PAT1029:Median

最新推荐文章于 2022-03-22 10:30:30 发布

转载最新推荐文章于 2022-03-22 10:30:30 发布 · 87 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/0kk470/p/7930065.html

本文介绍了一道关于寻找两个有序整数序列中位数的编程题。通过合并两个序列并排序的方法来找到中位数，特别指出当元素总数为偶数时，中位数取左侧的那个数。

1029. Median (25)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 nondecreasing 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.

Input

Each input file contains 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.

Output

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

Sample Input

4 11 12 13 14
5 9 10 15 16 17

Sample Output

思路

水题，将两个数组合并后排序就行，注意偶数个数字的数组的median是靠左的数。

代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
   int n;
   while(cin >> n)
   {
       vector<int> seq1(n);
       for(int i = 0;i < n;i++)
       {
           cin >> seq1[i];
       }
       int m;
       cin >> m;
       vector<int> seq2(m);
       for(int i = 0;i < m;i++)
       {
           cin >> seq2[i];
       }
       seq1.insert(seq1.end(),seq2.begin(),seq2.end());
       sort(seq1.begin(),seq1.end());
       cout << seq1[(m + n - 1)/2] << endl;
   }
}

转载于:https://www.cnblogs.com/0kk470/p/7930065.html