codeforces 676A Nicholas and Permutation

最新推荐文章于 2021-07-29 09:24:00 发布

原创最新推荐文章于 2021-07-29 09:24:00 发布 · 270 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#水题

ACM算法_水题专栏收录该内容

76 篇文章

订阅专栏

本文介绍了一个简单的编程问题：如何通过一次交换操作使数组中的最小值和最大值之间的距离最大化。文章提供了一段C++代码实现，并附有示例输入输出说明。

点击打开链接

Nicholas has an array a that contains n distinct integers from 1 to n. In other words, Nicholas has a permutation of size n.

Nicholas want the minimum element (integer 1) and the maximum element (integer n) to be as far as possible from each other. He wants to perform exactly one swap in order to maximize the distance between the minimum and the maximum elements. The distance between two elements is considered to be equal to the absolute difference between their positions.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100) — the size of the permutation.

The second line of the input contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n), where ai is equal to the element at the i-th position.

Output

Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap.

        Examples
      
          input
        
5
4 5 1 3 2

          output
        
3

          input
        
7
1 6 5 3 4 7 2

          output
        
6

          input
        
6
6 5 4 3 2 1

          output
        
5

Note

In the first sample, one may obtain the optimal answer by swapping elements 1 and 2.

In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap 7 and 2.

In the third sample, the distance between the minimum and the maximum elements is already maximum possible, so we just perform some unnecessary swap, for example, one can swap 5 and 2.

水题，直接找最大值和最小值的位置，判断把谁放到最边上。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int a[110];
    cin>>a[1];
    int mi=a[1],ma=a[1],p=1,q=1;
    for(int i=2;i<=n;i++)
    {
    	cin>>a[i];
    	if(a[i]<mi)
    	{
    		mi=a[i];
    		p=i;
    	}
    	if(a[i]>ma)
    	{
    		ma=a[i];
    		q=i;
    	}
    }
    int w1=p-1,w2=n-p,w3=q-1,w4=n-q;
    int w=max(w1,w2);
    w=max(w,w3);
    w=max(w,w4);
    cout<<w<<endl;
    return 0;
}