PAT.A1029. Median

最新推荐文章于 2022-11-06 17:45:16 发布

克拉拉123

最新推荐文章于 2022-11-06 17:45:16 发布

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本文链接：https://blog.youkuaiyun.com/mokena1/article/details/79967985

本文介绍了一种算法，该算法能够找到两个已排序整数序列的中位数。通过对两个序列进行比较并合并，文章实现了寻找合并后序列中位数的方法。此算法适用于计算机科学领域的数据处理和算法设计。

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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 <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <iostream>
using namespace std;
int a1[1000010], a2[1000010],a3[2000010];
int main(){
	int n, m;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d", &a1[i]);
	scanf("%d", &m);
	for (int i = 0; i < m; i++)
		scanf("%d", &a2[i]);
	int i = 0, j = 0,p=0;
	while (i < n&&j < m) {
		if (a1[i] <=a2[j]) {
			a3[p++] = a1[i];
			i++;
		}
		else {
			a3[p++] = a2[j];
			j++;
		}
	}
	while (i < n) a3[p++] = a1[i++];
	while (j < m) a3[p++] = a2[j++];
	int l = (n + m);
	if (l % 2 == 0) printf("%d", a3[l / 2 - 1]);
	else printf("%d", a3[(l - 1) / 2]);
	return 0;
	
}