A. Nicholas and Permutation 題目鏈接：http://codeforces.com/contest/676/problem/A

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.

?

題意：給定一個(gè)序列，只能交換1次[一次交換任意的2個(gè)數(shù)的位置]，問(wèn)最大值和最小值的距離最大是多少？

思路：因?yàn)橹荒芙粨Q一次。所以肯定把某一個(gè)[最大值/最小值]反正最邊是最優(yōu)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=100+5; int num[MAXN]; int main() { #ifdef kiritofreopen("in.txt","r",stdin);freopen("out.txt","w",stdout); #endifint n;while(~scanf("%d",&n)){int MAX_pos,MIN_pos;for(int i=1;i<=n;i++){scanf("%d",&num[i]);if(num[i]==1){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)-1));}return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/kirito520/p/5550331.html

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