# POJ 2081 Recaman’s Sequence C++版

POJ题解 2236浏览

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 …

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

Slyar:没难度，打表就行了，标记数组开到260W就差不多了，我才懒得搜索…

```#include

using namespace std;

char ext = {0};
int a;

void reset()
{
int i;
a = 0;
ext = 1;
for (i = 1; i <= 500000; i++)
{
a[i] = a[i-1] - i;
if (a[i] < 0 || ext[a[i]])
{
a[i] = a[i-1] + i;
}
ext[a[i]] = 1;
}
}

int main()
{
int k;
reset();
while(cin >> k && k != -1)
{
cout << a[k] << endl;
}
//system("pause");
return 0;
}
```