1029. Median (25)

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本文介绍了一种算法，该算法通过合并两个递增整数序列来找到它们的中位数。具体实现使用了标准模板库(STL)中的merge函数，首先读取两个序列的元素，然后将这两个序列合并成一个新的有序序列，并最终输出中位数。

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

本题用到merge函数

merge函数的作用是：将两个有序的序列合并为一个有序的序列。函数参数：merge(first1,last1,first2,last2,result,cmp);

//firs1t为第一个容器的首迭代器，last1为第一个容器的末迭代器，first2为第二个容器的首迭代器，last2为容器的末迭代器，result为存放结果的容器的一个迭代器（begin（）或任何一个位置），cmp为比较函数（可略写，默认为合并为一个升序序列）。

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std; 
vector<long long>a,b,c;
int main(){
	int m,n,tmp;
	cin>>m;
	for(int i=0;i<m;i++)
	{
		cin>>tmp;
		a.push_back(tmp);
	}
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>tmp;
		b.push_back(tmp);
	}
	c.resize(m+n);
	merge(a.begin(),a.end(),b.begin(),b.end(),c.begin());
	if((m+n)%2==0)	
		cout<<c[(m+n)/2-1]<<endl;
	else
		cout<<c[(m+n)/2]<<endl;
	return 0;
}