Description

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

Output

Sample Input

6
10
3
7
4
12
2

Sample Output

5

单调栈的入门题:单调栈就是维护一个栈，栈中的元素都保持着单调递增或递减的顺序。

题目意思：有n只牛站在一排，给出队伍中每只牛的高度，每只牛只能看到它右边比它矮的牛，求所有的牛能看到的牛数之和。

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，那这个牛肯定看不见这个高度的牛，就把这个元素弹栈。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”。

?

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       typedef long long ll; using namespace std; int main() { int n; ll heigh,ans; stack
      
       s; while(scanf("%d",&n)!=EOF) { ans=0; cin>>heigh; s.push(heigh);//入栈 for(int i=1;i
       
        >heigh; while(!s.empty()&&s.top()<=heigh) //比较栈顶元素和新加入元素的关系 { s.pop(); } ans+=s.size(); s.push(heigh); } cout<
        
         
?
         
?