The total fee for all \(N\) certificates is \(N \times A\) yen. If the available money \(B\) covers it, print Yes; otherwise print No. The program directly evaluates the condition \(N \cdot A \le B\) and outputs the result. Although the product fits in an int here, computing it with long long is a robust habit against overflow. Only one comparison is performed, so both the time and space complexities are \(O(1)\).