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.
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.
Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap.
5
4 5 1 3 2
3
7
1 6 5 3 4 7 2
6
6
6 5 4 3 2 1
5
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.
题意：给定一个序列，只能交换1次[一次交换任意的2个数的位置]，问最大值和最小值的距离最大是多少？
思路：因为只能交换一次。所以肯定把某一个[最大值/最小值]反正最边是最优的。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<bitset>
using namespace std;
#define INF 0x3f3f3f3f
#define PI 3.14159
const int MAXN=+;
int num[MAXN];
int main()
{
#ifdef kirito
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int n;
while(~scanf("%d",&n)){
int MAX_pos,MIN_pos;
for(int i=;i<=n;i++)
{
scanf("%d",&num[i]);
if(num[i]==){MIN_pos=i;}
if(num[i]==n){MAX_pos=i;}
}
printf("%d\n",max(n-min(MAX_pos,MIN_pos),max(MAX_pos,MIN_pos)-)); }
return ;
}