Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17614		Accepted: 5937

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

 1 //2016.8.23
 2 #include<cstdio>
 3
 4 const int N = 80005;
 5 int Stack[N], hi, top;
 6
 7 int main()
 8 {
 9     int n;
10     long long ans;
11     while(scanf("%d", &n)!=EOF)
12     {
13         ans = top = 0;
14         for(int i = 1; i <= n; i++)
15         {
16             scanf("%d", &hi);
17             while(top>0&&Stack[top-1]<=hi)top--;
18             ans += top;
19             Stack[top++] = hi;
20         }
21         printf("%lld\n", ans);
22     }
23
24     return 0;
25 }