一道简单的acm题目中的一个bug求调试DescriptionAs 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 ar

一道简单的acm题目中的一个bug求调试DescriptionAs 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 ar
一道简单的acm题目中的一个bug求调试
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
我写的程序是这样的：
#include
#include
int *a;
int length;
void output()
{
\x05for(int i = 0 ; i < length ; i ++)
\x05\x05printf("%d\n",a[i]);
}
int input()
{
\x05int temp;
\x05scanf("%d",&temp);
\x05for(length = 1 ; temp!=0 ; length++)
\x05{
\x05\x05if(a)
\x05\x05{
\x05\x05\x05a=(int *)realloc(a,sizeof(int)*length);
\x05\x05}
\x05\x05else
\x05\x05{
\x05\x05\x05a=(int *)malloc(sizeof(int));
\x05\x05}
\x05\x05a[length-1]=temp;
\x05\x05scanf("%d",&temp);
\x05}
\x05return a[0];
}
int solve()
{
\x05int count = 0,max = a[0];
\x05//首先查找最大值
\x05for(int i = 1 ; i< length ; i++)
\x05{
\x05\x05if(max < a[i])
\x05\x05\x05max = a[i];
\x05}
\x05for(int i = 0 ; i < length ; i++)
\x05{
\x05\x05for(int j = 0 ; j < length ; j++)
\x05\x05{
\x05\x05\x05if(a[i]

一道简单的acm题目中的一个bug求调试DescriptionAs 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 ar

总共最多15个数,为啥要动态分配内存啊