hdu Sumsets

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).

InputA single line with a single integer, N.OutputThe 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
题目大意就是将一个整数拆分为只有 1 或者 有 1 和偶数（2的次方数）组分的几个数的和

当于在f[n-1]的拆分结果上增加一个 1 即可；

当整数n为偶数时，分为两种情况，含有1 ，和不含有1;

含有1时，就是f[n-1]，不含1时，就是f[n/2]（拆分结果都除以2）的结果

综上结果，得出打表公式：

f[1]=1;   f[2]=2;

f[n]=f[n-1]              //n是奇数

f[n]=f[n-1]+f[n/2]   //n是偶数

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int a[];
int main()
{
    int n;
    a[] = ;
    a[] = a[] = ;
    a[] = a[] = ;
    for(int i=;i<=;i++)
    {
        if(i%==)
            a[i] = (a[i-] + a[i/])%;//这里取i/2的值
        else a[i] = a[i-]%;
    }
    while(cin >> n)
    {
        cout << a[n] << endl;
    }
    return ;
}