CodeForces - 676A Nicholas and Permutation (模拟) 水

CodeForces - 676A

Nicholas and Permutation

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Description

Nicholas has an array a that contains ndistinct 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, ..., a_n (1 ≤ a_i ≤ n), where a_i 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.

Sample Input

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

Sample Output

Hint

Source

Codeforces Round #354 (Div. 2)

//题意：

给你一个乱序数组，数组的元素为1--n,问现在可以交换两个数的位置（只有一次机会），问交换后最大值和最小值之间的最大距离是多少？

//思路：

直接模拟就行了。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
	int n,i,j,l,r,x;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<=n;i++)
		{
			scanf("%d",&x);
			if(x==1)
				l=i;
			else if(x==n)
				r=i;
		}
		if(l>r)
			printf("%d\n",l-r+max(n-l,r-1));
		else
			printf("%d\n",r-l+max(n-r,l-1));
	}
	return 0;
}