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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

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.

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

6

10

3

7

4

12

2

5

我本来用递归写的   现在发现好笨啊
简单的栈就可以
题目大意：

给你n个数，求每个数后面比他小的个数和

这个题是先把第一个a[0]入栈  再开始判断a[i]和入栈内的
如果栈顶的呢一个数比a[i]大  说明a[i]可以被栈内的每一个数看到然后就可以直接加上栈的长度。
然后再把a[i]入栈。

#include<stdio.h>

#include<stack>

#include<iostream>

#include<string.h>

using namespace std;

#define N 88000

stack<int >Q;

int a[N];

int main()

{

    int n,i;

    while(scanf("%d",&n)!=EOF)

    {

        for(i=;i<=n;i++)

        {

            scanf("%d",&a[i]);

        }

        Q.push(a[]);

        long long ans=;

        for(i=;i<=n;i++)

        {

            while(!Q.empty() && Q.top()<=a[i])

                Q.pop();

            ans+=Q.size();

            Q.push(a[i]);

        }

        printf("%lld\n",ans);

    }

    return ;

}