POJ-2081 Recaman's Sequence

Recaman’s Sequence
Time Limit: 3000MS Memory Limit: 60000K
Total Submissions: 22392 Accepted: 9614
Description

The Recaman’s sequence is defined by a0 = 0 ; for m > 0, am = am−1 − m if the rsulting am is positive and not already in the sequence, otherwise am = am−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 ak.
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 ak to the output.
Sample Input

7
10000
-1
Sample Output

20
18658

#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdlib.h>

using namespace std;
#define MAX 500000
long long int a[MAX+5];
bool tag[10000000];
void fun()
{
    a[0]=0;
    tag[0]=1;
    for(int i=1;i<=MAX;i++)
    {
        a[i]=a[i-1]-i;
        if(a[i]<0||tag[a[i]]==1)
            a[i]=a[i-1]+i;
        tag[a[i]]=1;
    }
}
int main()
{
    int n;
    memset(tag,0,sizeof(tag));
    fun();
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)
            break;
        printf("%d
",a[n]);

    }
}