Official
B - Failing Grade Editorial by en_translator
Whether a student failed or not can be determined by comparing which of \(a_i\) and \(P\) is larger with comparison operator >
or <
in an \(O(1)\) time.
Therefore, the problem can be solved in a total complexity of \(O(N)\) by preparing a variable \(\mathrm{ans}\) to store the answer, initialized to \(0\), and incrementing \(\mathrm{ans}\) by \(1\) if \(a_i \lt P\).
Sample codes in C++ and Python follow.
- C++
#include <iostream>
using namespace std;
int main() {
int N, P, ans = 0;
cin >> N >> P;
for (int i = 0, a; i < N; i++) {
cin >> a;
if (a < P) ans++;
}
cout << ans << "\n";
}
- Python
N, P = map(int, input().split())
print(sum(x < P for x in map(int, input().split())))
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