Sumset 递推 poj2299

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

 

 

递推

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

int ans[1000001];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        ans[1]=1;
        ans[2]=2;
        for(int i=3;i<=n;i++)
        {
            if(i%2==1)  ans[i]=ans[i-1];
            if(i%2==0)  ans[i]=(ans[i-1]+ans[i/2])%1000000000;
        }
        cout<<ans[n]<<endl;
    }
    return 0;
}

注意输出不超过十位数