POJ 2081 Recaman's Sequence

比较简单，加一个B数组判重即可

Recaman's Sequence

Time Limit: 3000MS		Memory Limit: 60000K
Total Submissions: 21743		Accepted: 9287

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

 

//oimonster
#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
int a[500001],b[10000000]={0};
int main(){
    int i,j,n,m;
    a[0]=0;
    b[0]=1;
    for(i=1;i<=500000;i++){
        m=a[i-1]-i;
        if((m<=0)||(b[m]==1)){
            a[i]=a[i-1]+i;
            b[a[i]]=1;
        }
        else{
            a[i]=m;
            b[m]=1;
        }
    }
    scanf("%d",&n);
    while(n!=-1){
        printf("%d
",a[n]);
        scanf("%d",&n);
    }
    return 0;
}