HackerEarth Monk and his Friend problem solution

In this HackerEarth Monk and his Friend, problem-solution Monk has a very good friend, Puchi. As weird as his name, are the games he plays.

One fine day, they decided to play a game to test how diverse their choices are. Both of them choose exactly one integer each. Monk chooses an integer M and Puchi chooses an integer P.

The diversity of their choices is defined as the number of bits whose status is different in the binary representation of M and P, i.e., count of bits that are, either set in M and unset in P or set in P and unset in M.

Find the answer to their game.

HackerEarth Monk and his Friend's problem solution.

#include<bits/stdc++.h>


using namespace std;

#define rep(i,n) for(i=0;i<n;i++)
#define ll long long
#define elif else if
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define CLEAR(array, value) memset(ptr, value, sizeof(array));
#define si(a)     scanf("%d", &a)
#define sl(a)     scanf("%lld", &a)
#define pi(a)     printf("%d", a)
#define pl(a)     printf("%lld", a)
#define pn        printf("\n")

ll int foo(ll int n)
{
  ll int count = 0;
  while(n)
  {
    count += n & 1;
    n >>= 1;
  }
  return count;
}

int main()
{
  freopen("in.txt","r",stdin);
  freopen("out","w",stdout);
  ios_base::sync_with_stdio(0);
  int t;
  cin>>t;
  assert(1<=t && t<=10000);
  while(t--)
  {
    ll int a,b;
    cin>>a>>b;
    assert(0<=a && a<=10000000000000000);
    assert(0<=b && b<=10000000000000000);
    a= a^b;
    cout<<foo(a);
    if(t>0)cout<<"\n";
  }
return 0;
}

Second solution

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

int main()
{
    int T; cin >> T; assert(T>=1 && T<=10000);
    for (int g=0; g<T; g++){
        long long a, b; cin >> a >> b;
        assert (a>=0 && a<=1e16); 
        assert (b>=0 && b<=1e16);
        cout << __builtin_popcountll (a^b) << '\n';
    }
    return 0;
}