hdu 1303 Doubles

Dear 丶 2022-06-16 12:42 132阅读 0赞

# Doubles #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 5231 Accepted Submission(s): 3551**

Problem Description

As part of an arithmetic competency program, your students will be given randomly generated lists of from 2 to 15 unique positive integers and asked to determine how many items in each list are twice some other item in the same list. You will need a program to help you with the grading. This program should be able to scan the lists and output the correct answer for each one. For example, given the list  
1 4 3 2 9 7 18 22  
  
your program should answer 3, as 2 is twice 1, 4 is twice 2, and 18 is twice 9.

Input

The input file will consist of one or more lists of numbers. There will be one list of numbers per line. Each list will contain from 2 to 15 unique positive integers. No integer will be larger than 99. Each line will be terminated with the integer 0, which is not considered part of the list. A line with the single number -1 will mark the end of the file. The example input below shows 3 separate lists. Some lists may not contain any doubles.

Output

The output will consist of one line per input list, containing a count of the items that are double some other item.

Sample Input

1 4 3 2 9 7 18 22 0
    2 4 8 10 0
    7 5 11 13 1 3 0
    -1

Sample Output

3
    2
    0
       
        
       
        
    
       
        
       
        
         
        题目大意：
       
        
       
        
         
        这个题目的意思很明确，就是让我们找出数串中可以用其他数字的二倍得到的数的数目；
       
        
       
        
    
       
        
       
        
         
        题目解析：
       
        
       
        
         
        这个题就是hdu最大的水题了，我们直接用包里就可以解决，很简单就不多说什么了，直接看代码吧；
       
        
       
        
    
       
        
       
        
         
        代码：
       
        
       
        
        
         #include<iostream>
    #include<algorithm>
    using namespace std;
    int main()
    {
        int a[1005];
        int len;
        int i,j,k;
        while(cin>>a[0]&&a[0]!=-1)
        {
            len=1;
            for(i=1; i<20; i++)
            {
                cin>>a[i];
                len++;
                if(!a[i])
                    break;
            }
            int cnt=0;
            sort(a,a+len);
            for(j=0; j<len; j++)
            {
                for(k=j+1; k<=len; k++)
                {
                    if(a[j]*2==a[k])
                        cnt++;
                }
            }
            cout<<cnt<<endl;
    
        }
        return 0;
    }