| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16425 | Accepted: 9404 |
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
your program should answer 3, as 2 is twice 1, 4 is twice 2, and 18 is twice 9.
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 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
Source
分析：
（1）题意：求每行中有二倍关系的数字的组数。
（2）模拟，因为每行最多不超过14个数字，数字的范围是99以内所以开辟一个100大小的数组，使用标记位的方法标记这个数字是否存在，这样判断二倍关系是就方便了许多
#include <stdio.h> int main() { int i,num[100],index,count; while(1) { //将数组初始化-1 for(i=0;i<100;i++)num[i] = -1; count = 0; scanf("%d",&index); if(index==-1)break; while(1) { num[index]=1; scanf("%d",&index); if(index==0)break; } for(i=0;i<100;i++) if(i*2<100&&num[i]==1&&num[i*2]==1) count++; printf("%d\n",count); } return 0; }
本文转自NewPanderKing51CTO博客，原文链接：http://www.cnblogs.com/newpanderking/archive/2012/10/05/2711995.html，如需转载请自行联系原作者
