C++ 求解：n 个不爽事连续 m 个刺痛值和的最小值

```cpp #include <iostream> using namespace std; int main() { int n, m; cin >> n >> m; int a[3005]; // 输入每个事件的刺痛值 for (int i = 0; i < n; i++) { cin >> a[i]; } // 计算前m个刺痛值的和作为初始最小值 int sum = 0; for (int i = 0; i < m; i++) { sum += a[i]; } int minSum = sum; // 遍历后续的连续m个刺痛值 for (int i = m; i < n; i++) { // 计算新的连续m个刺痛值的和 sum = sum - a[i - m] + a[i]; // 更新最小值 if (sum < minSum) { minSum = sum; } } cout << minSum << endl; return 0; } ``` ######[AI问答 | 241点数解答 | 2025-11-24 19:52:43]