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

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最新推荐文章于 2024-08-23 23:26:17 发布

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分类专栏：模拟 ACM 文章标签： codeforces

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本文介绍了一种计算数组中局部极值数量的方法。局部极值包括局部最大值和局部最小值，它们分别指比相邻元素大或小的元素。文章通过示例详细解释了如何通过遍历数组并比较每个元素与其邻居来确定局部极值的数量。

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You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1and ai < ai + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, ai > ai - 1 and ai > ai + 1). Since a1 and an 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, ..., an (1 ≤ ai ≤ 1000) — the elements of array a.

Output

Print the number of local extrema in the given array.

Examples

input
3
1 2 3

output
0

input
4
1 5 2 5

output
2

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;
}