Nicholas and Permutation

最新推荐文章于 2020-12-04 10:48:19 发布

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本文介绍了一个编程挑战，即通过一次交换操作使数组中最小和最大元素之间的距离最大化。提供了问题描述、输入输出示例及解决方案代码。

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, ..., 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.

Examples

Input

5
4 5 1 3 2

Output

Input

7
1 6 5 3 4 7 2

Output

Input

6
6 5 4 3 2 1

Output

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.

showing code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

struct node
{
    int v;
    int p;
};

node arr[110];

int main()
{
    int n;
    int d = 0;
    node ma, mi;

    while (scanf("%d", &n) != EOF)
    {
        d = 0;
        for (int i=0; i<n; i++)
        {
            arr[i].p = i;
            scanf("%d", &arr[i].v);
        }

        mi.v = ma.v = arr[0].v;
        mi.p = ma.p = arr[0].p;

        for (int i=1; i<n; i++)
        {
            if (ma.v < arr[i].v)
            {
                ma.v = arr[i].v;
                ma.p = arr[i].p;
            }


            if (mi.v > arr[i].v)
            {
                mi.v = arr[i].v;
                mi.p = arr[i].p;
            }
        }

        ma.p += 1;
        mi.p += 1;
        d = abs(ma.p - mi.p);
        d = max (abs(ma.p - n), d);
        d = max (abs(mi.p - n), d); 
        d = max (abs(ma.p - 1), d); //比赛时这里的ma写成了mi， 居然还可以过预测， 多么戏剧性的一幕呀
        d = max (abs(mi.p - 1), d);

        printf("%d\n", d);

    }
    return 0;
}

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