星际编码大赛终极对决：机械星AI与异星人激战“逆序对”统计难题

```cpp #include <iostream> using namespace std; long long merge(int arr[], int l, int m, int r) { int n1 = m - l + 1; int n2 = r - m; int L[n1], R[n2]; for (int i = 0; i < n1; i++) L[i] = arr[l + i]; for (int j = 0; j < n2; j++) R[j] = arr[m + 1 + j]; int i = 0, j = 0, k = l; long long inv_count = 0; while (i < n1 && j < n2) { if (L[i] <= R[j]) { arr[k++] = L[i++]; } else { arr[k++] = R[j++]; inv_count += n1 - i; } } while (i < n1) { arr[k++] = L[i++]; } while (j < n2) { arr[k++] = R[j++]; } return inv_count; } long long mergeSort(int arr[], int l, int r) { long long inv_count = 0; if (l < r) { int m = l + (r - l) / 2; inv_count += mergeSort(arr, l, m); inv_count += mergeSort(arr, m + 1, r); inv_count += merge(arr, l, m, r); } return inv_count; } int main() { int n; cin >> n; int arr[n]; for (int i = 0; i < n; i++) { cin >> arr[i]; } cout << mergeSort(arr, 0, n - 1) << endl; return 0; } // 额外注意逻辑点： // 1. 输入数据的边界处理。 // 2. 合并过程中逆序对数量的准确计算。 // 3. 确保代码在不同规模的输入下都能高效运行。 ``` ######[AI写代码神器 | 509点数解答 | 2025-04-19 17:33:00]