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
题目大意：第0项为0，如果第m-1项-m>0且该数未出现过，第m项为第m-1项-m，否则为第m-1项+m；
思路：谁把这题放在dp专题的........太不靠谱了，hash表标记一下出现状态直接模拟。
代码如下：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[500005]={0};
int p[10000000]={0};
int main()
{
int i;
for(i=1;i<=500000;i++)
{
if(a[i-1]-i>0&&p[a[i-1]-i]==0)
{
a[i]=a[i-1]-i;
}
else{
a[i]=a[i-1]+i;
p[a[i]]=1;
}
}
while(~scanf("%d",&i),i>=0)
printf("%d\n",a[i]);
return 0;
}