Educational Codeforces Round 32 A. Local Extrema（模拟水题）-蒲公英云

Educational Codeforces Round 32 A. Local Extrema（模拟水题）

A. Local Extrema

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array a. Some element of this array a**i is a local minimum iff it is strictly less than both of its neighbours (that is, a**i < a**i - 1and a**i < a**i + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, a**i > a**i - 1 and a**i > a**i + 1). Since a1 and a**n have only one neighbour each, they are neither local minima nor local maxima.

An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.

Input

The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.

The second line contains n integers a1, a2, …, a**n (1 ≤ a**i ≤ 1000) — the elements of array a.

Output

Print the number of local extrema in the given array.

Examples

input

3
1 2 3

output

input

4
1 5 2 5

output

题解：

今晚心血来潮打了一场cf，水几发博客hhhh，这题太水了不说

代码：

#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
#define lson k*2
#define rson k*2+1
#define M (t[k].l+t[k].r)/2
#define ll long long
int main()
{
    int n,i,j,ans=0;
    int a[1005];
    scanf("%d",&n);
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    for(i=1;i<n-1;i++)
    {
        if(a[i-1]>a[i]&&a[i+1]>a[i]||a[i-1]<a[i]&&a[i+1]<a[i])
            ans++;
    }
    printf("%d\n",ans);
    return 0;
}