>>

#define bug(...) __f(#__VA_ARGS__, __VA_ARGS__)

#define print(a)          \

    for (auto x : a)      \

        cout << x << " "; \

    cout << endl

#define print1(a)    \

    for (auto x : a) \

    cout << x.F << " " << x.S << endl

#define print2(a, x, y)         \

    for (int i = x; i < y; i++) \

        cout << a[i] << " ";    \

    cout << endl

void solve()

{

    int n;

    cin >> n;

    int ans = 0;

    while (n)

    {

        ans += (n & 1);

        n = n >> 1;

        //OR What you can do is this

        // n = n & (n-1);

        // ans++;

    }

    cout<<ans<<endl;

}

int32_t main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    clock_t z = clock();

    int t = 1;

    cin >> t;

    while (t--)

        solve();

    return 0;

}

Thank you for reading this.