山东理工大学ACM平台题答案关于C语言 1259 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 c1 through cN.

示例输入

6
10
3
7
4
12
2

示例输出

5

Bad Hair Day倒霉的日子Time Limit: 2000ms Memory limit: 65536K 有疑问？点这里^_^时间限制：2000ms内存限制：65536k有疑问点这里^ _ ^？题目描述题目描述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.一些农夫约翰的奶牛（1≤N≤80000）有一个糟糕的发型！由于每个奶牛对她凌乱的发型的自觉，福建要清点人数，其他的奶牛，可以看到其他牛的头上。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. 我有一个指定的每头奶牛的高度高（1≤嗨≤1000000000）和站立在一行牛所有面向东方（在图右）。因此，我可以看到牛在她面前的奶牛的头的顶端（即牛，我1，我2，等等），只要这些牛是严格小于牛岛Consider this example:考虑一下这个例子：　　　　　　　　＝＝＝　　　　　　　＝＝＝＝　　　＝　　　＝　　　　　　　Cows facing right－－＞＝＝＝牛朝右－－＞＝　　　＝　　　＝＝＝＝＝　＝　＝　＝　＝＝＝＝＝＝＝　＝　＝　＝　＝　＝＝＝＝＝＝＝１　２　３　４　５　６1 2 3 4 5 6Cow#1 can see the hairstyle of cows #2, 3, 4牛# 1可以看到牛# 2，3型，4Cow#2 can see no cow's hairstyle牛# 2可以看到没有牛的发型Cow#3 can see the hairstyle of cow #4牛# 3可以看到牛# 4发型Cow#4 can see no cow's hairstyle 牛# 4可以看到没有牛的发型Cow#5 can see the hairstyle of cow 6牛# 5可以看到牛6头发型Cow#6 can see no cows at all! 牛# 6可以看到牛都没有！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.让词表示的奶牛的发型是从牛我可见的数量；请通过CN计算C1的总和。在这个例子中，所需的回答“3 + 0 + 1 + 0 + 1 + 0 = 5。输入输入Line 1: The number of cows, N. 行1：牛的数量，N。Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.线2。n + 1 + 1：我行包含一个整数，牛岛的高度输出输出Line 1: A single integer that is the sum of c1 through cN.1行：一个整数，即通过CN C1的总和。示例输入示例输入66101033774412122示例输出2示例输出55
#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;
}