poj2388 更水
2024-08-29 13:46:58
Who's in the Middle
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 34974 | Accepted: 20396 |
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
#include<string.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int s[];
int main(){
int n;
while(scanf("%d",&n)!=EOF){
memset(s,,sizeof(s));
for(int i=;i<n;i++)
scanf("%d",&s[i]);
sort(s,s+n);
printf("%d\n",s[n/]);
}
return ;
}
最新文章
- 如何在openresty里解析域名
- wp8.1 Study16:网络之 使用Azure移动服务及利用Azure推送通知服务
- Agile Software Development ——敏捷开发
- 【转载】NIO客户端序列图
- iOS开发之Runloop(转)
- 与Jquery Mobile的第一次亲密接触
- android 显示特殊符号
- Ibatis collect select用法详解
- OC中extern,static,const的用法
- 【Centos7】hostnamectl 设置主机名
- Notepad++使用教程
- About Windows 10 SDK Preview Build 17110
- DSAPI多功能组件编程应用-网络相关(上)
- Java线程池ThreadPoolExecutor原理和用法
- 文件防删除保护(miniifiter)
- 金蝶CLOUD与EAS的区别
- Xgboost调参总结
- junit 测试quartz
- 思维题练习专场-DP篇(附题表)
- 【cocos2d-x 3.0-Mac配置篇】