1029. Median (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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思路：two pointer 需要注意的是中位数坐标的取值问题。当为奇数或者是偶数的时候取值不同。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 #define MAX 1000010
 6 long long  data1[MAX];
 7 long long  data2[MAX];
 8 long long  res[MAX];
 9 int main(int argc, char *argv[])
10 {
11     int N1,N2;
12     scanf("%d",&N1);
13     for(int i=0;i<N1;i++)
14     {
15         scanf("%lld",&data1[i]);
16     }
17     scanf("%d",&N2);
18     for(int i=0;i<N2;i++)
19     {
20         scanf("%lld",&data2[i]);
21     }
22     int i=0,j=0;
23     int index=0;
24     while(i<N1&&j<N2)
25     {
26         if(data1[i]<data2[j])
27         {
28             res[index++]=data1[i++];
29         }
30         else
31         {
32             res[index++]=data2[j++];
33         }
34     }
35     while(i<N1)
36        res[index++]=data1[i++];
37     while(j<N2)
38        res[index++]=data2[j++];
39     int pt=N1+N2;
40     if(pt%2==0)
41     {
42         pt=pt/2-1;
43     }
44     else
45     {
46         pt=pt/2;
47     }
48     printf("%d\n",res[pt]);
49     return 0;
50 }