2011年浙江大学计算机及软件工程研究生机试真题(3)

本文介绍了一种算法,该算法通过合并两个已排序的整数序列并找到合并后的中位数。输入包括两个递增序列,输出则是这两个序列合并后的中位数。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目描述:

    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 non-decreasing 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.

输入:

    Each input file may contain more than 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.

输出:

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

样例输入:
4 11 12 13 14
5 9 10 15 16 17
样例输出:
13

AC代码:

#include<iostream>
#include<algorithm>
#define maxsize 2000001
using namespace std;
long num[maxsize];

int main(){
	long i,n1,n2;

	while(cin>>n1){
		for(i=0;i<n1;i++)
			cin>>num[i];
		cin>>n2;
		for(i=n1;i<n2+n1;i++)
			cin>>num[i];
		sort(num,num+n1+n2);
		cout<<num[(n1+n2-1)/2]<<endl;
	}
}

将数组num定义为全局变量
评论 2
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值