| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15941 | Accepted: 5382 |
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 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
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
代码：
#include<iostream> #include<cstdio> #include<stack> #include<algorithm> using namespace std; typedef long long LL; int main(){ int n; while(~scanf("%d",&n)){ stack<int>S; int t; scanf("%d",&t); S.push(t); LL ans=0; for(int i=1;i<n;i++){ scanf("%d",&t); while(!S.empty()&&t>=S.top())S.pop(); ans+=S.size(); S.push(t); } printf("%lld\n",ans); } return 0; }
原文：http://www.cnblogs.com/handsomecui/p/4965767.html