Codeforce 486A – Calculating Function

Hi guys sorry for waiting too long, I just started to working at a startup in Jakarta. So this afternoon, at the office LOL, i just solved one question on codeforce.

It took me several minutes to find the pattern of the question, the pattern are :

1) By using even and odd number :
a) if the number is an even number then we should calculate by using (n/2).
b) then if we found out that the number is an odd one, just calculate it by using ((n+2)+1) *-1

Go and give it a try! the above pattern gave me a solid AC on my second try HAHA*

*beware of the questions that you should use high range variable, I tried using int at first and it gave me an error, then I changed it to long long int so that It can holds the result value.

Ok.. That’s for me now, See you on another post!

#include <iostream>

using namespace std;

int main(void)
{
    long long int n;
    cin >> n;
    long long int res = 0;

    if (n == 1){
        res = -1;
        cout << res << endl;
    }else if (n % 2 == 0){
        res = (n/2);
        cout << res << endl;
    }else{
        res = ((n/2) + 1) * -1;
        cout << res << endl;
    }
}