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, 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

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

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