Bad Hair Day

Bad Hair Day

Time Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^

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

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

Line 1: A single integer that is the sum of c ₁ through c _N.

示例输入

6
10
3
7
4
12
2

示例输出

5

来源

USACO 2006 November Silver

#include <stdio.h>  
int main()  
{  
    int n, i, j, count;  
    int a[80000];  
    while (scanf("%d", &n) == 1)  
        {  
        for (i=0; i<n; i++)  
            scanf("%d", a+i);  
        count = 0;  
        for (i=0; i<n; i++)  
        {  
            for (j=i+1; j<n; j++)  
                if (a[j] >= a[i]) break;  
            count += j-i-1;  
        }  
        printf("%d\n", count);  
    }  
    return 0;  
}