C++ 程序：依据机关门数字是否为 3 的倍数输出“stop”

```cpp #include <iostream> using namespace std; int main() { int n; cin >> n; for (int i = 0; i < n; i++) { int num; cin >> num; if (num % 3 == 0) { cout << "stop" << endl; } } return 0; } ``` ######[AI问答 | 95点数解答 | 2025-10-18 17:13:14]