poj--3250--Bad Hair Day(模拟)

Bad Hair Day

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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 h_i (1 ≤ h_i≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cowi 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 cowi.

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 c_i denote the number of cows whose hairstyle is visible from cowi; please compute the sum of c₁ through c_N.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 cowi.

Output

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

Sample Input

610374122

Sample Output

5

Source

USACO 2006 November Silver

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#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int num[101000];int main(){    int n;    while(scanf("%d",&n)!= EOF)    {        int top = 0;        long long ans = 0;        for(int i = 0; i < n; i++)        {            int h; scanf("%d",&h);//保证严格单调递减的序列             while(top > 0 && num[top] <= h)                --top;            ans += top;//记录每一个下标和             num[++top] = h;        }        printf("%lld\n",ans);    }    return 0;}