My code gives different results on different compilers

My code gives different results on different compilers, the following code gives 499999998352516354 when I enter 1,1000000000 as my input on vs code which is the desired results while it gives 499999998352516352 on codeforces

#include <bits/stdc++.h>  
using namespace std;
int main()
{
   cout<<fixed<<setprecision(90);
   long long x,y;
   long long sum=0;
   long long z=1;
   cin>>x;
   for (int i = 0; i < x; i++)
   {
      cin>>y;
      sum=y*(y+1)/2;
       z= log2(y);
       sum-=2*(pow(2,1+z)-1);
       cout<<sum<<"\n";
       sum=0;
   }
}

>Solution :

Use std::llround() function around pow() and your code will work.

It is because pow() gives floating value which can be incorrectly truncated to 1 less than needed. And llround() gives correct rounding to whole integer.

Below is fixed code, I also adjusted code formatting and changed to correct necessary C++ headers.

Try it online!

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main() {
    cout << fixed << setprecision(90);
    long long x, y;
    long long sum = 0;
    long long z = 1;
    cin >> x;
    for (int i = 0; i < x; i++) {
        cin >> y;
        sum = y * (y + 1) / 2;
        z = log2(y);
        sum -= 2 * (llround(pow(2, 1 + z)) - 1);
        cout << sum << "\n";
        sum = 0;
    }
}

Input:

1 1000000000

Output:

499999998352516354

As it is suggested in comments you may also use bit shifting 1LL << x instead of pow(2, x) if x is non-negative integer.

Try it online!

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main() {
    cout << fixed << setprecision(90);
    long long x, y;
    long long sum = 0;
    long long z = 1;
    cin >> x;
    for (int i = 0; i < x; i++) {
        cin >> y;
        sum = y * (y + 1) / 2;
        z = log2(y);
        sum -= 2 * ((1LL << (1 + z)) - 1);
        cout << sum << "\n";
        sum = 0;
    }
}

As @Bathsheba suggests in comments it is also needed to consider z = log2(y); line. Better to use z = llround(log2(y + 1e-5) - 0.5);, here 1e-5 is a small epsilon value that should be added. I re-checked, this new expression gives exactly same result.