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 hi (1 ≤ hi ≤ 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 ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.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 cN.
Sample Input
6 10 3 7 4 12 2
Sample Output
5
题意：一群牛往右看（大概右边有妹子 ?_? )，每头牛可以看见右边比他矮的，如果比他矮的牛，被比他高的牛挡到了，显然他看不见。 所以求这些牛一共能看见多少牛？
注意： 一样高的他也看不见。
第一感觉是单调栈，然后逆向思维一下，对于每头牛应该考虑他被多少头牛看见，然后建一个从左往右的单调减栈，得到答案；
题意转换成 每头牛能被在他左边比他矮的看见 应该才是突破；
我也不明白 int为什么会WA，，，
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <stack>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=2e6+7;
int n,m;
int main()
{
int t;
while(scanf("%d",&n)!=EOF&&n)
{
unsigned long long cnt=0;
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
stack<int> st;
///构造一个单调减的栈 从右往左
for(int i=1;i<=n;++i)
{
while(!st.empty()&&a[i]>=st.top())
{
st.pop();
}
cnt+=st.size();
st.push(a[i]);
}
cout<<cnt<<endl;
}
return 0;
}