1753: [Usaco2005 qua]Who's in the Middle

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 290 Solved: 242
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Description

FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. Given an odd number of cows N (1 <= N < 10,000) and their milk output (1..1,000,000), find the median amount of milk given such that at least half the cows give the same amount of milk or more and at least half give the same or less. 输入N个数,输出升序排列后中间那个数.

Input

* Line 1: A single integer N * Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.

Output

* Line 1: A single integer that is the median milk output.

Sample Input

5
2
4
1
3
5

INPUT DETAILS:

Five cows with milk outputs of 1..5

Sample Output

3

OUTPUT DETAILS:

1 and 2 are below 3; 4 and 5 are above 3.

HINT

Source

Gold

题解：难以想象这样的题目居然是usaco金组的= =，汗。。

其实直接排序就好啦，但是这样子太没意思了，于是来个简单的小优化（程序如下，很清晰，相信你们一定看得懂）

 1 /**************************************************************
 2     Problem: 1753
 3     User: HansBug
 4     Language: Pascal
 5     Result: Accepted
 6     Time:12 ms
 7     Memory:616 kb
 8 ****************************************************************/
 9  
10 var
11    i,j,k,l,m,n,x,y:longint;
12    a:array[0..100000] of longint;
13 procedure sort(l,r:longint);
14           var i,j,x,y:longint;
15           begin
16                if (l>m) or (r<m) then exit;   //here!!!^_^
17                i:=l;j:=r;x:=a[(l+r) div 2];
18                repeat
19                      while a[i]<x do inc(i);
20                      while a[j]>x do dec(j);
21                      if i<=j then
22                         begin
23                              y:=a[i];a[i]:=a[j];a[j]:=y;
24                              inc(i);dec(j);
25                         end;
26                until i>j;
27                if i<r then sort(i,r);
28                if l<j then sort(l,j);
29           end;
30  
31 begin
32      readln(n);m:=(1+n) div 2;
33      for i:=1 to n do readln(a[i]);
34      sort(1,n);
35      writeln(a[m]);
36      readln;
37 end.