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社区首页 >专栏 >Codeforces Round #483 (Div. 2) A. Game

Codeforces Round #483 (Div. 2) A. Game

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发布2018-08-30 16:25:17
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发布2018-08-30 16:25:17
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文章被收录于专栏:wym

A. Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Two players play a game.

Initially there are nn integers a1,a2,…,ana1,a2,…,an written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. n−1n−1 turns are made. The first player makes the first move, then players alternate turns.

The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.

You want to know what number will be left on the board after n−1n−1 turns if both players make optimal moves.

Input

The first line contains one integer nn (1≤n≤10001≤n≤1000) — the number of numbers on the board.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106).

Output

Print one number that will be left on the board.

Examples

input

Copy

代码语言:javascript
复制
3
2 1 3

output

Copy

代码语言:javascript
复制
2

input

Copy

代码语言:javascript
复制
3
2 2 2

output

Copy

代码语言:javascript
复制
2

Note

In the first sample, the first player erases 33 and the second erases 11. 22 is left on the board.

In the second sample, 22 is left on the board regardless of the actions of the players.

考排序的题目.给你n个数,第一次去掉最大的,第二次去掉最小的,一直到剩余一个数。

n为奇数,输出中间的,n为偶数,输出中位数中小的一个。

#include <iostream>

#include <algorithm> using namespace std; int a[1005]; int main() { int n;  while(scanf("%d",&n)!=EOF) {   for(int i=1;i<=n;i++)   { scanf("%d",&a[i]);  }   sort(a+1,a+n+1);  if(n&1)  printf("%d\n",a[(n+1)/2]);  else printf("%d\n",a[n/2]); } return 0;  } 

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原始发表:2018年05月16日,如有侵权请联系 cloudcommunity@tencent.com 删除

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