hdoj 1303 Doubles

最新推荐文章于 2021-03-24 09:24:54 发布

原创最新推荐文章于 2021-03-24 09:24:54 发布 · 649 阅读

0 ·

CC 4.0 BY-SA版权

|--hdoj 专栏收录该内容

134 篇文章

订阅专栏

本文解析了一道名为'Doubles'的ACM编程题，该题要求统计给定整数列表中有多少个数是列表内其他数的两倍。文章提供了完整的C语言实现代码，并详细解释了如何通过排序和遍历来解决此问题。

http://acm.hdu.edu.cn/showproblem.php?pid=1303

Doubles

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3501 Accepted Submission(s): 2450

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

题目大意就是求所给数有几个数的二倍包含在所给数中

#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a,const void *b)
{
	return *(int *)a-*(int *)b;
}
int main()
{
	int i,j,k,sum,a[200];
	while(scanf("%d",&a[0])&&a[0]!=-1)
	{
		for(i=1;a[i-1]!=0;i++)
		scanf("%d",&a[i]);
		qsort(a,i,sizeof(a[0]),cmp);
		
		for(j=0,sum=0;a[j]*2<=a[i-1];j++)
		  for(k=j+1;k<i;k++)
		  if(a[j]*2==a[k])
		  sum+=1;
        printf("%d\n",sum);
    }
    return 0;
}