1029

最新推荐文章于 2021-05-08 12:57:04 发布

比锅

最新推荐文章于 2021-05-08 12:57:04 发布

阅读量128

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT

本文链接：https://blog.youkuaiyun.com/czh1075208140/article/details/79789489

PAT 专栏收录该内容

43 篇文章

订阅专栏

本文介绍了一种算法，该算法接收两个递增整数序列作为输入，并将这两个序列合并为一个非递减序列。随后，算法计算并输出合并序列的中位数。示例演示了如何使用此算法来找出两个特定序列的中位数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

vector<long int> a;

int main()
{
    int n,m;
    long int t;
    cin >> n;
    for(int i=0;i<n;i++)
    {
        scanf("%ld", &t);
        a.push_back(t);
    }
    cin >> m;
    for(int i=0;i<m;i++)
    {
        scanf("%ld", &t);
        a.push_back(t);
    }
    sort(a.begin(), a.end());
    if((n+m)%2 == 0)
        printf("%ld", a[(n+m)/2-1]);
    else
        printf("%ld", a[(n+m)/2]);

    return 0;
}