CodeForces 676A Nicholas and Permutation（移动数字游戏）

http://codeforces.com/problemset/problem/676/A

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

66 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.

题意：

给定一串数字，在移动一个数字的情况，问最大值和最小值之间的距离最大可以是多少。

思路：

四种方案可以选择：

最大的移到序列两端，最小的移到序列两端，最后的四种距离存在的最大值就是答案。

AC  Code：

#include<cstdio>
#include<cstring>
const int MYDD=1103;

int MAX(int x,int y) {
return x>y? x:y;
}

int main() {
int n;
while(scanf("%d",&n)!=EOF) {
//		int a[MYDD];
int a;
int max=-1,min=1e9;
int locmax,locmin;
for(int j=1; j<=n; j++) {
scanf("%d",&a);
if(max<a) {
locmax=j;//最大值位置
max=a;
}
if(min>a) {
locmin=j;//最小值位置
min=a;
}
}
int max0=(locmin-locmax>0 ? locmin-locmax:locmax-locmin);//最初两者的距离
int max1=MAX(locmax-1,n-locmax);//移动最小值到两端
int max2=MAX(locmin-1,n-locmin);//移动最大值到两端
int ans=MAX(max0,max1);
ans=MAX(ans,max2);
printf("%d\n",ans);
}
return 0;
}
/*By: Shyazhut*/