ACM程序设计 -M（二倍数个数问题）

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

思路：输入一行 数字，如果 第一个数是-1 那就结束程序，如果 一行数字的末尾为0 那就结束 这一行数字的输入，输出这一行 ，所有 有二倍关系的数的 个数。

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

注意到主要是 输入 几个 循环之间 联系， 两个 判断条件 一个是 开头的 -1，一个是末尾的 0 。

Code:

#if    0                     //13
#include<bits/stdc++.h>
using namespace std;
int main()
{
vector <int> v;
int n;
cin>>n;
while(n!=-1)
{
 v.push_back(n);
 int x,sum=0;

 while( cin>>x&&x!=0)
 {
  
  v.push_back(x);
 
 }
 for(int j=0;j<v.size();j++)
 {    
  for(int jj=0;jj<v.size();jj++)
  if(v[jj]==2*v[j]) sum++;
 }
 cout<<sum<<endl;
 cin>>n;
 v.clear()                             //;
 } 
 
 
}

#endif