POJ1552 HDU1303 UVALive2787 ZOJ1760 Doubles【序列】

Doubles Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 23199 Accepted: 13094

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 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

Mid-Central USA 2003

Regionals 2003 >> North America - Mid-Central USA

问题链接：POJ1552 HDU1303 UVALive2787 ZOJ1760 Doubles

问题简述：

　　输入有若干组整数，最后以-1结束。每组是正整数，以０结束。求每组数中有多少对数，其中的一个是另外一个的两倍。

问题分析：

　　这个问题的关键是，题意中指出每个数不超过99。

　　这样问题就简单了，可以做一个标记数组来解决，读到一个数a，就标记数组flag[a]为1。

　　同时，判定a的倍数或其一半是否存在就可以了。

程序说明：

　　需要注意标记数组flag[]的长度。

　　标记数组清零是套路。

　　使用二重循环实现那是浮云！

题记：

　　时空可以互换，空间不仅可以换时间，也能换来程序的简洁性。

参考链接：（略）

AC的C语言程序如下：

/* POJ1552 HDU1303 UVALive2787 ZOJ1760 Doubles */ #include <stdio.h> #include <string.h> #define TRUE 1 #define FALSE 0 #define MAXN 99 int flag[MAXN * 2 + 1]; int main(void) { int a, count; while(~scanf("%d", &a) && a != -1) { memset(flag, 0, sizeof(flag)); count = 0; while(a) { if(flag[a * 2]) count++; if((a & 1) == 0 && flag[a / 2]) count++; flag[a] = TRUE; scanf("%d", &a); } printf("%d\n", count); } return 0; }