poj_2081

Recaman's Sequence

Time Limit: 3000MS		Memory Limit: 60000K
Total Submissions: 19016		Accepted: 7963

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

DP水题

#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 9999999;
int a[MAXN];
bool flag[MAXN];

void DP()
{
	for(int i=1;i<=500000;i++)
	{
		if(a[i-1] - i > 0 && flag[a[i-1] - i] == false)
		{
			a[i] = a[i-1] - i;
			flag[a[i-1] - i] = true;
		}
		else
		{
			a[i] = a[i-1] + i;
			flag[a[i-1] + i] = true;
		}
	}
}

int main()
{
	memset(flag,false,sizeof(false));
	a[0] = 0;
	DP();
	int k;
	while(cin>>k)
	{
		if(k == -1) break;
		cout<<a[k]<<endl;
	}
}