题目：
Who‘s in the Middle |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 2938 Accepted Submission(s): 1109 |
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 5 2 4 1 3 5 |
Sample Output 3 |
Source USACO 2004 November |
Recommend mcqsmall |
题目分析：
排序，然后输出中位数
代码如下：
/*
* i.cpp
*
* Created on: 2015年1月29日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 10005;
int a[maxn];
int main(){
int n;
while(scanf("%d",&n)!=EOF){
int i;
for(i = 0 ; i < n ; ++i){
scanf("%d",&a[i]);
}
sort(a,a+n);
printf("%d\n",a[n/2]);
}
return 0;
}
(hdu step 1.3.8)Who's in the Middle(排序)
原文：http://blog.csdn.net/hjd_love_zzt/article/details/43278469