文章作者：姜南(Slyar) 文章来源：Slyar Home (www.slyar.com) 转载请注明，谢谢合作。
水题，典型的空间换时间…
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
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[2600000] = {0};
int a[500001];
void reset()
{
int i;
a[0] = 0;
ext[0] = 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;
}