【BZOJ1677】[Usaco2005 Jan]Sumsets 求和 递推

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#include <iostream>
using namespace std;
int  f[1000010];
int n,i;
int main()
{
    cin>>n;
    f[1]=1;
    for (i=2;i<=n;i++)
    {
        f[i]=f[i-1];
        if (!(i&1)) f[i]+=f[i/2];
        f[i]%=1000000000;
    }
    cout<<f[n];
    return 0;
}

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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).

给出一个N(1≤N≤10^6)，使用一些2的若干次幂的数相加来求之．问有多少种方法

Input

   一个整数N.

Output

方法数．这个数可能很大，请输出其在十进制下的最后9位．

Sample Input

7

Sample Output

6

有以下六种方式
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

HINT

 

Source

Silver