I accepted this problem using long long variables. I know that is not optiomal solution but very interesting and I want to share it !!!
#include <iostream>
using namespace std;
typedef long long ll;
const ll mod = 10000000000ll;
void solve(int n) {
    ll s1 = 0, s2 = 0;
    for(int i = 1; i <= n; i ++) {
        // inc
        s2 = s2 + 1;
        s1 = s1 + s2 / mod;
        s2 = s2 % mod;
        //mult
        s2 *= i;
        s1 *= i;
        s1 = s1 + s2 / mod;
        s2 = s2 % mod;
    }
    // sub
    s2 -= n;
    if(s2 < 0) {
        s1 --;
        s2 += mod;
    }
    // print
    if(s1 > 0) {
        printf("%lld%010lld", s1, s2);
    } else {
        printf("%lld", s2);
    }
}
void test() {
    for(int n = 1; n <= 21; n ++) {
        cout << n << " ";
        solve(n);
        cout << endl;
    }
}
int main() {
    //test();
    int n;
    cin >> n;
    solve(n);
}

Edited by author 11.12.2018 11:39