poj 2081 Recaman's Sequence (dp)

Posted 2020-07-18 kk_kk

Recaman‘s Sequence

Time Limit: 3000MS		Memory Limit: 60000K
Total Submissions: 22566		Accepted: 9697

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

 

Java AC 代码

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        int n = 0;
        int[] result;
        boolean[] marked = new boolean[10000000];
        while((n = sc.nextInt()) != -1) {
            result = new int[n + 1];
            for(int i = 1; i < n + 1; i++) {
                int a = result[i - 1] - i;
                if(a > 0 && !marked[a]) {
                    result[i] = a;
                    marked[a] = true;
                }
                else {
                    result[i] = a + 2 * i;
                    marked[a + 2 * i] = true;
                }
            }
            System.out.println(result[n]);
            for(int i = 0; i < 10000000; i++ )
                marked[i] = false;
         }
    }
}