zoukankan      html  css  js  c++  java
  • HDU 3763 C D(二分)

    C D

    Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1090 Accepted Submission(s): 480

    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
    
    
    Source
    University of Waterloo Local Contest 2010.09.26
    
    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int a[1000010];
    int n,m,t;
    int fin(int p)
    {
        int left=0,right=n-1,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()
    {
        int k;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            if(n==0&&m==0)
                break;
            for(int i=0;i<n;i++)
                scanf("%d",&a[i]);
            k=0;
            while(m--)
            {
                scanf("%d",&t);
                if(fin(t)==1)
                    k++;
            }
            printf("%d
    ",k);
        }
        return 0;
    }
    
  • 相关阅读:
    js与jquery常用数组方法总结
    js 对象深复制,创建对象和继承
    Web前端面试常识
    jQuery插件开发之boxScroll与marquee
    jQuery插件开发之windowScroll
    《将博客搬至CSDN》
    蓝桥杯 翻硬币
    AcWing 756.蛇形矩阵
    货仓选址
    费解的开关
  • 原文地址:https://www.cnblogs.com/nanfenggu/p/7899982.html
Copyright © 2011-2022 走看看