湖大ACM-Recaman's Sequence

Recaman’s Sequence
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB
Total submit users: 1547, Accepted users: 1361
Problem 10010 : No special judgement
Problem 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

Problem Source
Shanghai-P 2004

解决方案：超大数组保存是否使用过该值

代码部分：

#include <stdio.h>#include <malloc.h>int main (){    int arr[500001],i;    int * b=(int *)malloc (4000000 *sizeof(int));    arr[0]=0;    for (i=0;i<500000;i++)       if ((arr[i-1]-i>0) &&((b[arr[i-1]-i])!=1) )       {arr[i]=arr[i-1]-i; b[arr[i-1]-i]=1;}       else       {arr[i]=arr[i-1]+i; b[arr[i-1]+i]=1;}    int key=0;    scanf("%d",&key);    while (key!=-1)    {        printf ("%d\n",arr[key]);        scanf("%d",&key);    }    return 0;}