codeforces 676A A. Nicholas and Permutation(水题)

题目链接：

A. Nicholas and Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

题意：

交换这个序列里面的两个数使最大的数和最小的数的距离最大,问这个最大的距离是多少;

思路：

把最大的和最小的分别和第一个和最后一个交换,最大的就是答案了;

AC代码：

#include <bits/stdc++.h>
/*
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
*/
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+4;
int n;
int a[110],flag[110];
int main()
{
       scanf("%d",&n);
       Riep(n)
       {
           scanf("%d",&a[i]);
           flag[a[i]]=i;
       }
       int ans=0;
           ans=max(ans,abs(n-flag[1]));
           ans=max(ans,abs(flag[n]-1));
           ans=max(ans,abs(n-flag[n]));
           ans=max(ans,abs(flag[1]-1));
       cout<<ans<<"
";



    return 0;
}