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  • poj 2388 -- Who's in the Middle

    Who's in the Middle
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 31287   Accepted: 18163

    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

    水题,sort排序即可。
     1 /*======================================================================
     2  *           Author :   kevin
     3  *         Filename :   WhoIsInTheMiddle.cpp
     4  *       Creat time :   2014-07-14 09:56
     5  *      Description :
     6 ========================================================================*/
     7 #include <iostream>
     8 #include <algorithm>
     9 #include <cstdio>
    10 #include <cstring>
    11 #include <queue>
    12 #include <cmath>
    13 #define clr(a,b) memset(a,b,sizeof(a))
    14 #define M 10005
    15 using namespace std;
    16 int s[M];
    17 int main(int argc,char *argv[])
    18 {
    19     int n;
    20     while(scanf("%d",&n)!=EOF){
    21         for(int i = 0; i < n; i++)
    22             scanf("%d",&s[i]);
    23         sort(s,s+n);
    24         printf("%d
    ",s[n/2]);
    25     }
    26     return 0;
    27 }
    View Code
    Do one thing , and do it well !
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  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/3842205.html
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