一日一九度之题目1004: Median (九度Online Judge)

一日一九度之题目1004: Median (九度Online Judge)

1，真题

题目1004：Median

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：9112

解决：2484

题目描述：

Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the non-decreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.
Given two increasing sequences of integers, you are asked to find their median.

输入：

Each input file may contain more than one test case.
Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (≤1000000) is the size of that sequence. Then N integers follow, separated by a space.
It is guaranteed that all the integers are in the range of long int.

输出：

For each test case you should output the median of the two given sequences in a line.

样例输入：

4 11 12 13 145 9 10 15 16 17

样例输出：

13

来源：

2011年浙江大学计算机及软件工程研究生机试真题

2，分析

题目的意思是有两个一维数组，由用户分时间给出各个序列，给出各个序列前，在每个序列前给出序列大小。然后对两个数组排序，得出中间值。

3，答案

import java.util.Arrays;import java.util.Scanner;/** * to test problem_1004 * @author Sunkun * Date: 2013.09.27 * Memory: 18836KB * Code Length: 1025B * Time Consuming: 120MS */public class problem_1004 {public static void main(String[] args) {Scanner input = new Scanner(System.in);while(input.hasNext()){long[] array1 = null;long[] array2 = null;// 第一个数组int len = input.nextInt();array1 = new long[len];for(int i = 0; i < len; i++){array1[i] = input.nextLong();}// 第二个数组len = input.nextInt();array2 = new long[len];for(int i = 0; i < len; i++){array2[i] = input.nextLong();}// 两数组合并long[] total = new long[array1.length + array2.length];int j = 0;for(int i = 0; i < array1.length; i++){total[j++] = array1[i];}for(int i = 0; i < array2.length; i++){total[j++] = array2[i];}// 合并后的数组排序Arrays.sort(total);// 输出合并后数组的中间值if(total.length%2 == 0){System.out.println(total[total.length/2-1]);}else{System.out.println(total[total.length/2]);}}}}

4，备注
上述代码直接复制是AC不了的，把类名problem_1004改为Main就可以直接上传通过了：）