要解决这个问题，我们需要对给出的排列 `p` 进行操作，寻找满足条件的下标 `i`。我们可以通过检查切分后的两个子数组是否为等差数列来实现。 下面是实现这个逻辑的 C++ 代码： ```cpp #include <iostream> #include <vector> #include <algorithm> using namespace std; // Function to check if a given array is an arithmetic sequence bool isArithmeticSequence(const vector<int>& arr) { if (arr.size() < 2) return true; // A single element or empty array is an arithmetic sequence int diff = arr[1] - arr[0]; // Calculate the common difference for (size_t i = 1; i < arr.size() - 1; ++i) { if (arr[i + 1] - arr[i] != diff) { return false; // If the difference doesn't match, it's not an arithmetic sequence } } return true; } int countArithmeticSplits(const vector<int>& p) { int n = p.size(); int count = 0; // Iterate through the possible split points for (int i = 1; i < n; ++i) { vector<int> left(p.begin(), p.begin() + i); // Left part [p1, ..., pi] vector<int> right(p.begin() + i, p.end()); // Right part [pi+1, ..., pn] // Sort both parts sort(left.begin(), left.end()); sort(right.begin(), right.end()); // Check if both sorted parts are arithmetic sequences if (isArithmeticSequence(left) && isArithmeticSequence(right)) { count++; } } return count; } int main() { int n; cout << "Enter n: "; cin >> n; vector<int> p(n); cout << "Enter the permutation: "; for (int i = 0; i < n; ++i) { cin >> p[i]; } int result = countArithmeticSplits(p); cout << "The number of valid splits is: " << result << endl; return 0; } ``` 代码说明 1. `isArithmeticSequence` 函数：用于检查给定的数组是否是等差数列。 2. `countArithmeticSplits` 函数：根据给定排列 `p` 计算合法切分数量。 - 遍历所有可能的切分点 `i`。 - 将 `p` 切分为左右两部分，分别排序后检查是否为等差数列。 3. ` [2024-08-17 16:52:29 | AI写代码神器 | 1345点数解答]