PAT-A-https://www.patest.cn/contests/pat-a-practise/1029
来源:互联网 发布:杭州莱茵矩阵国际 编辑:程序博客网 时间:2024/05/01 06:42
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 Input4 11 12 13 145 9 10 15 16 17Sample Output
13
给定两串序列求合并后的中间值,偶数个取第N/2个,基数取N/2+1;没难度直接上代码吧;
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<algorithm>
#include<string>
#include<map>
#include<queue>
#include<cstdio>
using namespace std;
long int s[2000100];
int main(){
int n1,n2;
cin>>n1;
for(int i=0;i<n1;i++)
scanf("%ld",&s[i]);
cin>>n2;
for(int i=n1;i<n1+n2;i++)
scanf("%ld",&s[i]);
sort(s,s+n1+n2);
if((n1+n2)%2==0)
cout<<s[(n1+n2)/2-1]<<endl;
else cout<<s[(n1+n2)/2]<<endl;
system("pause");
return 0;
}
- PAT-A-https://www.patest.cn/contests/pat-a-practise/1029
- http://pat.zju.edu.cn/contests/pat-practise/1029
- http://pat.zju.edu.cn/contests/pat-practise/1030
- http://pat.zju.edu.cn/contests/pat-practise/1034
- http://pat.zju.edu.cn/contests/pat-practise/1016
- http://pat.zju.edu.cn/contests/pat-practise/1032
- http://pat.zju.edu.cn/contests/pat-practise/1037
- http://pat.zju.edu.cn/contests/pat-practise/1021
- http://pat.zju.edu.cn/contests/pat-practise/1040
- http://pat.zju.edu.cn/contests/pat-practise/1038
- http://pat.zju.edu.cn/contests/pat-practise/1025
- http://pat.zju.edu.cn/contests/pat-practise/1023
- http://pat.zju.edu.cn/contests/pat-practise/1035
- http://pat.zju.edu.cn/contests/pat-practise/1024
- http://pat.zju.edu.cn/contests/pat-practise/1031
- http://pat.zju.edu.cn/contests/pat-practise/1028
- http://pat.zju.edu.cn/contests/pat-practise/1020
- http://pat.zju.edu.cn/contests/pat-practise/1015
- [转]c#中DateTime.Now函数的使用详解
- C语言基础练习
- EXTJS6.2 使用弹出确认框窗口Ext.Msg.show
- python找出列表间隐含的关联关系以及重复模式
- OpenGL -- DDS 加载
- PAT-A-https://www.patest.cn/contests/pat-a-practise/1029
- ACM DP Longest Run on a Snowboard
- 倒角距离匹配
- Red Hat Enterprise Linu 7 3.1.11系统时间管理
- 278. First Bad Version
- 软件开发模式简介
- 树莓派安装opencv打开摄像头实现实时传输
- linux initcall机制
- Django中render和render_to_response的区别