九度 题目1004：Median

利用给定的s1, s2是递增序列的性质，进行如归并排序(Merge Sort)中的遍历方法，找到第(n+m+1)/2个数即可。

代码：

#include <iostream>#include <vector>using namespace std;int main(){int n, m, num;while (cin >> n){vector<int> s1, s2;for (int i = 0; i < n; ++ i){cin >> num;s1.push_back(num);}cin >> m;for (int i = 0; i < m; ++ i){cin >> num;s2.push_back(num);}int cnt = (n + m + 1) >> 1;int i = 0, j = 0;int median = -1;for ( ; i<n && j<m && cnt>0; -- cnt){if (s1[i] < s2[j]){median = s1[i];++ i;} else{median = s2[j];++ j;}}for ( ; i<n && cnt>0; ++ i, -- cnt){median = s1[i];}for ( ; j<m && cnt>0; ++ j, -- cnt){median = s2[j];}cout << median << endl;}return 0;}