Azhar-Sheet1-G

#include <bits/stdc++.h>
#define endl "\n"
#define ll long long
#define IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
using namespace std;

int main() {
    IO;
    ll n;
    cin>>n;
    ll summation =  n *(1+n)*(1/2) ;
    ll summation1 =  n *(1+n)*(0.5) ;
    ll summation2 =  n *(1+n) /2;
    cout<<summation1;
    /*Integer vs. Floating-Point Context:

    In the first expression (n * (1 + n) * (0.5)), the multiplication by 0.5 promotes the calculation to
    floating-point arithmetic. If n and (1 + n) are large enough, the product might exceed the precision
    limits of floating-point representation.
    In the second expression (n * (1 + n) / 2),
    the entire calculation remains in integer arithmetic until the final division by 2,
    which is straightforward integer division. This means it maintains precision up to the limits of long long.*/
    return 0;
}