HDU - 3763 CD『二分』

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博客围绕Jack和Jill卖CD的问题展开，给出输入输出要求及样例。问题是求两人都拥有的CD数量，题解采用二分法，遍历一人的CD，对另一人进行二分查找相同的CD，找到则计数加一。

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CD

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1288 Accepted Submission(s): 558

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

Source

University of Waterloo Local Contest 2010.09.26

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notonlysuccess

题解：样例1，a,b每人有3，3张cd，a的cd编号是1 2 3,b的cd编号是1 2 4,问a b有几张相同的。

二分，遍历b的所以cd，同时对a进行二分，找到相同的，sum++；

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[100000010],s[100000010];
int m,n;
long long sum;
int check(int max,int p){
	int left=0;
	int right=max-1;
	int mid;
	while(left<=right){
		mid=(left+right)/2;
		if(a[mid]==p)
		    return 1;
		else{
			if(a[mid]>p){
				right=mid-1;
			}else{
				left=mid+1;
			}
		}
	}
	return 0;
}
int main(){
	while(scanf("%d%d",&n,&m)!=EOF&&(m+n)){
		sum=0;
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		sort(a,a+n);
		for(int j=0;j<m;j++){
			scanf("%d",&s[j]); 
			if(check(n,s[j])){
				sum=sum++;
			}
		}
		printf("%lld\n",sum);
	}
}