杭电 HDU 1157 Who's in the Middle

Who's in the Middle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10507    Accepted Submission(s): 5019

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

524135

 

 

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.
 

 

 

Source

USACO 2004 November

 

题目 没看懂 但放啥屁很明确，就是排序，输出中位数。中间出现了一次 非法使用内存，后来发现是scanf语句没加取地址符号。另外一次wa是因为用c++的输入出入流 结果超时。

#include<stdio.h>#include<algorithm>using namespace std;int ls[10001];int main(){    int N;    while(scanf("%d",&N)!=EOF){       for(int i=0;i<N;i++)          scanf("%d",&ls[i]);         sort(ls,ls+N);       printf("%d\n",ls[N/2]);}    return 0;}