Bad Hair Day（求数组中元素和它后面离它最近元素之间的元素个数）

题目链接：https://ac.nowcoder.com/acm/contest/984/A

链接：https://ac.nowcoder.com/acm/contest/984/A
来源：牛客网

Bad Hair Day

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

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.

输入描述:

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.

输出描述:

Line 1: A single integer that is the sum of c1 through cN.

示例1

输入

复制

6
10
3
7
4
12
2

输出

复制

5

题目大意：输入N  有N个数 求每个数与它后面的数之间有多少个数  （总和）
思路：利用栈求解，栈的用途还挺多的，具体看代码：

#include<iostream>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
using namespace std;
typedef long long LL;
const int maxn=1e9+5;
stack<int>s;
int main()
{
    int N,x;
    LL ans=0;
    cin>>N;
    for(int i=1;i<=N;i++)
    {
        cin>>x;
        while(!s.empty())
        {
            int y=s.top();
            if(y<=x) s.pop();//如果当前的数比栈顶元素大 出栈
            else//否则这个数可以对栈中所有元素都贡献一个
            {
                ans+=s.size();
                break;
            }
        }
        s.push(x);
    }
    cout<<ans<<endl;
    return 0;
}