1029. Median

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

13

本题是本意是考察归并排序， 两路归并， 但是由于题目的数据量比较大（$ N \le 10 $） 因此如果直接开两个long数组

int a[1000000], b[1000000]

会开崩掉， 接下来需要解决数据量大的问题, 一种办法是在全局区域定义a, b这样可以防止数组开崩掉，但是具体的没有试过, 这题用了一个比较奇怪的方法， 使用vector先装好一个数组，然后一边读取一边归并，这样可以省一个中间数组，虽然这样做的， 结果速度还行，但是内存占用很多，不知道为什么，有机会研究一下， Mark

#include <iostream>
#include <vector>
using namespace std;

void merge(vector<long> a, vector<long> & res)
{
    int N;
    cin >> N;
    int i = 0;
    while(N--)
    {
        long num;
        cin >> num;
        if(a[i] < num)
        {
            while(a[i] < num && i != a.size()) //一边读取一边归并
            {
                res.push_back(a[i]);
                i++;
            }
            res.push_back(num);
            if(i == a.size())
            {
                break;
            }
        }
        else
        {
            res.push_back(num);
        }
    }

    if(i == a.size())
    {
        while(N--)
        {
            long num;
            cin >> num;
            res.push_back(num);
        }
    }
    else{
        while( i != a.size())
        {
            res.push_back(a[i]);
            i++;
        }
    }
}

int main()
{
    vector<long> a;
    int N;
    cin >> N;
    int n = N;
    while(n--)
    {
        long num;
        cin >> num;
        a.push_back(num);
    }

    vector<long> res;
    merge(a, res);
    cout << res[(res.size() - 1) / 2] << endl;
    return 0;
}