C++ 破解 [jrksj r7] 铃音的第二分块难题：区间操作与计数实现

```c++ #include <iostream> #include <vector> using namespace std; int main() { int n, m; cin >> n >> m; vector<int> a(n); for (int i = 0; i < n; i++) { cin >> a[i]; } for (int i = 0; i < m; i++) { int opt, l, r, x_prime; cin >> opt >> l >> r >> x_prime; if (opt == 1) { for (int j = l - 1; j < r; j++) { if (a[j] > x_prime) { a[j] -= x_prime; } } } else { int count = 0; for (int j = l - 1; j < r; j++) { if (a[j] <= x_prime) { count++; } } cout << count << endl; } } return 0; } ``` ######[AI写代码神器 | 500点数解答 | 2024-07-18 21:05:15]