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Who's in the Middle

时间:2015-05-31 14:05:56      阅读:150      评论:0      收藏:0      [点我收藏+]

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import java.util.Scanner;

public class p1157 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();
			if(n==0){
				return;
			}
			int[] a = new int[n];
			for (int i = 0; i < a.length; i++) {
				a[i] = sc.nextInt();
			}
			mySort(a);
			System.out.println(a[a.length / 2]);
		}
	}

	public static void mySort(int[] a) {// 快排
		for (int i = 0; i < a.length - 1; i++) {
			int temp = i;
			for (int j = i + 1; j < a.length; j++) {
				if (a[temp] > a[j]) {
					temp = j;// 记下最小的数
				}
			}
			if (temp != i) {
				a[temp] = a[temp] ^ a[i];
				a[i] = a[temp] ^ a[i];
				a[temp] = a[temp] ^ a[i];
			}
		}
	}
}

题意:

首先:是无限输入,当输入零时结束(通过观看c代码总结出,非题意所给)

目标:求出输入母羊数的中间数(N是奇数)

方法:对输入的数据进行*快排*(其他的排序也行,但是不要超时)

Who‘s in the Middle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11096    Accepted Submission(s): 5248


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








Who's in the Middle

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原文地址:http://blog.csdn.net/hncu1306602liuqiang/article/details/46287875

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