题目链接：https://ac.nowcoder.com/acm/contest/984/A/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

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.

输入描述

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.

输出描述

Line 1: A single integer that is the sum of c1 through cN.

输入

6
10
3
7
4
12
2

输出

5

解题思路

题意:给定n头牛的高度，设d(i)为第i头牛可以看到(向右看)的其它牛的总数。
思路:反过来想，求第i头牛可以被左边多少头牛看到。维护一个单调栈即可。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n, a;
    long long ans = 0;
    stack <int> S;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a);
        while (!S.empty() && S.top() <= a)
            S.pop();
        ans += S.size();
        S.push(a);
    }
    printf("%lld\n", ans);
    return 0;
}