A. Warrior and Archer
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/594/problem/A
Description
In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this reason it was excluded from the contest. This mistake have been fixed and the current given problem statement and model solution corresponds to what jury wanted it to be during the contest.
Vova and Lesha are friends. They often meet at Vova's place and compete against each other in a computer game named The Ancient Papyri: Swordsink. Vova always chooses a warrior as his fighter and Leshac chooses an archer. After that they should choose initial positions for their characters and start the fight. A warrior is good at melee combat, so Vova will try to make the distance between fighters as small as possible. An archer prefers to keep the enemy at a distance, so Lesha will try to make the initial distance as large as possible.
There are n (n is always even) possible starting positions for characters marked along the Ox axis. The positions are given by their distinct coordinates x1, x2, ..., xn, two characters cannot end up at the same position.
Vova and Lesha take turns banning available positions, Vova moves first. During each turn one of the guys bans exactly one of the remaining positions. Banned positions cannot be used by both Vova and Lesha. They continue to make moves until there are only two possible positions remaining (thus, the total number of moves will be n - 2). After that Vova's character takes the position with the lesser coordinate and Lesha's character takes the position with the bigger coordinate and the guys start fighting.
Vova and Lesha are already tired by the game of choosing positions, as they need to play it before every fight, so they asked you (the developer of the The Ancient Papyri: Swordsink) to write a module that would automatically determine the distance at which the warrior and the archer will start fighting if both Vova and Lesha play optimally.
⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.
⋅2. Ignoring the buoys and relying on dogfighting to get point.
If you and your opponent meet in the same position, you can try to
fight with your opponent to score one point. For the proposal of game
balance, two players are not allowed to fight before buoy #2 is touched by anybody.
There are three types of players.
Speeder:
As a player specializing in high speed movement, he/she tries to avoid
dogfighting while attempting to gain points by touching buoys.
Fighter:
As a player specializing in dogfighting, he/she always tries to fight
with the opponent to score points. Since a fighter is slower than a
speeder, it's difficult for him/her to score points by touching buoys
when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.
There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.
Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting.
Since Asuka is slower than Shion, she decides to fight with Shion for
only one time during the match. It is also assumed that if Asuka and
Shion touch the buoy in the same time, the point will be given to Asuka
and Asuka could also fight with Shion at the buoy. We assume that in
such scenario, the dogfighting must happen after the buoy is touched by
Asuka or Shion.
The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?
Input
The first line on the input contains a single integer n (2 ≤ n ≤ 200 000, n is even) — the number of positions available initially. The second line contains n distinct integers x1, x2, ..., xn (0 ≤ xi ≤ 109), giving the coordinates of the corresponding positions.
Output
Print the distance between the warrior and the archer at the beginning of the fight, provided that both Vova and Lesha play optimally.
Sample Input
6
0 1 3 7 15 31
Sample Output
7
HINT
题意
给你n个点，两个人依次删点，使得只剩下2个点
第一个人想让最后剩下的两个点距离最近
第二个人想让最后剩下的两个点距离最远
问答案应该是多少
题解：
n一定是偶数咯，所以就贪心去做就好了
假设我们最后剩下的是A,B这两个数，那么AB之间的数，一定是被第二个人删除的，AB之外的数，一定是被第一个人删除的
所以就扫一遍就好了
代码
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std; long long a[];
int main()
{
int n;scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+,a++n);
long long ans = 2e9;
for(int i=;i<=n;i++)
{
if(i+n/>n)break;
//cout<<i+n/2<<" "<<i<<" "<<a[i+n/2]<<" "<<a[i]<<endl;
ans = min(ans,a[i+n/]-a[i]);
}
printf("%lld\n",ans);
}