北大ACM poj2081 Recaman's Sequence

最新推荐文章于 2020-05-30 03:37:51 发布

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最新推荐文章于 2020-05-30 03:37:51 发布

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文章标签： integer input table numbers each output

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Recaman's Sequence

Description

The Recaman's sequence is defined by a0 = 0 ; for m > 0, a_m = a_m−1 − m if the rsulting a_m is positive and not already in the sequence, otherwise a_m = a_m−1 + m.
The first few numbers in the Recaman's Sequence is 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9 ...
Given k, your task is to calculate a_k.

Input

The input consists of several test cases. Each line of the input contains an integer k where 0 <= k <= 500000.
The last line contains an integer −1, which should not be processed.

Output

For each k given in the input, print one line containing a_k to the output.

Sample Input

7
10000
-1

Sample Output

20
18658

/*解题思路：

       题目给出了一个序列的生成方法，求序列的第k个元素。

       k不大，直接递推生成第K个元素就可以了。

       生成过程中需要记录已经被生成的元素，可以用布尔数组打表记录

代码如下：
*/
#include<stdio.h>
int A[500005];
int table[3500000]={0};
void init()
{
      int i,t;
      A[0]=0;
      for(i=1;i<=500000;i++)
      {
            t=A[i-1]-i;
            if(t>0&&!table[t])
                  A[i]=t;
            else
                  A[i]=A[i-1]+i;
            table[A[i]]=1;
      }
}
int main()
{
      int n;
      init();//初始化，打表 
      while(scanf("%d",&n) && n!=-1)//输入-1结束 
      {
            printf("%d\n",A[n]);
      }
      return 0;
}