POJ 2081

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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

题意：
题意大概就是就是第m个位置的数是根据第m-1位置的数推出来的如果a[m-1]-m>0，并且a[m-1]-m在前面的序列中没有出现过那么a[m] = a[m-1]-m否则a[m] = a[m-1]+m

思路：
公式都已经写好了，只需要加一个bool值来标记当前a[m]是否出现过就可以了

代码：

import java.util.Arrays;import java.util.Scanner;public class Test1 {    private static int[] array = new int[500005];    private static boolean[] flag = new boolean[10000000];     public static void main(String[] args) {    Arrays.fill(flag, false);    Arrays.fill(array, 0);    for (int i = 1; i <=500000 ; i++) {        if (array[i-1]-i>0 && flag[array[i-1]-i]==false) {        array[i] = array[i-1]-i;        }else         array[i]= array[i-1]+i;        flag[array[i]] = true;    }    Scanner scanner = new Scanner(System.in);    while (scanner.hasNext()) {        int n = scanner.nextInt();        if (n<0)         break;        System.out.println(array[n]);    }    scanner.close();    }}

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