Print closer integer of a double solve by c++

#include<iostream>
#include<math.h>
using namespace std;

int main()
{
    double n;
    cout << "Enter the number : " ;
    cin >> n;
    double t = n - (int)n;
    if(t >= .5)
        cout << ceil(n) << endl;
    else
        cout << floor(n) << endl;

    return 0;
}

///ahmed shuvo969