POJ 2081

Recaman's Sequence

Time Limit: 3000MS		Memory Limit: 60000K
Total Submissions: 18575		Accepted: 7751

Description

The Recaman's sequence is defined by a0 = 0 ; for m > 0, a_m = a_m−1 − m if the rsulting 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 <string.h>
const int N=500010;
int ch[N];
bool vis[N*10];//状态数组一定要开得比数字数组大，否则测试时会ＲＥ 
void init()
{
    int i,j;
    memset(vis,false,sizeof(vis));
    memset(ch,0,sizeof(ch));
    vis[0]=vis[1]=vis[3]=true;
    ch[0]=0;ch[1]=1;
    for(i=2;i<N;i++)
    {
        ch[i]=ch[i-1]-i;
        if(ch[i]<1||vis[ch[i]])
            ch[i]=ch[i-1]+i;
        vis[ch[i]]=true;
    }
    return ;           
}
int main()
{
    int i,j,k;
    init();
    while(scanf("%d",&k),k!=-1)
        printf("%d\n",ch[k]);
    return 0;
}