Bad Hair Day

/*

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.

Input

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.

Output

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

Sample Input

6
10
3
7
4
12
2

Sample Output

5

*/
n个牛排成一列向右看，牛i能看到牛j的头顶，当且仅当牛j在牛i的右边并且牛i与牛j之间的所有牛均比牛i矮。设牛i能看到的牛数为Ci，求∑Ci

       单调栈-----所谓单调栈也就是每次加入一个新元素时，把栈中小于等于这个值的元素弹出。

　　求所有牛总共能看到多少牛，可以转化为：这n头牛共能被多少头牛看见。       

　　 当我们新加入一个高度值时，如果栈中存在元素小于新加入的高度值，那么这些小的牛肯定看不见这个高度的牛(那就看不见这头牛后边的所有牛)，所以就可以把这些元素弹出。每次加入新元素，并执行完弹出操作后，栈中元素个数便是可以看见这个牛的“牛数”。

       这道题要注意答案可能会超longint，要用int64。

#include <iostream>
#include <stack>
#include <stdio.h>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    while(cin >> n)
    {
        stack<int> s;
        int be_cow;
        cin >> be_cow;
        s.push(be_cow);
        long long ans = 0;
        for(int i = 1; i < n; i++)
        {
            int cow;
            cin >> cow;
            while(!s.empty() && cow >= s.top())
                s.pop();
            ans += s.size();
            s.push(cow);
        }
        cout << ans << endl;
    }
    return 0;
}