poj 2081【Recaman's Sequence】

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

用一个数组或容器标记数字是否已经存在过。map的话时间有点长，毕竟是树结构。

用数组吧。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<map>
#define N 500000
using namespace std;
int dp
;
map<int ,int >a;
void sq()
{
dp[0]=0;
a[0]++;
for(int i=1; i<N; i++)
{
if(dp[i-1]-i>=0&&!a[dp[i-1]-i])
{
dp[i]=dp[i-1]-i;
a[dp[i]]++;
}
else
{
dp[i]=dp[i-1]+i;
a[dp[i]]++;
}
}
}
int main()
{
int n;
sq();
while(scanf("%d", &n) && n != -1)
{
printf("%d\n", dp
);
a.clear();
}

}