Bad Hair Day

题目描述：

Some of Farmer John’s N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows’ heads.

Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.

Consider this example:

    =

= =
= - = Cows facing right -->
= = =
= - = = =
= = = = = =
1 2 3 4 5 6
Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow’s hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow’s hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

输入：

Line 1: The number of cows, N.
Lines 2…N+1: Line i+1 contains a single integer that is the height of cow i.

输出：

Line 1: A single integer that is the sum of c 1 through cN.

样例输入：

6
10
3
7
4
12
2

样例输出：

5

很经典的一个单调栈的使用，通过记录在该位置，能看到他的头发的牛的数量最终确定牛数

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<stack>
using namespace std;
int main()
{
	long long n;
	scanf("%lld",&n);
	long long gao;
	scanf("%lld",&gao);
	stack<long long> feng;
	feng.push(gao);
	long long ans=0;
	for(int i=1;i<n;i++)
	{
		scanf("%lld",&gao);
		while(!feng.empty()&&gao>=feng.top())
		{
			feng.pop();
		}
		ans+=feng.size();
		feng.push(gao);
	}
	printf("%lld\n",ans);
	return 0;
}

还是想要吐槽一下，有的时候翻译真的不是很好用