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问题描述
给定n个整数表示一个商店连续n天的销售量。如果某天之前销售量在增长，而后一天销售量减少，则称这一天为折点，反过来如果之前销售量减少而后一天销售量增长，也称这一天为折点。其他的天都不是折点。
给定n个整数a1, a2, …, an表示销售量，请计算出这些天总共有多少个折点。
为了减少歧义，我们给定的数据保证：在这n天中相邻两天的销售量总是不同的，即ai-1≠ai。注意，如果两天不相邻，销售量可能相同。
输入格式
输入的第一行包含一个整数n。
第二行包含n个整数，用空格分隔，分别表示a1, a2, …, an。
输出格式
输出一个整数，表示折点出现的数量。
样例输入
7
5 4 1 2 3 6 4
样例输出
2
评测用例规模与约定
所有评测用例满足：1 ≤ n ≤ 1000，每天的销售量是不超过10000的非负整数。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm>
#include<cmath> using namespace std; const int N=1000+10; int a[N],n; int
main() { scanf("%d",&n); scanf("%d%d",&a[0],&a[1]); int ans=0; for(int
i=1;i<n-1;i++) { scanf("%d",&a[i+1]);
if(a[i]<a[i-1]&&a[i]<a[i+1]||a[i]>a[i-1]&&a[i]>a[i+1]) ans++; }
printf("%d\n",ans); return 0; }
