arc108_a - Sum And Product

• Write both equations in terms of one variable, N or M: $N^2 - NS + P = 0$.
• The problem reduces to finding a positive root of that expression.
• This problem could be solved by doing a binary search, but it would result in integer overflow.
• Note, however, that $\max(N, M) <= \sqrt{max(S, P)}$.
• Since that value doesn’t exceed $10^6$, we can linear search the answer.

Time complexity: $O( \sqrt{ \max(S, P)})$

Memory complexity: $O(1)$

using namespace std;
using ll = long long;
using predicate = function<bool(ll)>;
bool solve(ll S, ll P)
{
    for (ll N = 1, MAX = 1e6; N <= MAX; ++N)
        if (N * (S - N) == P)
            return true;
    return false;
}
int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    ll S, P;
    cin >> S >> P;
    cout << (solve(S, P) ? "Yes" : "No") << endl;
    return 0;
}