poj--2229

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 15048		Accepted: 6000

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

解体思路:找规律题,当n为奇数时，a[n]=a[n-1];否则a[n]=a[n-1]+a[n/2];

代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int a[1000000+10];
int main(){
	int n;
	memset(a,0,sizeof(a));
	a[1]=1;
	for(int i=2;i<=1000000;i++){
	    if(i%2)a[i]=a[i-1]%1000000000;
	    else{
	    	a[i]=(a[i-1]+a[i/2])%1000000000;
	    }
	}
	while(scanf("%d",&n)!=EOF){
	     printf("%d\n",a[n]%1000000000);
	}
	return 0;
}