/*

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

610374122

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 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;int main(){    int n;    while(cin >> n)    {        stack
         
           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;}