POJ 2081 Recaman's Sequence

题目链接： http://poj.org/problem?id=2081

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

710000-1

Sample Output

2018658

注意：

比较大的数组一定要定义为全局变量。否则会出现栈溢出。原因是全局变量分配在数据段中，而如果在main()

函数中分配的话，则其为局部变量，而局部变量是分配在堆栈段中的，栈溢出不见得是递归调用，也可能是局部

变量过大。此题还有一个技巧是用bitset保存值 coll[i] 是否存在在向量中，而不用再依次查找，大大减小了时间复

杂度。

代码如下：

#include<iostream>#include<fstream>#include<vector>#include<string>#include<map>#include<iterator>#include<algorithm>#include<numeric>#include<cmath>#include<sstream>#include<bitset>using namespace std;const int N = 500010;vector<int> coll(N);bool b[N * 10];int main(){    #ifdef ONLINE_JUDGE    #else        freopen("D:\\in.txt", "r", stdin);        freopen("D:\\out.txt", "w", stdout);    #endif // ONLINE_JUDEG        coll[0] = 0;        b[0] = true;        for (int i = 1; i <= 500000; i++)        {            coll[i] = coll[i - 1] - i;            if (coll[i] > 0 && !b[coll[i]])            {                b[coll[i]] = true;                continue;            }            else            {                coll[i] = coll[i - 1] + i;                b[coll[i]] = true;            }                    }        int n(0);        while (cin >> n)        {            if (-1 == n)            {                break;            }            cout << coll[n] << endl;        }        return 0;}