Codeforces 768A Oath of the Night's Watch

A. Oath of the Night's Watch

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.

With that begins the watch of Jon Snow. He is assigned the task to support the stewards.

This time he has n stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.

Can you find how many stewards will Jon support?

Input

First line consists of a single integer n (1 ≤ n ≤ 10⁵) — the number of stewards with Jon Snow.

Second line consists of n space separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹) representing the values assigned to the stewards.

Output

Output a single integer representing the number of stewards which Jon will feed.

Examples

Input

2

1 5

Output

0

Input

3

1 2 5

Output

1

Note

In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.

In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2.

 题目链接：http://codeforces.com/problemset/problem/768/A

分析：把数字排下序，取最大值和最小值，然后分别进行比较，用一个数去计算其个数即可！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     int a[100005];
 7     while(scanf("%d",&n)!=EOF)
 8     {
 9         int ans=0;
10         for(int i=1;i<=n;i++)
11             scanf("%d",&a[i]);
12             sort(a+1,a+1+n);
13         if(n==1||n==2)
14             printf("0\n");
15         if(n>=3)
16         {
17             int minn=a[1];
18             int maxn=a[n];
19             for(int i=2;i<=n-1;i++)
20                 if(a[i]>minn&&a[i]<maxn)
21                     ans++;
22                     printf("%d\n",ans);
23         }
24     }
25     return 0;
26 }

 

您可以考虑给博主来个小小的打赏以资鼓励，您的肯定将是我最大的动力。thx.

微信打赏

支付宝打赏

作　　者： Angel_Kitty
出　　处：http://www.cnblogs.com/ECJTUACM-873284962/
关于作者：潜心机器学习以及信息安全的综合研究。如有问题或建议，请多多赐教！
版权声明：本文版权归作者和博客园共有，欢迎转载，但未经作者同意必须保留此段声明，且在文章页面明显位置给出原文链接。
特此声明：所有评论和私信都会在第一时间回复。也欢迎园子的大大们指正错误，共同进步。或者直接私信我
声援博主：如果您觉得文章对您有帮助，可以点击右下角【推荐】推荐一下该博文。您的鼓励是作者坚持原创和持续写作的最大动力！