poj 2081 Recaman's Sequence

Recaman's Sequence

Time Limit: 3000MS		Memory Limit: 60000K
Total Submissions: 17487		Accepted: 7246

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

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

bool f[10000000];
int r[500001];
int main()
{
    int n,i;
    memset(f,true,sizeof(f));
    r[0]=0;
    for(i=1;i<=500000;i++)
    {
        if(r[i-1]>i&&f[r[i-1]-i])
            r[i]=r[i-1]-i;
        else 
            r[i]=r[i-1]+i;
        f[r[i]]=false;
    }
    while(scanf("%d",&n)&&n!=-1)
        printf("%d\n",r[n]);
    return 0;
}