C++ 实现：巨龙喷火热量输出，避开倍数哑火情况

```cpp #include <iostream> using namespace std; int main() { int n; cin >> n; for (int i = 1; i <= 10; i++) { if (i % n != 0) { cout << i; if (i != 10 && (i + 1) % n != 0) { cout << " "; } } } cout << endl; return 0; } ``` ######[AI问答 | 119点数解答 | 2025-10-25 19:18:11]