[解题报告]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): 6458    Accepted Submission(s): 3158

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

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

 

 

Recommend

mcqsmall

 

 

 

 

 

 

题目很简单，就是排序然后输出中位数，但是依旧WA了好多次才过。。

为毛？

尼玛题目你倒是说一句是循环输入啊魂淡！！！

 

#include<stdio.h>
int main()
{
    int N;
    while(scanf("%d",&N)!=EOF)
    {
    int a[N];
    int i;
    for(i=0;i<N;i++)
    {
        scanf("%d",&a[i]);
    }
    int j,flag,temp;
    for(i=0;i<N-1;i++)//冒泡
    {
        flag=1;
        for(j=0;j<N-i-1;j++)
        {
            if(a[j]>a[j+1])
            {
                temp=a[j+1];
                a[j+1]=a[j];
                a[j]=temp;
                flag=0;
            }
        }
        if(flag==1) break;
    }
     printf("%d\n",a[(N-1)/2]);
    }
}