POJ 2081 Recaman's Sequence (递推)

Recaman's Sequence
Time Limit: 3000MS        Memory Limit: 60000K

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
Source

Shanghai 2004 Preliminary

题意：

      一组数列，从第一项开始，每一项根据前一项获得，对于第m项，如果am-1-m>0 && am在之前的数列中未出现过，则am=am-1-m;否则am=am-1+m;

思路：

        递推式已经很明确了，主要是判断当前数在之前的数列中是否出现过，需要设置标记数组flag[]，,，

以下AC代码：

#include<stdio.h>
#include<string.h>
int a[10000005];
int flag[10000005];
int main ()
{
int n;
int i;
while(scanf("%d",&n))
{
if(n==-1)
return 0;
a[0]=0;
memset(flag,0,sizeof(flag));
for(i=1;i<=n;i++)
{
if(a[i-1]-i>0 && !flag[a[i-1]-i])
a[i]=a[i-1]-i;
else
a[i]=a[i-1]+i;
flag[a[i]]=1;
}
printf("%d\n",a
);
}
return 0;
}