1411031037-hd-Doubles



Doubles

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3519    Accepted Submission(s): 2464

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
   给定多串数字，每串数字已0结束，所有数据以-1结束。要求出一串数字中存在几组2倍关系的数字
   例如：1 4 3 2 9 7 18 22
       2是1的2倍，4是2的2倍，18是9的2倍，所以输出3
解题思路
   要注意所有数据的输入存储方式。对给定的数据串进行排序。然后依次判断就好
代码
#include<stdio.h>
#include<algorithm>
using namespace std;
int num[20];
int main() 
{
	int i,j,k;
	int sum,len;
	while(scanf("%d",&num[0])&&num[0]!=-1)
	{
		len=1;
		for(i=1;;i++)
		{
			scanf("%d",&num[i]);
			if(!num[i])
			    break;
			else
			    len++;
		}
		sort(num,num+len);
		sum=0;
		for(i=0;i<len;i++)
		{
			k=num[i]*2;
			for(j=i+1;num[j]<=k&&j<len;j++)
			//注意j也需要小于len 
			    if(num[j]==k)
			        sum++;
		}
		printf("%d\n",sum);
	}
	return 0;
}