Recaman's Sequence（简单）

Recaman's Sequence

Time Limit: 3000MS
Memory Limit: 60000K

Total Submissions: 19147
Accepted: 8028

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

Source

Shanghai 2004 Preliminary

先预处理求出k<=500000的所有的ak，然后对于输入的k直接输出ak即可。

以下代码中使用STL map标记数字是否在序列中出现，用数组标记效率会更高，怎么知道数组要开多大？我的做法比较笨，就是先用map标记然后求出a[0:500000]中最大的ak。

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;

int a[500005] = {0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9};
int k;
map<int, bool> M;

int main()
{
     for(int i = 0; i < 15; i++) M[a[i]] = true;
     for(int i = 15; i < 500001; i++)
     {
        a[i] = a[i-1] - i;
        if(a[i] <= 0 || M[a[i]])
            a[i] = a[i-1] + i;
        M[a[i]] = true;
    }
    while(scanf("%d", &k) && k != -1)
        printf("%d\n", a[k]);
    return 0;
}