1029 Median (25 分)

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原文链接：http://www.cnblogs.com/moranzju/p/11158529.html

博客介绍了递增整数序列中位数的定义，即序列中间位置的数，两个序列的中位数是包含它们所有元素的非递减序列的中位数。要求根据给定的两个递增整数序列，找出它们的中位数，并说明了输入输出规范。

1029 Median (25 分)

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 Specification:

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 (

Output Specification:

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

注意点：1.要考虑到m个数字已经输入完但是k还没到count的情况
　　　　2.必须用cin.tie(0),否则超时。

#include<bits/stdc++.h>
using namespace std;

const int maxn = 200010;

int a[maxn];

int n,m;

int main(){
    
    ios::sync_with_stdio(false);
    cin.tie(0); 
    
    
    cin>>n;
    
    int i,j;
    
    for(i=0;i<n;i++)
        cin>>a[i];
        
    cin>>m;
    
    int count=0;
    
    bool flag=false;
    
    count=(m+n+1)/2;
    
    
    int k=0;
    
    
    int b;
    
    int ans;
    
    i=0;
    
    for(j=0;j<m&&k<count;j++){
        cin>>b;
        
        if(a[i]<=b){
            while(i<n&&a[i]<b){
            k++;
            if(k==count)
            {
                ans=a[i];
                break;
            }
            
            i++;
            }
            
            
        
        
    }
            
        k++;
    if(k==count)
    {
        ans=b;
        break;
    }
        
    
            
        
    }
    
    
    while(k<count&&i<n){
        ans=a[i];
        i++;
        k++;
    }
    
     
//    cout<<count<<endl;
    cout<<ans<<endl; 
    
    
        
            
}