HDU1157 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): 7034    Accepted Submission(s): 3432

Problem DescriptionFJ 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

3Hint INPUT DETAILS: Five cows with milk outputs of 1..5 OUTPUT DETAILS: 1 and 2 are below 3; 4 and 5 are above 3. 

 

SourceUSACO 2004 November 

Recommendmcqsmall

解題思路：好吧，我表示我很邪惡，拿了老師給的動歸題用庫函數直接排序做了，不知道用sort（）排序會不會逾時呢？恩，等會試一下。
思路，簡單啦，中位元嗎，直接把數列排序就好了，直接去中間的那一位元輸出即可啦！

01463876

2013-07-29 10:53:42

Accepted

1006

0 MS

352 KB

Visual C++

try it

效果不可思意的好，本來以為有可能逾時，沒想到竟然0ms飄過，STL強大啊！

#include<stdio.h>#include<queue>using namespace std;int main(){    int n,i,x;    priority_queue<int> q;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)        {            scanf("%d",&x);            q.push(x);   //直接進入優先隊列        }        n/=2;        for(i=0;i<n;i++)  //彈出前n/2個數            q.pop();        printf("%d\n",q.top()); //直接輸出即可，排序後第n/2+1個數，即中位元        q.pop();   //把隊列中得元素全部彈出，以免影響下一個測試資料{要是優先隊列能有個清空隊列的方法，就方便省時了}        for(i=0;i<n;i++)            q.pop();    }    return 0;}

8750179

2013-07-29 11:09:21

Accepted

1157

0MS

312K

286 B

C++

try it

#include<stdio.h>#include<algorithm>using namespace std;int main(){    int n,i;    int a[10000];    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)            scanf("%d",&a[i]);        sort(a,a+n);           //直接排序        printf("%d\n",a[n/2]);    }    return 0;}