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;
}