Poj 2081 Recaman's Sequence之解题报告

                                                                                                      Recaman's Sequence

Time Limit: 3000MS		Memory Limit: 60000K
Total Submissions: 22363		Accepted: 9605

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

28

18658

 

对于这题的正确做法就是模拟加暴力Dp；

在DP的时候一定记得考虑时间复杂度的问题；

代码:

 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int maxn = 500000+10;
int a[maxn];
bool used[3012500];

int  main(void)
{
     int k,i,j;
     
     a[0] = 0;
     memset(used,0,sizeof(used));
     used[0] = 1;
     for(i=1;i<=500000;++i)
     {
        if(a[i-1]-i>0&&!used[a[i-1]-i])  a[i] = a[i-1]-i;
        else    a[i] = a[i-1]+i;
        used[a[i]] = 1;
     }
     while(scanf("%d",&k),k!=-1)
     {
         
         cout<<a[k]<<endl;
     }
     
     return 0;
}