Posted on 2010-09-04 22:38
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ACM ( 水题 )
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋 
题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1157
题目描述:
Who's in the Middle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2451 Accepted Submission(s): 1204
Problem 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.
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
Sample Output
3
Hint
INPUT DETAILS:
Five cows with milk outputs of 1..5
OUTPUT DETAILS:
1 and 2 are below 3; 4 and 5 are above 3.
水题, 直接代码 :
/*
MiYuMiYu 原创, 转帖请注明 : 转载自 ______________白白の屋
http://www.cnblog.com/MiYu
Author By : MiYu
Test : 1
Program : 1157
*/
#include <iostream>
#include <algorithm>
using namespace std;
int cow[10010];
int main ()
{
int N;
while ( cin >> N ){
for ( int i = 0; i < N; ++ i ) cin >> cow[i];
sort ( cow, cow + N );
cout << cow[N/2] << endl;
}
return 0;
}