Codeforces Round #354 (Div. 2) A Nicholas and Permutation(水题)
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http://codeforces.com/problemset/problem/676/A
A. Nicholas and Permutation
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
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.
这个题很简单,只需要考虑三种情况就好了。
1.不交换
2.把位置靠前的放在最前面
3.把位置靠后的放在最后面
之后再比较这三种哪一个是符合条件的情况。
下面是AC代码:
#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int a[105];int main(){ int n; while(~scanf("%d",&n)) { int minx,maxx; for(int i=0;i<n;i++) { scanf("%d",&a[i]); if(a[i]==1) { minx=i+1; } else if(a[i]==n) { maxx=i+1; } } int s1=abs(maxx-minx); int maxxx=max(maxx,minx)-1; int minxx=n-min(maxx,minx); int re=max(s1,max(maxxx,minxx)); printf("%d\n",re); } return 0;}
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