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;

}