2081 Recaman's Sequence

Recaman's Sequence

Time Limit:

3000ms

Memory limit:

60000kB

题目描述

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.

输入

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.

输出

For each k given in the input, print one line containing a_k to the output.

样例输入

7
10000
-1

样例输出

20
18658

Global No.

1083

#include<iostream>
#include<cstring>
using namespace std;
int a[1000000],flag[10000000];
int main()
{
    //freopen("111.txt","w",stdout);
    for(int i=1;i<=500000;i++)
    {
        if(a[i-1]-i>0&&flag[a[i-1]-i]==0) {a[i]=a[i-1]-i;flag[a[i-1]-i]=1;}
        else {a[i]=a[i-1]+i;flag[a[i-1]+i]=1;}
    }  
    int n;
    while(cin>>n&&n!=-1)
    {
        cout<<a[n]<<endl;
    }
    return 0;
}