Sumsets
http://acm.hdu.edu.cn/showproblem.php?pid=2709
Problem 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
解题代码：
/*
设way[n]为和为 n 的种类数； 根据题目可知，加数为2的N次方，即 n 为奇数时等于它前一个数 n-1 的种类数 way[n-1] ，若 n 为偶数时分加数中有无 1 讨论，即关键是对 n 为偶数时进行讨论： 1.如果所求的n为奇数，那么所求的分解结果中必含有1，因此，直接将n-1的分拆结果中添加一个1即可 为way[n-1] 2.如果所求的n为偶数，那么n的分解结果分两种情况 (1).如果加数里含1，则一定至少有两个1，即对n-2的每一个加数式后面 +1+1，总类数为way[n-2]；
(2).不含有1 那么，分解因子的都是偶数，将每个分解的因子都除以2，刚好是n/2的分解结果，并且可以与之一一对应，这种情况有 s[n/2] 所以，状态转移方程为: 如果i为奇数 way[n] = way[n-1]
如果i为偶数 way[n] = way[n-2]+way[n/2];
*/
#include <stdio.h>
#include <math.h>
#include <string.h> #ifdef WINDOWS
#define LL __int64
#define LLd "%I64d"
#else
#define LL long long
#define LLD "%lld"
#endif const int max_n = 1e6+;
const LL MOD = 1e9;
LL way[max_n]; void deal()
{
way[] = way[] = ;
for (int i = ; i < max_n; i ++)
{
if (!(i&))
{
way[i] = way[i-] + way[i/];
way[i] %= MOD;
}
else
{
way[i] = way[i-];
way[i] %= MOD;
}
}
}
int main ()
{
int n;
deal();
while (~scanf ("%d", &n))
{
printf (LLD"\n", way[n]);
}
return ;
}