acm pku 2091 Recaman's Sequence

Recaman's Sequence
Time Limit:3000MS  Memory Limit:60000K
Total Submit:2885 Accepted:1071

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

---

#include <stdio.h>
#define A_SIZE 500000
#define REC_SIZE 5000000
long a[A_SIZE+10];
int rec[REC_SIZE+10];
int main()
...{
    long p;
    for (p=0;p<REC_SIZE;p++)
        rec[p]=0;
    a[0]=0;
    long m;
    for(m=1;m<A_SIZE;m++)
    ...{
        a[m]=a[m-1]-m;
        if( (a[m] <= 0) || rec[a[m]] == 1  )...{
            a[m]=a[m-1]+m;
        }
        rec[a[m]]=1;
    }
    long n=1;
    while( scanf("%d",&n) )
    ...{
        if( n == -1 )break;
        printf("%ld ",a[n]);
    }
}

#include <iostream>
using namespace std;
#define RSIZE 5000000
#define ASIZE 500000
bool rec[RSIZE];
long a[ASIZE];
int main()
...{

    long i;
    for (i=0;i<RSIZE;i++)
        rec[i]=false;
    for (i=0;i<ASIZE;i++)
        a[i]=0;
    a[0]=0;
    for (i=1;i<ASIZE;i++)
    ...{
        a[i]=a[i-1]-i;
        if(a[i]<=0 || rec[a[i]]==true)
                   a[i]=a[i-1]+i;
        rec[a[i]]=true;
    }
    long n;
    while(cin>>n)
    ...{
                 if(n==-1)return 0;
                 cout<<a[n]<<endl;    
    }    
    return 0;
    
}

注意大数组必须是全局的