题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3763

Problem Description

Jack and Jill have decided to sell some of their Compact Discs, while they still have some value. They have decided to sell one of each of the CD titles that they both own. How many CDs can Jack and Jill sell?

Neither Jack nor Jill owns more than one copy of each CD.

 

 

Input

The input consists of a sequence of test cases. The first line of each test case contains two non-negative integers N and M, each at most one million, specifying the number of CDs owned by Jack and by Jill, respectively. This line is followed by N lines listing the catalog numbers of the CDs owned by Jack in increasing order, and M more lines listing the catalog numbers of the CDs owned by Jill in increasing order. Each catalog number is a positive integer no greater than one billion. The input is terminated by a line containing two zeros. This last line is not a test case and should not be processed.

 

 

Output

For each test case, output a line containing one integer, the number of CDs that Jack and Jill both own.

 

 

Sample Input


 

3 3 1 2 3 1 2 4 0 0

 

 

Sample Output


 

2

题目大意：a和b分别有一些光盘，他们每个人对于每种光盘只有一个，他们准备卖出去他们共同拥有的光盘，问最多能卖多少

题解：便利一个二分一个O(nlogn)

符合二分总结中的第一种情况   1.对于查找目标值的二分查找

ac：

//#pragma comment(linker, "/STACK:1024000000,1024000000") 
 
#include<stdio.h>
#include<string.h>  
#include<math.h>  
  
#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<fstream>
#include<iostream>  
#include<algorithm>  
using namespace std;  
 
#define ll long long  
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);
const int MAXN=1e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);

int a[MAXN],b[MAXN];
int n,m;

int main()
{
	//std::ios::sync_with_stdio(false);
	while(scanf("%d%d",&n,&m))
	{
		if(n==0&&m==0)
			break;
		for(int i=1;i<=n;++i)
			scanf("%d",&a[i]);
		for(int i=1;i<=m;++i)
			scanf("%d",&b[i]);
		int ans=0;
		for(int i=1;i<=n;++i)
		{
			int l=1,r=m,mid;
			while(l<=r)
			{
				mid=(l+r)>>1;
				if(a[i]==b[mid])
				{
					ans++;
					break;
				}
				else if(b[mid]<a[i])//mid在key左边 
					l=mid+1;
				else
					r=mid-1;
			}
		}
		printf("%d\n",ans);
	}
}