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
我本来用递归写的 现在发现好笨啊
简单的栈就可以
题目大意：
给你n个数，求每个数后面比他小的个数和
#include<stdio.h> #include<stack> #include<iostream> #include<string.h> using namespace std; #define N 88000 stack<int >Q; int a[N]; int main() { int n,i; while(scanf("%d",&n)!=EOF) { for(i=1;i<=n;i++) { scanf("%d",&a[i]); } Q.push(a[1]); long long ans=0; for(i=2;i<=n;i++) { while(!Q.empty() && Q.top()<=a[i]) Q.pop(); ans+=Q.size(); Q.push(a[i]); } printf("%lld\n",ans); } return 0; }
原文：http://www.cnblogs.com/linliu/p/4964793.html