[动态规划]Recaman's Sequence

Recaman's Sequence

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Time Limit:5s	Memory limit:32M
Accepted Submit:383	Total Submit:1245

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The Recaman's sequence is defined by a(0)=0; for m>0, a(m)=a(m-1)-m if the resulting 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 <stdio.h>
#include <stdlib.h>
#include <string.h>

struct Entry
{
    int i;
    int val;
    int next;
};

int flag[1000000];
int out[500010];

int main()
{
    int i, val, t;

    memset(flag, 0, sizeof(flag));
    flag[0] = 1;
    out[0] = 0;
    for (i = 1; i <= 500000; ++i)
    {
        t = out[i - 1] - i;
        if ((t < 0) || (flag[t / 32] & (1 << (t & 0x1F))))
            t = out[i - 1] + i;

        out[i] = t;
        flag[t / 32] |= (1 << (t & 0x1F));
    }

    while (scanf("%d", &val) && (val >= 0))
        printf("%d\n", out[val]);

    return 0;
}