Doing calculation in the range 0f 10^30

Question

The following code is giving an error by saying the message "singed integer overflow". And the error is in this test case :

input : 208170109961052

Output : 4611790103482368430

Expected Output : 104085054980526

The input range is 1 ≤ n ≤ 10^15

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

int main()
{
    long long n;
    cin >> n;
    if (n % 2)
    {
        cout << (((n - 1) * (n - 1)) / 4 + (n - 1) / 2) - (((n + 1) / 2) * ((n + 1) / 2)) << '\n';
    }
    else
    {
        cout << ((n * n) / 4 + n / 2) - ((n / 2) * (n / 2)) << '\n';
    }

    return 0;
}

As far as I know long long is the largest datatype. How can I solve this ?

Answer 1

The solution will be to simplify your terms:

Term 1:

(((n - 1) * (n - 1)) / 4 + (n - 1) / 2) - (((n + 1) / 2) * ((n + 1) / 2)) 

n^2 - 2n + 1      n - 1            n+1      n+1 
------------  +  ------   -   (   ----- *  ----- )
      4             2               2        2


n^2 - 2n + 1      n - 1            (n+1)   *  (n+1) 
------------  +  ------   -   (   ---------------- )
      4             2                   4

n^2 - 2n + 1      2n - 2              n^2 + 2n -1 
------------  +  --------   -   (   ---------------- )
      4             4                      4
      
      
n^2 - 2n + 1  +    2n - 2    - ( n^2 + 2n -1 )
------------------------------------------------ 
                      4                      
                      
n^2 - 2n + 1  +    2n - 2    -n^2 - 2n + 1 
------------------------------------------------ 
                      4                      
                      
    -2n  
---------- 
     4    

 -n
----
  2
  

Term 2

  ((n * n) / 4 + n / 2) - ((n / 2) * (n / 2))
  
    n^2      n          n^2
   ----- +  ---   -  ( ----- )
     4       2           4
     
    n^2      2n         n^2
   ----- +  ---   -    ----- 
     4       4          4
     
    n^2   +   2n    -   n^2
   -------------------------- 
              4          
              
    2n
   ---- 
    4          
 
  n 
 --- 
  2

So, If I did no mistake, then the result is always n/2

Could be done with that:

#include <iostream> 

int main() {
    long long n{};
    if (std::cin >> n)
        std::cout << ((n%2)? (-n / 2):(n/2)) << '\n';
}