poj 2229

Sumsets

Time Limit: 2000MS
 
Memory Limit: 200000K
Total Submissions: 6980
 
Accepted: 2763
Description
Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 

1) 1+1+1+1+1+1+1 
2) 1+1+1+1+1+2 
3) 1+1+1+2+2 
4) 1+1+1+4 
5) 1+2+2+2 
6) 1+2+4 

Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). 
Input
A single line with a single integer, N.
Output
The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).
Sample Input
7
Sample Output
6
Source
USACO 2005 January Silver

难得写次递归，vs2008居然爆掉了》Stack overflow》，郁闷了下

据说是栈溢出，不清楚是什么情况

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int f[1000002];
int find(int num)
{
	if(f[num])
		return f[num];
	if(num==0)
		return 0;
	if(num%2==0)
		f[num]=(find(num-2)+find(num/2))%1000000000;
	else if(num%2==1)
		f[num]=find(num-1)%1000000000;
	return f[num];
}
int main()
{
	int n;
	memset(f,0,sizeof(f));
	f[1]=1;
	f[2]=2;
	while(scanf("%d",&n)!=EOF)
	{
		printf("%d\n",find(n));
	}
	return 0;
}