abc177_a - Dont Be Late
11 Dec 2020 — Tags: easy — URLTo arrive in time, this has to be true: \(T >= \frac{D}{S}\)
Time complexity: $O(1)$
Memory complexity: $O(1)$
Click to show code.
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
int main(void)
{
ios::sync_with_stdio(false), cin.tie(NULL);
int d, t, s;
cin >> d >> t >> s;
cout << (((d + s - 1) / s <= t) ? "Yes" : "No") << endl;
return 0;
}