abc103_c - Modulo Summation

Imagine there exist a value $m$ such that it maximizes the contribution of each term in the summation of $f$, meaning that $m \equiv a_i - 1 \mod a_i$ for all $i$.

If such a value exists, then $m + 1$ must be divisible by all $a_i$. We know an number with such properties exists, as the lcm of all $a_i$ behaves as an $m + 1$. Therefore, the answer is:

Time complexity: $O(n)$

Memory complexity: $O(n)$

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
ll solve(vi a) { return accumulate(begin(a), end(a), 0LL) - (int)(a).size(); }
int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    int n;
    cin >> n;
    vi a(n);
    for (auto &ai : a)
        cin >> ai;
    cout << solve(a) << endl;
    return 0;
}