题目：
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 c1through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
6 10 3 7 4 12 2
5
题目链接：https://cn.vjudge.net/contest/181259#problem/B
题意：前一头牛只能看到后面比它矮的牛。给出每头牛的高度，求总共能看到多少牛。
思路：这个题是一个比较简单的单调队列的问题。从前往后不好做，那就用当前牛去看之前的牛。看代码比较好理解吧。
结果因为数组开小了wa了一发。。心痛
源代码：
#include <iostream>
#include <cstring>
#include <algorithm>
#include <stdio.h>
#include <iomanip>
using namespace std;
long long n;
long long a[8000005];
int main()
{scanf("%lld",&n);long long temp = 0;long long sum = 0;long long h;memset(a,0,sizeof(a));for (int i = 1; i <= n; i++){scanf("%lld",&h);while (temp > 0 && h >= a[temp])temp--;sum += temp;temp++;a[temp] = h;}printf("%lld\n",sum);return 0;
}