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
#include<bits/stdc++.h>
using namespace std;
int s[105];
int b[105];
int changdu;
int main()
{
while(1)
{
int num = 0;
for(int i=0; ; i++)///循环输入
{
scanf("%d",&s[i]);
b[i] = s[i];///赋值给第二个数组
changdu = i;///记录长度
if(s[i] == 0)
{
break;///如果遇到0表示结尾,结束
}
if(s[i] == -1)
{
return 0;///遇到-1,表示程序结束
}
}
sort(b,b+changdu);///对b数组排序(从小到大)
for(int i=changdu-1; i>0; i--)///将b数组逆序枚举
{
for(int j=0; j<changdu; j++)///枚举s数组,查找有否有2倍的关系
{
if(b[i] == 2*s[j])
{
//printf("%d %d\n",s[j],b[i]);
num++;///如果有数量就+1
}
}
}
printf("%d\n",num);
}
}
本文介绍了一个用于检测列表中是否存在两倍关系元素的程序。该程序适用于教育场景中的算术能力评估,能处理2到15个唯一正整数的列表,并输出列表中存在两倍关系的元素数量。

被折叠的 条评论
为什么被折叠?



