题：3227. 折点计数

5 4 1 2 3 6 4

思路：

(a[i - 1] - a[i]) * (a[i + 1] - a[i]) > 0
(4 - 1) × (2 - 1) > 0
(3 - 6) × (4 - 6) > 0

或者

(a[i] - a[i - 1]) * (a[i + 1] - a[i]) < 0

#include <iostream>
using namespace std;
const int N = 1005;
int n, a[N], x;

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    int cnt = 0;
    for (int i = 1; i + 1 < n; i++) {
        if ((a[i - 1] - a[i]) * (a[i + 1] - a[i]) > 0) {
           cnt++;
        }
    }
    cout << cnt << endl;
    return 0;
}