1029. Median (25)

题目：

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 13 145 9 10 15 16 17

Sample Output

13

注意：

1、由于N可能会特别大，因此不能把两个直接合并然后排序，那样肯定会超时。

2、这里使用合并排序，并且只要合并到一半即找到了中值就直接结束。

3、需要特别小心的是可能会存在两个序列不重合的情况，需要特别处理。

代码：

#include<iostream>#include<vector>using namespace std;vector<long>s[2];long total=0;int main(){long n[2],t;for(int i=0;i<2;++i){cin>>n[i];total+=n[i];for(long j=0;j<n[i];++j){scanf( "%ld",&t);s[i].push_back(t);}}long index[2]={0},ss;for(long i=0;i<=(total-1)/2;++i){if(index[0]<n[0] && index[1]<n[1])ss = s[0][index[0]]<s[1][index[1]] ? s[0][index[0]++] : s[1][index[1]++];else if (index[0]>=n[0] || index[1]>=n[1])ss = index[0]>=n[0] ? s[1][index[1]++] : s[0][index[0]++];}cout<<ss;return 0;}