题目:Recaman's Sequence
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB
Total submit users: 204, Accepted users: 176
Problem 10010 : No special judgement
Problem description
The Recaman's sequence is defined by a0 = 0 ; for m > 0, am = am−1 − m if the rsulting am is positive and not already in the sequence, otherwise am = am−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 ak.

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 ak to the output.

Sample Input
7
10000
-1

Sample Output
20
18658
我自己用递归写的代码:#include<iostream>
#define N 500000
using namespace std;
long rectman(long n)
{long a[N]={0};
if(n==0)
return 0;

a[rectman(0)]=rectman(0);
if(n>0&&rectman(n-1)-n>0&&a[rectman(n-1)-n]==0)
{a[rectman(n-1)-n]=rectman(n-1)-n;
return rectman(n-1)-n;}
else {a[rectman(n-1)+n]=rectman(n-1)+n;
return rectman(n-1)+n;}
}
int main(){
long m;

while(1)
{ cin>>m;
if(m==-1)
break;
cout<<rectman(m)<<endl;}
return 0;
}

Maybe you do not understand the problem
And there are some errors in your program
code:

#include<iostream>
#define N 500000
using namespace std;

long a[N]={0};
bool find(long n,int index)
{
for(int i =0;i<index;i++)
{
if(a[i] == n)
return true;
}
return false;

}

long rectman(long m)
{
for(long i =1;i<=m;i++)
{
if(a[i-1]-i<0||find(a[i-1]-i,i))
a[i] = a[i-1]+i;
else
a[i] = a[i-1]-i;
}

return a[m];

}
int main()
{
long m;

while(1)
{
cin>>m;
if(m==-1)
return -1;
cout<<rectman(m)<<endl;
}
return 0;
}

谢谢了

[此贴子已经被作者于2006-12-11 15:45:34编辑过]